Solutions for exercises in chapter 1

E1.1 Verify that

S0 → ¬S1 = h2, 3, 1, 4i
and
(S0 → S1 ) → (¬S1 → ¬S0 ) = h2, 2, 3, 4, 2, 1, 4, 1, 3i.

S0 → ¬S1 = h2i⌢ S0⌢ ¬S1

= h2i⌢ h3i⌢ h1i⌢ S1
= h2, 3, 1, 4i;

(S0 → S1 ) → (¬S1 → ¬S0 ) = h2i⌢ (S0 → S1 )⌢ (¬S1 → ¬S0 )

= h2i⌢ h2i⌢ S0⌢ S1⌢ h2i⌢ ¬S1⌢ ¬S0
= h2, 2, 3, 4, 2i⌢h1i⌢ S1⌢ h1i⌢ S0
= h2, 2, 3, 4, 2, 1, 4, 1, 3i.

E1.2 Prove that there is a sentential formula of each positive integer length.
If m is a positive integer, then
m−1 times
z }| {
h1, 1, . . . , 1, S0 i
is a formula of length m, it is
m−1 times
z }| {
¬¬ · · · ¬ S0 .
E1.3 Prove that m is the length of a sentential formula not involving ¬ iff m is odd.
Proof. ⇒: We prove by induction on ϕ that if ϕ is a sentential formula not involving
¬, then the length of ϕ is odd. This is true of sentential variables, which have length 1.
Suppose that it is true of ϕ and ψ, which have length 2m + 1 and 2n + 1 respectively. Then
ϕ → ψ, which is h1i⌢ ϕ⌢ ψ, has length 1 + 2m + 1 + 2n + 1 = 2(m + n + 1) + 1, which is
again odd. This finishes the inductive proof.
⇐. We construct formulas without ¬ with length any odd integer by induction. hS0 i
is a formula of length 1. If ϕ has been constructed of length 2m + 1, then S0 → ϕ, which
is h1, S0 i⌢ ϕ, has length 2m + 3. This finishes the inductive construction.
E1.4 Prove that a truth table for a sentential formula involving n basic formulas has 2n
rows.
We prove this by induction on n. For n = 1, there are two rows. Assume that for n basic
formulas there are 2n rows. Given n + 1 basic formulas, let ϕ be one of them. For the
others, by the inductive hypothesis there are 2n rows. For each such row there are two
possibilities, 0 or 1, for ϕ. So for the n + 1 basic formulas there are 2n · 2 = 2n+1 rows.

1
E1.5 Use the truth table method to show that the formula

(ϕ → ψ) ↔ (¬ϕ ∨ ψ)

is a tautology.

ϕ ψ ϕ→ψ ¬ϕ ¬ϕ ∨ ψ (ϕ → ψ) ↔ (¬ϕ ∨ ψ)
1 1 1 0 1 1
1 0 0 0 0 1
0 1 1 1 1 1
0 0 1 1 1 1

E1.6 Use the truth table method to show that the formula

[ϕ ∨ (ψ ∧ χ)] ↔ [(ϕ ∨ ψ) ∧ (ϕ ∨ χ)]

is a tautology.
Let θ be the indicated formula.

ϕ ψ χ ϕ∨ψ ϕ∨χ (ϕ ∨ ψ) ∧ (ϕ ∨ χ) ψ∧χ ϕ ∨ (ψ ∧ χ) θ

1 1 1 1 1 1 1 1 1
1 1 0 1 1 1 0 1 1
1 0 1 1 1 1 0 1 1
1 0 0 1 1 1 0 1 1
0 1 1 1 1 1 1 1 1
0 1 0 1 0 0 0 0 1
0 0 1 0 1 0 0 0 1
0 0 0 0 0 0 0 0 1

E1.7 Use the truth table method to show that the formula

(ϕ → ψ) → (ϕ → ¬ψ)

is not a tautology. It is not necessary to work out the full truth table.

ϕ ψ ϕ→ψ ¬ψ ϕ → ¬ψ (ϕ → ψ) → (ϕ → ¬ψ)
1 1 1 0 0 0

2
E1.8 Determine whether or not the following is a tautology:

S0 → (S1 → (S2 → (S3 → S1 ))).

Suppose that f is an assignment making the indicated formula false; we work towards a
contradiction. Thus
(1) S0 [f ] = 1 and
(2) (S1 → (S2 → (S3 → S1 )))[f ] = 0.
From (2) we get
(3) S1 [f ] = 1 and
(4) (S2 → (S3 → S1 ))[f ] = 0.
From (4) we get
(5) S2 [f ] = 1 and
(6) (S3 → S1 )[f ] = 0.
From (6) we get S1 [f ] = 0, contradicting (3).
E1.9 Determine whether or not the following is a tautology; an informal method is better
than a truth table:

({[(ϕ → ψ) → (¬χ → ¬θ)] → χ} → τ ) → [(τ → ϕ) → (θ → ϕ)].

Suppose that f is an assignment which makes the given formula false; we want to get a
contradiction. Thus we have
(1) ({[(ϕ → ψ) → (¬χ → ¬θ)] → χ} → τ )[f ] = 1 and
(2) [(τ → ϕ) → (θ → ϕ)][f ] = 0.
By (2) we have
(3) (τ → ϕ)[f ] = 1 and
(4) (θ → ϕ)[f ] = 0.
By (4) we have
(5) θ[f ] = 1 and
(6) ϕ[f ] = 0.
By (3) and (6) we get
(7) τ [f ] = 0.
By (1) and (7) we get
(8) {[(ϕ → ψ) → (¬χ → ¬θ)] → χ}[f ] = 0.

3
It follows that
(9) [(ϕ → ψ) → (¬χ → ¬θ)][f ] = 1 and
(10) χ[f ] = 0.
Now by (6) we have
(11) (ϕ → ψ)[f ] = 1,
and hence by (9),
(12) (¬χ → ¬θ)[f ] = 1.
By (5) we have
(13) (¬θ)[f ] = 0,
and hence by (12),
(14) (¬χ)[f ] = 0.
This contradicts (10).
E1.10 Determine whether the following statements are logically consistent. If the contract
is valid, then Horatio is liable. If Horation is liable, he will go bankrupt. Either Horatio
will go bankrupt or the bank will lend him money. However, the bank will definitely not
lend him money.
Let S0 correspond to “the contract is valid”, S1 to “Horatio is liable”, S2 to “Horatio will
go bankrupt”, and S3 to “the bank will lend him money”. Then we want to see if there is
an assignment of values which makes the following sentence true:

(S0 → S1 ) ∧ (S1 → S2 ) ∧ (S2 ∨ S3 ) ∧ ¬S3 .

We can let f (0) = f (1) = f (2) = 1 and f (3) = 0, and this gives the sentence the value 1.
E1.11 Write out an actual proof for {ψ} ⊢ ¬ψ → ϕ. This can be done by following the
proof of Lemma 1.9, expanding it using the proof of the deduction theorem.
Following the proof of Lemma 1.9, the following is a {ψ, ¬ψ}-proof:
(a) ¬ψ
(b) ¬ψ → (¬ϕ → ¬ψ) (1)
(c) ¬ϕ → ¬ψ (a), (b), MP
(d) (¬ϕ → ¬ψ) → (ψ → ϕ) (3)
(e) ψ→ϕ (c), (d), MP
(f) ψ
(g) ϕ (e), (f), MP
Now applying the proof of the deduction theorem, the following is a {ψ}-proof:
(a) [¬ψ → [(¬ψ → ¬ψ) → ¬ψ]] → [[¬ψ → (¬ψ → ¬ψ)]
→ (¬ψ → ¬ψ)] (2)

4
(b) ¬ψ → [(¬ψ → ¬ψ) → ¬ψ] (1)
(c) [¬ψ → (¬ψ → ¬ψ)] → (¬ψ → ¬ψ) (a), (b), MP
(d) ¬ψ → (¬ψ → ¬ψ) (1)
(e) ¬ψ → ¬ψ (c), (d), MP
(f) [¬ψ → (¬ϕ → ¬ψ)] → [¬ψ → [¬ψ → (¬ϕ → ¬ψ)]] (1)
(g) ¬ψ → (¬ϕ → ¬ψ) (1)
(h) ¬ψ → [¬ψ → (¬ϕ → ¬ψ)]] (f), (g), MP
(i) [(¬ϕ → ¬ψ) → (ψ → ϕ)] → [¬ψ → [(¬ϕ → ¬ψ) → (ψ → ϕ)]] (1)
(j) (¬ϕ → ¬ψ) → (ψ → ϕ) (3)
(k) ¬ψ → [(¬ϕ → ¬ψ) → (ψ → ϕ)] (i), (j), MP
(l) [¬ψ → [(¬ϕ → ¬ψ) → (ψ → ϕ)]] → [[¬ψ → (¬ϕ → ¬ψ)]
→ [¬ψ → (ψ → ϕ)]] (2)
(m) [¬ψ → (¬ϕ → ¬ψ)] → [¬ψ → (ψ → ϕ)] (k), (l), MP
(n) ¬ψ → (ψ → ϕ) (g), (m), MP
(o) ψ → (¬ψ → ψ) (1)
(p) ψ
(q) ¬ψ → ψ (o), (p), MP
(r) [¬ψ → (ψ → ϕ)] → [(¬ψ → ψ) → (¬ψ → ϕ)] (2)
(s) (¬ψ → ψ) → (¬ψ → ϕ) (n), (r), MP
(t) ¬ψ → ϕ (q), (s), MP

Solutions for exercises in Chapter 2

E2.1 Give an exact definition of a language for the structure (ω, <).
The quadruple ({11}, ∅, ∅, rnk), where rnk is the function with domain {11} such that
rnk(11) = 2.
E2.2 Give an exact definition of a language for the set A (no individual constants, func-
tion symbols, or relation symbols).
The quadruple (∅, ∅, ∅, ∅). Note that the last ∅ is the empty function.
E2.3 Describe a term construction sequence which shows that + • v0 v0 v1 is a term in the
language for (R, +, ·, 0, 1, <).
hv0 , •v0 v0 , v1 , + • v0 v0 v1 i.
E2.4 In any first-order language, show that the sequence hv0 , v0 i is not a term. Hint: use
Proposition 2.2.
Suppose that hv0 , v0 i is a term. This contradicts Proposition 2.2(ii).
E2.5 In the language for (ω, S, 0, +, ·), show that the sequence h+, v0 , v1 , v2 i is not a term.
Here S(i) = i + 1 for any i ∈ ω. Hint: use Proposition 2.2.
Suppose it is a term. By Proposition 2.2(ii)(c), there are terms σ, τ such that h+, v0 , v1 , v2 i
is h+i⌢ σ ⌢ τ . Thus hv0 , v1 , v2 i = σ ⌢ τ . So the term v0 is an initial segment of the term
σ. By Proposition 2.2(iii) it follows that v0 = σ. Hence hv1 , v2 i = τ . This contradicts
Proposition 2.2(ii).

5
 E2.6 Prove Proposition 2.5.
We show by complete induction on i that ϕi ∈ Γ for all i < m. So, suppose that i < m
and ϕj ∈ Γ for all j < i. By the definition of formula construction sequence, we have the
following cases.
Case 1. ϕi is an atomic formula. Then ϕi ∈ Γ by (i).
Case 2. There is a j < i such that ϕi is ¬ϕj . By the inductive hypothesis, ϕj ∈ Γ.
Hence by (ii), ϕi ∈ Γ.
Case 2. There are j, k < i such that ϕi is ϕj → ϕk . By the inductive hypothesis,
ϕj ∈ Γ and ϕk ∈ Γ. Hence by (iii), ϕi ∈ Γ.
Case 4. There exist j < i and k ∈ ω such that ϕi is ∀vk ϕj . By the inductive
hypothesis, ϕj ∈ Γ. Hence by (iv), ϕi ∈ Γ.
This completes the inductive proof.
E2.7 Show how the structure (ω, S, 0, +, ·) can be put in the general framework of struc-
tures.
(ω, S, 0, +, ·) can be considered to be the structure (ω, Rel′ , F cn′ , Cn′ ) where Rel′ = ∅,
Cn′ is the function with domain {8} such that Cn′ (8) = 0, and F cn′ is the function with
domain {6, 7, 9} such that F cn′ (6) = S, F cn′ (7) = +, and F cn′ (9) = ·.
E2.8 Prove that in the language for the structure (ω, +), a term has length m iff m is
odd.
First we show by induction on terms that every term has odd length. This is true for
variables. Suppose that it is true for terms σ and τ . Then also σ + τ has odd length.
Hence every term has odd length.
Second we prove by induction on m that for all m, there is a term of length 2m + 1.
A variable has length 1, so our assertion holds for m = 0. Assume that there is a term σ
of length 2m + 1. Then σ + v0 has length 2m + 3. This finishes the inductive proof.
E2.9 Give a formula ϕ in the language for (Q, +, ·) such that for any a : ω → Q,
(Q, +, ·) |= ϕ[a] iff a0 = 1.
Let ϕ be the formula ∀v1 [v0 · v1 = v1 ].
E2.10 Give a formula ϕ which holds in a structure, under any assignment, iff the structure
has at least 3 elements.
∃v0 ∃v1 ∃v2 (¬(v0 = v1 ) ∧ ¬(v0 = v2 ) ∧ ¬(v1 = v2 )).
E2.11 Give a formula ϕ which holds in a structure, under any assignment, iff the structure
has exactly 4 elements.
∃v0 ∃v1 ∃v2 ∃v3 (¬(v0 = v1 ) ∧ ¬(v0 = v2 ) ∧ ¬(v0 = v3 ) ∧ ¬(v1 = v2 ) ∧ ¬(v1 = v3 ) ∧ ¬(v2 = v3 )
∧ ∀v4 (v0 = v4 ∨ v1 = v4 ∨ v2 = v4 ∨ v3 = v4 )).
E2.12 Write a formula ϕ in the language for (ω, <) such that for any assignment a,
(ω, <) |= ϕ[a] iff a0 < a1 and there are exactly two integers between a0 and a1 .
v0 < v1 ∧ ∃v2 ∃v3 [v0 < v2 ∧ v2 < v3 ∧ v3 < v1
∧ ∀v4 [v0 < v4 ∧ v4 < v1 → v4 = v2 ∨ v4 = v3 ]].

6
E2.13 Prove that the formula
v0 = v1 → (Rv0 v2 → Rv1 v2 )
is universally valid, where R is a binary relation symbol.
Let A be a structure and a : ω → A an assignment. Suppose that A |= (v0 = v1 )[a]. Then
a0 = a1 . Also suppose that A |= Rv0 v2 [a]. Then (a0 , a2 ) ∈ RA . Hence (a1 , a2 ) ∈ RA .
Hence A |= Rv1 v2 [a], as desired.
E2.14 Give an example showing that the formula
v0 = v1 → ∀v0 (v0 = v1 )
is not universally valid.
def
Consider the structure A = (ω, <), and let a : ω → ω be defined by a(i) = 0 for all i ∈ ω.
Then A |= (v0 = v1 )[a]. Now A 6|= (v0 = v1 )[a01 ] since 1 6= 0, so A 6|= ∀v0 (v0 = v1 )[a].
Therefore A 6|= (v0 = v1 → ∀v0 (v0 = v1 ))[a].
E2.15 Prove that ∃v0 ∀v1 ϕ → ∀v1 ∃v0 ϕ is universally valid.
Assume that a : ω → A and A |= ∃v0 ∀v1 ϕ[a]. Choose u ∈ A so that A |= ∀v1 ϕ[a0u ]. In
order to show that A |= ∀v1 ∃v0 ϕ[a], let w ∈ A be given. Then A |= ϕ01
uw . It follows that
A |= ∃v0 ϕ[u1w ]. Hence A |= ∀v1 ∃v0 ϕ[a], as desired.
Solutions to exercises in Chapter 3
E3.1 Do the case Rσ0 . . . σm−1 for some m-ary relation symbol and terms σ0 , . . . , σm−1
in the proof of Theorem 3.1, (L3).
We are assuming that vi does not occur in Rσ0 . . . σm−1 ; hence it does not occur in any
term σi .
A |= (Rσ0 . . . σm−1 )[a] iff hσ0A (a), . . . , σm−1
R
(a)i ∈ RA
iff hσ0A (b), . . . , σm−1
R
(b)i ∈ RA
(by Proposition 2.4)
iff A |= (Rσ0 . . . σm−1 )[b].
E3.2 Prove that (L6) is universally valid, in the proof of Theorem 3.1.
Assume that A |= (σ = τ )[a] and A |= (ρ = σ)[a]. Then σ A (a) = τ A (a) and ρA (a) = σ A (a),
so ρA (a) = τ A (a), hence A |= (ρ = τ )[a].
E3.3 Prove that (L8) is universally valid, in the proof of Theorem 3.1.
Assume that A |= (σ = τ )[a]. Then σ A (a) = τ A (a). Assume that
A |= (Rξ0 . . . ξi−1 σξi+1 . . . ξm−1 )[a]; hence
hξ0A (a), . . . , ξi−1
A
(a), σ A (a), ξi+1
A A
(a), . . . , ξm−1 (a)i ∈ RA ; hence
hξ0A (a), . . . , ξi−1
A
(a), τ A (a), ξi+1
A A
(a), . . . , ξm−1 (a)i ∈ RA ; hence
A |= (Rξ0 . . . ξi−1 τ ξi+1 . . . ξm−1 )[a];

7
hence (L8) is universally valid.
E3.4 Finish the proof of Proposition 3.9.
If i = 0 then ϕ itself is the desired segment, unique by Proposition 2.6(iii). If i > 0 then
actually i > 1 so that ϕi is within ψ, and the inductive hypothesis applies.
E3.5 Finish the proof of Proposition 3.11.
Suppose inductively that ϕ is ¬ψ. Thus ϕ is h1i⌢ ψ. It follows that i > 0, so that ϕi
appears in ψ; then the inductive hypothesis applies.
Suppose inductively that ϕ is ψ → χ. Thus ϕ is h2i⌢ ψ ⌢ χ. It follows that i > 0, so
that ϕi appears in ψ or χ; then the inductive hypothesis applies.
Finally, suppose that ϕ is ∀vs ψ with ψ a formula and s ∈ ω. Thus ϕ is h4, 5(s+1)i⌢ψ.
Hence i > 0. If i = 1, then h5(s + 1)i is the desired segment, unique by Proposition 2.6(iii).
Suppose that i > 1. So ϕi is an entry in ψ and hence by the inductive assumption, there
is a segment hϕi , ϕi+1 , . . . ϕm i which is a term; this is also a segment of ϕ, and it is unique
by Proposition 2.6(iii).
E3.6 Indicate which occurrences of the variables are bound and which ones free for the
following formulas.
∃v0 (v0 < v1 ) ∧ ∀v1 (v0 = v1 ).
v4 + v2 = v0 ∧ ∀v3 (v0 = v1 ).
∃v2 (v4 + v2 = v0 ).
First formula: the first and second occurrences of v0 are bound, and the third one is free.
The first occurrence of v1 is free, and the other two are bound.
Second formula: the occurrence of v3 is bound. All other occurrences of variables are free.
Third formula: the two occurrences of v2 are bound. The other occurrences of variables
are free.
E3.7 Prove Proposition 3.14.
Induction on ϕ. Suppose that ϕ is ρ = ξ. Then by Proposition 3.13, σ occurs in ρ or
ξ. Suppose that it occurs in ρ. Let ρ′ be obtained from ρ by replacing that occurrence
of σ by τ . Then ρ′ is a term by Proposition 3.14. Since ψ is ρ′ = ξ, ψ is a formula.
The case in which σ occurs in ξ is similar. Now suppose that ϕ is Rη0 . . . ηm−1 with
R an m-ary relation symbol and η0 , . . . , ηm−1 are terms. Then the occurrence of σ is
within some ηi . Let ηi′ be obtained from ηi by replacing that occurrence by τ . Now ψ is
Rη0 . . . ηi−1 ηi′ . . . ηm−1 , so ψ is a formula.
Now suppose that the result holds for ϕ′ , and ϕ is ¬ϕ′ . Then σ occurs in ϕ′ , so if
ψ is obtained from ϕ′ by replacing the occurrence of σ by τ , then ψ ′ is a formula by the
′

inductive assumption. Since ψ is ¬ψ ′ also ψ is a formula.

Next, suppose that the result holds for ϕ′ and ϕ′′ , and ϕ is ϕ′ → ϕ′′ . Then the
occurrence of σ is within ϕ′ or is within ϕ′′ . If it is within ϕ′ , let ψ ′ be obtained from ϕ′
by replacing that occurrence of σ by τ . Then ψ ′ is a formula by the inductive hypothesis.
Since ψ is ψ ′ → ϕ′′ , also ψ is a formula. If the occurrence is within ϕ′′ , let ψ ′′ be obtained

8
from ϕ′′ by replacing that occurrence of σ by τ . Then ψ ′′ is a formula by the inductive
hypothesis. Since ψ is ϕ′ → ψ ′′ , also ψ is a formula.
Finally, suppose that the result holds for ϕ′ , and ϕ is ∀vk ϕ′ . If i = 1, then σ is vk ,
and by hypothesis τ is some variable vl . Then ψ is ∀vl ϕ′ , which is a formula. If i > 1, then
σ occurs in ϕ′ , so if ψ ′ is obtained from ϕ′ by replacing the occurrence of σ by τ , then ψ ′
is a formula by the inductive assumption. Since ψ is ∀vk ψ ′ also ψ is a formula.
E3.8 Indicate all free and bound occurrences of terms in the formula v0 = v1 + v1 →
∃v2 (v0 + v2 = v1 ).
v0 is free in both of its occcurrences.
v1 is free in all three of its occurrences.
v2 is bound in both of its occurrences.
v1 + v1 is free in its occurrence.
v0 + v2 is bound in its occurrence.
E3.9 Prove Proposition 3.17
Induction on ϕ. If ϕ is atomic, then ψ is equal to ϕ, and θ is equal to χ and hence is
a formula. Suppose the result is true for ϕ′ and ϕ is ¬ϕ′ . If ψ = ϕ, again the desired
conclusion is clear. Otherwise the occurrence of ψ is within the subformula ϕ′ . If θ ′ is
obtained from ϕ′ by replacing that occurrence by χ, then θ ′ is a formula by the inductive
hypothesis. Since θ is ¬θ ′ , also θ is a formula.
Now suppose the result is true for ϕ′ and ϕ′′ , and ϕ is ϕ′ → ϕ′′ . If ψ = ϕ, again the
desired conclusion is clear. Otherwise the occurrence of ψ is within the subformula ϕ′ or
is within the subformula ϕ′′ . If it is within ϕ′ and θ ′ is obtained from ϕ′ by replacing that
occurrence by χ, then θ ′ is a formula by the inductive hypothesis. Since θ is θ ′ → ϕ′′ , also
θ is a formula. If it is within ϕ′′ and θ ′′ is obtained from ϕ′′ by replacing that occurrence
by χ, then θ ′′ is a formula by the inductive hypothesis. Since θ is ϕ′ → θ ′′ , also θ is a
formula.
Finally, suppose the result is true for ϕ′ and ϕ is ∀vi ϕ′ . If ψ = ϕ, again the desired
conclusion is clear. Otherwise the occurrence of ψ is within the subformula ϕ′ . If θ ′ is
obtained from ϕ′ by replacing that occurrence by χ, then θ ′ is a formula by the inductive
hypothesis. Since θ is ∀vi θ ′ , also θ is a formula.
E3.10 Show that the condition in Lemma 3.15 that the resulting occurrence of τ is free
is necessary. Hint: use Theorem 3.2; describe a specific formula of the type in Proposition
3.15, but with τ not free, such that the formula is not universally valid.
Consider the language for (ω, S), and the formula

v0 = v1 → (∃v1 (Sv0 = v1 ) ↔ ∃v1 (Sv1 = v1 )).

Taking an assignment a : ω → ω with a0 = a1 makes this sentence false; hence it is not

provable, by Theorem 3.2.
E3.11 Do the case of implication in the proof of Lemma 3.15.
Suppose inductively that ϕ is χ → θ.

9
 Case 1. The occurrence of σ in ϕ is within χ. Let χ′ be obtained from χ by replacing
that occurrence by τ , such that that occurrence is free in ψ, hence free in χ′ . By the
inductive hypothesis, ⊢ σ = τ → (χ ↔ χ′ ). Since ψ is χ′ → θ, a tautology gives the
desired result.
Case 2. The occurrence of σ in ϕ is within θ. Let θ ′ be obtained from θ by replacing
that occurrence by τ , such that that occurrence is free in ψ, hence free in θ ′ . By the
inductive hypothesis, ⊢ σ = τ → (θ ↔ θ ′ ). Since ψ is χ → θ ′ , a tautology gives the desired
result.
E3.12 Prove that the hypothesis of Theorem 3.25 is necessary.
Consider the formula
∀v0 ∃v1 (v0 < v1 ) → ∃v1 (v1 < v1 ).
This formula is not universally valid; it fails to hold in (ω, <), for example.
E3.13 Prove Proposition 3.29.
Proof. By definition, ∃vi ¬ϕ is ¬∀vi ¬¬ϕ. Now ⊢ ϕ ↔ ¬¬ϕ by a tautology. Hence
using generalization and (L2) we get ⊢ ∀vi ϕ ↔ ∀vi ¬¬ϕ. Hence another tautology yields
⊢ ¬∀vi ϕ ↔ ¬∀vi ¬¬ϕ, i.e., ⊢ ¬∀vi ϕ ↔ ∃vi ¬ϕ.
E3.14 Prove Proposition 3.30.
Proof. ¬∃vi ϕ is the formula ¬¬∀vi ¬ϕ, so a simple tautology gives the result.
E3.15 Prove Proposition 3.31.
Proof. By Theorem 3.25 we have ⊢ ∀vi ¬ϕ → Subfvσi (¬ϕ). Since clearly Subfvσi (¬ϕ)
is the same as ¬Subfvσi ϕ, a tautology gives ⊢ Subfvσi ϕ → ∃vi ϕ.
E3.16 Prove Proposition 3.33.
By Corollary 3.26 and Corollary 3.32.
E3.17 Prove Proposition 3.34.
Proof. ⊢ ¬ϕ ↔ ∀vi ¬ϕ. Now use a tautology.
E3.18 Prove Proposition 3.41.
Assume ⊢ ϕ ↔ ψ. By a tautology, ⊢ ϕ → ψ. Hence by generalization and (L2), ⊢ ∀vi ϕ →
∀vi ψ. Similarly, ⊢ ∀vi ψ → ∀vi ϕ. The exercise follows by a tautology.
E3.19 Prove Proposition 3.42
Assume ⊢ ϕ ↔ ψ. By a tautology, ⊢ ¬ϕ ↔ ¬ψ. Hence by exercise E3.18, ⊢ ∀vi ¬ϕ ↔
∀vi ¬ψ. Now a tautology gives the desired result.
E3.20 Prove that

⊢ ∀v0 ∀v1 (v0 = v1 ) → ∀v0 (v0 = v1 ∨ v0 = v2 ).

10
 ⊢ ∀v0 ∀v1 (v0 = v1 ) → v0 = v1 ; Cor. 3.26 twice, taut. (1)
⊢ ∀v1 (v0 = v1 ) → v0 = v2 ; Thm. 3.25 (2)
⊢ ∀v0 ∀v1 (v0 = v1 ) → v0 = v2 ; (2), Cor. 3.26, taut. (3)
⊢ ∀v0 ∀v1 (v0 = v1 ) → v0 = v1 ∨ v0 = v2 ; (1), (3), taut. (4)
⊢ ∀v0 ∀v0 ∀v1 (v0 = v1 ) → ∀v0 (v0 = v1 ∨ v0 = v2 ); (4), (L2), taut. (5)
⊢ ∀v0 ∀v1 (v0 = v1 ) → ∀v0 (v0 = v1 ∨ v0 = v2 ). (5), Prop. 3.27, taut.

E3.21 Prove that

⊢ ∃v0 (¬v0 = v1 ∧ ¬v0 = v2 ) → ∃v0 ∃v1 (¬v0 = v1 ).

⊢ ¬∀v0 (v0 = v1 ∨ v0 = v2 ) → ¬∀v0 ∀v1 (v0 = v1 ); E3.20, taut. (1)

⊢ ¬∀v0 (v0 = v1 ∨ v0 = v2 ) ↔ ∃v0 ¬(v0 = v1 ∨ v0 = v2 ); Prop. 3.29 (2)
⊢ ¬(v0 = v1 ∨ v0 = v2 ) ↔ (¬(v0 = v1 ) ∧ ¬(v0 = v2 )); taut. (3)
⊢ ∃v0 ¬(v0 = v1 ∨ v0 = v2 ) ↔ ∃v0 (¬(v0 = v1 ) ∧ ¬(v0 = v2 )); (3), Prop. 3.42 (4)
⊢ ¬∀v0 (v0 = v1 ∨ v0 = v2 ) ↔ ∃v0 (¬(v0 = v1 ) ∧ ¬(v0 = v2 )); (2), (4), taut. (5)
⊢ ¬∀v1 (v0 = v1 ) ↔ ∃v1 ¬(v0 = v1 ); Prop. 3.29 (6)
⊢ ∃v0 ¬∀v1 (v0 = v1 ) ↔ ∃v0 ∃v1 ¬(v0 = v1 ); (6), Prop. 3.42 (7)
⊢ ¬∀v0 ∀v1 (v0 = v1 ) ↔ ∃v0 ¬∀v1 (v0 = v1 ); Prop. 3.29 (8)
⊢ ¬∀v0 ∀v1 (v0 = v1 ) ↔ ∃v0 ∃v1 ¬(v0 = v1 ) (7), (8), taut. (9)
⊢ ∃v0 (¬v0 = v1 ∧ ¬v0 = v2 ) → ∃v0 ∃v1 (¬v0 = v1 ). (1), (5), (9), taut.

Solutions to exercises in Chapter 4

E4.1 Suppose that Γ ⊢ ϕ → ψ, Γ ⊢ ϕ → ¬ψ, and Γ ⊢ ¬ϕ → ϕ. Prove that Γ is
inconsistent.
The formula (¬ϕ → ϕ) → ϕ is a tautology. Hence by Lemma 3.3, Γ ⊢ (¬ϕ → ϕ) → ϕ.
Since also Γ ⊢ ¬ϕ → ϕ, it follows that Γ ⊢ ϕ. Hence Γ ⊢ ψ and Γ ⊢ ¬ψ. Hence by Lemma
4.1, Γ is inconsistent.
E4.2 Let L be a language with just one non-logical constant, a binary relation symbol R.
Let Γ consist of all sentences of the form ∃v1 ∀v0 [Rv0 v1 ↔ ϕ] with ϕ a formula with only
v0 free. Show that Γ is inconsistent. Hint: take ϕ to be ¬Rv0 v0 .
By Theorem 3.25 we have

(1) Γ ⊢ ∀v0 [Rv0 v1 ↔ ¬Rv0 v0 ] → [Rv1 v1 ↔ ¬Rv1 v1 ].

Now [Rv1 v1 ↔ ¬Rv1 v1 ] → ¬(v0 = v0 ) is a tautology, so from (1) we obtain

Γ ⊢ ∀v0 [Rv0 v1 ↔ ¬Rv0 v0 ] → ¬(v0 = v0 );

11
then generalization gives

Γ ⊢ ∀v1 [∀v0 [Rv0 v1 ↔ ¬Rv0 v0 ] → ¬(v0 = v0 )].

Then by Proposition 3.37 we get

Γ ⊢ ∃v1 ∀v0 [Rv0 v1 ↔ ¬Rv0 v0 ] → ¬(v0 = v0 ).

But the hypothesis here is a member of Γ, so we get Γ ⊢ ¬(v0 = v0 ). Hence by Lemma

4.1, Γ is inconsistent.
Alternate proof (due to a couple of students). Suppose that Γ is consistent. By
the completeness theorem let A be a model of Γ. Taking ϕ to be ¬Rv0 v0 , we get A |=
∃v1 ∀v0 [Rv0 v1 ↔ ¬Rv0 v0 ]. Let a : ω → A be any assignment. Then by Proposition 2.8(iv)
there is a b ∈ A such that A |= ∀v0 [Rv0 v1 ↔ ¬Rv0 v0 ][a1b ]. By the definition of satisfaction
of ∀, it follows that for any c ∈ A we have A |= [Rv0 v1 ↔ ¬Rv0 v0 ][a0c b1 ]. Hence (c, b) ∈ RA
/ RA , contradiction.
iff (c, b) ∈
E4.3 Show that the first-order deduction theorem fails if the condition that ϕ is a sentence
is omitted. Hint: take Γ = ∅, let ϕ be the formula v0 = v1 , and let ψ be the formula v0 = v2 .

{v0 = v1 } ⊢ v0 = v1
{v0 = v1 } ⊢ ∀v1 (v0 = v1 )
{v0 = v1 } ⊢ ∀v1 (v0 = v1 ) → v0 = v2 by Theorem 3.25
{v0 = v1 } ⊢ v0 = v2 .

On the other hand, let A be the structure with universe ω and define a = h0, 0, 1, 1, . . .i.
Clearly A 6|= [v0 = v1 → v0 = v2 ][a]. Hence 6⊢ v0 = v1 → v0 = v2 by Theorem 3.2.
def
E4.4 In the language for A = (ω, S, 0, +, ·), let τ be the term v0 + v1 · v2 and ν the
term v0 + v2 . Let a be the sequence h0, 1, 2, . . .i. Let ρ be obtained from τ by replacing the
occurrence of v1 by ν.
(a) Describe ρ as a sequence of integers.
(b) What is ρA (a)?
(c) What is ν A (a)?
(d) Describe the sequence a1 A as a sequence of integers.
ν (a)
A
(e) Verify that ρ (a) = τ A
(a1 A ) (cf. Lemma 4.4.)
ν (a)

(a) ρ is v0 + (v0 + v2 ) · v2 ; as a sequence of integers it is h7, 5, 9, 7, 5, 15, 15i.

(b) ρA (a) = 0 + (0 + 2) · 2 = 4.
(c) ν A (a) = 0 + 2 = 2.
(d) a1 A = h0, 2, 2, 3, . . .i.
ν (a)
(e) ρ (a) = 4, as above; τ A (a1 A
A
) = 0 + 2 · 2 = 4.
ν (a)

12
 def
E4.5 In the language for A = (ω, S, 0, +, ·), let ϕ be the formula ∀v0 (v0 · v1 = v1 ), let ν
be the formula v1 + v1 , and let a = h1, 0, 1, 0, . . .i.
(a) Describe Subfvν1 ϕ as a sequence of integers
(b) What is ν A (a)?
(c) Describe a1 A as a sequence of integers.
ν (a)
(d) Determine whether A |= Subfvν1 ϕ[a] or not.
(e) Determine whether A |= ϕ[a1 A ] or not.
ν (a)

(a) Subfvν1 ϕ is ∀v0 (v0 · (v1 + v1 ) = v1 + v1 ; as a sequence of integers it is

h4, 5, 3, 9, 5, 7, 10, 10, 7, 10, 10i.

(b) ν A (a) = (v1 + v1 )A (h1, 0, 1, 0, . . .i) = 0 + 0 = 0.

(c) a1 A = h1, 0, 1, 0, . . .i.
ν (a)
(d) A |= Subfvν1 ϕ[a] iff A |= [∀v0 (v0 · (v1 + v1 ) = v1 + v1 ][h1, 0, 1, 0, . . .i] iff for all a ∈ ω,
a · (0 + 0) = 0 + 0; this is true.
(e) A |= ϕ[a1 A ] iff A |= [∀v0 (v0 · v1 = v1 ][h1, 0, 1, 0, . . .i] iff for all a ∈ ω, a · 0 = 0;
ν (a)
this is true.
E4.6 Show that the condition in Lemma 4.6 that
no free occurrence of vi in ϕ is within a subformula of the form ∀vk µ with vk a variable
occurring in ν
is necessary for the conclusion of the lemma.
In the language for A = (ω, S, 0, +, ·), let ϕ be the formula ∃v1 [Sv1 = v0 ], ν = v1 ,
and a = h1, 1, . . .i. Note that the condition on v0 fails. Now Subfvv01 ϕ is the formula
∃v1 [Sv1 = v1 ], and there is no a ∈ ω such that Sa = a, and hence A 6|= Subfvv01 ϕ[a]. Also,
ν A (a) = v1A (a) = a1 = 1, and hence a0 A = h1, 1, . . .i. Since S0 = 1, it follows that
ν (a)
A |= ϕ[a0 A ].
ν (a)

E4.7 Let A be an L -structure, with L arbitrary. Define Γ = {ϕ : ϕ is a sentence and

A |= ϕ[a] for some a : ω → A}. Prove that Γ is complete and consistent.
Note by Lemma 4.4 that A |= ϕ[a] for some a : ω → A iff A |= ϕ[a] for every a : ω → A.
Let ϕ be any sentence. Take any a : ω → A. If A |= ϕ[a], then ϕ ∈ Γ and hence Γ ⊢ ϕ.
Suppose that A 6|= ϕ[a]. Then A |= ¬ϕ[a], hence ¬ϕ ∈ Γ, hence Γ ⊢ ¬ϕ.
This shows that Γ is complete. Suppose that Γ is not consistent. Then Γ ⊢ ¬(v0 = v0 )
by Lemma 4.1. Then Γ |= ¬(v0 = v0 ) by Theorem 3.2. Since A is a model of Γ, it is also
a model of ¬(v0 = v0 ), contradiction.
E4.8 Call a set Γ strongly complete iff for every formula ϕ, Γ ⊢ ϕ or Γ ⊢ ¬ϕ. Prove that
if Γ is strongly complete, then Γ ⊢ ∀v0 ∀v1 (v0 = v1 ).
Assume that Γ is strongly complete. Then Γ ⊢ v0 = v1 or Γ ⊢ ¬(v0 = v1 ). If Γ ⊢ v0 = v1 ,
then by generalization, Γ ⊢ ∀v0 ∀v1 (v0 = v1 ). Suppose that Γ ⊢ ¬(v0 = v1 ). Then by

13
generalization, Γ ⊢ ∀v0 ¬(v0 = v1 ). By Theorem 3.25, Γ ⊢ ∀v0 ¬(v0 = v1 ) → ¬(v1 = v1 ).
Hence Γ ⊢ ¬(v1 = v1 ). But also Γ ⊢ v1 = v1 by Proposition 3.4, so Γ is inconsistent by
Lemma 4.1, and hence again Γ ⊢ ∀v0 ∀v1 (v0 = v1 ).
E4.9 Prove that if Γ is rich, then for every term σ with no variables occurring in σ there
is an individual constant c such that Γ ⊢ σ = c.
By richness we have Γ ⊢ ∃v0 (v0 = σ) → c = σ for some individual constant c. Then using
(L4) it follows that Γ ⊢ c = σ.
E4.10 Prove that if Γ is rich, then for every sentence ϕ there is a sentence ψ with no
quantifiers in it such that Γ ⊢ ϕ ↔ ψ.
We proceed by induction on the number m of symbols ¬, →, ∀ in ϕ. (More exactly, by
the number of the integers 1,2,4 that occur in the sequence ϕ.) If m = 0, then ϕ is atomic
and we can take ψ = ϕ. Assume the result for m and suppose that ϕ has m + 1 integers
1,2,4 in it. Then there are three possibilities. First, ϕ = ¬ϕ′ . Let ψ ′ be a quantifier-free
sentence such that Γ ⊢ ϕ′ ↔ ψ ′ . Then Γ ⊢ ϕ ↔ ¬ψ ′ . Second, ϕ = (ϕ′ → ϕ′′ ). Choose
quantifier-free sentences ψ ′ and ψ ′′ such that Γ ⊢ ϕ′ ↔ ψ ′ and Γ ⊢ ϕ′′ ↔ ψ ′′ . Then
Γ ⊢ ϕ ↔ (ψ ′ → ψ ′′ ). Third, ϕ = ∀vi ϕ′ . By richness, let c be an individual constant such
that Γ ⊢ ∃vi ¬ϕ′ → Subfvc i ¬ϕ′ . Then by Theorem 3.31 we get
(1) Γ ⊢ ∃vi ¬ϕ′ ↔ Subfvc i ¬ϕ′ .
Now Subfvc i ϕ′ has only m integers 1,2,4 in it, so by the inductive hypothesis there is a
sentence ψ with no quantifiers in it such that Γ ⊢ Subfvc i ϕ′ ↔ ψ and hence
(2) Γ ⊢ Subfvc i ¬ϕ′ ↔ ¬ψ.
From (1) and (2) and a tautology we get Γ ⊢ ¬∃vi ¬ϕ′ ↔ ψ. Then by Proposition 3.31,
Γ ⊢ ∀vi ϕ′ ↔ ψ, finishing the inductive proof.
E4.11 Describe sentences in a language for ordering which say that < is a linear ordering
and there are infinitely many elements. Prove that the resulting set Γ of sentences is not
complete.
Let Γ consist of the following sentences:

¬∃v0 (v0 < v0 );

∀v0 ∀v1 ∀v2 [v0 < v1 ∧ v1 < v2 → v0 < v2 ];
∀v0 ∀v1 [v0 < v1 ∨ v0 = v1 ∨ v1 < v0 ];
^
¬(vi = vj ) for every positive integer n.
i<j<n

The following sentence ϕ holds in (Q, <) but not in (ω, <):

∀v0 ∀v1 [v0 < v1 → ∃v2 (v0 < v2 ∧ v2 < v1 )].

Since ϕ does not hold in (ω, <), we have Γ 6⊢ ϕ, by Theorem 4.2. But since ϕ holds in
(Q, <), we also have Γ 6⊢ ¬ϕ by Theorem 4.2. So Γ is not complete.

14
 E4.12 Prove that if a sentence ϕ holds in every infinite model of a set Γ of sentences,
then there is an m ∈ ω such that it holds in every model of Γ with at least m elements.
Suppose that ϕ holds in every infinite model of a set Γ of sentences, but for every m ∈ ω
there is a model M of Γ with at least m elements such that ϕ does not hold in M . Let ∆
be the following set:
 
 ^ 
Γ∪ ¬(vi = vj ) : n a positive integer ∪ {¬ϕ}.
 
i<j<n

Our hypothesis implies that every finite subset ∆′ of ∆ has a model; for if m is the
maximum of all n such that the above big conjunction is in ∆′ , then the hypothesis yields
a model of ∆′ . By the compactness theorem we get a model N of ∆. Thus N is an infinite
model of Γ in which ϕ does not hold, contradiction.
E4.13 Let L be the language of ordering. Prove that there is no set Γ of sentences whose
models are exactly the well-ordering structures.
Suppose there is such a set. Let us expand the language L to a new one L ′ by adding an
infinite sequence cm , m ∈ ω, of individual constants. Then consider the following set Θ of
sentences: all members of Γ, plus all sentences cm+1 < cm for m ∈ ω. Clearly every finite
subset of Θ has a model, so let A = (A, <, ai )i<ω be a model of Θ itself. (Here ai is the
0-ary function, i.e., element of A, corresponding to ci .) Then a0 > a1 > · · ·; so {ai : i ∈ ω}
is a nonempty subset of A with no least element, contradiction.
E4.14 Suppose that Γ is a set of sentences, and ϕ is a sentence. Prove that if Γ |= ϕ,
then ∆ |= ϕ for some finite ∆ ⊆ Γ.
We prove the contrapositive: Suppose that for every finite subset ∆ of Γ, ∆ 6|= ϕ. Thus
every finite subset of Γ ∪ {¬ϕ} has a model, so Γ ∪ {¬ϕ} has a model, proving that Γ 6|= ϕ.
E4.15 Suppose that f is a function mapping a set M into a set N . Let R = {(a, b) :
a, b ∈ M and f (a) = f (b)}. Prove that R is an equivalence relation on M .
If a ∈ M , then f (a) = f (a), so (a, a) ∈ R. Thus R is reflexive on M . Suppose that
(a, b) ∈ R. Then f (a) = f (b), so f (b) = f (a) and hence (b, a) ∈ R. Thus R is symmetric.
Suppose that (a, b) ∈ R and (b, c) ∈ R. Then f (a) = f (b) and f (b) = f (c), so f (a) = f (c)
and hence (a, c) ∈ R.
E4.16 Suppose that R is an equivalence relation on a set M . Prove that there is a function
f mapping M into some set N such that R = {(a, b) : a, b ∈ M and f (a) = f (b)}.
Let N be the collection of all equivalence classes under R. For each a ∈ M let f (a) = [a]R .
Then (a, b) ∈ R iff a, b ∈ M and [a]R = [b]R iff a, b ∈ M and f (a) = f (b).
E4.17 Let Γ be a set of sentences in a first-order language, and let ∆ be the collection of
all sentences holding in every model of Γ. Prove that ∆ = {ϕ : ϕ is a sentence and Γ ⊢ ϕ}.
For ⊆, suppose that ϕ ∈ ∆. To prove that Γ ⊢ ϕ we use the compactness theorem, proving
that Γ |= ϕ. Let A be any model of Γ. Since ϕ ∈ ∆, it follows that A is a model of Γ, as
desired.

15
 For ⊇, suppose that ϕ is a sentence and Γ ⊢ ϕ. Then by the easy direction of the
completeness theorem, Γ |= ϕ. That is, every model of Γ is a model of ϕ. Hence ϕ ∈ ∆.
Solutions, Chapter 6
E6.1 Prove that if f : A → B and hCi : i ∈ Ii is a system of subsets of A, then
S  S
f i∈I Ci = i∈I f [Ci ].

" # !
[ [
x∈f Ci iff x ∈ rng f ↾ Ci
i∈I i∈I
[
iff ∃y ∈ Ci [f (y) = x]
i∈I
iff ∃i ∈ I∃y ∈ Ci [f (y) = x]
iff ∃i ∈ I[x ∈ rng(f ↾ Ci )]
iff ∃i ∈ I[x ∈ f [Ci ]]
[
iff x∈ f [Ci ].
i∈I

E6.2 Prove that if f : A → B and C, D ⊆ A, then f [C ∩ D] ⊆ f [C] ∩ f [D]. Give an

example showing that equality does not hold in general.
Take any x ∈ f [C ∩ D]. Choose y ∈ C ∩ D such that x = f (y). Since y ∈ C, we have
x ∈ f [C]. Similarly, x ∈ f [D]. So x ∈ f [C] ∩ f [D]. Since x is arbitrary, this shows that
f [C ∩ D] ⊆ f [C] ∩ f [D].
For the required example, let dmn(f ) = {a, b} with a 6= b and with f (a) = a = f (b).
Let C = {a} and D = {b}. Then C ∩ D = ∅, so f [C ∩ D] = ∅, while f [C] = {a} = f [D]
and hence f [C] ∩ f [D] = {a} =
6 ∅. So f [C ∩ D] 6= f [C] ∩ f [D].
E6.3 Given f : A → B and C, D ⊆ A, compare f [C\D] and f [C]\f [D]: prove the
inclusions (if any) which hold, and give counterexamples for the inclusions that fail to
hold.
We claim that f [C]\f [D] ⊆ f [C\D]. For, suppose that x ∈ f [C]\f [D]. Choose c ∈ C
such that x = f (c). Since x ∈
/ f [D], we have c ∈
/ D. So c ∈ C\D and hence x ∈ f [C\D],
proving the claim.
The other inclusion does not hold. For, take the same f, C, D as for exercise E6.2.
Then C\D = {a} and so f [C\D] 6= ∅. But f [C] = {a} = f [D], so f [C]\f [D] = ∅.
E6.4 Prove that if f : A → B and hCi : i ∈ Ii is a system of subsets of B, then
S  S
f −1 i∈I Ci = i∈I f −1 [Ci ].
For any b ∈ B we have
" #
[ [
b ∈ f −1 Ci iff f (b) ∈ Ci
i∈I i∈I
iff ∃i ∈ I[f (b) ∈ Ci ]

16
 iff ∃i ∈ I[b ∈ f −1 [Ci ]]
[
iff b∈ f −1 [Ci ].
i∈I

E6.5 Prove that if f : A → B and hCi : i ∈ Ii is a system of subsets of B, then

T  T
f −1 i∈I Ci = i∈I f −1 [Ci ].
For any a,
" #
\ \
a ∈ f −1 Ci iff f (a) ∈ Ci
i∈I i∈I
iff ∀i ∈ I[f (a) ∈ Ci ]
iff ∀i ∈ I[a ∈ f −1 [Ci ]]
\
iff a∈ f −1 [Ci ].
i∈I

E6.6 Prove that if f : A → B and C, D ⊆ B, then f −1 [C\D] = f −1 [C]\f −1 [D].

For any a,

a ∈ f −1 [C\D] iff f (a) ∈ C\D

iff f (a) ∈ C and f (a) ∈
/D
iff a ∈ f −1 [C] and a ∈/ f −1 [D]
iff a ∈ f −1 [C]\f −1 [D].

E6.7 Prove that if f : A → B and C ⊆ A, then

{b ∈ B : f −1 [{b}] ⊆ C} = B\f [A\C].

First suppose that b is in the left side; but suppose also, aiming for a contradiction, that
b ∈ f [A\C]. Say b = f (a), with a ∈ A\C. Then a ∈ f −1 [{b}], so a ∈ C, contradiction.
Second, suppose that b is in the right side. Take any a ∈ f −1 [{b}]. Then f (a) = b,
and it follows that a ∈ C, as desired.
E6.8 For any sets A, B define A△B = (A\B) ∪ (B\A); this is called the symmetric
difference of A and B. Prove that if A, B, C are given sets, then A△(B△C) = (A△B)△C.
Let D = A ∪ B ∪ C, A′ = D\A, B ′ = D\B, and C ′ = D\C. Then

A△B = (A ∩ B ′ ) ∪ (B ∩ A′ );
(A△B)′ = ((A ∩ B ′ ) ∪ (B ∩ A′ ))′
= (A ∩ B ′ )′ ∩ (B ∩ A′ )′
= (A′ ∪ B) ∩ (B ′ ∪ A)
= (A′ ∩ B ′ ) ∪ (A ∩ B).

17
These equations hold for any sets A, B. Now

A△(B△C) = (A ∩ (B△C)′ ) ∪ ((B△C) ∩ A′

= (A ∩ ((B ′ ∩ C ′ ) ∪ (B ∩ C))) ∪ (((B ∩ C ′ ) ∪ (C ∩ B ′ )) ∩ A′ )
= (A ∩ B ′ ∩ C ′ ) ∪ (A ∩ B ∩ C) ∪ (A′ ∩ B ∩ C ′ ) ∪ (A′ ∩ B ′ ∩ C).

This holds for any sets A, B, C. Hence

(A△B)△C = C△(A△B)
= (C ∩ A′ ∩ B ′ ) ∪ (C ∩ A ∩ B) ∪ (C ′ ∩ A ∩ B ′ ) ∪ (C ′ ∩ A′ ∩ B)
= A△(B△C).

E6.9 For any set A let

IdA = {hx, xi : x ∈ A}.
Justify this definition on the basis of the axioms.

IdA = {y ∈ A × A : ∃x ∈ A[y = hx, xi]}.

E6.10 Suppose that f : A → B. Prove that f is surjective iff there is a g : B → A such

that f ◦ g = IdB . Note: the axiom of choice might be needed.
⇐: given b ∈ B, we have b = (f ◦ g)(b) = f (g(b)); so f is surjective.
⇒: Assume that f is surjective. Let

A = {{(b, a) : a ∈ A, f (a) = b} : b ∈ B}.

Each member of A is nonempty; for let x ∈ A . Choose b ∈ B such that x = {(b, a) : a ∈

A, f (a) = b}. Choose a ∈ A such that f (a) = b. So (b, a) ∈ x.
The members of A are pairwise disjoint: suppose x, y ∈ A with x 6= y. Choose b, c
so that x = {(b, a) : a ∈ A, f (a) = b} and y = {(b, a) : a ∈ A, f (a) = c}. If u ∈ x ∩ y, then
there exist a, a′ ∈ A such that u = (b, a), f (a) = b, and also (u = (c, a′ ), f (a′ ) = c. So by
Theorem 6.3, b = c. But then x = y, contradiction.
Now by the axiom of choice, let C have exactly one element in common with each
member of A . Then define

g = {(b, a) ∈ C : a ∈ A, f (a) = b}.

Now g is a function. For, suppose that (b, a), (b, a′) ∈ g. Let x = {(b, a′′ ) : a′′ ∈ A, f (a′′ ) =
b}. Then (b, a), (b, a′) ∈ C ∩ x, so (b, a) = (b, a′ ). Hence a = a′ .
Clearly g ⊆ B × A. Next, dmn(g) = B, for suppose that b ∈ B. Choose x ∈
C ∩ {(b, a′′ ) : a′′ ∈ A, f (a′′) = b}; say x = (b, a) with a ∈ A, f (a) = b. Then x ∈ g and so
b ∈ dmn(g).
Thus g : B → A. Take any b ∈ B, and let g(b) = a. So (b, a) ∈ g and hence f (a) = b.
So f ◦ g = IdB .

18
 E6.11 Let A be a nonempty set. Suppose that f : A → B. Prove that f is injective iff
there is a g : B → A such that g ◦ f = IdA .
First suppose that f is injective. Fix a ∈ A, and let

g = f −1 ∪ {(b, a) : b ∈ B\rng(f )}.

Then g is a function. In fact, suppose that (b, c), (b, d) ∈ g. If both are in f −1 , then
(c, b(, (d, b) ∈ f , so f (c) = b = f (d) and hence c = d since f is injective. If (b, c) ∈ f −1
/ f −1 , then
and b ∈ B\rng(f ), the (c, b) ∈ f , so b ∈ rng(f ), contradiction. If (b, c), (b, d) ∈
c = d = a.
Clearly then g : B → A. For any a ∈ A we have (a, f (a)) ∈ f , hence (f (a), a) ∈ f −1 ⊆
g, and so g(f (a)) = a.
Second, suppose that g : B → A and g ◦ f = IdA . Suppose that f (a) = f (a′ ). Then
a = (g ◦ f )(a) = g(f (a)) = g(f (a′)) = (g ◦ f )(a′ ) = a′ .
E6.12 Suppose that f : A → B. Prove that f is a bijection iff there is a g : B → A such
that f ◦ g = IdB and g ◦ f = IdA . Prove this without using the axiom of choice.
⇒: Assume that f is a bijection. By E6.11 there is a g : B → A such that g ◦ f = IdA .
We claim that f ◦ g = IdB . Since f is a bijection, the relation f −1 is also a bijection. Now
for any b ∈ B,

(f ◦ g)(b) = f (g(b)) = f (g(f (f −1(b)))) = f ((g ◦ f )(f −1 (b))) = f (f −1 (b)) = b.

So f ◦ g = IdB , as desired.
⇐: Assume that g is as indicated. Then f is injective, since f (a) = f (b) implies
that a = g(f (a)) = g(f (a′ )) = a′ . And f is surjective, since for a given b ∈ B we have
f (g(b)) = b.
E6.13 For any sets R, S define

R|S = {(x, z) : ∃y((x, y) ∈ R ∧ (y, z) ∈ S)}.

Justify this definition on the basis of the axioms.

R|S = {(x, z) ∈ dmn(R) × rng(S) : ∃y((x, y) ∈ R ∧ (y, z) ∈ S)}.

E6.14 Suppose that f, g : A → A. Prove that

(A × A)\[((A × A)\f )|((A × A)\g)]

is a function.
Suppose that (x, y), (x, z) are in the indicated set, with y 6= z. By symmetry say f (x) 6= y.
Then (x, y) ∈ [(A × A)\f ], so it follows that (y, z) ∈ g, as otherwise (x, z) ∈ [((A ×
A)\f )|((A × A)\g)]. Hence (y, y) ∈ / g, so (x, y) ∈ [((A × A)\f )|((A × A)\g)], contradiction.

19
 E6.15 Suppose that f : A → B is a surjection, g : A → C, and ∀x, y ∈ A[f (x) = f (y) →
g(x) = g(y)]. Prove that there is a function h : B → C such that h ◦ f = g. Define h as a
set of ordered pairs.
Let h = {(f (a), g(a)) : a ∈ A}. Then h is a function, for suppose that (x, y), (x, z) ∈ h.
Choose a, a′ ∈ A so that x = f (a), y = g(a), x = f (a′ ), and y = g(a′ ). Thus f (a) = f (a′ ),
so g(a) = g(a′ ), as desired.
Since f is a surjection it is clear that dmn(h) = B. Clearly rng(h) ⊆ C. So h : B → C.
If a ∈ A, then (f (a), b(a)) ∈ h, hence h(f (a)) = g(a). This shows that h ◦ f = g.
E6.16 The statement
S T
∀A ∈ A ∀B ∈ B(A ⊆ B) implies that A ⊆ B
is slightly wrong. Fix it, and prove the result.
If A has a nonempty member and B is empty, the implication does not hold. Add the
hypothesis B 6= ∅. S
Suppose that a ∈ A and B ∈ B; we want to show that a ∈ B. Choose A ∈ A such
that a ∈ A. Since A ⊆ B, we have a ∈ B.
S S
E6.17 Suppose that ∀A ∈ A ∃B ∈ B(A ⊆ B). Prove that A ⊆ B.
S
Suppose that a ∈ A ; we want to show that a ∈ B. ChooseSA ∈ A such that a ∈ A.
Then choose B ∈ B such that A ⊆ B. Then a ∈ B. Hence a ∈ B.
E6.18 The statement
T T
∀A ∈ A ∃B ∈ B(B ⊆ A) implies that B⊆ A.
is slightly wrong. Fix it, and prove the result.
T
If A is empty and B is nonempty, the statement is false. Fix it by adding the hypothesis
that A is nonempty. T
Suppose that b ∈ B and A T ∈ A ; we want to show that b ∈ A. Choose B ∈ B such
that B ⊆ A. Now b ∈ B since b ∈ B, so b ∈ A.
Solutions, exercises in Chapter 7
E7.1 Prove that if x is an ordinal, then x is transitive and (x, {(y, z) ∈ x × x : y ∈ z}) is
a well-ordered set.
By definition, x is transitive. Let R = {(y, z) ∈ x × x : y ∈ z}). Obviously R is a relation.
By definition, R ⊆ x × x. R is irreflexive on x by Theorem 7.5. R is transitive since x is
transitive. R is linear on x by Theorem 7.7. The final well-ordering property follows from
Theorem 7.13.
E7.2 Assume that x is transitive and (x, {(y, z) ∈ x × x : y ∈ z}) is a well-ordered set.
Prove that for all y, z ∈ x, either y = z or y ∈ z or z ∈ y.
This is obvious.
E7.3 Assume that x is transitve and for all y, z ∈ x, either y = z or y ∈ z or z ∈ y.
Prove that for all y, if y ⊂ x and y is transitive, then y ∈ x. Hint: apply the foundation
axiom to x\y.

20
Assume the hypothesis, and suppose that y ⊂ x and y is transitive. Choose z ∈ x\y such
that z ∩ (x\y) = ∅. If u ∈ y, then u ∈ x since y ⊂ x. So u, z ∈ x, so by hypothesis we have
u ∈ z, u = z, or z ∈ u. Now u 6= z since z ∈ / y and u ∈ y. And z ∈ / u, since z ∈ u would
imply, because y is transitive and u ∈ y, that z ∈ y, which is not true. Hence u ∈ z. This
is true for any u ∈ y. So y ⊆ z. Clearly also z ⊆ y, so y = z ∈ x.
E7.4 Assume that x is transitive and for all y, if y ⊂ x and y is transitive, then y ∈ x.
Show that x is an ordinal. Hint: let y = {z ∈ x : z is an ordinal}, and get a contradiction
from the assumption that y ⊂ x.
Assume the hypothesis. Let y = {z ∈ x : z is an ordinal}. So y ⊆ x. Suppose that y ⊂ x.
Now y is transitive, for assume that z ∈ y. Thus z ∈ x and z is an ordinal. Suppose that
w ∈ z. Then w ∈ x since x is transitive, and w is an ordinal since z is an ordinal. So
w ∈ y. Thus, indeed, y is transitive. So by assumption y ∈ x. Now y is a transitive set
of transitive sets, so y is an ordinal. It follows that y ∈ y, contradiction. This proves that
x = y. So x is a transitive set of transitive sets, and hence x is an ordinal.
E7.5 Show that if x is an ordinal, then the following two conditions hold:
(i) For all y ∈ x, either y ∪ {y} = x or y ∪ {y} ∈ x.
S S
(ii) For all y ⊆ x, either y = x or y ∈ x.
Assume S that x is an ordinal. Then (i) holds by Proposition 7.10. Now suppose that y ⊆ x.
If z ∈ y, choose S w ∈ y such that z ∈ w. Then S also w ∈ x, so z ∈ x since x is transitive.
This
S shows that y ⊆ x. By Proposition 7.3, y is an ordinal. Hence by Proposition 7.8,
y ≤ x.
E7.6 Assume the two conditions of exercise E7.5. Show that x is an ordinal. Hint: Show
that there is an ordinal α not in x. Taking such an ordinal α, show that there is a least
β ∈ α ∪ {α} such that β ∈/ x. Work with such a β to show that x is an ordinal.
By Theorem 7.6 there is an ordinal α not in x. Then by Theorem 7.13 there is a least
β ∈ α ∪ {α} such that β ∈
/ x. Now we have two possibilities:
S S
Case
S 1. β = β. Now β ⊆ x, so by (ii) second clause, since β =β ∈
/ x we have
x = β, hence x is an ordinal, as desired.
S S
Case S
2. β = ( β) + 1. Thus Sβ is an ordinal smaller than β, so it is in x. By (i),
since β = β + 1 ∈ / x we have x = ( β) + 1, hence x is an ordinal.
Solutions, exercises in Chapter 8
E8.1 Give an example of A, R such that R is not well-founded on A and is not set-like
on A.
We take On and R, where R = {(α, β) : α > β}. As shown after 8.2, R is not set-like on
On. It is also not well-founded on On, since ω is a nonempty set of ordinals, but if m ∈ ω
then (m +′ 1, m) ∈ R, so that ω does not have an R-minimal element.
E8.2 Give an example of A, R such that R is not well-founded on A but is set-like on
A. Give one example with R and A are proper classes, and one example where they are
sets.

21
Both are sets: let A = ω and R = {(m, n) : m, n ∈ ω and m > n. Then ω does not have
an R-minimal element, since for any m ∈ ω we have (m +′ 1, m) ∈ R.
Both are proper classes: let A = On and let
R = {(m, n) : m, n ∈ ω and m > n} ∪ {(α, β) : α < β}.
E8.3 Give an example of A, R such that R is well-founded on A but is not set-like on
A.
Let A = V and R = {(a, ∅) : a ∈ V, a 6= ∅}. Thus predAR (∅) = V, so R is not set-like
on V. Now let X be a nonempty set. If X = {∅}, then ∅ ∈ X and ∀a ∈ X[(a, ∅) ∈
/ R]. If
X 6= {∅}, take any a ∈ X\{∅}. Then ∀b ∈ X[(b, a) ∈
/ R].
E8.4 Suppose that R is a class relation contained in A × A, x ∈ A, and v ∈ predAR∗ (x).
Prove by induction on n that if n ∈ ω\1, f is a function with domain n +′ 1, ∀i <
n[(f (i), f (i +′ 1)) ∈ R] and f (n) = v, then f (0) ∈ predAR∗ (x).
Suppose that R is a class relation contained in A × A, x ∈ A, and v ∈ predAR∗ (x).
We take n = 1 in the condition to be proved. So, suppose that f is a function with
domain 2 such that ∀i < 1[(f (i), f (i +′ 1)) ∈ R] and f (1) = v. Thus (f (0), v) ∈ R, so
f (0) ∈ predAR (v). By Lemma 8.3(ii), f (0) ∈ predAR∗ (x).
Now suppose that if n ∈ ω\1, f is a function with domain n +′ 1, ∀i < n[(f (i), f (i +′
1)) ∈ R] and f (n) = v, then f (0) ∈ predAR∗ (x). Suppose also now that f is a function
with domain n +′ 2, ∀i < n +′ 1[(f (i), f (i +′ 1)) ∈ R] and f (n +′ 1) = v. Define g with
domain n +′ 1 by setting g(i) = f (i +′ 1) for all i < n +′ 1. Then ∀i < n[(g(i), g(i +′
1)) = (f (i +′ 1), f (i +′ 2)) ∈ R] and g(n) = f (n +′ 1) = v. Hence by the inductive
assumption, f (1) = g(0) ∈ predAR∗ (x). We also have (f (0), f (1)) ∈ R, so by Lemma
8.3(ii), f (0) ∈ predAR∗ (x).
E8.5 Suppose that R is a class relation contained in A×A, (u, v) ∈ R∗ , and (v, w) ∈ R∗ .
Show that (u, w) ∈ R∗ .
Assume that R is a class relation contained in A × A, (u, v) ∈ R∗ , and (v, w) ∈ R∗ .
Since (u, v) ∈ R∗ , there exist n ∈ ω\1 and a function f with domain n +′ 1 such that
∀i < n[(f (i), f (i +′ 1)) ∈ R, f (0) = u, and f (n) = v. From exercise E8.4 it follows that
(u, w) ∈ R∗ .
E8.6 Give an example of a proper class X which has a proper class of ∈-minimal elements.
Let X = {{α} : α ≥ 2}. We claim that all elements of X are ∈-minimal. Suppose that
α, β ≥ 2 and {α} ∈ {β}. Then {α} = β, Since β ≥ 2 we have 0, 1 ∈ β, so 0 = α = 1,
contradiction.
E8.7 Give an example of a proper class relation R contained in A × A for some proper
class A, and a class function G mapping A × V into V such that R is set-like on A
but not well-founded on A and there is no class function F mapping A into V such that
F(a) = G(a, FhpredAR (a)) for all a ∈ A.
Let A = On and
R = {(m, n) : m, n ∈ ω and m > n} ∪ {(α, β) : ω ≤ α < β}.

22
Thus R is a proper class relation contained in A × A. Clearly R is set-like on On but it
is not well-founded on On. Define G : On × V → V by setting

{a(α +′ 1)} if α ∈ ω and a is a function with domain {m ∈ ω : m > α},
G(α, a) =
∅ otherwise.

Suppose that F : A → V is such that F(α) = G(α, F ↾ predOnR (α)) for all α ∈ On. Let
f = F ↾ ω. Choose b ∈ rng(f ) such that b ∩ rng(f ) = ∅. Say b = f (m) with m ∈ ω. Now

f (m) = F(m) = G(m, F ↾ predOnR (m)) = G(m, F ↾ {n : n ∈ ω, n > m})

= {F(m +′ 1)} = {f (m +′ 1)},

so that f (m +′ 1) ∈ f (m) ∩ rng(f ), contradiction.

E8.8 Give an example of a proper class relation R contained in some A × A for some
proper class A and a class function G mapping A × V into V such that R is set-like on
A but not well-founded on A but still there is a class function F mapping A into V such
that F(a) = G(a, FhpredAR (a)) for all a ∈ A.
Let A and R be as in exercise E8.7, but define G(α, a) = α for all α ∈ On and all a ∈ V.
Then the function F : On → V such that F(α) = α for all α ∈ On is as desired.
Solutions to exercises in chapter 9
E9.1 Let (A, <) be a well order. Suppose that B ⊂ A and ∀b ∈ B∀a ∈ A[a < b → a ∈ B].
Prove that there is an element a ∈ A such that B = {b ∈ A : b < a}.
Let a be the least element of A\B. We claim that a is as desired. For, if b ∈ B, then it
cannot happen that a ≤ b, since this would imply that a ∈ B; so b < a. And if b < a, then
b ∈ B by the minimality of a.
E9.2 Let (A, <) be a well order. Suppose that B ⊂ A and ∀b ∈ B∀a ∈ A[a < b → a ∈ B].
Prove that (A, <) is not isomorphic to (B, <).
Suppose that f is such an isomorphism from (A, <) onto (B, <). By exercise E9.1, let
a ∈ A be such that B = {x ∈ A : x < a}. By Proposition 9.11, a ≤ f (a), contradicting
the assumption that f maps into B.
E9.3 Suppose that f is a one-one function mapping an ordinal α onto a set A. Define
a relation ≺ which is a subset of A × A such that (A, <) is a well-order and f is an
isomorphism of (α, <) onto (A, ≺).
Define ≺= {(a, b) ∈ A × A : f −1 (a) < f −1 (b)}. We check that (A, <) is a well-order.
If a ∈ A and a ≺ a, then f −1 (a) < f −a (a), contradiction. So ≺ is irreflexive. Suppose
that a ≺ b ≺ c. Then f −1 (a) < f −1 (b) < f −1 (c), so f −1 (a) < f −1 (c) and hence a ≺ c.
So ≺ is transitive. Now given a, b ∈ A, either f −1 (a) < f −1 (b) or f −1 (a) = f −1 (b) or
f −1 (b) < f −1 (a), so a ≺ b or a = b or b ≺ a. Thus (A, ≺) is a linear order. Finally,
suppose that ∅ 6= X ⊆ A. Then ∅ = 6 f −1 [X], so let ξ be the least element of f −1 [X].
Then f (ξ) ∈ X. Suppose that b ∈ X. Then f −1 (b) ∈ f −1 [X], so ξ ≤ f −1 (b). Hence

23
f (ξ)  b. This shows that f (ξ) is the ≺-least element of X. We have shown that (A, ≺) is
a well-order.
We are given that f is a bijection from α onto A. If ξ, η ∈ α and ξ < η, then
f (ξ) ≺ f (η). If f (ξ) ≺ f (η), then ξ < η. Thus f is an isomorphism.
E9.4 Prove that 1 + m = m + 1 for any m ∈ ω.
(Ordinary) induction on m. 0 + 1 = 1 = 1 + 0 using Theorem 9.21(vi). Assume that
1 + m = m + 1. Then 1 + (m + 1) = (1 + m) + 1 = (m + 1) + 1.
E9.5 Prove that m + n = n + m for any m, n ∈ ω.
With m fixed, induction on n. 0 + m = m = m + 0 using Theorem 9.21(vi). Assume
that m + n = n + m. Then (n + 1) + m = n + (1 + m) = n + (m + 1) (by exercise E9.4)
= (n + m) + 1 = (m + n) + 1 = m + (n + 1).
E9.6 Prove that ω ≤ α iff 1 + α = α.
S S
First note that 1 + ω = m∈ω (1 + m) = m∈ω (m + 1) = ω, using Theorem 9.21(vi).
⇒: Assume that ω ≤ α. By Theorem 9.21(vii) let δ be such that ω + δ = α. Then
1 + α = 1 + (ω + δ) = (1 + ω) + δ = ω + δ = α.
⇐: It suffices to show that if m < ω then 1 + m 6= m. This is true by Theorem
9.21(vi).
E9.7 For any ordinals α, β let

α ⊕ β = (α × {0}) ∪ (β × {1}).

We define a relation ≺ as follows. For any x, y ∈ α ⊕ β, x ≺ y iff one of the following

three conditions holds:
(i) There are ξ, η < α such that x = (ξ, 0), y = (η, 0), and ξ < η.
(ii) There are ξ, η < β such that x = (ξ, 1), y = (η, 1), and ξ < η.
(ii) There are ξ < α and η < β such that x = (ξ, 0) and y = (η, 1).
Prove that (α ⊕ β, ≺) is a well order which is isomorphic to α + β.
Clearly ≺ is a well-order. We show by transfinite induction on β, with α fixed, that
(α ⊕ β, ≺) is order isomorphic to α + β. For β = 0 we have α + β = α + 0 = α, while
α ⊕ β = α ⊕ 0 = α × {0}. Clearly ξ 7→ (ξ, 0) defines an order-isomorphism from α onto
(α × {0}, ≺). So our result holds for β = 0. Assume it for β, and suppose that f is an
order-isomorphism from α + β onto (α ⊕ β, ≺). Now the last element of α ⊕ (β + 1) is
(β, 1), and the last element of α + (β + 1) is α + β, so the function

f ∪ {(α + β, (β, 1))}

is an order-isomorphism from α + (β + 1) onto α ⊕ (β + 1).

Now assume that β is a limit ordinal, and for each γ < β, the ordinal α + γ is
isomorphic to α ⊕ γ. For each such γ let fγ be the unique isomorphism from α + γ onto

24
α ⊕ γ. Note that if γ < δ < β, then fδ ↾ γ is an isomorphism from α + γ onto α ⊕ γ; hence
fδ ↾ γ = fγ . It follows that [
fγ
γ<β

is an isomorphism from α + β onto α ⊕ β, finishing the inductive proof.

E9.8 Given ordinals α, β, we define the following relation ≺ on α × β:

(ξ, η) ≺ (ξ ′ , η ′ ) iff ((ξ, η) and (ξ ′ , η ′ ) are in α × β and:

η < η ′ , or (η = η ′ and ξ < ξ ′ ).

We may say that this is the anti-dictionary or anti-lexicographic order.

Show that the set α × β under the anti-lexicographic order is a well order which is
isomorphic to α · β.
We may assume that α 6= 0. It is straightforward to check that ≺ is a well-order.
Now we define, for any (ξ, η) ∈ α × β,

f (ξ, η) = α · η + ξ.

We claim that f is the desired order-isomorphism from α × β onto α · β. If (ξ, η) ∈ α × β,

then
f (ξ, η) = α · η + ξ < α · η + α = α · (η + 1) ≤ α · β.
Thus f maps into α · β.
To show that f is one-one, suppose that (ξ, η), (ξ ′, η ′ ) ∈ α × β and f (ξ, η) = f (ξ ′ , η ′ ).
Then by Theorem 9.26, (ξ, η) = (ξ ′ , η ′ ). So f is one-one.
To show that f maps onto α · β, let γ < α · β. Choose ξ and η so that γ = α · η + ξ
with ξ < α. Now η < β, as otherwise

γ = α · η + ξ ≥ α · η ≥ α · β.

It follows that f (ξ, η) = α · η + ξ = γ. so f is onto.

Finally, we show that the order is preserved. Suppose that (ξ, η) ≺ (ξ ′ , η ′ ). Then one
of these cases holds:
Case 1. η < η ′ . Then

f (ξ, η) = α · η + ξ < α · η + α = α · (η + 1) ≤ α · η ′ ≤ α · η ′ + ξ ′ = f (ξ ′ , η ′ ),

as desired.
Case 2. η = η ′ and ξ < ξ ′ . Then f (ξ, η) < f (ξ ′ , η ′ ).
Now it follows that f is the desired isomorphism.
E9.9 Suppose that α and β are ordinals, with β 6= 0. We define
α w
β = {f ∈ α β : {ξ < α : f (ξ) 6= 0} is finite}.

25
For f, g ∈ α β w we write f ≺ g iff f 6= g and f (ξ) < g(ξ) for the greatest ξ < α for which
f (ξ) 6= g(ξ).
Prove that (α β w , ≺) is a well-order which is order-isomorphic to the ordinal exponent
β α . (A set X is finite iff there is a bijection from some natural number onto X.)
If α = 0, then β α = 1, and α β w also has only one element, the empty function (= the
emptyset). So, assume that α 6= 0. If β = 1, then α β w has only one member, namely the
function with domain α whose value is always 0. This is clearly order-isomorphic to 1, as
desired. So, suppose that β > 1.
Now we define a function f mapping β α into α β w . Let f (0) be the member of α β w
which takes only the value 0. Now suppose that 0 < ε < β α . By Theorem 9.29 write

ε = β γ(0) · δ(0) + β γ(1) · δ(1) + · · · + β γ(m−1) · δ(m − 1),

where ε ≥ γ(0) > γ(1) > · · · > γ(m − 1) and 0 < δ(i) < β for each i < m. Note that
β γ(0) ≤ ε < β α , so γ(0) < α. Then we define, for any ζ < α,

0 if ζ ∈
/ {γ(0), . . . , γ(m − 1)},
(f (ε))(ζ) =
δ(i) if ζ = γ(i) with i < m.

Clearly f (ε) ∈ α β w . To see that f maps onto α β w , suppose that x ∈ α β w . If x takes only
the value 0, then f (0) = x. Suppose that x takes on some nonzero value. Let

{ξ < α : x(ξ) 6= 0} = {γ(0), γ(1), . . . , γ(m − 1)},

where γ(0) > γ(1) > · · · > γ(m − 1). Let δ(i) = x(γ(i)) for each i < m, and let

ε = β γ(0) · δ(0) + β γ(1) · δ(1) + · · · + β γ(m−1) · δ(m − 1).

Clearly then f (ε) = x.

Now we complete the proof by showing that for any ε, θ < β α , ε < θ iff f (ε) < f (θ).
This equivalence is clear if one of ε, θ is 0, so suppose that both are nonzero. Write

ε = β γ(0) · δ(0) + β γ(1) · δ(1) + · · · + β γ(m−1) · δ(m − 1),

where α ≥ γ(0) > γ(1) > · · · > γ(m − 1) and 0 < δ(i) < β for each i < m, and
′ ′ ′
θ = βγ (0)
· δ ′ (0) + β γ (1)
· δ ′ (1) + · · · + β γ (n−1)
· δ ′ (n − 1),

where α ≥ γ ′ (0) > γ ′ (1) > · · · > γ ′ (n − 1) and 0 < δ ′ (i) < β for each i < n.
By symmetry we may suppose that m ≤ n. Note that N (β, m, γ, δ), k(β, m, γ, δ) = ε,
N (β, n, γ ′ , δ ′ ), and k(β, n, γ ′ , δ ′ ) = θ. We now consider several possibilities.
Case 1. ε = θ. Then clearly f (ε) = f (θ).
Case 2. γ ⊆ γ ′ , δ ⊆ δ ′ , and m < n. Thus ε < θ. Also, γ ′ (m) is the largest ξ < α such
that (f (ε))(ξ) 6= (f (θ))(ξ), and (f (ε))(ξ) = 0 < δ ′ (m) = (f (θ))(γ ′(m)), so f (ε) < f (θ).

26
 Case 3. There is an i < m such that γ(j) = γ ′ (j) and δ(j) = δ ′ (j) for all j < i, while
γ(i) 6= γ ′ (i). By symmetry, say that γ(i) < γ ′ (i). Then we have ε < θ. Since γ ′ (i) is the
largest ξ < α such that (f (ε))(ξ) 6= (f (θ))(ξ), and (f (ε))(γ ′(i)) = 0 < δ ′ (i) = (f (θ))(γ ′(i)),
we also have f (ε) < f (θ).
Case 4. There is an i < m such that γ(j) = γ ′ (j) and δ(j) = δ ′ (j) for all j < i, while
γ(i) = γ ′ (i) and δ(i) 6= δ ′ (i). By symmetry, say that δ(i) < δ ′ (i). Then we have ε < θ.
Since γ(i) is the largest ξ < α such that (f (ε))(ξ) 6= (f (θ))(ξ), and (f (ε))(γ ′ (i)) = δ(i) <
δ ′ (i) = (f (θ))(γ ′(i)), we also have f (ε) < f (θ).
E9.10 Show that for every nonzero ordinal α there are only finitely many ordinals β such
that α = γ · β for some γ.
Suppose there are infinitely many such β; let hβi : i ∈ ωi be a one-one enumeration of
infinitely many of them. For each i ∈ ω let γi be such that α = γi · βi . Clearly βi < βj
iff γj < γi . We define hij : j ∈ ωi by recursion. Let i0 be such that βi0 is the smallest
element of {βk : k ∈ ω}. Having defined i0 , . . . , is , let is+1 be such that βis+1 is the smallest
element of
{βk : k ∈ ω}\{βik : k ≤ s}
Clearly βi0 < βi1 < · · ·, and hence γi0 > γi1 > · · ·, contradiction.
ω ω
E9.11 Prove that n(ω )
= ω (ω )
for every natural number n > 1.
Note that nω = ω by an easy argument. Hence
ω ω
ω (ω )
= (nω )(ω )
ω
= n(ω·(ω ))
ω
= n(ω ) . by Theorem 9.32

E9.12 Show that the following conditions are equivalent for any ordinals α, β:
(i) α + β = β + α.
(ii) There exist an ordinal γ and natural numbers k, l such that α = γ · k and β = γ · l.
⇒: Assume that α + β = β + α. The desired conclusion is clear if α = 0 or β = 0, so
assume that α, β 6= 0. Write α = ω δ · k + ε with δ ≤ α, 0 < k ∈ ω, and ε < ω δ , and write
β = ω ρ · l + σ with ρ ≤ β, 0 < l ∈ ω, and σ < ω ρ . If δ < ρ, then

α + β = β < β + α,

contradiction. A similar contradiction is reached if ρ < δ. So δ = ρ. Now

α + β = ω δ · (k + l) + σ = β + α = ω δ · (k + l) + ε,

so σ = ε. Hence α = (ω δ + ε) · k and β = (ω δ + ε) · l, as desired.

⇐: Obvious.
E9.13 Suppose that α < ω γ . Show that α + β + ω γ = β + ω γ .

27
Suppose that α, β, γ are ordinals and α < ω γ . If also β < ω γ , then α + β < ω γ by Theorem
9.31, and also by Theorem 9.31 α + β + ω γ = ω γ and β + ω γ = ω γ .
Now suppose that ω γ ≤ β. Write β = ω γ · δ + ε with δ > 0 and ε < ω γ .
(1) α + ω γ · ϕ = ω γ · ϕ for every positive ϕ.
We prove (1) by induction on ϕ. It is true for ϕ = 1 by Theorem 9.31. Assume that it
holds for ϕ. Then

α + ω γ · (ϕ + 1) = α + ω γ · ϕ + ω γ = ω γ · ϕ + ω γ = ω γ · (ϕ + 1),

as desired. Finally, assume that ϕ is limit and (1) holds for all ψ < ϕ. Let F (ϕ) = α + ϕ
for all ϕ, and G(ϕ) = ω γ · ϕ. Both of these are normal functions. Hence
[ [ [
α + ω γ · ϕ = F (G(ϕ)) = F (G(ψ)) = (α + ω γ · ψ) = (ω γ · ψ) = ω γ · ϕ,
ψ<ϕ ψ<ϕ ψ<ϕ

finishing the inductive proof of (1).

Now by (1) we have

α + β + ωγ = α + ωγ · δ + ε + ωγ = ωγ · δ + ε + ωγ = β + ωγ .

E9.14 Show that the following conditions are equivalent:

(i) α is a limit ordinal
(ii) α = ω · β for some β 6= 0.
(iii) For every m ∈ ω\1 we have m · α = α, and α 6= 0.
(i)⇒(ii): Assume (i). By Theorem 9.26 write α = ω · β + n with n < ω. If β = 0, then
α = n, contradiction. If n 6= 0, then α = ω · β + (n − 1) + 1, contradiction.
(ii)⇒(iii):
S Assume (ii). By Theorem 9.23(iii), α 6= 0. Suppose that m ∈ ω\1. Then
m · ω = n∈ω (m · n) = ω by Theorem 9.23(iii), so m · α = α.
(iii)⇒(i): Assume (iii), but suppose that α = β + 1. Then α = 2 · α = 2 · (β + 1) =
2 · β + 2 > α, contradiction.
E9.15 Show that (α + β) · γ ≤ α · γ + β · γ for any ordinals α, β, γ.

Assume that α, β.γ 6= 0. Write α = ω δ · k + ε with δ ≤ α, 0 6= k ∈ ω, ε < ω δ , and

β = ω ρ · l + σ with ρ ≤ β, 0 6= l ∈ ω, σ < ω ρ . Also, write γ = ω · ξ + m with m ∈ ω. Now
we consider some cases.
Case 1. δ < ρ. Then α + β = β, and the desired conclusion follows.
Case 2. δ = ρ. Note that if m > 0, then

α · m = ω δ · k · m + ε;
β · m = ω δ · l · m + σ;
(α + β) · m = ω δ · (k + l) · m + σ.

28
If ξ = 0 it is then clear that (α + β) · γ = α · γ + β · γ. Hence assume that ξ > 0. Then

α·γ =α·ω·ξ+α·m
= ω δ+1 · ξ + α · m;
β · γ = ω δ+1 · ξ + β · m;
α · γ + β · γ = ω δ+1 · ξ · 2 + β · m;
(α + β) · γ = ω δ+1 · ξ + (α + β) · m,

and the desired conclusion is clear.

Case 3 ρ < δ. Then if m > 0 we have

α · m = ω δ · k · m + ε;
β · m = ω ρ · l · m + σ;
α · m + β · m = ω δ · k · m + ε + ω ρ · l · m + σ;
(α + β) · m = ω δ · k · m + ε + ω ρ · l + σ.

Hence the desired conclusion follows if ξ = 0. Assume now that ξ 6= 0. Then

α · γ = ω δ+1 · ξ + α · m;
β · γ = ω ρ+1 · ξ + β · m;
α · γ + β · γ = ω δ+1 · ξ + α · m + ω ρ+1 · ξ + β · m;
(α + β) · γ = ω δ+1 · ξ + ω δ · k · m + ε + ω ρ · l + σ.

Again the desired conclusion holds.

Solutions to exercises in chapter 10
E10.1 Show that any vector space over a field has a basis (possibly infinite).
Let V be any vector space over F . Let A = {X ⊆ V : X is linearly independent}, partially
ordered by ⊆. Then A 6= ∅, since trivially S ∅ ∈ A. Now suppose that B is a subset of A
simply ordered by ⊆. We claim that S B ∈ A; this will verify the hypothesis of Zorn’s
lemma. Suppose that v1 , . . . , vn ∈ B, a1 , . . . , an ∈ F , and a1 v1 + · · · + an vn = 0; we
want to show that all ai are 0. For each i = 1, . . . , n choose Xi ∈ B such that vi ∈ Xi .
Now {Xi : i = 1, . . . , n} has a largest member Xj under ⊆, since B is simply ordered.
[Easy proof by induction on n.] Clearly vi ∈ Xj for all i = 1, . . . , n. Since Xj is linearly
independent, it follows that each ai = 0, as desired.
Now we apply Zorn’s lemma to obtain a maximal member Y of A under ⊆. We claim
that Y is a basis for V . Since Y is linearly independent, it suffices to show that Y spans
V . Suppose that w ∈ V . If w ∈ Y , then obviously w is in the span of Y . Suppose that
w ∈ / Y . Then Y ⊂ Y ∪ {w} so by the maximality of Y , Y ∪ {w} is linearly dependent.
Hence there is a natural number n, distinct elements v1 , . . . , vn ∈ Y ∪ {w}, and elements
a1 , . . . , an ∈ F , not all 0, such that a1 v1 + · · · + an vn = 0. Since Y is linearly independent,

29
not all vi are in Y ; say that vj = w. Then again because Y is linearly independent, we
must have aj 6= 0. So
       
a1 aj−1 aj+1 an
w = − v1 + · · · + − vj−1 + − vj+1 + · · · + − vn ,
aj aj aj aj

so that w is in the span of Y , as desired.

E10.2 A subset C of R is closed iff the following condition holds:
For every sequence f ∈ ω C, if f converges to a real number x, then x ∈ C.
Here to say that f converges to x means that

∀ε > 0∃M ∀m ≥ M [|fm − x| < ε].

Prove that if hCm : m ∈ ωi is a sequence of nonempty closed

T subsets of R, ∀m ∈ ω∀x, y ∈
Cm [|x − y| < 1/(m + 1)], and Cm ⊇ Cn for m < n, then m∈ω Cm is nonempty. Hint:
use the Cauchy convergence criterion.
Let c be a choice function for P(R)\{∅}. For each m ∈ ω let fm = c(Cm ). We claim that
f is a Cauchy sequence, and hence it converges to some point x. For, let ε > 0 be given.
1
Choose m ∈ ω such that m+1 < ε. Then for any n, p ≥ m we have fn , fp ∈ Cm and hence
1
by a hypothesis of the exercise, |fn − fp | < m+1 < ε, as desired.
T Now for any m ∈ ω we
have fn ∈ Cm for all n ≥ m, and hence x ∈ Cm . Thus x ∈ m∈ω Cm .
E10.3 Prove that every nontrivial commutative ring with identity has a maximal ideal.
Nontrivial means that 0 6= 1. Only very elementary definitions and facts are needed here;
they can be found in most abstract algebra books. Hint: use Zorn’s lemma.
Let R be a nontrivial commutative ring with identity. Let A be the collection of all proper
ideals, partially ordered under ⊂. Obviously
S A 6= ∅. Suppose that B is a nonempty subset
of A simply ordered by ⊂. Let I = B. We claim that I is a proper ideal, so that it is
an upper bound for B. In fact, if a, b ∈ I, choose J, K ∈ B such that a ∈ J and b ∈ K.
Since B is simply ordered by ⊂, by symmetry say J ⊂ K. Then a, b ∈ K, hence a + b and
a − b are also in K, and hence they are in I too. Also, if a ∈ I and b ∈ R, then a · b ∈ I
by an even easier argument. Thus I is an ideal. Clearly 1 ∈/ I, so I is proper.
By Zorn’s lemma, A has a maximal element L. Clearly L is a maximal ideal.
E10.4 A function g : R → R is continuous at a ∈ R iff for every sequence f ∈ ω R which
converges to a, the sequence g ◦ f converges to g(a). (See Exercise E10.2.) Show that g is
continuous at a iff the following condition holds:

∀ε > 0∃δ > 0∀x ∈ R[|x − a| < δ → |g(x) − g(a)| < ε].

Hint: for →, argue by contradiction.

→: Suppose that g is continuous at a but the indicated condition fails. Thus

(∗) ∃ε > 0∀δ > 0∃x ∈ R[|x − a| < δ and |g(x) − g(a)| ≥ ε].

30
Let c be a choice function for R. For each m ∈ ω let
  
1
fm = c x ∈ R |x − a| < and |g(x) − g(a)| ≥ ε .
m+1

Then f converges to a. In fact, given ξ > 0, choose M such that M1−1 < ξ. Then for
1
any m ≥ M , |fm − a| < m+1 ≤ M1−1 < ξ. Since f converges to a and g is continuous
at a, it follows that g ◦ f converges to g(a). Hence we can choose N such that ∀n ≥
N [|g(fm ) − g(a)| < ε]. But by the definition of f , |g(fN ) − g(a)| ≥ ε, contradiction.
←: Assume the indicated condition, and suppose that f ∈ ω R converges to a. In
order to show that g ◦ f also converges to a, let ε > 0 be given. By the condition, choose
δ > 0 such that ∀x ∈ R[|x − a| < δ → |g(x) − g(a)| < ε]. Since f converges to a, choose
M such that ∀m ≥ M [|fm − a| < δ]. Then for any m ≥ M we have |g(fm ) − g(a)| < ε, as
desired.
E10.5 Show by induction on m, without using the axiom of choice, that if m ∈ ω and
hAi : i ∈ mi is a system of nonempty sets, then there is a function f with domain m such
that f (i) ∈ Ai for all i ∈ m.
For m = 0, the system itself is empty, and the desired function f is the empty set.
Now suppose that hAi : i ∈ m + 1i is a system of nonempty sets, and we know our
result for a system of m nonempty sets. So, let f be a function with domain m such that
f (i) ∈ Ai for all i ∈ m. Pick a ∈ Am , and let g = f ∪ {(m, a)}. Clearly g is as desired,
completing the inductive proof.
E10.6 Using AC, prove the following, which is called the Principle of Dependent Choice
(which is also weaker than the axiom of choice, but cannot be proved in ZF). If A is a
nonempty set, R is a relation, R ⊆ A × A, and for every a ∈ A there is a b ∈ A such that
aRb, then there is a function f : ω → A such that f (i)Rf (i + 1) for all i ∈ ω.
Let c be a choice function for nonempty subsets of A. We define f : ω → A by recursion,
as follows. Fix a ∈ A. For any m ∈ ω let
(
a if m = 0,
f (m) = c({x ∈ A : f (n)Rx}) if m = n + 1 and {x ∈ A : f (n)Rx} =
6 0,
a otherwise.
By induction, f (i)Rf (i + 1) for every i ∈ ω, as desired.
E10.7 Show that the axiom of choice implies (1), where (1) is
(1) If < is a partial ordering and ≺ is a simple ordering which is a subset of <, then
there is a maximal (under ⊆) simple ordering ≪ such that ≺ is a subset of ≪, which in
turn is a subset of <.
Let
A = {≪:≪ is a simple ordering and ≺⊆≪⊆<}.
Note that A is nonempty, since ≺∈ A . We partially order A by inclusion. To check the
hypothesis of Zorn’s lemma, suppose that B is a nonempty subset of A simply ordered by

31
 S
claim that B ∈ A ; this is clear, by checking
inclusion. We S S all the necessary conditions.
For example, B is transitive since if (a, b), (b, c) ∈ B, then there are R, S ∈ B with
(a, b) ∈ R andS(b, c) ∈ S; by symmetry R ⊆ S, hence (a, b), (b, c) ∈ S, hence (a, c) ∈ S,
hence (a, c) ∈ B.
So we apply Zorn’s lemma to obtain a maximal member ≪ of A ; this is as desired.
E10.8 Prove that (1) implies (2). [Given sets A and B, define f < g iff f and g are
one-one functions which are subsets of A × B, and f ⊂ g. Apply (1) to < and the empty
simple ordering.] Here (2) is
(2) For any two sets A and B, either there is a one-one function mapping A into B or
there is a one-one function mapping B into A.
S
Following the hint, we get a maximal simple ordering ≺ such that ≺⊆<. Let f = (≺).
Since ≺ is a simply ordered collection of one-one functions, it is clear that f is a one-one
function. It suffices to show that dmn(f ) = A or rng(f ) = B. Suppose that this is not
true, and choose a ∈ A\dmn(f ) and b ∈ B\rng(f ). Let g = f ∪ {(a, b)}. Clearly g is a
one-one function contained in A × B. Thus if we define ≺′ as an extension of ≺ with g ≺′ f
for all g in the domain of ≺, we get a proper extension of ≺, contradiction.
E10.9 Prove that (2) implies (3). [Easy] Here (3) is
(3) For any two nonempty sets A and B, either there is a function mapping A onto B or
there is a function mapping B onto A.
Assume (2), and let A and B be nonempty sets. By (2) and symmetry, say that f is a
one-one function mapping A into B. Fix a ∈ A, and define g with domain B by setting,
for each b ∈ B, 
−1
g(b) = f (b) if b ∈ rng(f ),
a otherwise.
Clearly g maps B onto A, as desired.
E10.10 Show in ZF that for any set A there is an ordinal α such that there is no one-one
function mapping α into A. Hint: consider all well-orderings contained in A × A.
Let X be the set of all well-orderings
S contained in A × A. Now each ≺∈ X is isomorphic
to an ordinal β≺ . Let α = ≺∈X (β≺ + 1). Suppose that f is a one-one function mapping
α into A. Let ≺= {(f (ξ), f (η)) : ξ < η}. Then ≺ is a well-ordering contained in A × A,
and so βX = α; consequently α ∈ α, contradiction.
E10.11 Prove that (3) implies the axiom of choice. [Show that any set A can be well-
ordered, as follows. Use exercise E10.10 to find an ordinal α which cannot be mapped
one-one into P(A). Show that if f : A → α maps onto α, then hf −1 [{β}] : β < αi is a
one-one function from α into P(A).]
We follow the hint. Suppose that f : A → α maps onto α. Let g = hf −1 [{β}] : β < αi.
Clearly g maps α into P(A). Suppose that g(β) = g(γ). Thus f −1 [{β}] = f −1 [{γ}].
Choose a ∈ A such that f (a) = β; this is possible because f maps onto α. thus a ∈
f −1 [{β}] = f −1 [{γ}], so f (a) ∈ {γ}, hence β = f (a) = γ. So g is one-one, contradicting
the choice of α.

32
 Now it follows from (3) that there is a function f mapping α onto A. Define a ≺ b
iff the least element of f −1 [{a}] is less than the first element of f −1 [{b}]. Clearly ≺ is a
linear order on A. To show that it is a well-order, let B be a nonempty subset of A. Let
β be the least element of f −1 [B]. Then f (β) is clearly the ≺-least element of B.
E10.12 Show that the axiom of choice implies (4). [Use Zorn’s lemma.] Here (4) is
(4) A family F of subsets of a set A has finite character if for all X ⊆ A, X ∈ F iff every
finite subset of X is in F . Principle (4) says that every family of finite character has a
maximal element under ⊆.
Let F , a nonempty family of subsets of A, have finite character. We consider F as a
partially ordered set under inclusion. It is nonempty by assumption. Now suppose S that
G is a nonempty subset of F linearly ordered by inclusion. To show that G ∈ F , it
suffices to show that every finite F subset of it is in F , by the definition of finite character.
For each a ∈ F choose Xa ∈ G such that a ∈ Xa . Since G is linearly ordered by inclusion,
choose a ∈ F such that Xb ⊆ Xa for all b ∈ F . Now Xa ∈ F since G ⊆ F , and F is a
finite subset of Xa , so F ∈ F by the definition of finite character.
Thus we have verified the hypotheses of Zorn’s lemma, and it gives the desired maximal
element.
E10.13 Show that (4) implies (5). Here (5) is
(5) For any relation R there is a function f ⊆ R such that dmn R = dmn f .
[Given a relation R, let F consist of all functions contained in R.]
Taking F as indicated, we verify that F has finite character. It is obviously nonempty,
since ∅ ∈ F . Of course, if f ∈ F , then every finite subset of f is in F . Now suppose that
f ⊆ R and every finite subset of f is in F . We just need to show that f is a function.
Suppose that (a, b), (a, c) ∈ f . Then {(a, b), (a, c)} is a finite subset of f , and so it is in F ,
which means that it is a function, and so b = c. Thus f is a function.
Now by (4), let f be a maximal member of F under inclusion. So, f is a function
included in R. Suppose that a ∈ dmn(R)\dmn(f ). Choose b such that (a, b) ∈ R. Then
f ⊂ f ∪ {(a, b)} ∈ F , contradiction. Therefore, dmn(R) = dmn(f ), as desired.
E10.14 Show that (5) implies the axiom of choice. [Given a family hAi : i ∈ Ii of
nonempty sets, let R = {(i, x) : i ∈ I and x ∈ Ai }.]
We follow the hint. Let f be a function such that dmn(f ) = dmn(R). Thus dmn(f ) = I
and f (i) ∈ Ai for all i ∈ I.
Solutions to exercises in chapter 12
E12.1 Define sets A, B with |A| = |B| such that there is a one-one function f : A → B
which is not onto.
Let A = B = ω and define f (m) = m + 1 for all m ∈ ω. Then f is not onto, since
0∈/ rng(f ). Suppose that f (m) = f (n) and m 6= n. Say m < n. By Proposition 4.10,
m + 1 ≤ n < n + 1 = m + 1, contradiction.

33
E12.2 Define sets A, B with |A| = |B| such that there is an onto function f : A → B
which is not one-one.
Let A = B = ω. Define f (m + 1) = m for any m ∈ ω and f (0) = 0.
E12.3 Show that the restriction λ 6= 0 is necessary in Proposition 12.43(ix).
Let κ = λ = 0, µ = 0, ν = 1. Then κµ = 00 = 1 by Theorem 12.43(i), and κν = 01 = 0 by
Theorem 12.43(ii).
E12.4 Let F : P(A) → P(A), and assume that for all X, Y ⊆ A, ifS X ⊆ Y , then
F (X) ⊆ F (Y ). Let A = {X : X ⊆ A and X ⊆ F (X)}, and set X0 = X∈A X. Then
X0 ⊆ F (X0 ).
For any Y ∈ A we have Y ⊆ X0 , and hence Y ⊆ F (Y ) ⊆ F (X0 )), so X0 ⊆ F (X0 ).
E12.5 Under the assumptions of exercise E12.4 we actually have X0 = F (X0 ).
By exercise E12.4, X0 ⊆ F (X0 ), so F (X0 ) ⊆ F (F (X0 )), hence F (X0 ) ∈ A , hence F (X0 ) ⊆
X0 ; together with exercise E12.4 this proves that X0 = F (X0 ).
E12.6 Suppose that f : A → B is one-one and g : B → A is also one-one. For every X ⊆
A let F (X) = A\g[B\f [X]]. Show that for all X, Y ⊆ A, if X ⊆ Y then F (X) ⊆ F (Y ).
We have f [X] ⊆ f [Y ], hence B\f [Y ] ⊆ B\f [X], hence g[B\f [Y ]] ⊆ g[B\f [X]], hence
F (X) = A\g[B\f [X]] ⊆ A\g[B\f [Y ]] = F (Y ).
E12.7 Prove the Cantor-Schröder-Bernstein theorem as follows. Assume that f and g
are as in exercise E12.6, and choose F as in that exercise. Let X0 be as in exercise E12.4.
Show that A\X0 ⊆ rng(g). Then define h : A → B by setting, for any a ∈ A,

f (a) if a ∈ X0 ,
h(a) =
g −1 (a) if a ∈ A\X0 .

Show that h is one-one and maps onto B.

A\X0 = A\F (X0 ) = g[B\f [X]] ⊆ rng(g). Now note that h ↾ X0 maps X0 onto f [X0 ] and
is one-one, and h ↾ (A\X0 ) maps A\X0 onto g −1 [A\X0 ] = g −1 [g[B\f [X0]]] = B\f [X0 ] and
is one-one. So h is the union of two functions with disjoint domains and disjoint ranges,
so h is a one-one function, and it maps A onto B.
E12.8 Show that if α and β are ordinals, then |α ∔ β| = |α| + |β|, where ∔ is ordinal
addition and + is cardinal addition.
This is immediate from Proposition 12.21.
E12.9 Show that if α and β are ordinals, then |α ⊙ β| = |α| · |β|, where ⊙ is ordinal
multiplication and · is cardinal multiplication.
This is immediate from Proposition 12.29.
E12.10 Show that if α and β are ordinals, 2 ≤ α, and ω ≤ β, then |· αβ | = |α| · |β|. Here
the dot to the left of the first exponent indicates that ordinal exponentiation is involved.

34
First note that |α| ≤ α ≤ · αβ and |β| ≤ β ≤ · αβ , so |α| · |β| ≤ |· αβ |. Hence it suffices
to prove the other direction, which
S we do by induction on β, starting with β = ω. First,
β = ω: If α < ω, then |· αω | = m∈ω · αm = ω = |α| · |ω|. If α ≥ ω, then
[ X X
|· αω | = | · m
α |≤ |· αm | ≤ |α| ≤ ω · |α|,
m∈ω m∈ω m∈ω

as desired.
Now we assume the result for β ≥ ω. Then

|· αβ+1 | = |· αβ ⊙ α| = |· αβ | · |α| = |α| · |β| · |α| = |α| · |β|.

using the inductive hypothesis. Finally, if β is a limit ordinal > ω and the result is true
for all γ < β, then

[ [ X
|· αβ | = · γ
α = · γ
α ≤ |· αγ |
γ<β ω≤γ<β ω≤γ<β
X X
= |α| · |γ| ≤ |α| · |β| ≤ |α| · |β| · |β| = |α| · |β|.
ω≤γ<β ω≤γ<β

E12.11 Prove that if |A| ≤ |B| then |P(A)| ≤ |P(B)|.

Let f be a one-one function mapping A into B. For each X ∈ P(A) let g(X) = f [X]. So
g maps P(A) into P(B). We claim that g is one-one. For, suppose that X, Y ∈ P(A)
and X 6= Y . Say by symmetry that x ∈ X\Y . Then f (x) ∈ f [X] but f (x) ∈/ f [Y ] by
one-oneness.
E12.12 Prove the following general distributive law:

c X
Y c
XY
κij = κi,f (i) ,
i∈I j∈Ji f ∈P i∈I

Q
where P = i∈I Ji .
The left side is the number of elements of
 
Y [
(1)  (κij × {j}) ,
i∈I j∈Ji

and the right side is the number of elements of

c
!
[ Y
(2) κi,f (i) × {f } .
f ∈P i∈I

35
 Q Qc
For each f ∈ P let Ff be a bijection from i∈I κi,f (i) onto i∈ISκi,f (i) . Now given x in
(1) we define G(x) in (2) as follows. For each i ∈ I we have xi ∈ j∈Ji (κij × {j}), and so
there is a unique j ∈ Ji such that x ∈ κij × {j}; Q let fx (i) be this j. Thus fx ∈ P . Now
st st
1 (xi ) ∈ κi,fx (i) for all i ∈ I, so h1 (xi ) : i ∈ Ii ∈ i∈I κi,fx (i) . Now we define

G(x) = (Ffx (h1st (xi ) : i ∈ Ii), fx).

Clearly G(x) is in (2).

Suppose that G(x) = G(y). Now fx = 2nd (G(x)) = 2nd (G(y) = fy . Write fx = g.
Then for any i ∈ I,

1st (xi ) = (Fg−1 (1st (G(x))))i

= (Fg−1 (1st (G(y))))i
= 1st (yi ),

and 2nd (xi ) = g(i) = 2nd (yi ). So xi = yi . Hence x = y. So G is one-one.

Qc that G maps onto (2), suppose that zQis a member of (2). Choose h ∈ P such
To show
that z ∈ ( i∈I κi,h(i) ) × {h}. Now Fh−1 (1st (z)) ∈ i∈I κi,h(i) , so for each i ∈ I we can let

xi = ((Fh−1 (1st (z))i , h(i)).

Then x is in (1). Moreover, clearly fx = h. Then 1st (xi ) = (Fh−1 (1st (z))i , hence h1st (xi ) :
i ∈ Ii = Fh−1 (1st (z)), and so

G(x) = (Ffx (h1st (xi ) : i ∈ Ii), fx)

= (Fh (h1st (xi ) : i ∈ Ii), h)
= (Fh (Fh−1 (1st (z))), h)
= (1st (z), 2nd (z))
= z,

as desired.
E12.13 Show that for any cardinal κ we have κ+ = {α : α is an ordinal and |α| ≤ κ}.
First suppose that α < κ+ . Then |α| ≤ α, so |α| ≤ κ. Now suppose that |α| ≤ κ. Thus
there is a one-one function from α into κ. If κ+ ≤ α, then we could also get a one-one
function from κ+ into κ, so κ+ = |κ+ | ≤ |κ| = κ, contradiction. So α < κ+ , as desired.
E12.14 For every infinite cardinal λ there is a cardinal κ > λ such that κλ = κ.
Let κ = 2λ . Then κλ = (2λ )λ = 2λ·λ = 2λ = κ.
E12.15 For every infinite cardinal λ there is a cardinal κ > λ such that κλ > κ.
Let λ = ℵα . Note that cf(ℵα+ω ) = ω ≤ λ. Let κ = ℵα+ω . Then κλ > κ.
E12.16 Prove that for every n ∈ ω, and every infinite cardinal κ, ℵκn = 2κ · ℵn .

36
We prove this by induction on n. n = 0: ℵκ0 = 2κ = 2κ · ℵ0 . Assume it for n. Then by
Hausdorff’s theorem,
ℵκn+1 = ((ℵn )+ )κ = ℵκn · (ℵn )+ = 2κ · ℵn · ℵn+1 = 2κ · ℵn+1 .
Q Q Q
E12.17 Prove that ci∈I (κi · λi ) = ci∈I κi · ci∈I λi .
We have
c
Y Y
(1) (κi · λi ) = (κi · λi )
i∈I i∈I

and
c c c
! c
!
Y Y Y Y
(2) κi · λi = κi × λi
i∈I i∈I i∈I i∈I

For each i ∈ Q
Q I let fi be a bijection from κi · λi onto Qc κi × λi . Let
Q g be a bijection from
c
κ
i∈I i onto κ , and let h be a bijection from λ onto i∈I λi . Now we define a
Q i
i∈I Qc
Qc
i∈I i Q
function F from i∈I (κi · λi ) to i∈I κi × i∈I λi by setting, for any x ∈ i∈I (κi · λi ),

F (x) = (g −1 (h1st (fi (xi )) : i ∈ Ii), h−1 (2nd (fi (xi )) : i ∈ Ii)).
Qc
Qc

This does map into i∈I κi × i∈I λi , since for any i ∈ I we haveQ xi ∈ κi · λi , hence
st st
fi (xi ) ∈ κi × λi ; so 1 (fi (xi )) ∈ κi . Thus h1 (fi (xi )) : i ∈ Ii ∈ i∈I κi and hence
Qc Qc
g −1 (h1st (fi (xi ))
: i ∈ Ii) ∈
i∈I κi . Similarly h−1 (2nd (fi (xi )) : i ∈ Ii) ∈ i∈I λi , so that
Qc Qc
F (x) ∈ i∈I κ i × i∈I λi .
F is one-one; for suppose that F (x) = F (y). Then for any i ∈ I,
1st (F (x)) = g −1 (h1st (fi (xi )) : i ∈ Ii);
g(1st(F (x)) = h1st (fi (xi )) : i ∈ Ii;
(g(1st(F (x))i = 1st (fi (xi )).
Similarly, (g(1st (F (y))i = 1st (fi (yi )). Then 1st (fi (xi )) = 1st (fi (yi )) follows from the
assumption that F (x) = F (y). Similarly, 2nd (fi (xi )) = 2nd (fi (yi )). So fi (xi ) = fi (yi ),
and hence xi = yi . This being true for allQi ∈ I, we
have
Qc x =
y, as desired. Qc
c
F maps onto; for suppose that z ∈ i∈I κ i × i∈I λi . Then 1st (z) ∈ i∈I κi , so
Q
g(1st(z)) ∈ i∈I κi . Hence for any i ∈ I, (g(1st(z)))i ∈ κi . Similarly, (h(2nd (z))i ∈ λi . It
follows that ((g(1st(z)))i , (h(2nd (z))i ) ∈ κi × λi , and hence
fi−1 (((g(1st(z)))i , (h(2nd (z))i )) ∈ κi · λi .
Q
We let xi = fi−1 (((g(1st(z)))i , (h(2nd (z))i )). So x ∈ i∈I (κi · λi ), and

fi (xi ) = ((g(1st(z)))i , (h(2nd (z))i );

1st (fi (xi )) = (g(1st (z)))i ;
h1st (fi (xi )) : i ∈ Ii = g(1st (z));
g −1 (h1st (fi (xi )) : i ∈ Ii) = 1st (z).

37
Similarly, h−1 (h2nd (fi (xi )) : i ∈ Ii) = 2nd (z). So F (x) = z.
By (1) and (2) this completes the exercise.
E12.18 Prove that ℵℵω1 = 2ℵ1 · ℵℵω0 .
P P
Note that ℵω = m∈ω ℵm , since ℵn ≤ m∈ω ℵm for each n ∈ ω, hence
X X
ℵω ≤ ℵm ≤ ℵω = ω · ℵω = ℵω .
m∈ω m∈ω
Qc Qc
Hence by Theorem 12.41 we have ℵω < m∈ω ℵm+1 ≤ m∈ω ℵm . So

c
!ℵ1
Y
ℵℵω1 ≤ ℵm
m∈ω
c
Y
= ℵℵm1
m∈ω
Yc
= (2ℵ1 · ℵm ) by exercise E12.16
m∈ω
Yc c
Y
ℵ1
= 2 · ℵm by exercise E12.17
m∈ω m∈ω
c
Y
= 2ℵ1 ·ℵ0 · ℵm
m∈ω
c
Y
ℵ1
=2 · ℵm
m∈ω
Yc
≤ 2ℵ1 · ℵω
m∈ω

= 2ℵ1 · ℵℵω0
≤ ℵℵω1 .

Q
E12.19 Prove that ℵℵω0 = n∈ω ℵn .
P
By the argument at the beginning of the solution of exercise E12.18, ℵω =
Q m∈ω ℵm <
c
m∈ω ℵm . Hence

c
!ℵ0 c c c c c
Y Y Y Y Y Y
ℵℵω0 ≤ ℵm = ℵℵm0 = ℵ0
(2 ·ℵm ) = 2 · ℵ0
ℵm = ℵm ≤ ℵω = ℵℵω0 .
m∈ω m∈ω m∈ω m∈ω m∈ω m∈ω

E12.20 Prove that for any infinite cardinal κ, (κ+ )κ = 2κ .

By Hausdorff’s theorem, (κ+ )κ = κκ · κ+ = 2κ · κ+ = 2κ .

38
 E12.21 Show that if κ is an infinite cardinal and C is the collection of all cardinals less
than κ, then |C| ≤ κ.
Let κ = ℵα . Thus C ⊆ ω ∪ {ℵβ : β < α}. Hence |C| ≤ ω + |α|. Now ω ≤ κ, and
|α| ≤ α ≤ ℵα = κ since ℵ is a normal function. Hence |C| ≤ κ.
E12.22 Show that if κ is an infinite cardinal and C is the collection of all cardinals less
than κ, then
!cf(κ)
X
2κ = 2ν .
ν∈C

First suppose that κ is a successor cardinal λ+ . Then

!κ !cf(κ) !κ
X X X
2κ ≤ 2ν = 2ν ≤ 2λ = (|C| · 2λ )κ ≤ (λ+ · 2λ )κ ≤ (2κ )κ = 2κ ,
ν∈C ν∈C ν∈C

as desired.
Now suppose that κ is a limit cardinal. Let hµξ : ξ < cf(κ)i be a strictly increasing
sequence of cardinals with supremum κ. Then

P c
!cf(κ)
µξ Y X
2κ = 2 ξ<cf (κ) = 2µξ ≤ 2λ ≤ (2κ )cf(κ) = 2κ .
ξ<cf (κ) ν∈C

Q
E12.23 Prove that for any limit ordinal τ , cξ<τ 2ℵξ = 2ℵτ .
P Yc
ℵτ ℵξ
2 =2 ξ<τ = 2ℵξ .
ξ<τ

E12.24 Assume that κ is an infinite cardinal, and 2λ < κ for every cardinal λ < κ. Show
that 2κ = κcf(κ) .
If κ is a successor cardinal, then cf(κ) = κ and the desired conclusion is clear. Suppose
that κ is a limit cardinal. Let hµξ : ξ < cf(κ)i be a strictly increasing sequence of cardinals
with supremum κ. Then
P c
Y c
Y
µξ
κ
2 =2 ξ<cf(κ) = 2 µξ
≤ κ = κcf(κ) ≤ κκ = 2κ .
ξ<cf (κ) ξ<cf(κ)

E12.25 Suppose that λ is a singular cardinal, cf(λ) = ω, and 2κ < λ for every κ < λ.
Prove that 2λ = λω .
Let hκn : n ∈ ωi be a system of cardinals less than λ with supremum λ. Then
P c
Y c
Y
λ κn κn
2 =2 n∈ω = 2 ≤ λ = λω ≤ λλ = 2λ .
n∈ω n∈ω

39
 Solutions to exercises in Chapter 13
E13.1 Let (A, +, ·, −, 0, 1) be a Boolean algebra. Show that (A, △, ·, 0, 1) is a ring with
identity in which every element is idempotent. This means that x · x = x for all x.
Obviously △ is commutative, and it is associative by Proposition 13.11(iii). Clearly x△0 =
x for all x. Clearly x△x = 0, so each element x has itself as additive inverse. Hence
(A, △, 0) is an abelian group.
Clearly · is associative. The distributive law holds by Proposition 13.11(ii). Clearly
x · 1 = x for all x, and clearly x · x = x for all x.
Hence (A, △, ·, 0, 1) is a ring with identity in which every element is idempotent.
E13.2 Let (A, +, ·, 0, 1) be a ring with identity in which every element is idempotent.
Show that A is a commutative ring, and (A, ⊕, ·, −, 0, 1) is a Boolean algebra, where for
any x, y ∈ A, x ⊕ y = x + y + xy and for any x ∈ A, −x = 1 + x. Hint: expand (x + y)2 .
x + y = (x + y)2 = x2 + xy + yx + y 2 = x + xy + yx + y, and hence 0 = xy + yx for any
x, y. Setting x = y, we get 0 = x + x, and so x is its own additive inverse. Then from
0 = xy + yx we see that yx is the additive inverse of xy, hence xy = yx. Thus the ring is
commutative.
To show that (A, ⊕, ·, −, 0, 1) is a Boolean algebra, we need to check all of the axioms.
(C): Clear.
(A): For any x, y, z,

x ⊕ (y ⊕ z) = x + (y ⊕ z) + x(y ⊕ z)
= x + y + z + yz + x(y + z + yz)
= x + y + z + yz + xy + xz + xyz;

Hence, using (C),

(x ⊕ y) ⊕ z = z ⊕ (x ⊕ y)
= z + x + y + xy + zy + zxy
= above.

(A′ ): obvious.
(C′ ): obvious.
(L):
x ⊕ xy = x + xy + xxy = x + xy + xy = x.
(L′ ):
x(x ⊕ y) = x(x + y + xy) = xx + xy + xxy = x + xy + xy = x.
(D):

x(y ⊕ z) = x(y + z + yz) = xy + xz + xyz;

xy ⊕ xz = xy + xz + xyxz = xy + xz + xyz.

40
(D′ ): See Proposition 13.12.
(K): x + (−x) = x + 1 + x = 1.
(K’): x(1 + x) = x + xx = x + x = 0.
Thus we have a BA.
E13.3 Show that the processes described in exercises E2.1 and E2.2 are inverses of one
another.
For each BA (A, +, ·, −, 0, 1) let R(A, +, ·, −, 0, 1) = (A, △, ·, 0, 1) be the associated ring,
and for each ring (A, +, ·, 0, 1) with identity in which every element is idempotent let
B(A, +, ·, 0, 1) = (A, ⊕, ·, −, 0, 1) be the associated Boolean algebra. We want to show
that R and B are inverses of each other.
First suppose that (A, +, ·, −, 0, 1) is a BA. Let R((A, +, ·, −, 0, 1)) = (A, △, ·, 0, 1) be
the associated ring, and let B(R((A, +, ·, −, 0, 1)) = (A, ⊕, ·, −′ , 0, 1) be the BA associated
with that ring; we want to show that + = ⊕ and − = −′ . We have

x ⊕ y = x△y△(x · y)
= x△(y · −(x · y) + x · y · −y)
= x△(y · −x)
= x · −(y · −x) + y · −x · −x
= x + y · −x
= x + y · x + y · −x
= x + y.

Also, −′ x = 1△x = −x.

Second, suppose that (A, +, ·, 0, 1) is a ring with identity in which every element
is idempotent, let B((A, +, ·, 0, 1)) = (A, +′ , ·, −′ , 0, 1) be the associated BA, and let
R(B((A, +, ·, 0, 1))) = (A, △′ , ·, 0, 1) be the ring associated with it. We want to show
that + = △′ . We have

x△′ y = (x · −′ y) +′ (y · −′ x)
= (x · (1 + y)) +′ (y · (1 + x))
= (x + xy) +′ (y + xy)
= x + xy + y + xy + (x + xy)(y + xy)
= x + y + xy + xy + xy + xxy + xyy + xyxy
= x + y + xy + xy + xy + xy + xy + xy
= x + y.

E13.4 Prove that a filter F is an ultrafilter iff F is maximal among the set of all filters
G such that 0 ∈
/ G.

41
⇒: Assume that F is an ultrafilter. Hence by definition 0 ∈ / F . Suppose that F ⊂ G with
G a filter. Choose x ∈ G\F . Since x ∈ / F it follows that −x ∈ F , and hence −x ∈ G. So
0 = x · −x ∈ G. So F is maximal among the set of filters G such that 0 ∈ / G.
⇐: Suppose that F is maximal among the set of filters G such that 0 ∈ / G. Suppose
that a ∈ A and a ∈ / F . Let G = {x ∈ A : a · y ≤ x for some y ∈ F }. Then G is a filter on
A. In fact, obviously conditions (1) and (2) hold. For (3), suppose that x, z ∈ G. Choose
y, w ∈ F such that x = a · y and z = a · w. Now y · w ∈ F , and x · z = a · y · w. So x · z ∈ G.
Thus, indeed, G is a filter on A. Clearly also F ⊆ G. Clearly a ∈ G (taking y = 1), so
F ⊂ G.
It follows by supposition that 0 ∈ G. Say 0 = a · y, with y ∈ F . then y ≤ −a, so
−a ∈ F . Thus F is an ultrafilter.
E13.5 Prove that for any nonzero a ∈ A there is an ultrafilter F such that a ∈ F .
Let A = {G : G is a filter in A, a ∈ G, and 0 ∈/ G}. We consider A as a partially ordered
set under ⊆. To verify the hypothesis of Zorn’s lemma, suppose that B is a subset of A
linearly ordered by ⊆. Now {xS∈ A : a ≤ x} is clearly a member of A , so we may assume
that B is nonempty. Let H = B. Since B is nonempty, it is clear that a ∈ H. Suppose
that x ∈ H and x ≤ y. Choose G ∈ B such that x ∈ G. Then y ∈ G since G is a filter.
So y ∈ H. Suppose that x, y ∈ H. Choose G, G′ ∈ B such that x ∈ G and y ∈ G′ . By
symmetry say G ⊆ G′ . Then x, y ∈ G′ , so x · y ∈ G′ , hence x · y ∈ H. Thus we have shown
that H is a filter on A. Clearly 0 ∈
/ H. So H is a member of A which is an upper bound
for B.
Thus by Zorn’s lemma, A has a maximal member G. By exercise E13.4, G is as
desired.
E13.6 Prove that any BA is isomorphic to a field of sets. (Stone’s representation theorem)
Hint: given a BA A, let X be the set of all ultrafilters on A and define f (a) = {F ∈ X :
a ∈ F }.
Let X be the collection of all ultrafilters, and let F, G ∈ X.
F ∈ f (−a) iff −a ∈ F iff a ∈/ F , so f (−a) = X\f (a).
Suppose that F ∈ f (a + b). Then a + b ∈ F . Suppose that F ∈ / f (a). Then a ∈
/ F , so
−a ∈ F , hence −a · (a + b) ∈ F . Since −a · (a + b) ≤ b, also b ∈ F , so F ∈ f (b). This shows
that f (a + b) ⊆ f (a) ∪ f (b). On the other hand, if F ∈ f (a), then a ∈ F ; but a ≤ a + b,
so also a + b ∈ F ; hence F ∈ f (a + b). Altogether this shows that f (a + b) = f (a) ∪ f (b).
Suppose that a 6= b. Then a△b 6= 0, so a△b ∈ F for some ultrafilter F , by exercise
E13. Hence F ∈ f (a△b) = [f (a)\f (b)] ∪ [f (b)\f (a)], and so f (a) 6= f (b). So f is one-one.
E13.7 Suppose that F is an ultrafilter on a BA A. Let 2 be the two-element BA. (This
is, up to isomorphism, the BA of all subsets of 1.) For any a ∈ A let

1 if a ∈ F ,
f (a) =
0 if a ∈
/ F.

Show that f is a homomorphism of A into 2.

f (a · b) = 1 iff a · b ∈ F iff a, b ∈ F iff f (a) · f (b) = 1. Hence f (a · b) = f (a) · f (b).

42
 f (−a) = 1 iff −a ∈ F iff a ∈ / F iff f (a) = 0. Hence f (−a) = −f (a).
f (a + b) = f (−(−a · −b)) = −(−f (a) · −f (b)) = f (a) + f (b).
f (0) = f (a · −a) = f (a) · −f (a) = 0.
f (1) = f (a + −a) = f (a) + −f (a) = 1.
E13.8 (A knowledge of logic is assumed.) Suppose that L is a first-order language and T
is a set of sentences of L . Define ϕ ≡T ψ iff ϕ and ψ are sentences of L and T |= ϕ ↔ ψ.
Show that this is an equivalence relation on the set S of all sentences of L . Let A be the
collection of all equivalence classes under this equivalence relation. Show that there are
operations +, ·, − on A such that for any sentences ϕ, ψ,

[ϕ] + [ψ] = [ϕ ∨ ψ];

[ϕ] · [ψ] = [ϕ ∧ ψ];
−[ϕ] = [¬ϕ].

Finally, show that (A, +, ·, −, [∃v0(¬(v0 = v0 ))], [∃v0 (v0 = v0 )]) is a Boolean algebra.
≡T is reflexive: T |= ϕ ↔ ϕ for any sentence ϕ.
≡T is symmetric: If T |= ϕ ↔ ψ, then T |= ψ ↔ ϕ.
≡T is transitive: If T |= ϕ ↔ ψ and T |= ψ ↔ χ, then T |= ϕ ↔ χ.
+ is well-defined: If T |= ϕ ↔ ϕ′ and T |= ψ ↔ ψ ′ , then T |= (ϕ ∨ ψ) ↔ (ϕ′ ∨ ψ ′ ).
· is well-defined: If T |= ϕ ↔ ϕ′ and T |= ψ ↔ ψ ′ , then T |= (ϕ ∧ ψ) ↔ (ϕ′ ∧ ψ ′ ).
− is well-defined: If T |= ϕ ↔ ϕ′ , then T |= ¬ϕ ↔ ¬ϕ′ .
Finally, we need to check the axioms for BAs:
(A) holds since

[ϕ] + ([ψ] + [χ]) = [ϕ ∨ (ψ ∨ χ)] = [(ϕ ∨ ψ) ∨ χ] = ([ϕ] + [ψ]) + [χ];

(A′ ) holds since

[ϕ] · ([ψ] · [χ]) = [ϕ ∧ (ψ ∧ χ)] = [(ϕ ∧ ψ) ∧ χ] = ([ϕ] · [ψ]) · [χ];

(C) holds since

[ϕ] + [ψ] = [ϕ ∨ ψ] = [ψ ∨ ϕ] = [ψ] + [ϕ];
(C′ ) holds since
[ϕ] · [ψ] = [ϕ ∧ ψ] = [ψ ∧ ϕ] = [ψ] · [ϕ];
(L) holds since
[ϕ] + [ϕ] · [ψ]) = [ϕ ∨ (ϕ ∧ ψ)] = [ϕ];
(L′ ) holds since
[ϕ] · [ϕ] + [ψ]) = [ϕ ∧ (ϕ ∨ ψ)] = [ϕ];
(D) holds since

[ϕ] · ([ψ] + [χ]) = [ϕ ∧ (ψ ∨ χ)] = [(ϕ ∧ ψ) ∨ (ϕ ∧ χ)] = [ϕ] · [ψ] + [ϕ] · [χ];

43
for (D′ ) see Proposition 2.12; (K) holds since

[ϕ] + −[ϕ] = [ϕ ∨ ¬ϕ] = [∃v0 (v0 = v0 )];

(K′ ) holds since

[ϕ] · −[ϕ] = [ϕ ∧ ¬ϕ] = [∃v0 (¬(v0 = v0 ))].

E13.9 (A knowledge of logic is assumed.) Show that every Boolean algebra is isomorphic
to one obtained as in exercise E13.8. Hint: Let A be a Boolean algebra. Let L be the
first-order language which has a unary relation symbol Ra for each a ∈ A. Let T be the
following set of sentences of L :

∀x∀y(x = y);
∀x[R−a (x) ↔ ¬Ra (x)] for each a ∈ A;
∀x[Ra·b (x) ↔ Ra (x) ∧ Rb (x)] for all a, b ∈ A;
∀xR1 (x).

We follow the hint, and consider ≡T . Define f (a) = [∀xRa (x)] for any a ∈ A. To show
that f preserves ·, suppose that a, b ∈ A. Note that

T |= ∀xRa·b (x) ↔ ∀xRa (x) ∧ ∀xRb (x);

hence f (a · b) = f (a) · f (b).

To proceed we need the following fact
(1) T |= ∀xϕ ↔ ϕ for any variable x and any formula ϕ.
In fact, trivially T |= ∀xϕ → ϕ, and T |= ϕ → ∃xϕ. Since T |= x = y, clearly T |= ∃xϕ →
∀xϕ. So (1) holds.
Now to show that f preserves −, suppose that a ∈ A. Then T |= ∀xR−a (x) ↔
∀x¬Ra (x). By (1), T |= ∀x¬Ra (x) ↔ ¬Ra (x) and T |= ¬Ra (x) ↔ ¬∀xRa (x). Putting
these statements together we have T |= ∀xR−a (x) ↔ ¬∀xRa (x), and it follows that f
preserves −.
To show that f is one-one, suppose that a, b ∈ A and a 6= b; say a · −b 6= 0. Let F
be an ultrafilter on A such that a · −b ∈ F . We now define an L -structure A. Let A = 1.
For each a ∈ A, let n
1 if a ∈ F ,
RaA =
0 otherwise.
Clearly A is a model of T . Also, A |= Ra·−b (x). It follows that [∀xRa·−b (x)] 6= [∃v0 (¬(v0 =
v0 ))], and so f (a) = [∀xRa (x)] 6= [∀xRb (x)] = f (b), as desired.
It remains only to show that f maps onto.
(2) For any formula ϕ there is an a ∈ A such that T |= ϕ ↔ Ra (x).
Condition (2) is easily proved by induction on ϕ, using (1). Hence f is onto.

44
 def
E13.10 Let A be the collection of all subsets X of Y = {r ∈ Q : 0 ≤ r} such that there
exist an m ∈ ω and a, b ∈ m (Y ∪ {∞}) such that a0 < b0 < a1 < b1 < · · · < am−1 <
bm−1 ≤ ∞ and
X = [a0 , b0 ) ∪ [a1 , b1 ) ∪ . . . ∪ [am−1 , bm−1 ).
Note that ∅ ∈ A by taking m = 0, and Y ∈ A since Y = [0, ∞).
(i) Show that if X is as above, c, d ∈ Y ∪ {∞} with c < d, c ≤ a0 , then X ∪ [c, d) ∈ A,
and c is the first element of X ∪ [c, d).
(ii) Show that if X is as above and c, d ∈ Y ∪ {∞} with c < d, then X ∪ [c, d) ∈ A.
(iii) Show that (A, ∪, ∩, \, ∅, Y ) is a Boolean algebra.
(i): Assume the hypothesis. If m = 0 the desired conclusion is clear, so suppose that
m > 0. We consider several cases.
Case 1. bm−1 ≤ d. Then X ∪ [c, d) = [c, d) ∈ A.
Case 2. There is an i < m − 1 such that bi ≤ d < ai+1 . Then

X ∪ [c, d) = [c, d) ∪ [ai+1 , bi+1 ) ∪ . . . ∪ [am−1 , bm−1 ) ∈ A.

Case 3. There is an i < m such that ai ≤ d < bi . Then

X ∪ [c, d) = [c, bi ) ∪ [ai+1 , bi+1 ) ∪ . . . ∪ [am−1 , bm−1 ) ∈ A.

Case 4. d < a0 . Then

X ∪ [c, d) = [c, d) ∪ [a0 , b0 ) ∪ . . . ∪ [am−1 , bm−1 ) ∈ A.

(ii): Again we consider several cases.

Case 1. c ≤ a0 . Then X ∪ [c, d) ∈ A by (i).
Case 2. There is an i < m such that ai ≤ c ≤ bi . Let X ′ = [ai , bi ) ∪ . . .∪ [am−1 , bm−1 ).
Then by (i) applied to X ′ and [ai , d) we get X ′ ∪ [ai , d) ∈ A, and ai is the least element of
X ′ ∪ [ai , d). Clearly

X ∪ [c, d) = [a0 , b0 ) ∪ . . . ∪ [ai−1 , bi−1 ) ∪ X ′ ∪ [ai , d) ∈ A.

Case 3. There is an i < m − 1 such that bi < c < ai+1 . Then we can apply (i) to
[ai+1 , bi+1 ) ∪ . . . ∪ [am−1 , bm−1 ) and [c, d) to get the desired result as in Case 2.
Case 4. c = bm−1 . Then

X ∪ [c, d) = [a0 , b0 ) ∪ . . . ∪ [am−1 , d) ∈ A.

Case 5. bm−1 < c. This case is clear.

(iii): From (ii) it is clear that A is closed under ∪. Now suppose that X is given as
above. To show that also Y \X ∈ A, we consider several cases.
Case 1. m = 0. So X = ∅, and Y = [0, ∞) ∈ A.
Case 2. m > 0, 0 < a0 , and bm−1 < ∞. Then

Y \X = [0, a0 ) ∪ [b0 , a1 ) ∪ . . . ∪ [bm−2 , am−1 ) ∪ [bm−1 , ∞) ∈ A.

45
 Case 3. m > 0, a0 = 0, and bm−1 < ∞. Then

Y \X = [b0 , a − 1) ∪ . . . ∪ [bm−2 , am−1 ) ∪ [bm−1 , ∞) ∈ A.

Case 4. m > 0, 0 < a0 , and bm−1 = ∞. Then

Y \X = [0, a0 ) ∪ [b0 , a1 ) ∪ . . . ∪ [bm−2 , am−1 ) ∈ A.

Case 5. m > 0, 0 = a0 , and bm−1 = ∞. Then

Y \X = [b0 , a1 ) ∪ . . . ∪ [bm−2 , am−1 ) ∈ A.

Thus (iii) holds.

E13.11 (Continuing
P exercise E13.10) For each n ∈ ω let xn = [n, n + 1), an interval in
Q. Show that n∈ω x2n does not exist in A.
P
Suppose that the sum does exist. Let X = n∈ω x2n , and assume that X is as in exercise
E13.10.
We claim that bm−1 = ∞. In fact, if bm−1 < ∞, then there is an m ∈ ω such that
bm−1 < 2m; then x2m = [2m, 2m + 1) is disjoint from X according to the form of X, but
x2m ≤ X by definition, contradiction. So our claim holds.
Now choose m ∈ ω so that am−1 < 2m + 1. Then [2m + 1, 2m + 2) ∩ x2n = ∅ for all n,
hence [2m + 1, 2m + 2) ∩ X = ∅. But [2m + 1, 2m + 2) ⊆ [am−1 , bm−1 ) ⊆ X, contradiciton.
E13.12 Let A be the Boolean algebra of all subsets of some nonempty set X, under the
natural set-theoretic operations. Show that if hai : i ∈ Ii is a system of elements of A, then
Y X Y ε(i)
(ai + −ai ) = 1 = ai ,
i∈I ε∈I 2 i∈I

where for any y, y 1 = y and y 0 = −y.

First note that the big products and sums are just the ordinary intersections and unions.
Obviously ai + −ai = ai ∪ (X\ai ) = X = 1, giving the first equality. Now suppose that
x ∈ X. We define n
1 if x ∈ ai ,
ε(i) =
0 otherwise.
ε(i)
Clearly then x ∈ ai for each i ∈ I, and hence x is in the right side of the second equality,
as desired.
E13.13 Let M be the set of all finite functions f ⊆ ω × 2. For each f ∈ M let

Uf = {g ∈ ω 2 : f ⊆ g}.

Let A consist of all finite unions of sets Uf .

(i) Show that A is a Boolean algebra under the set-theoretic operations.

46
 (ii) For each i ∈ ω, let xi = U{(i,1)} . Show that
Y
ω
2= (xi + −xi )
i∈ω

while X Y ε(i)
xi = ∅,
ε∈ω 2 i∈ω

where for any y, y 1 = y and y 0 = −y.

(i): Obviously A is closed
S under ∪. Now suppose that a ∈ A; we want to show that
ω
( 2)\a ∈ A. Say a = f ∈N Uf , where N is a finite subset of M . Let
 
 [ 
P = g ∈ M : dmn(g) ⊆ dmn(f ) and ∀f ∈ N ∃i ∈ dmn(g) ∩ dmn(f )[f (i) 6= g(i)] .
 
f ∈N

S
Clearly P is a finite subset
S of M . We claim that (ω 2)\a = g∈P Ug . First suppose that
h ∈ (ω 2)\a. Let g = h ↾ f ∈N dmn(f ). So g ∈ M and h ∈ Ug . We claim that g ∈ P . For,
suppose that f ∈ N . Then Uf ⊆ a, so it follows that h ∈ / Uf . So we can choose i ∈ dmn(f )
such that f (i) 6= h(i). Clearly i ∈ dmn(g) and f (i) 6= g(i). This shows that g ∈ P , proving
⊆ of our claim.
For ⊇, suppose that g ∈ P and h ∈ Ug . Suppose that h ∈ a. Choose f ∈ N such
that h ∈ Uf , hence f ⊆ h. But g ⊆ H too, so there is an i ∈ dmn(g) ∩ dmn(f ) such that
f (i) 6= g(i). But this means that f (i) 6= h(i), contradicting f ⊆ h. We have now shown
(i). Q
Clearly xi ∪ ((ω 2)\xi = ω 2 for any i ∈ ω. Hence ω 2 = i∈ω (xi + −xi ).
Q ε(i)
Now suppose that ε ∈ ω 2; we want to show that i∈ω xi = 0, i.e., that there is no
ε(i)
nonzero element a of A such that a ≤ xi for all i ∈ ω. suppose that a is such an element.
Then there is a g ∈ M such that Ug ⊆ a. Take any i ∈ / dmn(g), and let h ∈ ω 2 be any
ε(i)
function such that g ⊆ h and h(i) 6= ε(i). Then h ∈ Ug but h ∈ / xi , contradiction.
E13.14 Suppose that (P, ≤, 1) is a forcing order. Define

p≡q iff p, q ∈ P, p ≤ q, and q ≤ p.

Show that ≡ is an equivalence relation, and if Q is the collection of all ≡-classes, then
there is a relation  on Q such that for all p, q ∈ P , [p]≡  [q]≡ iff p ≤ q. Finally,
show that (Q, ) is a partial order, i.e.,  is reflexive on Q, transitive, and antisymmetric
(q1  q2  q1 implies that q1 = q2 ); moreover, q ≤ [1] for all q ∈ Q.
Since ≤ is reflexive on P , clearly also ≡ is reflexive on P . Clearly ≡ is symmetric. Now
suppose that p ≡ q ≡ r. Thus p ≤ q, q ≤ p, q ≤ r, and r ≤ q. Then p ≤ r and r ≤ p, so
p ≡ r. So ≡ is an equivalence relation on P .
Let
= {(a, b) : ∃p, q ∈ P [p ≤ q, a = [p], and b = [q]]}.

47
Obviously then p ≤ q implies that [p]  [q]. Now suppose that [p]  [q]. Choose p′ , q ′ ∈ P
such that p′ ≤ q ′ , [p] = [p′ ], and [q] = [q ′ ]. Then p ≤ p′ and q ′ ≤ q, so p ≤ q.
To show that  is a partial order on Q, first suppose that a ∈ Q. Write a = [p]. Then
p ≤ p, so a  a. Thus  is reflexive on Q. Now suppose that a  b  c. Then there exist
p, q, q ′ , r such that p ≤ q, a = [p], b = [q], q ′ ≤ r, b = [q ′ ], and c = [r]. Then q ≤ q ′ since
[q] = [q ′ ]. So p ≤ q ≤ q ′ ≤ r, hence p ≤ r. So a = [q]  [r] = c. This shows that  is
transitive. Finally, suppose that a  b  a. Then there exist p, q, q ′ , r such that p ≤ q,
a = [p], b = [q], q ′ ≤ r, b = [q ′ ], and a = [r]. Then q ≤ q ′ since [q] = [q ′ ]. Also r ≤ p since
[p] = [r], so q ≤ q ′ ≤ r ≤ p, hence q ≤ p. But also p ≤ q, so a = [p] = [q] = b. So  is a
partial order. Clearly a ≤ [1] for all a ∈ Q.
E13.15 We say that (P, <) is a partial order in the second sense iff < is transitive and
irreflexive. (Irreflexive means that for all p ∈ P , p 6< p.) Show that if (P, <) is a partial
order in the second sense and if we define  by p  q iff (p, q ∈ P and p < q or p = q),
then A (P, ) is a partial order. Furthermore, show that if (P, ≤) is a partial order, and
we define p ≺ q by p ≺ q iff (p, q ∈ P , p ≤ q, and p 6= q), then B(P, ≺) is a partial order
in the second sense.
Also prove that A and B are inverses of one another.
Clearly  is reflexive on P . Now suppose that x  y  z. If x = y or y = z, then x  z
by supposition. If x < y < z, then x < z, and so x  z. Thus  is transitive. Suppose
that x  y  x, but x 6= y. Then x < y < x, hence x < x, contradiction. So  is
antisymmetric. Hence (P, ) is a partial order.
Now suppose that (P, ≤) is a partial order, and define p ≺ q by p ≺ q iff (p, q ∈ P ,
p ≤ q, and p 6= q). Clearly ≺ is irreflexive. Suppose that p ≺ q ≺ r. Then p ≤ q ≤ r, so
p ≤ r. Suppose that p = r. Then p ≤ q ≤ p, so p = q by antisymmetry, contradiction.
Thus p 6= r, and so p ≺ r. So (P, ≺) is a partial order in the second sense.
Next, suppose that (P, <) is a partial order in the second sense, and let A (P, <) =
(P, ). Furthermore, let B(A (P, <)) = (P, <′ ). Then

p <′ q iff (p  q and p 6= q) iff ((p < q or p = q) and p 6= q) iff p < q.

Thus B(A (P, <)) = (P, <).

Finally, suppose that (P, ≤) is a partial order. Let B(P, ≤) = (P, ≺), and let
A (B(P, ≤)) = (P, ≤′ ). Then

p ≤′ q iff (p ≺ q or p = q) iff ((p ≤ q and p 6= q) or p = q) iff p ≤ q.

E13.16 Show that if (P, ≤, 1) is a forcing order and we define ≺ by p ≺ q iff (p, q ∈ P ,
p ≤ q and q 6≤ p), then (P, ≺) is a partial order in the second sense. Give an example
where this partial order is not isomorphic to the one derived from (P, ≤, 1) by the procedure
of exercise E13.14.
≺ is irreflexive, since x ≺ x would imply that x 6≤ x, a contradiction. For transitivity,
suppose that x ≺ y ≺ z. Then x ≤ y and y ≤ z, so x ≤ z. Also, y 6≤ x and z 6≤ y. Suppose
that z ≤ x. Then y ≤ z ≤ x and hence y ≤ x, contradiction. Hence z 6≤ x, and so x ≺ z.
Thus (P, ≺) is a partial order in the second sense.

48
 For the example, let X be any infinite set, and let ≤ be X ×X. Fix 1 ∈ X. So (X, ≤, 1)
is a quasiorder. The partial order constructed in exercise E13.14 has only one element,
while the partial order of the present exercise has X, an infinite set, as its underlying set.
Note that ≺ is empty.
E13.17 Prove that if (P, , 1) is a forcing order such that the mapping e from P into
RO(P ) is one-one, then (P, ) is a partial order. Give an example of a forcing order such
that e is not one-one. Give an example of an infinite forcing order Q such that e is not
one-one, while for any p, q ∈ Q, p ≤ q iff e(p) ⊆ e(q).
Suppose that P is a forcing order such that e is one-one, and p ≤ q ≤ p. Then P ↓ p =
P ↓ q, and hence e(p) = e(q). So p = q. Hence (P, ≤) is a partial order.
For an example of a forcing order such that e is not one-one, take any simple ordering
with greatest element; see the remarks preceding Proposition 13.21.
For the final example, take any infinite set Q, and take the forcing order (Q, Q × Q, q)
for any element q ∈ Q. So e is the constant function with value Q. For any p, q ∈ Q we
have p ≤ q and e(p) ⊆ e(q), so these statements are equivalent trivially.
E13.18 (Continuing E13.14.) Let P = (P, ≤, 1) be a forcing order, and let Q = (Q, , [1])
be as in exercise E13.14. Show that there is an isomorphism f of RO(P) onto RO(Q) such
that f ◦ eP = eQ ◦ π, where π : P → Q is defined by π(p) = [p] for all p ∈ P .
We will apply Theorem 13.22. For any p ∈ P let j(p) = eQ (π(p)). Thus j : P → RO(Q).
Suppose that 0 6= X ∈ RO(Q). By Theorem 13.20(i), choose q ∈ Q such that
eQ (q) ≤ X. Say q = [p]. Then j(p) = eQ (π(p)) ≤ X. So j[P ] is dense in RO(Q).
Suppose that p, q ∈ P and p ≤ q. Then [p]  [q], and so j(p) ≤ j(q) by Theorem
13.20(ii).
Suppose that p, q ∈ P . If p 6⊥ q, choose r ≤ p, q. Then j(r) ≤ j(p), j(q), so j(p)∩j(q) 6=
∅. If j(p) ∩ j(q) 6= ∅, then by Theorem 13.20(iii), π(p) 6⊥ π(q). So there is an r ∈ Q such
that r ≤ π(p), π(q). Say r = π(s). Then s ≤ p, q, so p 6⊥ q.
This verifies the hypotheses of Theorem 13.22, and the desired conclusion follows.
Solutions to exercises in chapter 14
E14.1 Write out all the elements of Vα for α = 0, 1, 2, 3, 4.
V0 = ∅.
V1 = P(V0 ) = P(∅) = {∅} = 1.
V2 = P(V1 ) = P({∅}) = {∅, {∅}} = 2.
V3 = P(V2 ) = P({∅, {∅}}) = {∅, {∅}, {{∅}}, {∅, {∅}}}. Note that V3 has four elements,
but it is not equal to 4, since, for example, {{∅}} ∈ V3 \4.
For V4 , it helps to use the usual abbreviations for natural numbers. Thus V3 = {0, 1, 2, {1}}.
We list out the subsets of V3 with 0,1,2,3,4 elements:

V4 = P(V3 ) = {0 0 elements; 1 of these

{0}, {1}, {2}, {{1}} 1 element, 4 of these

49
 {0, 1}, {0, 2}, {0, {1}}, {1, 2}, {1, {1}}, {2, {1}} 2 elements, 6 of these
{0, 1, 2}, {0, 1, {1}}, {0, 2, {1}}, {1, 2, {1}} 3 elements, 4 of these
{0, 1, 2, {1}} 4 elements, 1 of these.}

E14.2 Define by recursion

[
S(α) = P(S(β))
β<α

for every ordinal α. Prove that Vα = S(α) for every ordinal α.

First we apply the recursion theorem 8.7. Define G : On × V → V by setting, for any
ordinal α and any set x,
S
G(α, x) = β<α P(x(β)) if x is a function with domain α,
∅ otherwise.

S F : On → V by Theorem 6.7: for any ordinal α, F(α) = G(α, F ↾ α). Thus

Then obtain
F(α) = β<α P(F(β)).
We prove that Vα = S(α) for all α by induction:
[
S(0) = P(S(β)) = ∅ = V0 ;
β<0
[
S(α + 1) = P(S(β))
β<α+1
[
= P(S(β)) ∪ P(S(α))
β<α
= Vα ∪ P(Vα ) (inductive hypothesis)
= Vα+1 (using Theorem 14.5(ii));
[
S(α) = P(S(β)) (with α limit)
β<α
[ [
= P(S(β))
γ<α β<γ
[
= Vγ (inductive hypothesis)
γ<α
= Vα .

E14.3 Determine exactly the ranks of the following sets in terms of the ranks of the sets
entering into their definitions. In some cases the rank is not completely determined by the
ranks of the constituents; in such cases, describe all possibilities.
S
(i) {x} (iv) x ∪ y (vii) x (x) R−1
(ii) {x, y} (v) x ∩ y (viii) dmn(R)
(iii) (x, y) (vi) x\y (ix) P(x)

50
(i): Let α = rank(x). Thus x ∈ Vα+1 . Hence {x} ⊆ Vα+1 , so {x} ∈ P(Vα+1 ) = Vα+2 . So
rank({x}) ≤ α + 1. Suppose that rank({x}) < α + 1. Then {x} ∈ Vα+1 = P(Vα ), hence
{x} ⊆ Vα , so x ∈ Vα , and so rank(x) < α, contradiction. Thus rank({x}) = α + 1.
(ii): Let α = rank(x) and β = rank(y). We claim that rank({x, y}) = max(α, β) + 1.
To prove this, by symmetry it suffices to assume that α ≤ β. Hence x, y ∈ Vβ+1 , so
{x, y} ⊆ Vβ+1 , so {x, y} ∈ P(Vβ+1 ) = Vβ+2 . Hence rank({x, y}) ≤ β + 1. Suppose that
rank({x, y}) < β + 1. Then {x, y} ∈ Vβ+1 = P(Vβ ), so {x, y} ⊆ Vβ . Hence y ∈ Vβ ,
contradiction.
(iii): rank((x, y)) = max(rank(x), rank(y)) + 2 by (i) and (ii).
(iv): Let rank(x) = α and rank(y) = β. We claim that rank(x ∪ y) = max(α, β). To
prove this, by symmetry we may assume that α ≤ β. Thus x, y ∈ Vβ+1 , so x, y ⊆ Vβ , hence
x ∪ y ⊆ Vβ and so x ∪ y ∈ Vβ+1 . Hence rank(x ∪ y) ≤ β. Suppose that rank(x ∪ y) < β.
Then x ∪ y ∈ Vβ , so by exercise E14.2, x ∪ y ⊆ Vγ for some γ < β. So y ⊆ Vγ , hence
y ∈ Vγ+1 ⊆ Vβ , contradiction.
(v): Let α and β be as in (iv). We claim that rank(x ∩ y) ≤ min(rank(x), rank(y)),
and actually can be anything ≤ this minimum. To prove this, by symmetry assume that
α ≤ β. Then x ∩ y ⊆ x ⊆ Vα , so rank(x ∩ y) ≤ α.
For the second assertion, suppose that γ is any ordinal, and suppose that δ ≤ γ. We
define two sets x and y such that min(rank(x), rank(y)) = γ while rank(x ∩ y) = δ. Let
x = δ ∪ {γ} and y = γ. Then rank(x) = γ + 1, rank(y) = γ, and rank(x ∩ y) = rank(δ) = δ.
(vi): Let α = rank(x). We claim that rank(x\y) ≤ α, and it can be anything ≤ α. In
fact, x\y ⊆ x ⊆ Vα , so x\y ∈ Vα+1 , and so rank(x\y) ≤ α.
For the second assertion, let β ≤ α; we define x, y so that rank(x) = α while
rank(x\y) = β. Let x = α and y = α\β. S S
(vii): Let rank(x) = α. We claim that rank ( x) = α. (Thus S it is α if α is limit or
0, and it is β if α = β + 1.) To prove this, first we show that x ⊆ VS α . For, suppose
S
that y ∈ x. Say y ∈ z ∈ x. Now x ∈ Vα+1 , so x ⊆ V Sα . Hence z S∈ Vα . Hence by exercise
E14.2, z ⊆ Vβ for some β < α. So y ∈ Vβ . Now β ⊆ α, so β ≤ α. Hence y ∈ VS α , as
desired. S S S S
It follows that x ∈ V(S α)+1 , and hence rank ( x) ≤ α. Suppose that x ∈ VS α .
S S
Then by exercise E14.2 Sx ⊆ Vγ for some γ < α. Say γ < δ < α. Then x ⊆ Vδ . In
fact, if y ∈ x, then y ⊆ x ⊆ Vγ , hence y ∈ Vγ+1 S ⊆ Vδ . S So, indeed, x ⊆ Vδ . Hence
x ∈ Vδ+1 ⊆ Vα , contradiction. This proves that rank ( x) = α. SS
(viii): Let α = rank(R). We claim, first of all, that rank(dmn(R)) ≤ α. For, take
any Sx∈ S that (x, y) ∈ R. So x ∈ {x} ∈ S
S dmn(R). Choose yS such (x,Sy) ∈ R;SitSfollows that
x∈ R. So dmn(R) ⊆ R,
SSand so rank(dmn(R)) ≤ rank( R) = α by (vii).
Next we claim that if β ≤ α, then there is a set R such that rank(R) = α while
rank(dmn(R)) = β. To give the examples here, we consider two cases.
Case 1. β = 0. Then we take R = α. We use the easy fact that no ordinal is an ordered
pair. [(a, b) has at most two elements, and is a nonempty set. So the only possibilities for
(a, b) to be an ordinal are (a, b) = 1 or (a, b) = 2. Since (a, b) = {{a}, {a, b}}, neither case
is really possible, since the members of (a, b) are nonempty.]
Case 2. 0 < β. Let R = {(ξ, 0) : ξ < β} ∪ α. To see that this works, note that
dmn(R) = β = rank(β). But we also need to see that rank(R) ≤ α. For this we consider

51
several subcases. SS
Subcase 2.1. α is a limit ordinal. Then α = α, rank({(ξ, 0) : ξ < β}) ≤ α
Subcase 2.2. α = γ + 1 for some limit ordinal γ. Since β ≤ γ, clearly rank({(ξ, 0) :
ξ < β}) ≤ γ < α
Subcase 2.3. α = γ + 2 for some limit ordinal γ. Similar to Subcase 2.2.
Subcase 2.4. α = γ + n for some limit ordinal γ and some n ∈ ω\3. Then
β ≤ γ + n − 2, and so rank({(ξ, 0) : ξ < β}) ≤ γ + n
Subcase 2.5. 0 < β ≤ α − 2, with α ∈ ω\3. Again clearly ok.
These are all of the possibilities.
(ix): Let rank(x) = α. We claim that rank(P(x)) = α + 1. Now x ∈ Vα+1 , so x ⊆ Vα .
Hence y ⊆ Vα for every y ⊆ x. So P(x) ⊆ P(Vα ) = Vα+1 ; hence P(x) ∈ Vα+2 . This
shows that rank(P(x)) ≤ α + 1. Suppose that P(x) ∈ Vα+1 . Hence x ∈ P(x) ⊆ Vα , so
x ∈ Vα , contradiction. Hence rank(P(x)) = α + 1.
(x): We claim that for all ordinals α, β, ∃R[rank(R) = α and rank(R−1 ) = β] iff β ≤ α
and one of the following conditions holds:
(1) β = γ + 3 for some ordinal γ.
(2) β is a limit ordinal.
To prove this, we first note that if a and b have ranks ξ, η respectively, then (a, b) has
rank max(ξ, η) + 2 by (iii). Hence if S is a collection of ordered pairs, then rank(S) =
sup{rank(s)+1 : s ∈ S}, and hence rank(S) is either a limit ordinal (if {rank(s)+1 : s ∈ S}
does not have a greatest element) or it is of the form γ + 3. It follows that if rank(R) = α
and rank(R−1 ) = β, then β ≤ α and (1) or (2) holds.
Now suppose that β ≤ α. If β = γ + 3 for some γ, let R = {(γ, γ)} ∪ α; then
rank(R) = α and rank(R−1 ) = β. If β is a limit ordinal, let R = {(γ, γ) : γ < β} ∪ α; then
rank(R) = α and rank(R−1 ) = β.
E14.4 Define xRy iff (x, 1) ∈ y. Show that R is well-founded and set-like on V.
Suppose that X is a nonempty set. Choose x ∈ X of smallest rank. Suppose that y ∈ X
and yRx. Thus (y, 1) ∈ x, so y ∈ {y} ∈ (y, 1) ∈ x, and hence rank(y) < rank(x),
contradiction. Hence R is well-founded on V.
For any y ∈ V we have predVR (y) =S{x S : (x, 1) ∈ y}. Note that if S
(x,S1) ∈ y then x ∈
{x} ∈ {{x}, {x, 1}} = (x, 1) ∈ y, so x ∈ y. So predVR (y) = {x ∈ y : (x, 1) ∈ y},
and hence predVR (y) is a set. Thus R is set-like on V.
E14.5 (Continuing E14.4) By recursion let y̌ = {(x̌, 1) : x ∈ y} for any set y. Let F be
the Mostowski collapsing function for R, V in exercise E14.4. Prove that F(y̌) = y for
every set y.
First we show that the function ˇ exists. Let S = {(x, y) : x ∈ y}. So S is well-founded
and set-like on V. Define G : V × V → V by setting, for any sets y, f ,

{(f (x), 1) : x ∈ y} if f is a function with domain predVS (y),
G(y, f ) =
∅ otherwise.

52
Then let H be obtained from G by Theorem 5.7: for any set y, H(y) = G(y, H ↾
predVS (y)). Hence for any set y, H(y) = {(H(x), 1) : x ∈ y}, as desired.
Now for the main part of the exercise, suppose that it is not true. So there is a y such
that F(y̌) 6= y. Let W = {w ∈ trcl(y ∪ {y}) : F(w̌) 6= w}. Thus y ∈ W , so W 6= ∅. By the
foundation axiom, choose z ∈ W with z ∩ W = ∅. Then

F(ž) = {F(s) : sRž} (definition of F)

= {F(s) : (s, 1) ∈ ž} (definition of R)
= {F(ť) : t ∈ z} (definition of ž)
= {t : t ∈ z} (since z ∩ W = ∅)
= z,

contradiction.
E14.6 Define xRy iff x ∈ trcl(y). Show that R is well-founded and set like on V.
Clearly xRy implies that rank(x) < rank(trcl(y)) = rank(y), so R is well-founded. For
any a ∈ V, the class {b ∈ V : bRa} = {b : b ∈ trcl(a)} = trcl(a), a set.
E14.7 Let F be the Mostowski collapsing function for R, V. Show that F(x) = rank(x)
for every set x.
Suppose not, and by the foundation axiom let x be a set such that F(x) 6= rank(x) while
F(y) = rank(y) for every y ∈ x. Note that F(x) is transitive: if u ∈ v ∈ F(x), choose
y ∈ trcl(x) such that v = F (y). Then choose z ∈ trcl(y) such that u = F(z). Then
z ∈ trcl(x), and so u ∈ F(y). It follows that F(x) is an ordinal. Now

F(x) = {F(y) : y ∈ V and yRx}

= {F(y) : y ∈ trcl(x)}
= {rank(y) : y ∈ trcl(x)}.

Now if y ∈ trcl(x), then rank(y) < rank(trcl(x)) = rank(x). So F(x) ≤ rank(x). Con-
versely, if α ∈ rank(x), then α ∈ rank(trcl(x)) and hence α ≤ rank(y) for some y ∈ trcl(x).
Since F(x) is an ordinal, it follows that α ∈ F(x).
Thus F(x) = rank(x), contradiction.
E14.8 Prove that if a is transitive, then {rank(b) : b ∈ a} is an ordinal.
Let S = {rank(b) : b ∈ a}, and let α be the least ordinal not in S. Thus α ⊆ S. We
claim that α = S. Suppose not, and let β be the least member of S\α; so α < β. Thus
there is a b ∈ a such that β = rank(b). Now a is transitive, so if c ∈ b then c ∈ a and
hence rank(c) ∈ S with rank(c) < β, so rank(c) < α. Hence β = supc∈b (rank(c) + 1) ⊆ α,
contradiction.
E14.9 Show that for any set a we have rank(trcl(a)) = rank(a).

53
We use the notation in the proof of Theorem 14.6. First we prove that rank(dm ) = rank(a)
for every m ∈ ω, by induction on m. Obviously rank(d0 ) = rank(a). For the inductive
step we use Theorem 14.8:

rank(a) ≤ rank(dm+1 )
[
= max(rank(dm ), rank( dm ))
[
= max(rank(a), rank(b))
b∈dm
[
≤ max(rank(a), rank(dm ))
b∈dm
= max(rank(a), rank(a))
= rank(a).

Now rank(trcl(a)) = rank(a) by Exercise E14.3(vii).

E14.10 For any infinite cardinal κ, let H(κ) be the set of all x such that |trcl(x)| < κ.
Prove that Vω = H(ω). (H(ω) is the collection of all hereditarily finite sets.) Hint:
Vω ⊆ H(ω) is easy. For the other direction, suppose that x ∈ H(ω), let t = trcl(x), and let
S = {rank(y) : y ∈ t}. Show that S is an ordinal.
First we show Vω ⊆ H(ω). Suppose that x ∈ Vω . Say x ∈ Vn with n ∈ ω. Then n = m + 1
for some m, and so x ⊆ Vm . Hence by Theorem 14.6, trcl(x) ⊆ Vm . So |trcl(x)| ≤ |Vm | < ω.
It follows that x ∈ H(ω), as desired.
Conversely, suppose that x ∈ H(ω). Let t = trcl(x). So t is finite. Let S = {rank(y) :
y ∈ t}. Thus S is finite, since t is. S is a set of ordinals; we claim that it actually is
an ordinal. It suffices to show that S is transitive. Suppose that β ∈ α ∈ S; we may
assume that α is minimum with this property. Write α = rank(y) with y ∈ t. Now by
Theorem 14.8(iv), rank(y) = supz∈y (rank(z) + 1). It follows that there is a z ∈ y such
that β < rank(z) + 1. Now z ∈ t since t is transitive. Thus with γ = rank(z), we have
γ ∈ S, and hence we cannot have β < γ, as this would contradict the minimality of α. So
β = γ ∈ S, proving our claim: S is a finite ordinal.
For each y ∈ t we have y ∈ VS+1 , hence x ⊆ t ⊆ VS+1 , so x ∈ P(VS+1 ) = VS+2 ⊆ Vω ,
as desired.
E14.11 Which axioms of ZFC are true in On?
Extensionality holds, by Theorem 14.11.
Comprehension fails. Let ϕ(x, z) be the formula ∃y(y ∈ x). Suppose that [∃w∀x(x ∈
w ↔ x ∈ 2 ∧ ∃y(y ∈ x))]On . So, choose an ordinal w such that ∀x ∈ On(x ∈ w ↔ x ∈
2 ∧ ∃y ∈ On(y ∈ x)). Thus ∀x(x ∈ w ↔ x ∈ 2 ∧ x 6= ∅). So 1 ∈ w but 0 ∈ / w, contradicting
w being an ordinal.
Pairing holds, by Theorem 14.13.
Union holds, by Theorem 14.14
Power set holds, by Theorem 14.15. In fact, given an ordinal x, any subset of x which
is an ordinal is ≤ x itself, so we can take y = x.

54
 Replacement holds. To prove this, we use Theorem 14.16. So, suppose that ϕ is a
formula with free variables among x, y, A, w1, . . . , wn , and suppose that A, w1 , . . . , wn are
ordinals such that

∀x ∈ A∃!y[y ∈ On ∧ ϕOn (x, y, A, w1, . . . , wn )].

For each x ∈ A, let yx be the unique ordinal such that ϕOn (x, yx , A, w1 , . . . , wn ). Let
α = supx∈A yx . Then clearly {y ∈ On : ∃x ∈ AϕOn (x, y, A, w1, . . . , wn )} ⊆ α, as desired.
Foundation holds, by Theorem 14.17.
Infinity holds. This is clear if we write the axiom out:

∃x[∃y(y ∈ x ∧ ∀z(z ∈
/ y)∧
∀y ∈ x∃z ∈ x∀w(w ∈ z ↔ w ∈ y ∨ w = y)]

Clearly ω works for x.

Axiom of choice. This holds trivially. Given an ordinal A , if A = 0, then the
conclusion holds if we take B to be any ordinal; and if A 6= 0, the hypothesis of the
implication in the axiom is false.
E14.12 Show that the power set operation is absolute for Vα for α limit.
We first show that the power set operation is defined in Vα . We take the following
formula as the official definition for this operation:

∀z ∈ y[z ⊆ x] ∧ ∀z[z ⊆ x → z ∈ y].

Now given x ∈ Vα , we have rank(P(x)) = rank(x)+1 by the solution for exercise E14.3(ix).
Hence P(x) ∈ Vα . Clearly y = P(x) satisfies the above formula. The uniqueness condition
is clear.
Since every subset of x is in Vα by Theorem 14.8(iii), clearly P Vα (x) = P(x). See
Proposition 14.21.
E14.13 Let M be a countable transitive model of ZFC. Show that the power set operation
is not absolute for M .
Note that ω ∈ M by Theorem 14.28. If the power set operation is absolute for M , then
P M (ω) = P(ω) ∈ M . But |P(ω)| = 2ω and M is transitive, so M is uncountable,
contradiction.
E14.14 Show that Vω is a model of ZFC − Inf.
Extensionality: true by Theorem 14.11.
Comprehension: assume that ϕ has free variables among x, z, w1 , . . . , wn and we are
given z, w1 , . . . , wn ∈ Vω . Let

A = {x ∈ z : ϕVω (x, z, w1 , . . . , wn )}.

Choose m ∈ ω so that z, w1 , . . . , wn ∈ Vm . Then A ⊆ z ⊆ Vm , so A ∈ P(Vm ) = Vm+1 ⊆

Vω . So A ∈ Vω . Hence the comprehension axiom holds in Vω , by Theorem 14.12.

55
 Pairing: if x, y ∈ Vω , choose n such that x, y ∈ Vn . Then {x, y} ⊆ Vn , so {x, y} ∈
P(Vn ) = Vn+1 ⊆ Vω . So {x, y} ∈ Vω . By Theorem 14.13, the S pairing axiom S holds in Vω .
Union: if x ∈ Vω , choose n such that x ∈ Vn . Then x ⊆ Vn , so x ∈ P(Vn ) =
Vn+1 ⊆ Vω . By Theorem 14.14, the union axiom holds in Vω .
Power set: if x ∈ Vω , choose m such that x ∈ Vn . Then y ⊆ Vn for each y ⊆ x, so
P(x) ⊆ P(Vn ) = Vn+1 ∈ Vn+1 ⊆ Vω . So the power set axiom holds by Theorem 14.15.
Replacement: preparing to use Theorem 14.16, let ϕ be a formula with free variables
among x, y, A, w1, . . . , wn , suppose that A, w1 , . . . , wn ∈ Vω , and also assume that

∀x ∈ A∃!y[y ∈ Vω ∧ ϕVω (x, y, A, w1, . . . , wn )].

Choose m ∈ ω such that A, w1 , . . . , wn ∈ Vm . For each x ∈ A, let yx be the unique set

such that
yx ∈ Vω ∧ ϕVω (x, yx , A, w1 , . . . , wn )],
and let px ∈ ω be such that yx ∈ Vpx . Now A is finite, so there is a q ∈ ω such that m < q
and px < q for every x ∈ A. It follows that

{y ∈ Vω : ∃x ∈ Aϕ(x, y, A, w1, . . . , wn )} = {yx : x ∈ A} ⊆ Vq ∈ Vω ,

as desired.
Finally, foundation holds by Theorem 14.17.
E14.15 Show that the formula ∃x(x ∈ y) is not absolute for all nonempty sets, but it is
absolute for all nonempty transitive sets.
Let A = {{∅}}. Then ∃x(x ∈ {∅}) holds in V , but not in A, since there is no a ∈ A such
that a ∈ {∅}.
Now suppose that B is a nonempty transitive set, and y ∈ B. Then y has an element
iff it has an element in B, and so ∃x(x ∈ y) iff ∃x ∈ B(x ∈ y) iff (∃x(x ∈ y))B . So the
formula is absolute for B. (Note that ∃x(x ∈ y) is not quite a ∆0 formula.)
E14.16 Show that the formula ∃z(x ∈ z) is not absolute for every nonempty transitive
set.
Take the transitive set 2. Then ∃z(1 ∈ z), but this does not hold in 2, since there is no
z ∈ 2 such that 1 ∈ z.
E14.17 A formula is Σ1 iff it has the form ∃xϕ with ϕ a ∆0 formula; it is Π1 iff it has
the form ∀xϕ with ϕ a ∆0 formula.
(i) Show that “X is countable” is equivalent on the basis of ZF to a Σ1 formula.
(ii) Show that “α is a cardinal” is equivalent on the basis of ZF to a Π1 formula.
(i) Basically the following statement proves this:

X is countable iff ∃f [f is a one-one function and rng(f ) ⊆ ω].

(ii) Basically the following statement proves this:

α is a cardinal iff ∀f [α is an ordinal and ∀β ∈ α(f : β → α → f is not onto)].

56
However, in each case we have more work to do, since the “insides” were not proved to be
∆0 in this chapter; all we can really use are the definitions. For brevity, we say “is ∆0 ”
instead of “is equivalent under ZF to a ∆0 formula”.
(1) “x is an ordinal” is ∆0 . For,

x is an ordinal iff x is transitive and ∀y ∈ x[y is transitive].

(2) “x is a successor ordinal” is ∆0 . For,

x is a successor ordinal iff x is an ordinal and ∃y ∈ x(x = y ∪ {y})].

(3) “n is a natural number” is ∆0 . For,

n is a natural number iff (n = ∅ or n is a successor ordinal)

and (∀y ∈ n[y = ∅ or y is a successor ordinal]).

(4) “a is an ordered pair” is ∆0 . For,

a is an ordered pair iff ∃x ∈ a∃y ∈ a∃u ∈ x∃v ∈ y[x = {u} and y = {u, v} and a = {x, y}]

(5) “R is a relation” is ∆0 . For,

R is a relation iff ∀a ∈ R[a is an ordered pair]

S
(6) If ϕ(a, f ) is ∆0 , then so is ∀a ∈ f ϕ(a, f ). For,
[
∀a ∈ f ϕ(a, f ) iff ∀x ∈ f ∀a ∈ xϕ(a, f ).

(7) “f is a function” is ∆0 . For,

S S S
f is a function iff f is a relation and ∀a ∈ f ∀b ∈ f ∀x ∈ a∀y ∈ a∀z ∈ b
[a = (x, y) and b = (x, z) → y = z].

(8) “f is a one-one function” is ∆0 . For,

S S S
f is a one-one function iff f is a function and ∀a ∈ f ∀b ∈ f ∀x ∈ a∀y ∈ a∀z ∈ b
[a = (x, y) and b = (z, y) → x = z].

(9) “x is in the range of f ” is ∆0 . For,

S
x is in the range of f iff ∃a ∈ f ∃y ∈ a[a = (y, x)].

(10) “x is in the domain of f ” is ∆0 . For,

S
x is in the domain of f iff ∃a ∈ f ∃y ∈ a[a = (x, y)].

57
(11) x = dmn(f ) is ∆0 . For,

x = dmn(f ) iff ∀y ∈ x(y is in the domain of f ) and

∀a ∈ f ∀y ∈ a∀u ∈ a(u is in the domain of f implies that u ∈ x).

Now we can give the details on (i) and (ii):

X is countable iff ∃f [f is a one-one function and X = dmn(f )

S
and ∀a ∈ f ∀x ∈ a[x is in the range of f implies that x is a natural number]].

α is a cardinal iff ∀f [α is an ordinal and

∀β ∈ α[f is a function and β = dmn(f ) implies that ∃x ∈ α(x is not in the range of f )]

E14.18 Prove that if κ is an infinite cardinal, then H(κ) ⊆ Vκ .

Let a ∈ H(κ). Let α = {rank(b) : b ∈ trcl(a)}. Thus α is an ordinal by exercise E14.8. By
exercise E14.9 and Theorem 14.8(iv) rank(a) = rank(trcl(a)) ≤ α + 1. Now |trcl(a)| < κ,
and hrank(b) : b ∈ trcl(a)i maps trcl(a) onto α, so |rank(a)| ≤ |trcl(a)| < κ. Hence
rank(a) < κ, as desired.
E14.19 Prove that for κ regular, H(κ) = Vκ iff κ = ω or κ is inaccessible.
⇒: Assume that H(κ) = Vκ and κ is uncountable. Suppose that λ < κ ≤ 2λ . Thus
λ ∈ Vλ+1 , and so P(λ) ∈ Vλ+2 ⊆ Vκ . But |P(λ) = 2λ ≥ κ, so P(λ) ∈/ Vκ .
⇐: κ = ω implies that H(κ) = Vκ by exercise E14.14. Now suppose that κ is
inaccessible. By exercise E14.18 we have H(κ) ⊆ Vκ . Now suppose that S ∈ Vκ . Let
α = rank(S). Then also α = rank(trcl(S)) by exercise E14.14. So trcl(S) ∈ Vα+1 , hence
trcl(S) ⊆ Vα+1 and so |trcl(S)| ≤ |Vα+1 | = iβ with ω + β = α + 1. Now β < κ and
iβ < iκ = κ, so S ∈ H(κ).
E14.20 Assume that κ is an infinite cardinal. Prove the following:
(a) H(κ) is transitive.
(b) H(κ) ∩ On = κ. S
(c) If x ∈ H(κ), then x ∈ H(κ).
(d) If x, y ∈ H(κ), then {x, y} ∈ H(κ).
(e) If y ⊆ x ∈ H(κ), then y ∈ H(κ).
(f ) If κ is regular and x is any set, then x ∈ H(κ) iff x ⊆ H(κ) and |x| < κ.
(a): Suppose that x ∈ y ∈ H(κ). Then |trcl(y)| < κ. Moreover, x ∈ trcl(y), hence
x ⊆ trcl(y), hence trcl(x) ⊆ trcl(y). So |trcl(x)| < κ.
(b): If α ∈ H(κ) ∩ ON, then |α| = |trcl(α)| < κ, and so α < κ. Conversely, if α ∈ κ,
then |trcl(α)| =S|α| < κ, so α ∈ H(κ).
S (c): If y ∈ x, then S ∃z[y ∈ z ∈ x], henceS ∃z[y ∈ z ∈ trcl(x)], hence y ∈ trcl(x). Thus
x ⊆ trcl(x), hence trcl( x) ⊆ trcl(x). So x ∈ H(κ).
(d): Note that trcl(x) ∪ {x} is transitive and contains {x}. If b is any transitive set
containing {x}, then x ∈ b, hence trcl(x) ⊆ b, so that trcl(x) ∪ {x} ⊆ b. This proves that
trcl({x}) = trcl(x) ∪ {x}. Hence |trcl({x})| < κ.

58
 Similarly, trcl({x, y}) = trcl({x}) ∪ trcl({y}), so |trcl({x, y})| < κ. So {x, y} ∈ H(κ).
(e): obvious.
(f): Suppose that κ is regular and x is any set. If x ∈ H(κ), then x ⊆ H(κ) by
(a), and |x| ≤S |trcl(x)| < κ. Now suppose that Px ⊆ H(κ) and |x| < κ. Now clearly
trcl(x) = x ∪ y∈x trcl(y). Hence |trcl(x)| ≤ 1 + y∈x |trcl(y)| < κ; so x ∈ H(κ).

E14.21 Show that if κ is regular and uncountable, then H(κ) is a model of all of the ZFC
axioms except possibly the power set axiom.
Extensionality: true since H(κ) is transitive.
Comprehension: using Theorem 14.12, it suffices to take a formula ϕ with free vari-
ables among x, z, w1 , . . . , wn , assume that z, w1 , . . . , wn ∈ H(κ), and prove that {x ∈
z : ϕH(κ) (x, z, w1 , . . . , wn ) ∈ H(κ). This is true since the indicated set is a subset of z;
hence its transitive closure is a subset of trcl(z), which has size less than κ.
Pairing: Given x, y ∈ H(κ), clearly {x, y} ∈ H(κ).
S S S
Union: given x ∈ H(κ), clearly x ⊆ trcl(x), hence trcl( x) ⊆ trcl(x), and so x ∈
H(κ).
Replacement: Suppose that ϕ is a formula with free variables among those in the list
x, y, A, w1, . . . , wn , A, w1 , . . . , wn ∈ H(κ), and

∀x ∈ A∃!y[y ∈ H(κ) ∧ ϕH(κ) (x, y, A, w1, . . . , wn )].

So for each x ∈ A let yx ∈ SH(κ) such that ϕH(κ) (x, y, A, w1, . . . , wn ) holds. Let Y = {yx :
x ∈ A}. Then trcl(Y ) = x∈A trcl(yx ), and this has size less than κ since κ is regular.
Clearly {z ∈ H(κ) : ∃x ∈ AϕH(κ) (x, y, A, w1, . . . , wn )} ⊆ Y .
Infinity: obviously ω ∈ H(κ); see Theorem 14.27.
Foundation: true since H(κ) is transitive.
Axiom of choice: obvious.
Solutions to exercises in chapter 15
E15.1 Let I and J be sets with I infinite and |J| > 1, and let P = (P, ≤, ∅), where P is
the collection of all finite functions contained in I × J and ≤ is ⊇ restricted to P . Show
that P satisfies the condition of Lemma 15.2.
Suppose that p ∈ P . Pick any i ∈ I\dmn(p), and let j, k be distinct elements of J. Then
p ⊆ p ∪ {(i, j)}, p ⊆ p ∪ {(i, k)}, and these two extensions of p are incompatible.
E15.2 Show that if the condition in the hypothesis of Lemma 15.2 fails, then there is a
P-generic filter G over M such that G ∈ M , and G intersects every dense subset of P (not
only those in M ). [Cf. Lemma 15.1.]
Let p be such that for all q, r, if q and r are ≤ p then they are compatible. Define

G = {q : ∃r[r ≤ q and (r ≤ p or p ≤ r)].

We claim that G is as desired. Clearly G ∈ M .

59
 Note that p ∈ G, by taking r = p.
To check (1), suppose that q, r ∈ G; we want to find s ∈ G with s ≤ q, r. Choose
t ≤ q such that t ≤ p or p ≤ t, and choose u ≤ r such that u ≤ p or p ≤ u. If t ≤ p and
u ≤ p, then by the choice of p there is a v ≤ t, u. Then v ≤ u ≤ p, and v ≤ t ≤ q, hence
v ∈ G, and v ≤ q and v ≤ u ≤ r, as desired.
If t ≤ p and p ≤ u, then t ≤ u ≤ r, so t ∈ G and t ≤ q, r, as desired.
If p ≤ t and u ≤ p, then u ≤ r implies that u ∈ G, and u ≤ p ≤ t ≤ q, as desired.
Finally, if p ≤ t, u, then p ≤ q, r and p ∈ G, as desired. So (1) holds.
(2) is obvious.
Now suppose that D is dense. Choose q ∈ D such that q ≤ p. Then q ∈ G, as desired.
E15.3 Assume the hypothesis of Lemma 15.2. Show that there does not exist a P-generic
filter over M which intersects every dense subset of P (not only those which are in M ).
Hint: Take G generic, and show that {p ∈ P : p ∈ / G} is dense. Thus in the definition of
generic filter, the condition on dense sets being in M is necessary.
Let D be the set indicated in the hint. Let p be any element of P , and choose incompatible
q, r ≤ p by the hypothesis of Lemma 15.2. Then it is not true that both q, r ∈ G, as desired;
this checks that D is dense. Obviously G ∩ D = ∅, proving the assertion of the exercise.
E15.4 Show that if P satisfies the condition of Lemma 15.2, then it has uncountably many
dense subsets.

(Solution due to Josh Sanders) For each p ∈ P let p(0), p(1) be members of P such that
p(0), p(1) ≤ p and p(0) ⊥ p(1); thus p(0), p(1) < p. We now define an element pf for
each finite sequence f of 0s and 1s, by recursion on the domain of f . Let p∅ be any
element of P . Having defined pf , let pf 0 = pf (0) and pf 1 = pf (1). For each f ∈ ω 2
let Kf = {pf ↾m : m ∈ ω}. Now if f, g ∈ ω 2 and f 6= g, choose m minimum such that
f (m) 6= g(m). Clearly then pf ↾(m+1) ∈ Kf \Kg . Thus Kf 6= Kg for distinct f, g ∈ ω 2. For
each f ∈ ω 2 let Df = P \Kf . So Df 6= Dg for f 6= g.
We claim that each Df is dense; this will prove the statement of the exercise. To see
this, take any q ∈ P . If q ∈ Df , then there is nothing to prove. Suppose that q ∈ / Df .
Thus q ∈ Kf . Say q = pf ↾m . Let ε = 1 − f (m). Then p(f ↾m)ε ∈ Df and p(f ↾m)ε ≤ q, as
desired.
E15.5 Assume the hypothesis of Lemma 15.2. Show that there are 2ω filters which are
P-generic over M .
Let D0 , D1 , . . . list all of the dense subsets of P which are in M . We now define elements
rf and pf in P for f a finite sequence of 0’s and 1’s, by recursion on the length of f . Let
p∅ = 1 and choose r∅ ∈ D0 so that r∅ ≤ p∅ . Now suppose that pf and rf have been
defined, with f having domain n ∈ ω. Choose pf 0 and pf 1 both ≤ rf so that pf 0 ⊥ pf 1 .
Then choose rf 0 ≤ pf 0 with rf 0 ∈ Dn+1 , and choose rf 1 ≤ pf 1 with rf 1 ∈ Dn+1 .
For any f ∈ ω 2 let

Gf = {q ∈ P : pf ↾n ≤ q for some n ∈ ω}.

Clearly Gf is P-generic; and there are 2ω such filters.

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E15.6 Let P = ({1}, ≤, 1). Prove that the collection of all P-names is a proper class.
Clearly the following two facts hold with no assumption about P .
(1) If σ is a P -name, then so is {(σ, 1)}.
(2) If A is a set of P -names, then {(σ, 1) : σ ∈ A} is a P -name.
Now let A be the collection of all P -names, and suppose that A is a set. By (2), also
def def
τ = {(σ, 1) : σ ∈ A} is a P -name, and by (1), {(τ, 1)} is a P -name. So σ = {(τ, 1)} ∈ A.
Thus

τ ∈ {τ } ∈ {{τ }, {τ, 1}} = (τ, 1) ∈ {(τ, 1)} = σ ∈ {σ} ∈ {{σ}, {σ, 1}} = (σ, 1) ∈ τ,

contradiction.
E15.7 Show that p σ = τ iff the following two conditions hold.
(i) For every (ξ, q) ∈ σ and every r ≤ p, q one has r ξ ∈ τ .
(ii) For every (ξ, q) ∈ τ and every r ≤ p, q one has r ξ ∈ σ.
First assume that p σ = τ . For (i), suppose that (ξ, q) ∈ σ and r ≤ p, q. Let G be
P-generic over M with r ∈ G. Then also p, q ∈ G, so ξG ∈ σG and σG = τG . Hence
ξG ∈ τG , as desired. (ii) is similar.
Second assume the two conditions. To show that p σ = τ , let G be P-generic over
M with p ∈ G. Suppose that x ∈ σG . Then there is a (ξ, q) ∈ σ such that q ∈ G and
x = ξG . Choose r ∈ G such that r ≤ p, q. By (i) we have ξG ∈ τG . This shows that
σG ⊆ τG . The other inclusion is similar.
E15.8 Assume that P ∈ M , p, q ∈ P , and p ⊥ q. Show that {τ ∈ M P : p τ = ˇ∅} is a
proper class in M .
In M we define members τα of M P by recursion, such that p τα = 0̌, and such that
the ranks increase. let τ0 = 0̌. Having defined τα , let τα+1 = {(τα , q)}. Note that if G is
generic and p ∈ G, then q ∈ / G, and so (τα+1 )G = ∅; so p τα+1 = ∅. For λ a limit ordinal,
let τλ = {(τα , q) : α < λ}. Clearly p τλ = 0̌.
E15.9 A forcing order is separative iff it is antisymmetric (p ≤ q ≤ p implies that p = q),
and for all p, q, if p 6≤ q then there is an r ≤ p such that r ⊥ q. Show that the forcing
order of exercise E15.1 is separative.
If p ⊇ q ⊇ p, then p = q. Now suppose that p ⊇ 6 q. Choose (i, j) ∈ q\p. If i ∈ dmn(p),
then p ⊥ q, as desired. If i ∈
/ dmn(p), let k ∈ J\{j} and define r = p ∪ {(i, k)}. Then
r ⊇ p and r ⊥ q.
E15.10 Assume that P ∈ M is separative and p, q, r ∈ P . Prove that the following two
conditions are equivalent:
(i) p {({(∅, q)}, r)} = 1̌.
(ii) p ≤ r and p ⊥ q.
⇒: Assume that

(1) p {({(0, q)}, r)} = 1̌.

61
Suppose that p 6≤ r. By the definition of separative, choose s such that s ≤ p and s ⊥ r.
Let G be P-generic over M with s ∈ G. Then {({(0, q)}, r)}G = 0, contradiction. Thus
p ≤ r. Suppose that p and q are compatible; say t ≤ p, q. Let G be P-generic over M
with t ∈ G. Then q ∈ G, so {(0, q)}G = {0}. Also r ∈ G, so {({(0, q)}, r)}G = {{0}} 6= 1,
contradiction.
⇐: Suppose that p ≤ r and p ⊥ q. Suppose that G is P-generic over M and p ∈ G.
Then q ∈ / G, so {(0, q)}G = 0. Also, r ∈ G, so {({(0, q)}, r)}G = {{(0, q)}G } = {0} = 1, as
desired.
E15.11 Suppose that f : A → M with f ∈ M [G]. Show that there is a B ∈ M such that
f : A → B. Hint: let f = τG and B = {b : ∃p ∈ P [p b̌ ∈ rng(τ )]}.
In the hint, the definition of B takes place in M ; so B ∈ M . Suppose that b is in the range
of f . Thus b̌G = b ∈ rng(τG ), so we can choose p ∈ B such that p b̌ ∈ rng(τ ). So b ∈ B,
as desired.
E15.12 Assume that P ∈ M and α is a cardinal of M . Then for any P-generic G over
M the following conditions are equivalent:
(1) For all B ∈ M , α B ∩ M = α B ∩ M [G].
(2) α M ∩ M = α M ∩ M [G].
(1)⇒(2): Assume (1). Obviously ⊆ in (2) holds. Now suppose that f ∈ α M ∩ M [G]. By
exercise E15.11 choose B ∈ M such that f : α → B. So by (1), f ∈ M , as desired.
(2)⇒(1): Assume (2). Then ⊆ in (1) is clear. Suppose that f ∈ α B ∩ M [G]. Then
f ∈ α M ∩ M [G] since M is transitive, so f ∈ M by (2).
E15.13 Suppose that P ∈ M is a forcing order satisfying the condition of Lemma 15.2.
Assume that
M = M0 ⊆ M1 ⊆ M2 ⊆ · · · ⊆ Mn ⊆ · · · (n ∈ ω),
where Mn+1 = Mn [Gn ] for someSGn which is P-generic over Mn , for each n ∈ ω. Show
that the power set axiom fails in n∈ω Mn .
S
Assume that R = n∈ω Mn does satisfy the power set axiom. Then R |= ∃y∀z(z ⊆
P → z ∈ y). Choose y ∈ R so that R |= ∀z(z ⊆ P → z ∈ y). Say y ∈ Mn . Then
R |= Gn ⊆ P → z ∈ y. By absoluteness, R |= Gn ⊆ P . So R |= Gn ∈ y, hence
Gn ∈ y ∈ Mn . This contradicts Lemma 15.2.
E15.14 Prove that the following conditions are equivalent:

[[ϕ(σ0 , . . . , σm−1 ) ↔ ψ(σ0 , . . . , σm−1 )]] = 1

[[ϕ(σ0 , . . . , σm−1 )]] = [[ψ(σ0 , . . . , σm−1 )]].

We omit the parameters σ0 , . . . , σm−1 . First assume that [[ϕ ↔ ψ]] = 1. Then

0 = −[[ϕ ↔ ψ]]

62
 = −[[(ϕ → ψ) ∧ (ψ → ϕ)]]
= −[[¬(ϕ ∧ ¬ψ) ∧ ¬(ψ ∧ ¬ϕ)]]
= −([[¬(ϕ ∧ ¬ψ)]] · [[¬(ψ ∧ ¬ϕ)]]
= −(−[[ϕ ∧ ¬ψ) · −[[ψ ∧ ¬ϕ]])
= [[ϕ ∧ ¬ψ]] + [[ψ ∧ ¬ϕ]]
= ([[ϕ]] · −[[ψ]]) + ([[ψ]] · −[[ψ]].

It follows that [[ϕ]] = [[ψ]].

Conversely, if [[ϕ]] = [[ψ]], then we can reverse the above equations to get −[[ϕ ↔ ψ]] = 0,
so that [[ϕ ↔ ψ]] = 1.
E15.15 Prove that [[σ = τ ]] · [[τ = ρ]] ≤ [[σ = ρ]].

Suppose not. Then [[σ = τ ]] · [[τ = ρ]] · −[[σ = ρ]] 6= 0. By Theorem 9.20(i) choose p so
that e(p) ≤ [[σ = τ ]] · [[τ = ρ]] · −[[σ = ρ]]. Then (p ∗ σ = τ )M , (p ∗ τ = ρ)M , and
p ∗ (¬σ = ρ)M . Let G be P-generic over M . Then by Theorem 15.19, σG = τG , τG = ρG ,
and σG 6= ρG , contradiction.
E15.16 Prove that if ZFC |= ϕ then [[ϕ]] = 1, for any sentence ϕ.
Suppose that [[ϕ]] 6= 1. Thus −[[ϕ]] 6= 0, so by Theorem 13.20(i) choose p so that e(p) ≤
−[[ϕ]] = [[¬ϕ]]. Thus (p ∗ ¬ϕ)M . Let G be P-generic over M . Then by Theorem 15.19 we
have ¬ϕM [G] .
Solutions to exercises in chapter 16
E16.1 Show that Fn(ω1 , 2, ω1 ) preserves cardinals ≥ ω2 .

Clearly |Fn(ω1 , 2, ω1 )| = ω1 , so Fn(ω1 , 2, ω1 ) is ω2 -cc. Hence the result follows by Propo-

sition 16.5.
E16.2 A system hAi : i ∈ Ii of sets is an indexed ∆-system iff there is a set r (again
called the root such that Ai ∩ Aj = r for all distinct i, j ∈ I. Note that in an indexed
system hAi : i ∈ Ii it is possible to have distinct i, j ∈ I such that Ai = Aj ; in fact, all of
the Ai ’s could be equal, in which case the system is already an indexed ∆-system.
Prove that if κ is an uncountable regular cardinal and hAi : i ∈ Ii is a system of finite
sets with |I| ≥ κ, then there is a J ∈ [I]κ such that hAi : i ∈ Ji is an indexed ∆-system.
We may assume that |I| = κ. Define i ≡ j iff i, j ∈ I and Ai = Aj . Thus ≡ is an
equivalence relation on I. If some equivalence class K has κ elements, then hAi : i ∈ Ki
is an indexed ∆ system with |K| = κ, as desired; the kernel is Ai for any i ∈ K. Suppose
that every equivalence class has size less than κ. Then there are κ equivalence classes. Let
J ⊆ I have one element from each equivalence class, and let A = {Aj : j ∈ J}. Then
A is a collection of finite sets, and |A | = κ. Hence by Theorem 16.6 let B ∈ [A ]κ be a
∆-system. For each B ∈ B there is an iB ∈ J such that AiB = B. Let K = {iB : B ∈ B}.
Then hAj : j ∈ Ki is an indexed ∆-system, and |K| = κ since the function i is clearly
one-one.

63
E16.3 Here we work only in ZFC (or in a fixed model of it). Suppose that (X, <) is a linear
order. Let P be the set of all pairs (p, n) such that n ∈ ω and p ⊆ X ×n is a finite function.
Define (p, n) ≤ (q, m) iff m ≤ n, dmn(q) ⊆ dmn(p), ∀x ∈ dmn(q)[p(x) ∩ m = q(x), and

∀x, y ∈ dmn(q), if x < y then p(x)\p(y) ⊆ m.

Show that P has ccc.

Suppose that A is an uncountable subset of P . By the ∆-system theorem, we may assume
that hdmn(p) : (p, n) ∈ A i is a ∆-system, say with root r. We may also assume that
p ↾ r = q ↾ r whenever (p, n), (q, m) ∈ A . Now suppose that (p, n), (q, m) ∈ A . Let s be
the maximum of m and n. Clearly p ∪ q is a function, and so (p ∪ q, s) ∈ P . We claim that
(p∪q, s) ≤ (p, n), (q, m), as desired. By symmetry it suffices to show that (p∪q, s) ≤ (q, m).
Suppose that x ∈ dmn(q). Then (p ∪ q)(x) ∩ m = q(x) ∩ m = q(x). If x, y ∈ dmn(q) and
x < y, then (p ∪ q)(x)\(p ∪ q)(y) = q(x)\q(y) ⊆ m.
E16.4 Continuing exercise E16.3, suppose that we are working in a c.t.m. M of ZFC.
Let G be P-generic over M . For each x ∈ X let
[
ax = {p(x) : (p, n) ∈ G for some n ∈ ω, with x ∈ dmn(p)}.

Thus ax ⊆ ω. Show that if x < y, then ax \ay is finite.

For each z ∈ X let Dz = {(p, n) : z ∈ dmn(p)}. Given any (q, m) ∈ P , if z ∈ / dmn(q)
clearly (q ∪ {(z, 0)}, m) ∈ P , (q ∪ {(z, 0)}, m) ∈ Dz , and (q ∪ {(z, 0)}, m) ≤ (q, m). So Dz
is dense.
Choose (p, n) ∈ Dx ∩ G and (q, m) ∈ Dy ∩ G. Say (p, n), (q, m) ≥ (r, s) ∈ G. We claim
then that ax \ay ⊆ s. Let i ∈ ax \ay . Say i ∈ u(x) with (u, t) ∈ G and x ∈ dmnu. Say
(u, t), (r, s) ≥ (v, z) ∈ G. Thus v(x)\v(y) ⊆ s. Now i ∈ u(x), so i ∈ v(x). Also, i ∈ / v(y)
since i ∈/ ay . So i ∈ s, as desired.
E16.5 Continuing exercises E16.3 and E16.4, show that if x < y, then ay \ax is infinite.
Hint: for each i < ω let

E i = {(p, n) : x, y ∈ dmn(p) and |p(y)\p(x)| ≥ i},

and show that E i is dense.

For each i < ω let

E i = {(p, n) : x, y ∈ dmn(p) and |p(y)\p(x)| ≥ i}.

We claim that E i is dense. Let (q, n) be given. Wlog x, y ∈ dmn(q). Say dmn(q) is

u0 < · · · < uj = x < · · · < um−1 .

Let dmn(r) = dmn(q), r(ut ) = q(ut ) for t ≤ j; choose w > n with |w − n| = i, and let
r(ut ) = q(ut ) ∪ (w\n) for j < t. Then (q, n) ≥ (r, w) ∈ E i , as desired.

64
 Now for any i ∈ ω we show that |ay \ax | ≥ i. Choose (p, n) ∈ E i ∩ G. We claim
that p(y)\p(x) ⊆ ay \ax (as desired). Let j ∈ p(y)\p(x). So j ∈ ay , and j < n since
p(y) ⊆ n. Suppose that j ∈ ax . Say y ∈ q(x), with (q, v) ∈ G and x ∈ dmn(q). Say
(p, n), (q, v) ≥ (r, s) ∈ G. Then j ∈ r(x) since j ∈ q(x). Hence j ∈ r(x) = p(x) ∩ n, so
j ∈ p(x), contradiction.
E16.6 Define a set A of finite sets with |A | = ω while there is no ∆-system B ∈ [A ]ω .
Let A = ω. Suppose that B ∈ [A ]ω is a ∆-system, say with kernel r. Choose m ∈ ω with
n < m for all n ∈ r. Then m ∩ (m + 1) = m 6= r, contradiction.
E16.7 Let κ be singular. Define a set A of finite sets with |A | = κ while there is no
∆-system B ∈ [A ]κ .
Suppose that κ is uncountable and singular. Let hλξ : ξ < cf(κ)i be a strictly in-
creasing continuous sequence of cardinals with supremum κ. Let A = {{λξ , α} : ξ <
cf(κ), λξ < α < λξ+1 }. Clearly |A | = κ. Suppose that B ∈ [A ]κ is a ∆-system, say with
kernel G. If G = ∅, then for each ξ < cf(κ), B has at most one member in [λξ , λξ+1 ), and
so |B| ≤ cf(κ) < κ, contradiction. Let α ∈ G. Say λξ ≤ α < λξ+1 . Now B has some
member F not in [λξ , λξ+1 ), as otherwise |B| ≤ λξ+1 . Then G 6⊆ F , contradiction.
Solutions to exercises in Chapter 17
E17.1 Show that Theorem 17.2 does not extend to ω1 . Hint: consider ω1 × Q and ω1∗ × Q,
both with the lexicographic order, where ω1∗ is ω1 under the reverse order (α <∗ β iff
β < α).
Clearly each of these linear orders is dense with no first or last element. In fact, concerning
the first linear order, suppose that (ξ, r) < (η, s). If ξ < η, then (ξ, r) < (ξ, r + 1) < (η, s).
If ξ = η, then r < s, and so
 
r+s
(ξ, r) < ξ, < (ξ, s).
2

Thus ω1 × Q is dense. It does not have a first or last element, since if (ξ, r) is given, then
(0, r − 1) < (ξ, r) < (ξ, r + 1). The other order is treated similarly.
We claim that these two linear orders are not isomorphic. Suppose to the contrary
that f is an isomorphism of ω1 × Q onto ω1∗ × Q. For all ξ < ω1 let f (ξ, 0) = (α(ξ), q(ξ)).
Now if ξ < η < ω1 , then α(ξ) ≥ α(η). Hence
(1) There is a ρ < ω1 such that for all ξ ∈ [ρ, ω1 ) we have α(ξ) = α(ρ).
In fact, suppose not. Thus for every ρ < ω1 there is a ξ ∈ (ρ, ω1 ) such that α(ξ) < α(ρ).
Define hρn : n ∈ ωi by recursion as follows. Let ρ0 = 0. If ρn has been defined, choose
ρn+1 > ρn such that α(ρn+1 ) < α(ρn ). Then hα(ρn ) : n ∈ ωi is a strictly decreasing
sequence of ordinals, contradiction.
Thus (1) holds. Now hq(ξ) : ρ ≤ ξ < ω1 i is a strictly increasing sequence of rationals,
contradicting the fact that |Q| = ω.
E17.2 For any infinite cardinal κ, consider κ 2 under the lexicographic order, as for Hα .
Show that it is a complete linear order.

65
Let A ⊆ κ 2; we want to find a lub for A. We define f ∈ κ 2 by recursion, as follows: for
any α < κ,
n
f (α) = 1 if there is a g ∈ A such that g ↾ α = f ↾ α and g(α) = 1,
0 otherwise.
Suppose that g ∈ A and f < g. Then f ↾ χ(f, g) = g ↾ χ(f, g), f (χ(f, g)) = 0, and
g(χ(f, g)) = 1. This contradicts the definition of f . (We are using χ(f, g) in the same way
as for Hα .) Thus g ≤ f for all g ∈ A.
Suppose that h is an upper bound for A and h < f . Thus h ↾ χ(h, f ) = f ↾ χ(h, f ),
h(χ(h, f )) = 0, and f (χ(h, f )) = 1. By the definition of f , there is a g ∈ A such that
χ(g, f ) = χ(h, f ) and g(χ(g, f ) = 1. But then χ(g, h) = χ(h, f ) too, and it follows that
h < g. This contradicts h being an upper bound for A.
Thus f is the desired lub of A.
E17.3 Suppose that κ and λ are cardinals, with ω ≤ λ ≤ κ. Let µ be minimum such that
κ < λµ . Take the lexicographic order on µ λ, as for Hα . Show that this gives a dense linear
order of size λµ with a dense subset of size κ.
Note that obviously µ ≤ κ. Clearly µ λ is a linear order. Let

D = {f ∈ µ λ : there is a ξ < µ such that f (ξ) = 1 and f (η) = 0 for all η ∈ (ξ, µ)}.

Clearly |D| ≤ κ. We can show that µ λ is dense, and D is dense in it, by finding an element
of D between any two elements f < g of µ λ. For brevity let α = χ(f, g). Now define h by:

 f (β) if β ≤ α;

f (α + 1) + 1 if β = α + 1;
h(β) =
1
 if β = α + 2;
0 if β ∈ (α + 2, µ).
Clearly h is as desired.
If |D| < κ, we can simply add κ elements to it to satisfy the requirement that |D| = κ.
E17.4 Show that P(ω) under ⊆ contains a chain of size 2ω . Hint: remember that
|ω| = |Q|.
Let f : ω → Q be a bijection. For each real number r, let ar = f −1 [{s ∈ Q : s < r}].
Clearly r < t implies that ar ⊆ at ; and in fact ar ⊂ at since there is a rational number
u such that r < u < t, and so f −1 (u) ∈ at \ar . Now a is an isomorphic embedding by
Proposition 4.14.
E17.5 A subset S of a linear order L is weakly dense iff for all a, b ∈ L, if a < b then
there is an s ∈ S such that a ≤ s ≤ b. Show that the following conditions are equivalent
for any cardinals κ, λ such that ω ≤ κ ≤ λ:
(i) There is a linear order of size λ with a weakly dense subset of size κ.
(ii) P(κ) has a chain of size λ.
⇒: Clearly we may assume that κ < λ. Assume that L is a linear order with a weakly
dense subset D of size κ. Let f be a bijection from D to κ. For each r ∈ L\D let

66
ar = f [{d ∈ D : d < r}]. thus ar ∈ P(κ), and ar ⊆ as if r, s ∈ L\D with r < s. Also,
by the weak denseness of D, there is a d ∈ D such that r ≤ d ≤ s. Since r, s ∈ L\D, we
have r < d < s, and so d ∈ as \ar . Thus r < s implies that ar ⊂ as . The other implication
follows from this one. Note that |L\D| = λ.
⇐: Again we may assume that κ < λ. Let L be a chain in P(κ) of size λ. For each
α < κ let [
xα = {a : a ∈ L, α ∈ / a}.

(1) If α, β < κ, then xα ⊆ xβ or xβ ⊆ xα .

For, suppose that γ ∈ xα \xβ and δ ∈ xβ \xα . Say γ ∈ a ∈ L with α ∈/ a and δ ∈ b ∈ L
with β ∈/ b. Also, β ∈ a and α ∈ b, since γ ∈
/ xβ and δ ∈/ xα . Say by symmetry a ⊆ b.
Then β ∈ b, contradiction.
(2) If α ∈ κ and a ∈ L, then a ⊆ xα or xα ⊆ a.
For, suppose that α ∈ κ and a ∈ L. If α ∈ / a, then a ⊆ xα . Suppose that α ∈ a. If
α∈/ b ∈ L, we then must have b ⊂ a, since a and b are comparable. Thus xα ⊆ a in this
case. So (2) holds.
def
By (1) and (2), the set M = L∪{xα : α ∈ κ} is a chain. Its size is clearly λ. We claim
that {xα : α < κ} is weakly dense in it, which will finish the proof (expanding {xα : α < κ}
to a set of size κ if necessary). For, suppose that a, b ∈ L and a ⊂ b. Choose α ∈ b\a.
Then clearly a ⊆ xα . Also, xα ⊆ b by the proof of (2).
E17.6 Suppose that Li is a linear order with at least two elements, for each i ∈ ω. Let
Q
i∈ω Li have the lexicographic order. Show that it is not a well-order.

For each i ∈ ω let ai < bi be elements of Li . For each j ∈ ω define f j by setting, for each
i ∈ ω, 
j ai if i ≤ j or j + 1 < i;
f (i) =
bi if i = j + 1.

Then f 0 > f 1 > · · ·.

E17.7 Suppose that L is a ccc dense linear order. Show that L has a dense subset of size
≤ ω1 . Hint: let ≺ be a well-order of L, and let

N = {p ∈ L : there is an open set U in L such that p is the ≺-first element of U },

and show that N is dense in L and has size at most ω1 .

We may assume that |L| ≥ ω1 . Let ≺ be a well-order of L, and define

N = {p ∈ L : there is an open U ⊆ L such that p is the ≺-first element of U }.

Then N is dense in L. For, if a, b ∈ L and a < b, let p be the ≺-first element of (a, b).
Then p ∈ N and a < p < b, as desired.
Thus it suffices to show that |N | ≤ ω1 . Suppose, to the contrary, that |N | > ω1 .

67
 For each p ∈ N let
[
Ip = {(a, b) : a, b ∈ L, a < b, and p is the ≺-first element of (a, b)}.

Then
(1) For each p ∈ N , the set Ip is a nonempty convex open subset of L.
In fact, it is clearly nonempty and open. If u < w < v with u, v ∈ Ip , choose a, b, c, d ∈ L
such that u ∈ (a, b), v ∈ (c, d), and p is the ≺-first element of both (a, b) and (c, d). Clearly
then p is the ≺-first element of (min(a, c), max(b, d)), and w ∈ (min(a, c), max(b, d)), so
w ∈ Ip . Thus (1) holds.
(2) If p ≺ q, then Ip ∩ Iq = ∅ or Iq ⊂ Ip .
In fact, suppose that Ip ∩ Iq 6= ∅; say x ∈ Ip ∩ Iq . Choose a, b, c, d ∈ L such that x ∈ (a, b),
p is the ≺-first element of (a, b), x ∈ (c, d), and q is the ≺-first element of (c, d). Since
x ∈ (a, b) ∩ (c, d), it follows that (a, b) ∪ (c, d) = (min(a, c), max(b, d)), and p is the ≺-first
element of this interval. Thus
(min(a, c), max(b, d)) ⊆ Ip .
Now let t be any member of Iq . Say that t ∈ (e, f ), with q the ≺-first element of (e, f ).
Now q is also the ≺-first element of (c, d), so (c, d) ∪ (e, f ) = (min(c, e), max(d, f )), and
q is the ≺-first element of this interval. q is also a member of (min(a, c), max(b, d)), so
(min(a, c), max(b, d)) ∪ (min(c, e), max(d, f )) = (min(a, c, e), max(b, d, f )), and p is the ≺-
first element of this interval. Since t is in this interval, it follows that t ∈ Ip . This proves
(2).
Let hpα : α < ω2 i list the first ω2 elements of N under ≺. Let M = {Ip : p ∈ N }. Now
we construct by recursion two sequences hAα : α < ω1 i and hβα : α < ω1 i. Let β0 = 0,
and let A0 be a subset of ω2 with 0 ∈ A0 such that {Ipγ : γ ∈ A0 } is a maximal collection
of pairwise disjoint members of M . Thus A0 is countable. Suppose now that we have
constructed Aγ and βγ for all γ < α, where α < ω1 , so that each Aγ is a countable
S subset
of ω2 . Let βα be any ordinal less than ω2 and greater than each ordinal in γ<α Aγ . Then
let Aα be maximal subject to the following conditions:
(3) Aα ⊆ [βα , ω2 ).
(4) βα ∈ Aα .
(5) {Ipγ : γ ∈ Aα } is pairwise disjoint.
S
This finishes the construction. S Now take any δ greater than each ordinal in α<ω1 Aα ,
and with pε < pδ for all ε ∈ α<ω1 Aα . For each α < ω1 there is a γα ∈ Aα such
that Ipδ ∩ Ipγα 6= ∅. Hence by (2) we have Ipδ ⊂ Ipγα . Hence if α < β < ω1 then
Ipγβ ⊂ Ipγα . For brevity let q(α) = pγα for each α < ω1 . So Iq(β) ⊂ Iq(α) if α < β < ω1 .
Now for each limit ordinal µ < ω1 choose ξ(µ), η(µ), ρ(µ) such that ξ(µ) ∈ Iq(µ) \Iq(µ+1) ,
η(µ) ∈ Iq(µ+1) \Iq(µ+2) , and ρ(µ) ∈ Iq(µ+2) \Iq(µ+3) . Then ξ(µ), η(µ), ρ(µ) are distinct
elements all in Iq(µ) \Iq(µ+3) . Hence two of them, say εµ and θµ , with εµ < θµ , are on the

68
same side of Iq(µ+3) , say the left side. It follows that h(εµ , θµ ) : µ limit < ω1 i is a system
of pairwise disjoint sets, contradiction.
E17.8 Let hLi : i ∈ Ii be a system of linear orders, with I itself
P an ordered set. Show that
if each Li is dense without first or last elements, then also i∈I Li is dense without first
or last elements.
Suppose that (i, a) < (j, b). If i < j, let c be an element of Li such that a < c. Then
(i, a) < (i, c) < (j, b). If i P
= j, let c be an element of Li such that a < c < b. Then
(i, a) < (i, c) < (i, b). Thus i∈I LPi is dense. Given (i, a), choose c, d ∈ Li with c < a < d.
Then (i, c) < (i, a) < (i, d). Thus i∈I Li is does not have a first or last element.
E17.9 Let κ be any infinite cardinal number. Let L0 be a linear order similar to ω ∗ +ω +1;
specifically, let it consist of a copy of Z followed by one element a greater than every integer,
and let L1 be a linear order similar to ω ∗ + ω + 2; say it consists of a copy of Z followed
by two elements a < b greater than every integer. For any f ∈ κ 2 let
X
Mf = Lf (α) .
α<κ
κ
Show that if f, g ∈ 2 then Mf and Mg are not isomorphic.
Conclude that there are exactly 2κ linear orders of size κ up to isomorphism.
Let f, g ∈ κ 2. We assume that Mf is isomorphic to Mg and show that f = g. Let F be an
isomorphism of Mf onto Mg . Clearly
(1) For all x ∈ Mf ∪ Mg the following conditions are equivalent:
(a) x does not have an immediate predecessor.
(b) x = (a, ξ) for some ξ < κ.
Now for any x ∈ Mf , x has an immediate predecessor iff F (x) has an immediate prede-
cessor, as is easily seen. We claim then that F (a, ξ) = (a, ξ) for all ξ < κ. We prove this
by transfinite induction. Suppose that F (a, η) = (a, η) for all η < ξ. Now F (a, ξ) does not
have an immediate predecessor, so by (1) it has the form (a, ρ) for some ρ. We cannot have
ρ < ξ, since this would contradict F being one-one, by the supposition. If ξ < ρ, then we
would have F −1 (a, ξ) < F −1 (a, ρ) = (a, ξ), again contradicting the inductive assumption.
Thus F (a, ξ) = (a, ξ), finishing the inductive proof.
Next we claim
(2) For any x ∈ Mf the following conditions are equivalent:
(a) x does not have an immediate predecessor, but it has an immediate successor y
which in turn does not have an immediate successor.
(b) x = (a, ξ) for some ξ such that f (ξ) = 1.
This is obvious, and a similar condition for Mg holds.
Now the property given in (2)(a) is preserved under isomorphisms, so by the above,
for any ξ < κ,
f (ξ) = 1 iff (a, ξ) satisfies (2)(a)
iff F (a, ξ) satisfies (2)(a)
iff g(ξ) = 1.

69
Thus f = g, as desired.
The required conclusion in the exercise is clear.
E17.10 Let κ be an uncountable cardinal. Let L0 be a linear order similar to η + 1 + η · ω1∗ ;
specifically consisting of a copy of the rational numbers in the interval (0, 1] followed by
Q × ω1 , where Q × ω1 is ordered as follows: (r, α) < (s, β) iff α > β, or α = β and r < s.
Let L1 be a linear order similar to η · ω1 + 1 + η · ω1∗ ; specifically, we take L1 to be the set

{(q, α, 0) : q ∈ Q, α < ω1 } ∪ {(0, 0, 1)} ∪ {(q, α, 2) : q ∈ Q, α < ω1 },

with the following ordering:

(q, α, 0) < (r, β, 0) iff α < β, or α = β and q < r;

(q, α, 0) < (0, 0, 1) < (r, β, 2) for all relevant q, r, α, β;
(q, α, 2) < (r, β, 2) iff α > β, or α = β and q < r.

For each f ∈ κ 2 let X

Mf = Lf (α) .
α<κ

Show that each Mf is a dense linear order without first or last elements, and if f, g ∈ κ 2
and f 6= g, then Mf and Mg are not isomorphic.
Conclude that for κ uncountable there are exactly 2κ dense linear orders without first
or last elements, of size κ, up to isomorphism.
For L0 , if q is an element in the initial (0, 1] part, then q −1 < q; so L0 has no least element.
If (q, α) is an element in the second part, then (q, α) < (q + 1, 0); so L0 has no greatest
element. For denseness, suppose that x < y in L0 . If both are in the first part, clearly
there is a z such that x < z < y. Suppose that x is in the first part and y = (q, α) is in
the second part. Then x < (q − 1, α) < y. Finally, suppose that x = (r, β) and y = (s, γ)
are both in the second part. If β > γ, then x < (r + 1, β) < y. If β = γ, then r < s and
the desired element is clear.
For L1 , given any element (q, α, 0) in the first part, we have (q − 1, α, 0) < (q, α, 0),
so L1 does not have a least element. If (q, α, 2) is any element in the third part, then
(q, α, 2) < (q + 1, α, 2), so L2 does not have a greatest element. Suppose that x < y in L1 .
We can consider several cases.
Case 1. x = (q, α, 0), y = (r, β, 0), and α < β. Then x < (q + 1, α, 0) < y.
Case 2. x = (q, α, 0), y = (r, α, 0). Then q < r; so with q < t < r we have
x < (t, α, 0) < y.
Case 3. x = (q, α, 0), y = (0, 0, 1). Then x < (q + 1, α, 0) < y.
Case 4. x = (q, α, 0), y = (r, β, 2). Then x < (0, 0, 1) < y.
Case 5. x = (0, 0, 1), y = (q, α, 2). Then x < (q − 1, α, 2) < y.
Case 6. x = (q, α, 2), y = (r, β, 2), and α > β. Then x < (q + 1, α, 2) < y.
Case 7. x = (q, α, 2), y = (r, α, 2). Thus q < r. Then with q < t < r we have
x < (t, α, 2) < y.
Thus L1 is dense.

70
 Before beginning the main part of the exercise, we note the following facts.
(1) Each q in the first part of L0 , except for the final 1, has character (ω, ω).
(2) The final 1 has character (ω, ω1 ). A strictly decreasing sequence with limit 1 is

h(0, α) : α < ω1 i.

(3) Every element in the second part of L0 has character (ω, ω).
(4) Every element in the first part of L1 has character (ω, ω).
(5) (0, 0, 1) has character (ω1 , ω1 ).
In fact, a strictly increasing sequence with limit (0, 0, 1) is

h(0, α, 0) : α < ω1 i.

A strictly decreasing sequence with limit (0, 0, 1) is

h(0, α, 2) : α < ω1 i.

Now we suppose that f ∈ κ 2. We give some properties of Mf :

(6) The character of an element (x, ξ) of Mf is equal to the character of x in Lf (ξ) .
This is true because Lf (ξ) is a convex set in Mf , i.e., if u < v < w with u, w ∈ Lf (ξ) , then
v ∈ Lf (ξ) . Also, the fact that Lf (ξ)) does not have a least or greatest element is needed to
see (6).
(7) For each ξ < κ there is a unique element of Mf of the form (xξ , ξ) which has character
(ω, ω1 ) if f (ξ) = 0 and has character (ω1 , ω1 ) if f (ξ) = 1; all other elements of the form
(y, ξ) have character (ω, ω).
κ
Now we can treat the main part of the exercise. Suppose that f, g ∈ 2 and Mf is
isomorphic to Mg ; say F is an isomorphism. The sequence

hF (xξ , ξ) : ξ < κi,

with xξ given in (7), is an increasing sequence of elements of Mg such that all other elements
of Mg have character (ω, ω). In Mg there is only one sequence of order type κ consisting
of elements which do not have character (ω, ω), by (7) for Mg . Hence F (xξ , ξ) = (yξ , ξ),
where yξ is defined for Mg like xξ was for Mf in (7). But xξ and yξ then have the same
characters, and so f (ξ) = g(ξ). Thus f = g.
The final statement of the exercise is clear.
Solutions to exercises in Chapter 18
E18.1 Let κ be an uncountable regular cardinal, and suppose that there is a κ-Aronszajn
tree. Show that there is one which is a normal subtree of <κ 2. Hint: for each α < κ let gα
be an injection of Levα (T ) into |Levα (T )| 2 and glue these maps together.

71
Let T be a κ-Aronszajn tree. By Theorem 18.7 we may assume that it is well-pruned. For
s ∈ T and α < ht(s, T ) let sα be the unique element of height α below s. For each α < κ,
let gα be an injection from Levα (T ) into |Levα (T )| 2.
We define by recursion sequences hµα : α < κi and hFα : α < κi. Let µ0 = |Lev0 (T )|,
and let F0 = g0 . Now suppose that µα , Fα have been defined so that the following
conditions hold:
(1α ) µα < κ.
S
(2α ) Fα is a function with domain β≤α Levβ (T ).
(3α ) for all β, γ < α, if β < γ then µβ ≤ µγ and Fβ ⊆ Fγ .
(Clearly these conditions hold for α = 0.) Now let µα+1 = µα + |Levα+1 (T )| (ordinal addi-
tion). Let Fα+1 be the extension of Fα such that for every t ∈ Levα (T ), every immediate
successor s of t, and every β < µα+1 ,

(Fα (t))(β) if β < µα ,
(Fα+1 (s))(β) =
(gα (s))(ξ) if β = µα + ξ with ξ < |Levα+1 (T )|.

Clearly (1α+1 )–(3α+1 ) hold.

Now suppose that α is a limitS ordinal andSµβ , Fβ have been defined for all β < α so
that (1β )–(3β ) hold. Let ν = β<α µβ , G = β<α Fβ , and set µα = ν + |Levα (T )|. Let
Fα be the extension of G such that for every s ∈ Levα (T ) and every β < µα ,

(G(sγ ))(β) if β < µγ with γ < α,
(Fα (s))(β) =
(gα (s))(ξ) if β = ν + ξ with ξ < |Levα (T )|.

Clearly (1α )–(3

Sα ) hold.
Let H = α<κ Fα . Clearly
(4) If s ∈ Levα (T ), then H(s) ∈ µα 2,
(5) If u < s then H(u) ⊂ H(s).
We prove (5) by induction on the level of s. It is vacuously true for level 0. Now suppose
inductively that s has level α + 1. Say that t is the immediate predecessor of s. Let γ be
the level of u. Then for any β < µγ we have

(H(s))(β) = (Fα+1 (s))(β) = (Fα (t))(β) = (H(t))(β) = (H(u))(β).

Finally, suppose inductively that s has limit level α. Then for any β < γ we have

(H(s))(β) = (Fγ (sγ ))(β) = (H(u))(β).

Hence (5) holds.

(6) If s, t ∈ T have the same height and s 6= t, then H(s) 6= H(t).
We prove (6) by induction on the common height α of s and t. If α = 0 the conclusion is
clear since g0 is one-one. Suppose inductively that they both have height α + 1. Let s′ , t′

72
be their immediate predecessors. If s′ 6= t′ , then H(s′ ) 6= H(t′ ), so H(s) 6= H(t) by (5).
Suppose that s′ = t′ . Then H(s) 6= H(t) since gα is one-one. Finally, suppose inductively
that α is limit. Then H(s) 6= H(t) since gα is one-one. So (6) holds.
Now let T ′ = {h ∈ <κ 2 : h ⊆ H(s) for some s ∈ T }. We claim that T ′ is as desired.
Clearly it is a normal subtree of <κ 2. Now consider any α < κ. Choose β minimum such
that α ≤ µβ .
(7) If h ∈ T ′ with dmn(h) = α, then there is an s ∈ T of height β such that h ⊆ H(s).
In fact, choose t ∈ T such that h ⊆ H(t). Then dmn(H(t)) ≥ α ≥ µβ , so t has height ≥ β.
Let s ∈ T of height β with s ≤ t. Then H(s), h ⊆ H(t), so h ⊆ H(s), as desired.
It follows from (7) that each level of T ′ has size less than κ. From (5) and (7) it
follows that T ′ does not have a chain of size κ.
E18.2 Do exercise E18.1 for κ-Suslin trees.
We use the same construction as in exercise 18.1. Thus our new tree T ′ does not have any
chain of size κ. Suppose that A is an antichain of size κ. For each a ∈ A choose sa ∈ T
such that a ⊆ H(sa ). Since T is a Suslin tree, choose distinct a, b ∈ A such that sa and sb
are comparable. Say sa ≤ sb . Then H(sa ) ⊆ H(sb ). Since a, b ⊆ H(sb ), it follows that a
and b are comparable, contradiction.
E18.3 Suppose that T and T ′ are κ-Aronszajn trees. Define an order < on T × T ′ by
(s, s′ ) < (t, t′ ) iff s < t and s′ < t′ . Show that T × T ′ is not a tree.
Clearly T × T ′ is a partial order. Let x0 < x1 < x2 in T and y0 < y1 < y2 in T ′ . Then
(x1 , y0 ) < (x2 , y2 ) and (x0 , y1 ) < (x2 , y2 ), but (x1 , y0 ) and (x0 , y1 ) are incomparable.
E18.4 Suppose that T and T ′ are κ-Aronszajn trees. Let
[
T ×′ T ′ = Levα (T ) × Levα (T ′ );
α<κ

(s, s ) < (t, t ) iff (s, s′ ), (t, t′ ) ∈ T ×′ T ′ , s < t, and s′ < t′ ;

′ ′

Show that (T ×′ T ′ , <) is a κ-Aronszajn tree.

Clearly T ×′ T ′ is a partial order. Suppose that (s, t) ∈ T ×′ T ′ . Say that s and t have height
α, and let haξ : σ < αi and hbξ : σ < αi are the systems of predecessors of s, t respectively.
Clearly h(aξ , bξ ) : ξ < αi is the system of predecessors of (s, t). So we have a tree. Each
level α has size |Levα (T ) × Levα (T ′ )| < κ; so T ×′ T ′ is a κ-tree. If h(xξ , yξ ) : ξ < βi is
a chain in T ×′ T ′ , then hxξ : ξ < βi is a chain in T , and so β < κ. Thus T ×′ T ′ is a
κ-Aronszajn tree.
E18.5 Assume that κ is regular and uncountable. Suppose that T is a κ-Suslin tree. With
the order on T ×′ T given in exercise E18.4, show that T ×′ T is not a κ-Suslin tree. Hint:
first show that for every α < κ there is an element s of T at level α such that there are
incomparable t, u > s.
First we claim

73
(1) For every α < κ there is an s of height α such that there are incomparable t, u > s.
In fact, otherwise we get an α < κ such that for every s of height α, the set {t ∈ T : s < t}
is a chain. Since all chains have size less than κ, for each s of level α choose βs greater
than the level of all t with s < t. Let γ < κ be such that βs < γ for all s of level α.
Let β be any element of T of level γ, and let s be its predecessor at level α. But γ > βs ,
contradiction. Thus (1) holds.
Now we construct elements sα , tα , uα for α < κ by recursion. Suppose that they have
been constructed for all β < α. Let

γ = max(sup{ht(tβ , T ) : β < α}, sup{ht(uβ , T ) : β < α}) + 1.

By (1), choose sα of height γ with incomparable tα , uα > sα of some common level > γ.
We claim that h(tα , uα ) : α < κi is an antichain in T ×′ T . For, suppose that β < α
and (tβ , uβ ) < (tα , uα ). So tβ < tα and uβ < uα . Since tβ , sα < tα , we have tβ < sα
(since sα is at a higher level than tβ ). Similarly, uβ < sα . So tβ and uβ are comparable,
contradiction.
E18.6 A tree T is everywhere branching iff every t ∈ T has at least two immediate
successors. Show that every everywhere branching tree has at least 2ω branches.
We define a branch bf for every f ∈ ω 2 by defining elements ah for every h ∈ <ω 2 by
recursion on dmn(h). Let a∅ be a root of the tree. Suppose that ah has been defined for
every h ∈ n 2. For each h ∈ n 2, let ah0 and ah1 be two immediate successors of ah . This
finishes the definition of the ah ’s. Now let bf be an extension of haf ↾n : n ∈ ωi to a branch.
Clearly this is as desired.
E18.7 Show that the hypothesis that all levels are finite is necessary in König’s theorem.
For each n ∈ ω let fn ∈ n+1 ω be any function such that fn (0) = n, and let T be the tree
consisting of all g ∈ <ω ω such that g ⊆ fn for some n. Then two elements g, h ∈ T are
comparable iff they are both contained in the same fn . If C is a maximal chain in T , it
must be a subset of some fn , and hence is finite.
E18.8 Show that if κ is singular with cf(κ) = ω, then there is no κ-Aronszajn tree with
all levels finite.
Let hαn : n ∈ ωi be a strictly increasing sequence of ordinals with supremum κ. Suppose
that T is a κ-Aronszajn tree with all levels finite. Define

T ′ = {t ∈ T : there is an n ∈ ω such that t has height αn }.

Then T ′ , with the order induced by T , is a tree of height ω with all levels finite. Hence
by König’s theorem it has an infinite branch B. Let B ′ = {t ∈ T : t ≤ s for some s ∈ T ′ .
Then B ′ is a branch in T of size κ, contradiction.
E18.9 Prove that if κ is singular and there is a cf(κ)-Aronszajn tree, then there is a
κ-Aronszajn tree with all levels of power less than cf(κ).

74
Let T be a cf(κ)-Aronszajn tree. Let hµα : α < cf(κ)i be a strictly increasing continuous
sequence of cardinals with supremum κ, and with µ0 = 0. We define

T ′ = {(t, β) : there is an α < cf(κ) such that t ∈ Levα (T ) and µα ≤ β < µα+1 ;
(t, β) < (t′ , β ′ ) iff t < t′ , or t = t′ and β < β ′ .

Clearly this gives a partial order. To show that it is a tree, suppose that (t, β) ∈ T ′ . We
define a function f from β into the set of predecessors of (t, β) as follows. Let ht(t) = α.
Suppose that γ < β. then there is a δ such that µδ ≤ γ < µδ+1 . Clearly δ ≤ α. We define
f (γ) = (t′ , γ), where t′ is the predecessor of t at level δ. Clearly then f (γ) < (t, β). Clearly
also f is order preserving and maps onto the set of all predecessors of (t, β). Thus T ′ is a
tree. For each β < κ, say with µα ≤ β < µβ+1 we have

Levβ (T ′ ) = {(t, β) : t ∈ Levα (T )}.

Thus each level of T ′ has size less than cf(κ). If B is a branch of size κ, then for each
β < κ it has an element of height at least β, and hence for each α < cf(κ) it has an element
whose first coordinate has height at least α. These first coordinates are linearly ordered.
This contradicts T being a cf(κ)-Aronszajn tree.
Thus T ′ is as desired.
E18.10 Show that for every infinite cardinal κ there is an eventually branching tree T of
height κ such that for every subset S of T , if S is a tree under the order induced by T and
every element of S has at least two immediate successors, then S has height ω.
The idea is to put a copy of <ω 2 on top of longer and longer chains. More precisely, define

T = {(α, ξ, ∅) : α < κ, ξ < α} ∪ {(α, α, f ) : α < κ, f ∈ <ω 2};

(α, ξ, f ) < (β, η, g) iff α = β and either ξ < η, or ξ = η = α and f ⊂ g.

Clearly T is a tree. The height of an element (α, ξ, ∅) is ξ, and the height of an element
(α, α, f ) is α + n, where f ∈ n 2. In particular, T has height κ.
Now suppose that S is as indicated in the exercise, and take any element (α, ξ, f ) of
S. (α, ξ, f ) is a root of S iff α = ξ and f = ∅. It follows that all the non-root elements of
S have the form (α, α, f ), and so in S the height of every element is finite.
E18.11 Show that if κ is an uncountable regular cardinal and T is a κ-Aronszajn tree, then
T has a subset S such that under the order induced by T , S is a well-pruned κ-Aronszajn
tree in which every element has at least two immediate successors.
By Theorem 18.7, we may assume that T is well-pruned. Now we construct a strictly
increasing sequence hαξ : ξ < κi of ordinals less than κ. Let α0 = 0. Suppose that αξ
has been defined. Now T is eventually branching. (See the remark before Theorem 18.7.)
Hence for each t ∈ Levαξ (T ) there is an ordinal βt > αξ such that t has at least two
successors at level βt . Let αξ+1 be any ordinal less than κ such that βt < αξ+1 for all
t ∈ Levαξ (T ). Note by the well-prunedness condition, each t ∈ Levαξ has at least two

75
successors at level αξ+1 . Finally, suppose that η is a limit ordinal less than κ, and αξ has
been constructed
S for all ξ < η. Let αη = supξ<η αξ .
Let S = ξ<κ Levαξ (T ). Clearly S is as desired.
Solutions to exercises in Chapter 19
E19.1 Assume that κ is an uncountable regular cardinal and hAα : α < κi is a sequence
of subsets of κ. Let D = △α<κ Aα . Prove the following:
(i) For all α < κ, the set D\Aα is nonstationary.
(ii) Suppose that E ⊆ κ and for every α < κ, the set E\Aα is nonstationary. Show
that E\D is nonstationary.
(i): For any β < κ,

D\Aβ = {α < κ : ∀γ < α(α ∈ Aγ ) and α ∈

/ Aβ }
⊆ {α < κ : α ≤ β};

Hence D\Aβ is nonstationary.

(ii): For each β < κ let Cβ be club in κ such that (E\Aβ ) ∩ Cβ = ∅. Let F be the
diagonal intersection of the Cβ ’s; thus

F = {γ < κ : ∀α < γ(γ ∈ Cα )}.

Thus F is club. We claim that F ∩ (E\D) = ∅ (as desired). For, suppose that γ ∈
F ∩ (E\D). Since γ ∈
/ D, there is a β < γ such that γ ∈
/ Aβ . Since γ ∈ F , we have γ ∈ Cβ .
Since (E\Aβ ) ∩ Cβ = ∅, this is a contradiction.
E19.2 Let κ > ω be regular. Show that there is a sequence hSα : α < κi of stationary
subsets of κ such that Sβ ⊆ Sα whenever α < β < κ, and △α<κ Sα = {0}. Hint: use
Theorem 19.12.
By Theorem 19.12, let hAα : α < κi be pairwise
S disjoint stationary subsets of κ. Then
Aα \(α + 1) is also stationary. Let Sα = β>α (Aβ \(β + 1)). Then Sα is stationary and
α < β implies that Sβ ⊆ Sα . Let D be the diagonal intersection of the Sα ’s:

D = {γ < κ : ∀α < γ(γ ∈ Sα )}.

Thus 0 ∈ D. Suppose that 0 6= γ ∈ D. Then 0 < γ, so γ ∈ S0 . Hence there is a β > 0 such

that γ ∈ Aβ \(β + 1). Hence β < γ. So γ ∈ Sβ . Choose δ > β such that γ ∈ Aδ \(δ + 1).
So γ ∈ Aβ ∩ Aδ = ∅, contradiction.
E19.3 Suppose that κ is uncountable and regular, and for each limit ordinal α < κ we are
given a function fα ∈ ω α. Suppose that S is a stationary subset of κ. Let n ∈ ω. Show
that there exist a t ∈ n κ and a stationary S ′ ⊆ S such that for all α ∈ S ′ , fα ↾ n = t.
We define sequences hS0 , S1 , . . . , Sn i of stationary subsets of S and hβ0 , . . . , βn−1 i of ordi-
nals less than κ. Let S0 = S. Suppose that Si has been defined, i < n. Let g(α) = fα (i) for
all α ∈ Si . Then g is a regressive function, and hence there exist a stationary subset Si+1
of Si and an ordinal βi such that g(α) = βi for all α ∈ Si+1 . This finishes the construction.

76
 If α ∈ Sn , then for any i < n we have α ∈ Si+1 , and hence fα (i) = βi . Hence we can
let t(i) = βi for all i < n, and the property of the exercise holds.

E19.4 Suppose that cf(κ) > ω, C ⊆ κ is club of order type cf(κ), and hcβ : β < cf(κ)i is
the strictly increasing enumeration of C. Let X ⊆ κ. Show that X is stationary in κ iff
{β < cf(κ) : cβ ∈ X} is stationary in cf(κ).

Assume X ⊆ κ. Let X ′ = {β < cf(κ) : cβ ∈ X}.

⇒: Assume that X is stationary in κ. We want to show that X ′ is stationary in
cf(κ). Let D′ be club in cf(κ). Define D = {cβ : β ∈ D′ }. We claim that D is club in
S closure, suppose that α < κ is a limit ordinal and D ∩ α is unbounded in α. Let
κ. For
γ = {β ∈ D′ : cβ < α}.

(1) γ < cf(κ).

For, since C is unbounded in κ, there is a β < cf(κ) such that α < cβ . Now if ε ∈ D′ and
cε < α we have cε < cβ , hence ε < β. Therefore γ ≤ β < cf(κ), proving (1).

(2) γ is a limit ordinal, and D′ ∩ γ is unbounded in γ.

For, suppose that δ < γ. Choose β ∈ D′ such that cβ < α and δ < β. Since D ∩ α is
unbounded in α, choose cε ∈ D ∩ α such that cβ < cε . Then δ < β < ε ≤ γ. Since β ∈ D′ ,
this proves (2).

(3) cγ = α.
S
For, suppose that δ < cγ . Now cγ = β<γ cβ by (2), so there is a β < γ such that
δ < cβ . By the definition of γ, there is a β ′ ∈ D′ such that β < β ′ and cβ ′ < α. Thus
δ < cβ < cβ ′ < α, so δ < α. This proves that cγ ≤ α. On the other hand, suppose that
δ < α. Since D ∩ α is unbounded in α, there is a β ∈ D′ such that δ < cβ < α. Thus
β ≤ γ, so δ < cγ . This proves that α ≤ cγ , and finishes the proof of (3).
Now by (2) we have γ ∈ D′ , and hence (3) yields α ∈ D, as desired; we have now
proved that D is closed in κ.
To show that D is unbounded in κ, let α < κ be arbitrary. Choose β < cf(κ) such
that α < cβ . Since D′ is unbounded in cf(κ), choose β ′ ∈ D′ such that β < β ′ . Thus
α < cβ < cβ ′ ∈ D, as desired.
So we have shown that D is club in κ. Since X is stationary in κ, choose β ∈ D′ such
that cβ ∈ X. thus β ∈ D′ ∩ X ′ , as desired.
⇐: Assume that X ′ is stationary in cf(κ). Let D be club in κ, and let D′ = {β <
cf(κ) : cβ ∈ D}. We claim that D′ is club in cf(κ). To show that it is closed, suppose
that α < cf(κ) is a limit ordinal and D′ ∩ α is unbounded in α. We claim that D ∩ cα
is unbounded in cα . For, suppose that γ < cα . Then there is a β < α such that γ < cβ .
Choose δ ∈ D′ ∩ α such that β < δ; this is possible since D′ ∩ α is unbounded in α. Thus
γ < cβ < cδ ∈ D ∩ cα , as desired. So D′ is closed in cf(κ). To show that it is unbounded,
suppose that α < cf(κ). Now C ∩ D is club in κ, so there is a β such that cα < cβ ∈ D.
So α < β ∈ D′ . This shows that D′ is unbounded in cf(κ). Hence D′ is club in cf(κ).
Choose β ∈ D′ ∩ X ′ . Then cβ ∈ D ∩ X, as desired.

77
 E19.5 Suppose that κ is regular and uncountable, and S ⊆ κ is stationary. Also, suppose
that every α ∈ S is an uncountable regular cardinal. Show that

def
T = {α ∈ S : S ∩ α is non-stationary in α}

is stationary in κ. Hint: given a club C in κ, let C ′ be the set of all limit points of C and
let α be the least element of C ′ ∩ S; show that α ∈ T ∩ C.
Assume the notation of the exercise. Let C be club in κ; we want to show that T ∩ C 6= ∅.
Let C ′ be the set of all limit points of C, i.e., the set of all limit ordinals α ∈ κ such that
C ∩ α is unbounded in α. Clearly C ′ ⊆ C, and C ′ is club in κ. Since S is stationary in
κ, let α be the least element of S ∩ C ′ . Clearly C ′ ∩ α is closed in α; we claim that it is
also unbounded in α. For, suppose that β < α. Now C ∩ α is unbounded in α, so we can
construct a sequence hγi : i < ωi of members of C such that β < γ0 < γ1 < · · · < α. Let
δ = supi∈ω γi . Then δ ∈ C ′ , and δ < α since α is uncountable and regular. So C ′ ∩ α is
club in α. Now S ∩ C ′ ∩ α = ∅ by the minimality of α, so C ′ ∩ α is a club in α which shows
that S ∩ α is non-stationary in α. So α ∈ T ∩ C, as desired.
E19.6 Suppose that κ is uncountable and regular, and κ ≤ |A|. Suppose that C is a
closed subset of [A]<κ and D is a directed subset of C with |D| < κ. S(Directed means that
if x, y ∈ D then there is a z ∈ D such that x ∪ y ⊆ z.) Show that D ∈ C. Hint: use
induction on |D|.
S S
Proof. If D is finite, then D ∈ D; so D ∈ C. Suppose that |D| = ω; say D =
{xn : nS∈ ω}. For each n ∈ ω choose yn ∈S S {xm : m ≤ n} ∪ {ym : m < n} ⊆ yn .
D so that
Then n∈ω yn ∈ C since C is closed, and D = n∈ω yn .
Now suppose inductively that |D| > ω. Let |D| = κ and write D = {xα : α < κ}.
For all y, z ∈ D let f (y, z) ∈ D be such that y, z ⊆ f (y, z). We now define hEα : α < κi
by recursion. Suppose that Eβ has been defined forS all β < α so that Eβ ⊆ D, Eβ
is directed, and |Eβ | ≤ |β| + ω. Let F0 = {xα } ∪ β<α Eβ . So |F0 | S ≤ |α| + ω. Let
Fn+1 = Fn ∪ {f (x, y) : x, y ∈ Fn }. Then |Fn+1 ≤ |α| + ω. Let Eα = n∈ω Fn . Then
Eα ⊆ D, Eα is directed, and |Eα | ≤ |α| + ω.
def S
S BySthe inductive hypothesis, y α = Eα is in C, and yα ⊆ yβ for α < β. Hence
D = α<κ yα ∈ C.

E19.7 Let κ be uncountable and regular, and κ ≤ |A|. If f : [A]<ω → [A]<κ let Cf =
{x ∈ [A]<κ : ∀s ∈ [x]<ω [f (s) ⊆ x]}. Show that Cf is club in [A]<κ .
S
Suppose that xξ ∈ Cf for all ξ < α, with α < κ and xξ ⊆ xη for ξ < η. Clearly ξ<α xξ ∈
hS i<ω
Cf if α is a successor ordinal. Suppose that α is a limit ordinal. Take any s ∈ x
ξ<α ξ .
S
Then there is a ξ < α such that s ∈ [xξ ]<ω , and hence f (s) ∈ xξ ⊆ η<ζ xη . Thus Cf is
closed.
To show that Cf is unbounded, let y ∈ [A]<κ . DefineSz0 = y and zn+1 = zn ∪ {f (s) :
s ∈ [zn ]<ω }. By induction, zn ∈ [A]<κ for all n ∈ ω. Now n∈ω zn ∈ Cf , showing that Cf
is unbounded.

78
 E19.8 (Continuing Exercise E19.7) Let κ be uncountable and regular, and κ ≤ |A|. Let D
be club in [A]<κ . Show that there is an f : [A]<ω → [A]<κ such that Cf ⊆ D. Hint: show
that there is an f : [A]<ω → D such that ∀e ∈ [A]<ω [e ⊆ f (e)] and ∀e1 , e2 ∈ [A]<ω [e1 ⊆
e2 → f (e1 ) ⊆ f (e2 )].
We claim that there is an f : [A]<ω → D such that ∀e ∈ [A]<ω [e ⊆ f (e)] and ∀e1 , e2 ∈
[A]<ω [e1 ⊆ e2 → f (e1 ) ⊆ f (e2 )]. We define f by induction on |e|. Let f (∅) be any
member of D. Suppose that f (e) has been defined for all e ∈ [A]<ω such that |e| < m, and
suppose that e ∈ [A]<ω with |e| = m. Let f (e) be a member of D such that e ⊆ f (e) and
f (e\{a}) ⊆ f (e) for all a ∈ e. Clearly f is as desired.
Now we show that Cf ⊆ D. Let x ∈ Cf . Note that {f (e) : e ∈ [x]<ω } is directed and
has union x. Hence x ∈ D by Exercise 19.6.
E19.9 Let κ be uncountable and regular, κ ≤ |A|, and A ⊆ B. If Y ∈ [A]<κ , let
Y B = {x ∈ [B]<κ : x ∩ A ∈ Y }. Show that if Y is club in [A]<κ , then Y B is club in [B]<κ .
Assume the hypotheses. Suppose that α < κ and hxξ : ξ < αi is a sequence of members
of Y B with xξS⊆ xη for ξ < η. ThenSxξ ∩ A ∈ Y for all ξ < α, and xξ ∩ A ⊆ xη ∩ A for
ξ < η. Hence ξ<α (xξ ∩ A) ∈ Y . So ξ<α xξ ∈ Y B . Thus Y B is closed.
To show that Y B is unbounded, let b ∈ [B]<κ . Then b ∩ A ∈ [A]<κ , so there is a c ∈ Y
such that b ∩ A ⊆ c. Then b ⊆ c ∪ (b\A), and (c ∪ (b\A)) ∩ A = c ∈ Y . So c ∪ (b\A) ∈ Y B .
E19.10 Let κ be uncountable and regular, κ ≤ |A|, and A ⊆ B. If Y ∈ [B]<κ , let
Y ↾ A = {y ∩ A : y ∈ Y }. Show that if Y is stationary in [B]<κ then Y ↾ A is stationary
in [A]<κ .
Assume the hypotheses. Suppose that C is club in [A]<κ . Then by Exercise E19.9, C B is
club in [B]<κ . Choose y ∈ Y ∩ C B . Then y ∩ A ∈ (Y ↾ A) ∩ C.
E19.11 With κ, A, B as in exercise E19.9, suppose that f : [B]<ω → [B]<κ . For each
e ∈ [A]<ω define

x0 (e) = e;
xn+1 (e) = xn (e) ∪ {f (s) : s ∈ [xn (e)]<ω };
[
w(e) = xn (e).
n∈ω

S
Also, for each y ∈ [A]<κ let v(y) = {w(e) : e ∈ [y]<ω }.
Prove that w(e) ∈ Cf for all e ∈ [A]<ω and v(y) ∈ Cf for all y ∈ [A]<κ .
If z ∈ [w(e)]<ω , then z ∈ [xn (e)]<ω for some n, and hence f (z) ∈ xn+1 (e) ⊆ w(e). Thus
w(e) ∈ Cf . Note that if e1 ⊆ e2 then xn (e1 ) ⊆ xn (e2 ) for all n (by induction), and so
<ω
w(e1S) ⊆ w(e2 ). Now supposeS that z ∈ [v(y)]
S . Then there is a finite F ⊆ [y]<ω such that
′
z ⊆ e∈F w(e). Let e = e∈F e. Then e∈F w(e) ⊆ w(e′ ). So z ∈ [w(e′ )]<ω . It follows
from the first part of this solution that f (z) ∈ w(e′ ) ⊆ v(y). Thus v(y) ∈ Cf .
E19.12 With κ, A, B as in exercise E19.9, suppose that S is stationary in [A]<κ . Show
that S B is stationary in [B]<κ . Hint: use exercises E19.8 and E19.11.

79
Assume the hypotheses. Suppose that D is club in [B]<κ . By Exercise 19.8 there is an
f : [B]<ω → [B]<κ such that Cf ⊆ D. For any e ∈ [A]<ω let g(e) = w(e) ∩ A. We claim
that Cf ↾ A = Cg . Suppose that y ∈ Cf , so that y ∩ A ∈ Cf ↾ A. To show that y ∩ A ∈ Cg ,
let e ∈ [y ∩ A]<ω . Then xn (e) ⊆ y by induction on n. Since xn (e) = e, it is true for n = 0.
Suppose that xn (e) ⊆ y. If s ∈ [xn (e)]<ω , then s ∈ [y]<ω , so f (s) ⊆ y since y ∈ Cf . Hence
xn+1 (e) ⊆ y. It follows that w(e) ⊆ y, and so g(e) ⊆ y ∩ A. This shows that y ∩ A ∈ Cg ,
and proves that Cf ↾ A ⊆ Cg . Now suppose that y ∈ Cg . Hence ∀e ∈ [y]<ω [g(e) ⊆ y]. We
claim that v(y)∩A = y. For, if e ∈ [y]<ω , then g(e) ⊆ y, i.e., w(e)∩A ⊆ y. So v(y)∩A ⊆ y.
If a ∈ y, then a ∈ w({a}) ⊆ v(y); so v(y) ∩ A = y. This shows that Cg ⊆ Cf ↾ A. Thus
Cg = Cf ↾ A.
Choose z ∈ Cg ∩ S. Then z ∈ (Cf ↾ A) ∩ S, so there is a y ∈ Cf such that z = y ∩ A.
Thus y ∩ A ∈ S, so y ∈ S B ∩ D.
Solutions to exercises in Chapter 20
E20.1 Suppose that κω > κ. Show that there is a family A of subsets of κ, each of size
ω, with |A | = κ+ and the intersection of any two members of A is finite.
Let K = <ω κ. Thus |K| = κ. Let F be a bijection from K to κ. For each f ∈ ω κ let

Xf = F [{f ↾ m : m ∈ ω}].

Clearly each Xf has size ω. If f, g ∈ ω κ and f 6= g, choose p ∈ ω such that f (p) 6= g(p).
Then

Xf ∩ Xg = F [{f ↾ m : m ∈ ω}] ∩ F [{g ↾ m : m ∈ ω}]

= F [{f ↾ m : m ∈ ω} ∩ {g ↾ m : m ∈ ω}]
⊆ F [{f ↾ m : m ≤ p}]

and hence Xf ∩ Xg is finite. Since κω ≥ κ+ , the desired result follows.

E20.2 Suppose that κ is any infinite cardinal, and λ is minimum such that κλ > κ. Show
that there is a family A of subsets of κ, each of size λ, with the intersection of any two
members of A being of size less than λ, and with |A | = λ+ .
Let K = <λ κ. Thus |K| = κ. Let F be a bijection from K to κ. For each f ∈ λ κ let

Xf = F [{f ↾ m : m ∈ λ}].

Clearly each Xf has size λ. If f, g ∈ λ κ and f 6= g, choose p ∈ λ such that f (p) 6= g(p).
Then

Xf ∩ Xg = F [{f ↾ m : m ∈ λ}] ∩ F [{g ↾ m : m ∈ λ}]

= F [{f ↾ m : m ∈ λ} ∩ {g ↾ m : m ∈ λ}]
⊆ F [{f ↾ m : m ≤ p}]

and hence Xf ∩ Xg has size less than λ. Since κλ > κ, the desired result follows.

80
E20.3 Suppose that κ is uncountable and regular. Show that there is a family A of subsets
of κ, each of size κ with the intersection of any two members of A of size less than κ, and
with |A | = κ+ . Hint: (1) show that there is a partition of κ into κ subsets, each of size
κ; (2) Use Zorn’s lemma to start from (1) and produce a maximal almost disjoint set; (3)
Use a diagonal construction to show that the resulting family must have size > κ.
First of all, recall that κ can be partitioned into κ sets, each of size κ. Namely, if f :
κ × κ → κ is a bijection, let Xα = f [{(α, β) : β < κ}]; then clearly hXα : α < κi is as
claimed.
Thus we can apply Zorn’s lemma to get a maximal collection A ⊆ [κ]κ such that the
members of A are pairwise almost disjoint, and |A | ≥ κ.
Hence we just have to get a contradiction from the assumption that |A | = κ. Making
this assumption, let hYα : α < κi be a one-one enumeration of A . Note that for any α < κ,
[ [
Yα \ Yβ = Yα \ (Yα ∩ Yβ )
β<α β<α

has size κ. This enables us to define by recursion a sequence hzα : α < κi like this: having
defined zβ for all β < α, choose
 
[
zα ∈ Yα \ {zβ : β < α} ∪ Yβ  .
β<α

def
Then Z = {zα : α < κ} is a set of size κ, and for any α < κ,

Z ∩ Yα ⊆ {zβ : β ≤ α},

so that |Z ∩ Yα | < κ. This contradicts the maximality of A .

E20.4 Prove that if F is an uncountable family of finite functions each with range ⊆ ω,
then there are distinct f, g ∈ F such that f ∪ g is a function.
We apply Theorem 20.4 to the indexed system hdmn(f ) : f ∈ F i and get an uncountable
subset G of F such that hdmn(f ) : f ∈ G i is an indexed ∆-system; say that dmn(f ) ∩
dmn(g) = D for all distinct f, g ∈ G . Then
[
G = {f ∈ G : f ↾ D = h};
h∈D ω

since the index set D ω is countable and G is uncountable, there exist an h ∈ D ω for which
there are two distinct f, g ∈ G such that f ↾ D = g ↾ D = h. Then f ∪ g is a function.
E20.5 (Double ∆-system theorem) Suppose that κ is a singular cardinal with cf(κ) > ω.
Let hλα : α < cf(κ)i be a strictly increasing sequence of successor
Pcardinals+ with supremum
κ, with cf(κ) < λ0 , and such that for each α < cf(κ) we have ( β<α λβ ) ≤ λα . Suppose
that hAξ : ξ < κi is a system of finite sets. Then there exist a set Γ ∈ [cf(κ)]cf(κ) , a

81
sequence h Ξα : α ∈ Γi of subsets of κ, a sequence hFα : α ∈ Γi of finite sets, and a finite
set G, such that the following conditions hold:
(i) h Ξα : α ∈ Γi is a pairwise disjoint system, and |Ξα | = λα for every α ∈ Γ.
(ii) hAξ : ξ ∈ Ξα i is a ∆-system with root Fα for every α ∈ Γ.
(iii) hFα : α ∈ Γi is a ∆-system with root G.
(iv) If ξ ∈ Ξα , η ∈ Ξβ , and α 6= β, then Aξ ∩ Aη = G.
S
Let κ = α<cf(κ) Ξ′α where the Ξ′α ’s are pairwise disjoint, with |Ξ′α | = λα for every
α < cf(κ). For each α < cf(κ) let Ξ′′α ∈ [Ξ′α ]λα be such that hAη : η ∈ Ξ′′α i is a ∆-system,
say with root Fα . Choose Γ ∈ [cf(κ)]cf(κ) such that hFα : α ∈ Γi is a ∆-system, say with
root G. For each α ∈ Γ let
 
[ [ 
Bα = Aξ : β ∈ Γ, β < α .
 ′′ 
ξ∈Ξβ

We claim
(1) |Bα | < λα .
In fact,
X [ X
|Bα | ≤ Aξ ≤ λβ < λα .
β<α ξ∈Ξ′′
β
β<α
β∈Γ β∈Γ

So (1) holds. Now for any α ∈ Γ,

[
Ξ′′α = {ξ ∈ Ξ′′α : Aξ ∩ Bα = J},
J∈[Bα ]<ω

def
so by (1) there is a Cα ∈ [Bα ]<ω such that Ξ′′′ ′′
α = {ξ ∈ Ξα : Aξ ∩ Bα = Cα } has size λα .
Note that Cα ⊆ Fα , since for distinct ξ, η ∈ Ξ′′′
α we have Cα = Aξ ∩ Aη ∩ Bα ⊆ Fα . Next
′′′
note that hAξ \Fα : ξ ∈ Ξα i is a system of pairwise disjoint sets; hence for each β ∈ Γ, the
set {ξ ∈ Ξ′′′
α : (Aξ \Fα ) ∩ Fβ 6= 0} is finite. Since |Γ| = cf(κ) < λα , it follows that the set

def
Ξα = Ξ′′′ ′′
α \{ξ ∈ Ξα : (Aξ \Fα ) ∩ Fβ 6= ∅ for some β ∈ Γ}.

has size λα .
Now we can verify the conditions of the exercise. Conditions (i)–(iii) are clear. Now
suppose that ξ ∈ Ξα , η ∈ Ξβ , and α 6= β. Say β < α. Suppose that γ ∈ Aξ ∩ Aη \G; we
want to get a contradiction. Since Fα ∩ Fβ = G, we have two possibilities.
Case 1. γ ∈
/ Fα . But γ ∈ Aξ ∩ Bα = Cα ⊆ Fα , contradiction.
Case 2. γ ∈ Fα \Fβ . Thus γ ∈ (Aη \Fβ ) ∩ Fα , contradicting the definition of Ξβ .
E20.6 Suppose that F is a collection of countable functions, each with range ⊆ 2ω , and
with |F | = (2ω )+ . Show that there are distinct f, g ∈ F such that f ∪ g is a function.

82
Let κ = ω1 , λ = (2ω )+ , and apply Theorem 20.4 with hAi : i ∈ Ii replaced by hdmn(f ) :
f ∈ F i. We get J ∈ [F ]λ such that hdmn(f ) : f ∈ Ji is an indexed ∆-system, say with
root r. Now [
J= {f ∈ J : f ↾ r = h},
h:r→2ω

def
and |r (2ω )| = 2ω , so there is an h : r → 2ω such that K = {f ∈ J : f ↾ r = h} has size
(2ω )+ . For any two f, g ∈ K, the set f ∪ g is a function.
E20.7 For any infinite cardinal κ, any linear order of size at least (2κ )+ has a subset of
order type κ+ or one similar to (κ+ , >).
Let L be a linear order of size (2κ )+ , and let ≺ be a well-order of L. Define f : [L]2 → 2
by setting, for any {a, b} ∈ [L]2 , say with a < b,
n
0 if a ≺ b,
f ({a, b}) =
1 if b ≺ a.
+
By the Erdös-Rado theorem, i.e., Corollary 8.7, let A ∈ [L]κ such that A is homogeneous
for f . If f takes the value 0 on [A]2 , then A is well-ordered under <, and since its size is
κ+ , it has a subset of order type κ+ . Similarly if f takes the value 1 on [A]2 .
E20.8 For any infinite cardinal κ, any tree of size at least (2κ )+ has a branch or an
antichain of size at least κ+ .
Suppose that T is a tree of size at least (2κ )+ . Let S be a subset of T of size (2κ )+ . Define
f : [S]2 → 2 by setting, for distinct s, t ∈ S,
n
1 if s and t are comparable,
f ({s, t}) =
0 otherwise.

By the Erdös-Rado theorem, let X ⊆ S be homogeneous for f of size κ+ . So if f has

constant value 1 on [X]2 , then X is a chain of size κ+ , hence extends to a branch of size
at least κ+ , while if f has constant value 0 on [X]2 , then X is an antichain of size κ+ .
E20.9 Any uncountable tree either has an uncountable branch or an infinite antichain.
Suppose that T is an uncountable tree. Let S ∈ [T ]ℵ1 . We define f : [S]2 → 2 by setting,
for any distinct s, t ∈ S,
n
1 if s and t are comparable,
f ({s, t}) =
0 otherwise.

Then the desired conclusion follows from the Dushnik-Miller theorem.

E20.10 Suppose that m is a positive integer. Show that any infinite set X of positive
integers contains an infinite subset Y such that one of the following conditions holds:
(i) The members of Y are pairwise relatively prime.
(ii) There is a prime p < m such that for any two a, b ∈ Y , p divides a − b.

83
 (iii) If a, b are distinct members of Y , then a, b are not relatively prime, but the
smallest prime divisor of a − b is at least equal to m.
Let p0 , . . . , pi−1 list all of the primes < m, in order. Thus i = 0 if m = 1. Define
f : [X]2 → i + 2 by setting, for any distinct x, y ∈ X,
(
i if x and y are relatively prime,
f ({x, y}) = j if j < i and pj is the smallest prime dividing x − y,
i+1 otherwise.

Applying Ramsey’s theorem in the form

ω → (ω, . . . , ω)2 ,
| {z }
i+1 times

we get an infinite homogeneous subset Y of X. If f [[Y ]2 ] = {i}, then any two members
of Y are relatively prime. If f [[Y ]2 ] = {j} with j < i, then pj divides x − y for any two
members x, y of Y . If f [[Y ]2 ] = {i + 1}, then for any two members x, y of Y , the least
prime dividing x − y is at least as big as m.
E20.11 Suppose that X is an infinite set, and (X, <) and (X, ≺) are two well-orderings
of X. Show that there is an infinite subset Y of X such that for all y, z ∈ Y , y < z iff
y ≺ z.
Define f : [X]2 → 2 by setting, for any distinct x, y ∈ X, say with x < y,
n
1 if x ≺ y
f ({x, y}) =
0 otherwise.

By Ramsey’s theorem, let Y be an infinite subset of X which is homogeneous for f . If

f [[Y ]2 = {0}, then x < y implies x ≻ y for all distinct x, y ∈ Y . Now since Y is infinite
and < is a well-order of Y , the order type of Y under < is an infinite ordinal. Hence there
is a system hyn : n ∈ ωi of elements of Y such that y0 < y1 < · · ·. Hence y0 ≻ y1 ≻ · · ·,
contradicting the fact that ≺ is a well-order.
Hence f [[Y ]2 = 1. This means that for any distinct x, y ∈ Y we have x < y iff x ≺ y,
as desired.
E20.12 Let S be an infinite set of points in the plane. Show that S has an infinite subset
T such that all members of T are on the same line, or else no three distinct points of T
are collinear.
Define f : [S]3 → 2 by
n
1 if s, t, u are on a line,
f ({s, t, u}) =
0 otherwise.

Let T be an infinite subset of S homogeneous for f . If f [[T ]2 ] = {1}, then all points of T
are on a line. If f [[T ]2 ] = {0}, then no three points of T are on a line.

84
E20.13 We consider the following variation of the arrow relation. For cardinals κ, λ, µ, ν,
we define
κ → [λ]µν
to mean that for every function f : [κ]µ → ν there exist an α < ν and a Γ ∈ [κ]λ such
that f [[Γ]µ ] ⊆ ν\{α}. In coloring terminology, we color the µ-element subsets of κ with ν
colors, and then there is a set which is anti-homogeneous for f , in the sense that there is
a color α and a subset of size λ all of whose µ-element subsets do not get the color α.
Prove that for any infinite cardinal κ,

κ 6→ [κ]κ2κ .

Hint: (1) Show that there is an enumeration hXα : α < 2κ i of [κ]κ in which every member
of [κ]κ is repeated 2κ times. (2) Show that |[κ]κ | = 2κ . (3) Show that there is a one-one
hYα : α < 2κ i such that Yα ∈ [Xα ]κ for all α < 2κ . (4) Define f : [κ]κ → 2κ so that for all
α < 2κ one has
f (Yα ) = o.t.({β < α : Xβ = Xα }).
We follow the hint.
(1) There is an enumeration hXα : α < 2κ i of [κ]κ in which every member of [κ]κ is
repeated 2κ times.
In fact, let f : 2κ → [κ]κ be a surjection and let g : 2κ × 2κ → 2κ be a bijection. For each
α < 2κ let Xα = f (1st(g −1 (α))). Then for all α, β < 2κ ,

Xg(α,β) = f (1st (g −1 (g(α, β)))) = f (1st (α, β)) = f (α),

and (1) follows.

(2) |[κ]κ | = 2κ .
To prove (2), let f : κ × κ → κ be a bijection. For each g ∈ κ 2 let
[
Zg = f [κ × {α}].
α<κ,
g(α)=1

Then |Zg | = κ provided that g is not identically 0, and Zg 6= Zh if g 6= h. (If g(α) 6= h(α),
say g(α) = 1 and h(α) = 0; then f [κ × α] ⊆ Zg but f [κ × {α}] ∩ Zh = ∅.) Thus
2κ ≤ |[κ]κ | + 1 ≤ 2κ + 1 with cardinal addition, and so (2) follows.
(3) There is a one-one hYα : α < 2κ i such that Yα ∈ [Xα ]κ for all α < 2κ .
We construct Yα by recursion. If Yβ has been constructed for all β < α, where α < 2κ ,
choose Yα ∈ [Xα ]κ \{Yβ : β < α}; this is possible by (2). So (3) holds.
Now we define f : [κ]κ → 2κ so that for all α < 2κ one has

f (Yα ) = o.t.({β < α : Xβ = Xα }).

85
This defines f on {Yα : α < 2κ }, and it can be extended to all of [κ]κ in any fashion.
Now we show that f is the desired counterexample. For, suppose that β < 2κ , Γ ∈ [κ]κ ,
and f [[Γ]κ ] ⊆ 2κ \{β}. Choose α < 2κ such that Xα = Γ and {γ < α : Xγ = Γ} has order
type β. Then Yα ∈ [Γ]κ and f (Yα ) = β, contradiction.
Solutions to exercises in Chapter 21
E21.1 Assume MA(κ). Suppose that X is a compact Hausdorff space, and any pairwise
disjoint collection of open
T sets in X is countable. Suppose that Uα is dense open in X for
each α < κ. Show that α<κ Uα 6= ∅.
Let P consist of all nonempty open subsets of X, with ⊆ as the order. Note that for
U, V ∈ P , U is compatible with V iff U ∩ V 6= ∅. Hence ccc holds for P . For each α < κ
let Dα = {p ∈ P : p ⊆ Uα }. We claim that Dα is dense in the sense of P . For, suppose
that p ∈ P . Since Uα is (topologically) dense, we have p ∩ Uα 6= ∅. By regularity of the
space there is a nonempty open set q such that q ⊆ p ∩ Uα . Thus q ∈ Dα and q ⊆ p, as
desired.
So, we apply MA(κ) and obtain a filter G intersecting each Dα .TBecause G is a filter,
it has the fip as a collection of open sets. Hence by compactness, T p∈G p 6=T∅. For any
α < κ there is a p ∈ G ∩ Dα , and hence p ⊆ Uα . This implies that p∈G p ⊆ α<κ Uα .
E21.2 A partial order P is said to have ω1 as a precaliber iff for every system hpα : α < ω1 i
of elements of P there is an X ∈ [ω1 ]ω1 such that for every finite subset F of X there is a
q ∈ P such that q ≤ pα for all α ∈ F .
Show that MA(ω1 ) implies that every ccc partial order P has ω1 as a precaliber.
Hint: for each α < ω1 let

Wα = {q ∈ P : ∃β > α(q and pα are compatible)}.

Show that there is an α < ω1 such that Wα = Wβ for all β > α, and apply MA(ω1 ) to Wα .
Let hpα : α < ω1 i be a system of elements of P ; we want to come up with a set X as
indicated. For each α < ω1 let

Wα = {q ∈ P : ∃β > α(q and pα are compatible)}.

Clearly if α < β < ω1 then Wβ ⊆ Wα . Now we claim

(1) ∃α∀β ∈ (α, ω1 )[Wα = Wβ ].

In fact, otherwise we get a strictly increasing sequence hαξ : ξ < ω1 i of ordinals such that
Wαξ+1 ⊂ Wαξ for all ξ < ω1 . Choose qξ ∈ Wαξ \Wαξ+1 for all ξ < ω1 . Then there is an
ordinal βξ such that αξ < βξ ≤ αξ+1 and qξ and pβξ are compatible; say rξ ≤ qξ , pβξ . We
claim that rξ and rη are incompatible for ξ < η < ω1 (contradicting ccc for P ). For, if
s ≤ rξ , rη , then qξ and pβη are compatible, and hence qξ ∈ Wαξ+1 , contradiction. Thus (1)
holds.
We are going to apply MA(ω1 ) to Wα . The dense sets are as follows. For each
β ∈ (α, ω1 ), let
Dβ = {q ∈ Wα : ∃γ ∈ (β, ω1 )[q ≤ pγ ]}.

86
To prove density, suppose that r ∈ Wα . Then, since Wα = Wβ it follows that r and pγ are
compatible for some γ > β, as desired.
So, let G be a filter on Wα intersecting each set Dβ . It follows that there exist a
strictly increasing sequence hβξ : ξ < ω1 i and a sequence hqξ : ξ < ω1 i such that qξ ≤ pβξ
with qξ ∈ G for all ξ < ω1 . Clearly then {pβξ : ξ < ω1 } has the desired property.
E21.3 Call a topological space
Q X ccc iff every collection of
Qpairwise disjoint open sets in
<ω
X is countable. Show that i∈I Xi is ccc iff ∀F ∈ [I] [ i∈F Xi is ccc]. Hint: use the
∆-system theorem.
Q
⇒: Suppose that i∈I XiQ is ccc and F ∈ [I]<ω . Also suppose that A is a pairwise disjoint
collection of open sets in i∈F Xi . Then
(( ) )
Y
′ def
A = x∈ Xi : hxi : i ∈ T i ∈ U :U ∈A
i∈I

Q
is a collection of pairwise disjoint open sets in i∈I Xi , and hence A ′ is countable. Clearly
this implies that A is countable. Q
<ω
⇐: Suppose
Q that ∀F ∈ [I] [ i∈F Xi is ccc], and B is a collection of pairwise disjoint
open sets in i∈I Xi . Suppose that B is uncountable; we want to get a contradiction. We
may assume that each Q U ∈ B is basic open, which means
Q that there exist a finite set FU
and an open set VU in i∈F Xi such that U = {x ∈ i∈I Xi : hxi : i ∈ FU i ∈ VU }.
E21.4 Assuming MA(ω1 ), show that any product of ccc spaces is ccc.
By exercise E21.3 it suffices to show that any product of two ccc spaces X, Y is ccc. Suppose
that A is an uncountable collection of pairwise disjoint open subsets of X × Y ; we want
to get a contradiction. We may assume that each member of A has the form U × V , with
U open in X and V open in Y . Let hUα × Vα : α < ω1 i be a one-one enumeration of a
subset of A . Let P be the partially ordered set consisting of all nonempty open subsets of
X, ordered by ⊆. Thus by exercise 21.2, P has ω1 as a precaliber. Hence let M ∈ [ω1 ]ω1
be such that for every finite subset F of M there is a V ∈ P such that V ⊆ Uα for all
α ∈ F . Take any distinct α, β ∈ M . Then Uα ∩ Uβ 6= ∅, while (Uα × Vα ) ∩ (Uβ × Vβ ) = ∅,
so Vα ∩ Vβ = ∅. This contradicts ccc for Y .
E21.5 Assume MA(ω1 ). Suppose that P and Q are ccc partially ordered sets. Define ≤
on P × Q by setting (a, b) ≤ (c, d) iff a ≤ c and b ≤ d. Show that < is a ccc partial order
on P × Q. Hint: use exercise E21.2.
Suppose that h(pα , qα ) : α < ω1 i is a system of elements of P × Q; we want to find distinct
α, β < ω1 such that (pα , qα ) and (pβ , qβ ) are compatible. By exercise 21.2, let Γ ∈ [ω1 ]ω1
be such that for every finite subset F of Γ there is an r ∈ P such that r ≤ pα for all α ∈ F .
Since Q has ccc, there exist distinct α, β ∈ Γ such that qα and qβ are compatible. Also,
pα and pβ are compatible. So (pα , qα ) and (pβ , qβ ) are compatible.
E21.6 We define <∗ on ω ω by setting f <∗ g iff f, g ∈ ω ω and ∃n∀m > n(f (m) < g(m).
Suppose that MA(κ) holds and F ∈ [ω ω]κ . Show that there is a g ∈ ω ω such that f <∗ g

87
for all f ∈ F . Hint: let P be the set of all pairs (p, F ) such that p is a finite function
mapping a subset of ω into ω and F is a finite subset of F . Define (p, F ) ≤ (q, G) iff
q ⊆ p, G ⊆ F , and
∀f ∈ G∀n ∈ dmn(p)\dmn(q)[p(n) > f (n)].
P is ccc: assume that X is an uncountable subset of P. There are only countably many
finite functions from ω to ω, so there are distinct (p, F ), (p, G) ∈ P. Then (p, F ∪ G) ≤
(p, F ), (p, G). So X is not an incompatible set.
For each h ∈ F let Dh = {(p, F ) ∈ P : h ∈ F }. Then Dh is dense, since for any
(p, F ) ∈ P we have (p, F ∪ {h}) ≤ (p, F ).
For each n ∈ ω let En = {(p, F ) : n ∈ dmn(p)}. Then En is dense. For, suppose that
(p, F ) ∈ P and n ∈ / dmn(p). Choose m ∈ ω greater than each member of {f (n) : f ∈ F }.
Then clearly (p ∪ {(n, m)}, F ) ≤ (p, F ). S
Let G be a filter on P intersecting each Dh and En . Let g = (p,F )∈F p. Then g is a
function since G is a filter. Moreover, g ∈ ω ω since G ∩ En 6= ∅ for all n. Now let f ∈ F .
Choose (p, F ) ∈ G ∩ Df . Thus p ∈ F . We claim that if m is greater than each member of
dmn(p) then g(m) > f (m) (as desired).
Since m ∈ dmn(g), choose (q, H) ∈ G such that m ∈ dmn(q). Choose (r, K) ∈ G
with (r, K) ≤ (p, F ), (q, H). Then m ∈ dmn(q) ⊆ dmn(r), so m ∈ dmn(r). Hence
m ∈ dmn(r)\dmn(p), and (r, K) ≤ (p, F ), so g(m) = r(m) > f (m).
E21.7 Let B ⊆ [ω]ω be almost disjoint of size κ, with ω ≤ κ < 2ω . Let A ⊆ B with A
countable. Assume MA(κ). Show that there is a d ⊆ ω such that |d ∩ x| < ω for all x ∈ A ,
and |x\d| < ω for all x ∈ B\A . Hint: Let hai : i ∈ ωi enumerate A . Let
P = {(s, F, m) : s ∈ [ω]<ω , F ∈ [B\A ]<ω , and m ∈ ω};
(s′ , F ′ , m′ ) ≤ (s, F, m) iff s ⊆ s′ , F ⊆ F ′ , m ≤ m′ , and
" ! #
[
∀x ∈ F x\ ai ∩ s′ ⊆ s .
i∈m

Show that P satisfies ccc. To apply MA(κ), one needs various dense sets. The most
complicated is defined as follows. Let D = {(s, F, m, i, n) : (s, F, m) ∈ P, i < m, and
n ∈ ai \s}. Clearly |D| = κ. For each (s, F, m, i, n) ∈ D let
E(s,F,m,i,n) = {(s′ , F ′ , m′ ) ∈ P : (s, F, m) and (s′ , F ′ , m′ ) are incompatible
or (s′ , F ′ , m′ ) ≤ (s, F, m) and n ∈ s′ }.
P is ccc: assume that X is an uncountable subset of P. There are only countably many
finite functions from ω to ω, so there are distinct (p, F ), (p, G) ∈ P. Then (p, F ∪ G) ≤
(p, F ), (p, G). So X is not an incompatible set.
For each x ∈ B\A let Dx = {(s, F, m) ∈ P : x ∈ F }. Then Dx is dense, since clearly
(s, F ∪ {x}, m) ≤ (s, F, m) for any (s, F, m) ∈ P.
Let D = {(s, F, m, i, n) : (s, F, m) ∈ P, i < m, and n ∈ ai \s}. Clearly |D| = κ. For
each (s, F, m, i, n) ∈ D let
E(s,F,m,i,n) = {(s′ , F ′ , m′ ) ∈ P : (s, F, m) and (s′ , F ′ , m′ ) are incompatible
or (s′ , F ′ , m′ ) ≤ (s, F, m) and n ∈ s′ }.

88
Now E(s,F,m,i,n) is dense for each (s, F, m, i, n) ∈ D. For, suppose that (s′ , F ′ , m′ ) is
given. We may assume that (s, F, m) and (s′ , F ′ , m′ ) are compatible; say (s′′ , F ′′ , m′′ ) ≤
(s, F, m), (s′, F ′ , m′ ). We may assume that n ∈ / s′′ . We claim that (s′′ ∪ {n}, S F ′′ , m′′ ) ≤
(s′′ , F ′′ , m′′ ) (as desired). This is true since for any x ∈ F ′′ we have n ∈ / (x\ j∈m′′ aj ), by
virtue of n ∈ ai and i ∈ m ≤ m′′ .
Next, for any i < ω let Hi = {(s, F, m) ∈ P : i < m}. Then Hi is dense, since
(s, F, i + 1) ≤ (s, F, m) for any m ≤ i. S
Now let G be a filter on P intersecting all of these dense sets. Let d = (s,F,m)∈G s.
Take any x ∈ B\A , and choose (s, F, m) ∈ G ∩ Dx ; so x ∈ F . We claim that x ∩ d ⊆
S ′ ′ ′
i∈m (x ∩ ai ) ∪ s, so that x ∩ d is finite. For, suppose that n ∈ x ∩ d. choose (s , F , m ) ∈ G
′ ′′ ′′ ′′ ′′ ′′ ′′ ′ ′ ′
such that n ∈ s . Take S (s , F ,′′m ) ∈ G with (s , F′′ , m ) ≤ (s, F, m), (s S , F , m ). Then
′′
n ∈ s . Since (x\ i∈m ai ) ∩ s ⊆ s and n ∈ x ∩ s , it follows that n ∈ i∈m ai ∪ s, as
desired.
Next, take any i < ω. Choose (s, F, m) ∈ G∩Hi . Thus i < m. We claim that ai \s ⊆ d,
so that a\d is finite. To prove this, take any n ∈ ai \s. Then (s, F, m, i, n) ∈ D, so we
can choose (s′ , F ′ , m′ ) ∈ G ∩ E(s,F,m,i,n) . Since (s, F, m) and (x′ , F ′ , m′ ) are compatible as
elements of G, it follows that (s′ , F ′ , m′ ) ≤ (s, F, m) and n ∈ s′ . Thus n ∈ d, as desired.
This proves the claim.
E21.8 [The condition that A is countable is needed in exercise E21.7.] Show that there
exist A , B such that B is an almost disjoint family of infinite subsets of ω, A ⊆ B,
|A | = |B\A | = ω1 , and there does not exist a d ⊆ ω such that |x\d| < ω for all x ∈ A ,
and |x ∩ d| < ω for all x ∈ B\A . Hint: construct A = {aα : α < ω1 } and B\A = {bα :
α < ω1 } by constructing aα , bα inductively, making sure that the elements are infinite and
pairwise almost disjoint, and also aα ∩ bα = ∅, while for α 6= β we have aα ∩ bβ 6= ∅.
First we show that the hint works. Suppose that we have constructed haα : α < ω1 i and
hbα : α < ω1 i, so that they are infinite and pairwise almost disjoint, with the additional
indicated property. Suppose that d exists as indicated. Wlog ∀α < ω1 (aα \d = F and
bα ∩ d = G). Choose m ∈ a0 ∩ b1 . If m ∈ d, then m ∈ b1 ∩ d = G ⊆ b0 , so m ∈ a0 ∩ b0 ,
contradiction. If m ∈/ d, then m ∈ a0 \d = F ⊆ a1 , so m ∈ a1 ∩ b1 , contradiction.
Now we do the construction indicated in the hint. Let hci : i < ωi be a system of
pairwise disjoint infinite subsets of ω. Let hci,j : j < ωi be a one-one enumeration of ci .
Then we define for each m ∈ ω

am = c2m ∪ {c2n+1,0 : n < m};

bm = c2m+1 ∪ {c2n,0 : n < m}.

Clearly this defines infinite pairwise almost disjoint subsets of ω, am ∩bm = ∅ for all m ∈ ω,
and for n < m we have c2n,0 ∈ an ∩ bm and c2n+1,0 ∈ am ∩ bn .
Now suppose that aβ and bβ have been defined for all β < α, with ω ≤ α < ω1 . Let
f be a function from ω onto α. By recursion, choose
!
[ [
xm ∈ bf (m) \ bf (n) ∪ af (n) ∪ {xn : n < m} ∪ {yn : n < m} ;
n<m n<m

89
 !
[ [
ym ∈ af (m) \ bf (n) ∪ af (n) ∪ {xn : n ≤ m} ∪ {yn : n < m} .
n<m n<m

Let aα = {xm : m ∈ ω} and bα = {ym : m ∈ ω}. Then aα ∩ bα = ∅ and

xm ∈ aα ∩ bf (m) ⊆ {x0 , . . . , xm };
ym ∈ bα ∩ af (m) ⊆ {y0 , . . . , ym };
aα ∩ af (m) ⊆ {x0 , . . . , xm };
bα ∩ bf (m) ⊆ {y0 , . . . , ym }.

Hence the construction is complete.

T
E21.9 Suppose that A is a family of infinite subsets of ω such that F is infinite for
every finite subset F of A . Suppose that |A | ≤ κ. Assuming MA(κ), show that there is
an infinite X ⊆ ω such that X\A is finite for every A ∈ A . Hint: use Theorem 21.5.
Let A ′ = {X ⊆ ω : ω\X ∈ A }, and let B = {ω}. Clearly the hypothesis of Theorem 21.5
holds for A ′ and B. Hence by 21.5 there is a d ⊆ ω such that |d ∩ X| < ω for all X ∈ A ′
and |d| = |d ∩ ω| = ω. Clearly d is as desired.
E21.10 Show that MA(κ) is equivalent to MA(κ) restricted to ccc partial orders of car-
dinality ≤ κ. Hint: Assume the indicated special form of MA(κ), and assume given a ccc
partially ordered set P and a family D of at most κ dense sets in P ; we want to find a
filter on P intersecting each member of D. We introduce some operations on P . For each
D ∈ D define fD : P → P by setting, for each p ∈ P , fD (p) to be some element of D
which is ≤ p. Also we define g : P × P → P by setting, for all p, q ∈ P ,

p if p and q are incompatible,
g(p, q) =
r with r ≤ p, q if there is such an r.

Here, as in the definition of fD , we are implicitly using the axiom of choice; for g, we
choose any r of the indicated form.
We may assume that D 6= ∅. Choose D ∈ D, and choose s ∈ D. Now let Q be the
intersection of all subsets of P which have s as a member and are closed under all of the
operations fD and g. We take the order on Q to be the order induced from P . Apply the
special form to Q.
We assume the indicated special form of MA(κ), and assume given a ccc partially ordered
set P and a family D of at most κ dense sets in P ; we want to find a filter on P intersecting
each member of D. We introduce some operations on P . For each D ∈ D define fD : P →
P by setting, for each p ∈ P , fD (p) to be some element of D which is ≤ p. Also we define
g : P × P → P by setting, for all p, q ∈ P ,

p if p and q are incompatible,
g(p, q) =
r with r ≤ p, q if there is such an r.

90
Here, as in the definition of fD , we are implicitly using the axiom of choice; for g, we
choose any r of the indicated form.
We may assume that D 6= ∅. Choose D ∈ D, and choose s ∈ D. Now let Q be the
intersection of all subsets of P which have s as a member and are closed under all of the
operations fD and g. We take the order on Q to be the order induced from P .
(1) |Q| ≤ κ.
To prove this, we give an alternative definition of Q. Define

R0 = {s};
Rn+1 = Rn ∪ {g(a, b) : a, b ∈ Rn } ∪ {fD (a) : D ∈ D and a ∈ Rn }.
S
Clearly n∈ω Rn = Q. By induction, |Rn | ≤ κ for all n ∈ ω, and hence |Q| ≤ κ, as desired
in (1).
We also need to check that Q is ccc. Suppose that X is a collection of pairwise
incompatible elements of Q. Then these elements are also incompatible in P , since x, y ∈ X
with x, y compatible in P implies that g(x, y) ≤ x, y and g(x, y) ∈ Q, so that x, y are
compatible in Q. It follows that X is countable. So Q is ccc.
Now if D ∈ D, then D ∩ Q is dense in Q. In fact, take any q ∈ Q. Then fD (q) ∈ Q
and fD (q) ≤ q, as desired.
Now we can apply our special case of MA(κ) to Q and {D ∩ Q : D ∈ D}; we obtain a
filter G on Q such that G ∩ D ∩ Q 6= ∅ for all D ∈ D. Let

G′ = {p ∈ P : q ≤ p for some q ∈ G}.

We claim that G′ is the desired filter on P intersecting each D ∈ D.

Clearly if p ∈ G′ and p ≤ r, then r ∈ G′ .
Suppose that p1 , p2 ∈ G′ . Choose q1 , q2 ∈ G such that qi ≤ p1 for each i = 1, 2. Then
there is an r ∈ G such that r ≤ q1 , q2 . Then r ∈ G′ and r ≤ p1 , p2 . So G′ is a filter on P .
Now take any D ∈ D. Then as proved above, D ∩ Q is dense in Q. It follows that
G ∩ D ∩ Q 6= ∅; say q ∈ G ∩ D ∩ Q. Then q ∈ G′ ∩ D, as desired.
E21.11 Define x ⊂∗ y iff x, y ⊆ ω, x\y is finite, and y\x is infinite. Assume MA(κ),
and suppose that L, <) is a linear ordering of size at most κ. Show that there is a system
hax : x ∈ Li of infinite subsets of ω such that for all x, y ∈ L, x < y iff ax ⊂∗ ay . Hint: let
P consist of all pairs (p, n) such that n ∈ ω, p is a function whose domain is a finite subset
of L, and ∀x ∈ dmn(p)[p(x) ⊆ n]. Define (p, n) ≤ (q, m) iff m ≤ n, dmn(q) ⊆ dmn(p),
∀x ∈ dmn(q)[p(x) ∩ m = q(x)], and ∀x, y ∈ dmn(q)[x < y → p(x)\p(y) ⊆ m].
First note that it is enough to do the construction so that ∀x, y ∈ L[x < y → ax ⊂∗ ay ].
In fact, knowing this, if ax ⊂∗ ay , then we must have x < y, as otherwise y ≤ x and hence
ay ⊂∗ ax or ay = ax , both of which are ruled out by ax ⊂∗ ay .
Now we show that P has ccc. Suppose that A is an uncountable subset of P . By the
indexed ∆-system theorem, Theorem 16.4, there is an uncountable subset B of A such that

91
hdmn(p) : (p, n) ∈ Bi is an indexed ∆-system. Say M ∈ [L]<ω with dmn(p) ∩dmn(q) = M
for any distinct (p, n), (q, m) ∈ B. Now let Q =M ω. Then
[
B= {(p, n) ∈ B : ∀x ∈ M [p(x) = f (x)]}.
f ∈Q

Since Q is countable, there exist a C ∈ [B]<ω and an f ∈ Q such that ∀(p, n) ∈ C ∀x ∈

M [p(x) = f (x)]. Now take two distinct members (p, n) and (q, m) of C . Say m ≤ n. We
now define a function r. dmn(r) = dmn(p) ∪ dmn(q). For any x ∈ dmn(r),

p(x) if x ∈ dmn(p),
r(x) =
q(x) if x ∈ dmn(q)\dmn(p).

Clearly (r, n) ∈ P . We claim (r, n) ≤ (p, m), (q, n); this will show that P has ccc. First we
show that (r, n) ≤ (p, m). We have m ≤ n and dmn(p) ⊆ dmn(r). Take any x ∈ dmn(p).
Then r(x) ∩ m = p(x) ∩ m = p(x). Suppose that x, y ∈ dmn(p) and x < y. Then
r(x)\r(y) = p(x)\p(y) ⊆ m. Therefore (r, n) ≤ (p, m). Second we show that (r, n) ≤ (q, n).
We have n ≤ n and dmn(q) ⊆ dmn(r). Take any x ∈ dmn(q). If also x ∈ dmn(p) then
x ∈ M , and r(x) ∩ n = p(x) ∩ n = f (x) ∩ n = q(x) ∩ n = q(x). Suppose that x, y ∈ dmn(q)
and x < y.
Case 1. x, y ∈ dmn(p). Then r(x)\r(y) = p(x)\p(y) ⊆ m ≤ n.
Case 2. x ∈ dmn(p) and y ∈ / dmn(p). Then x ∈ M , and so r(x)\r(y) = p(x)\q(y) =
f (x)\q(y) = q(x)\q(y) ⊆ n.
Case 3. x ∈/ dmn(p) and y ∈ dmn(p). Similar to Case 2.
Case 4. x, y ∈/ dmn(p). The conclusion is clear.
This finishes the proof that P has ccc.
Now after defining certain dense sets we are going to take a filter G with respect to
them and then define [
ax = p(x)
(p,n)∈G
x∈dmn(p)

for each x ∈ L.
def
To show that we can define ax for each x ∈ L we consider Dx = {(p, n) ∈ P : x ∈
dmn(p)}. Then Dx is dense, since given (p, n) ∈ P with x ∈ / dmn(p), we may assume that
n > 0, and we have (p ∪ {(x, 0)}, n) ∈ P and (p ∪ {(x, 0)}, n) ≤ (p, n), as is easily checked.
To show that for a given x ∈ L the set ax is infinite, we consider for each i ∈ ω the
set
def
Eix = {(p, n) : x ∈ dmn(p) and i < n and p(x) 6⊆ i}.
To show that this set is dense, let (q, m) ∈ P . By the argument for Dx we may assume that
x ∈ dmn(q). Let n = max(i + 1, m). Let p have domain dmn(q), and for each y ∈ dmn(q)
let p(y) = q(y) ∪ {n}. Clearly (p, n + 1) ∈ P . We have m < n + 1 and dmn(q) = dmn(p).
For any y ∈ dmn(q) we have p(y) ∩ m = q(y). Now suppose that y, z ∈ dmn(q) and y < z.
Then p(y)\p(z) = q(y)\q(z) ⊆ m. So (p, n + 1) ≤ (q, m), and clearly (p, n + 1) ∈ Eix .

92
 Next, for any x, y ∈ L such that x < y and any i ∈ ω let

Fixy = {(p, n) ∈ P : x, y ∈ dmn(p) and ∃j ≥ i[j ∈ p(y)\p(x)]}.

To show that this set is dense, let (q, m) be given. Wlog x, y ∈ dmn(q). Let n = max(i +
1, m). Let dmn(p) = dmn(q), and for z ∈ dmn(q) let

q(z) if z 6= y,
p(z) =
q(z) ∪ {n} if y ≤ z.

Clearly (p, n + 1) ∈ P . Since p(x) = q(x) ⊆ m and n ≥ m, we have n ∈ p(y)\p(x).

If z ∈ dmn(q), then p(z) ∩ m = q(z). Suppose that u, v ∈ dmn(q) and u < v. Then
p(u)\p(v) ⊆ m; this is only questionable if y ≤ u, and this case is clear. Thus (p, n + 1) ≤
(q, m).
Now we take a filter G with respect to all of these dense sets. So we can define ax for
x ∈ L as above, and each ax is infinite. Now suppose that x, y ∈ L and x < y.
Using Dx and Dy , let (p, n) ∈ G with x, y ∈ dmn(p). We claim that ax \ay ⊆ n. We
prove that ax \n ⊆ ay . Suppose that i ∈ ax \n. Choose (q, m) ∈ G such that x ∈ dmn(q)
and i ∈ q(x). Then choose (r, s) ∈ G such that (r, s) ≤ (p, n), (q, m). Now i ∈ q(x), so
i ∈ r(x). Since x < y and x, y ∈ dmn(p), we have r(x)\r(y) ⊆ n. Now i ≥ n, so it follows
that i ∈ r(y). Hence i ∈ ay , as desired.
It remains only to show that ay \ax is infinite. Suppose it is finite; say ay \ax ⊆ i
with i ∈ ω. Choose (p, n) ∈ Fixy ∩ G. Thus x, y ∈ dmn(p) and there is a j ≥ i such that
j ∈ p(y)\p(x). Thus j ∈ ay . Since ay \i ⊆ ax , we have j ∈ ax . Choose (q, m) ∈ G such
that x ∈ dmn(q) and j ∈ q(x). Choose (r, s) ≤ (p, n), (q, m). Then j ∈ q(x), so j ∈ r(x).
Now r(x) ∩ n = p(x), so j ∈ p(x), contradiction.
E21.12 If A , B are nonempty countable subsets of [ω]ω and a ⊆∗ b whenever a ∈ A and
b ∈ B, then there is a c ∈ [ω]ω such that a ⊆∗ c ⊆∗ b whenever a ∈ A and b ∈ B.
Write A = {an : n ∈ ω} and B = {bn : n ∈ ω}. Let
  
[ [ \
c=  am  ∩ bm  .
n∈ω m≤n m≤n

Now suppose that p ∈ ω. Then

  
\ \ [
ap \c = ap ∩  (ω\am ) ∪ (ω\bm )
n∈ω m≤n m≤n
  
\ \ [
= ap ∩  (ω\am ) ∪ (ω\bm )
n<p m≤n m≤n
  
\ \ [
∩ ap ∩  (ω\am ) ∪ (ω\bm )
n≥p m≤n m≤n

93
   
\ \ [
= ap ∩  (ω\am ) ∪ (ω\bm )
n<p m≤n m≤n
 
\ [
∩ ap ∩ (ω\bm )
n≥p m≤n
[
⊆ ap ∩ (ω\bm ),
m≤p

and this last set is finite.

Furthermore,
  
[ [ \
c\bp =  am  ∩ bm ∩ (ω\bp )
n<p m≤n m≤n
!
[
⊆ am \bp ,
m<p

and this last set is finite.

The set c is infinite, as otherwise a0 = (a0 ∩ c) ∪ (a0 \c) would be finite.
E21.13 Suppose that A is a nonempty countable family of members of [ω]ω , and ∀a, b ∈
A [a ⊆∗ b or b ⊆∗ a]. Also suppose that ∀a ∈ A [a ⊂∗ d], where d ∈ [ω]ω . Then there is a
c ∈ [ω]ω such that ∀a ∈ A [a ⊆∗ c ⊂∗ d].
If ∃a ∈ A ∀b ∈ A [b ⊆∗ a], then the conclusion is obvious. So suppose that no such a exists.
Then there is a sequence han : n ∈ ωi of elements of A such that an ⊂∗ am for n < m, and
the sequence is cofinal in A in the ⊆∗ -sense. Let C = {a0 } ∪ {am+1 \am : m ∈ ω} ∪ {ω\d}.
Then C is an almost disjoint family, except that possibly ω\d is finite. By Theorem 11.1
and Corollary 11.6, let e ⊆ ω be infinite and almost disjoint from each member of C . Let
c = d\e. Then for any n ∈ ω,

an+1 \c = (an+1 \d) ∪ (an+1 ∩ e)

 
[
⊆ (an+1 \d) ∪  (ai+1 \ai ) ∪ a0  ∩ e,
i≤n

and the last set is finite. Thus an+1 ⊆∗ c, hence b ⊆∗ c for all b ∈ A .
Since c ⊆ d, we have c ⊆∗ d. Also, d\c = d ∩ e, and this is infinite since e\d is finite.
Thus c ⊂∗ d
Note that c is infinite, since a ⊆∗ c for all a ∈ A .
E21.14 If a, b ∈ [ω]ω and a ⊂∗ b, then there is a c ∈ [ω]ω such that a ⊂∗ c ⊂∗ b.
Write b\a = d ∪ e with d, e infinite and disjoint. Let c = a ∪ d.

94
E21.15 Suppose that A and B are nonempty countable subsets of [ω]ω , ∀x, y ∈ A [x ⊆∗ y
or y ⊆∗ x], ∀x, y ∈ B[x ⊆∗ y or y ⊆∗ x], and ∀x ∈ A ∀y ∈ B[a ⊂∗ b]. Then there is a
c ∈ [ω]ω such that a ⊂∗ c ⊂∗ b for all a ∈ A and b ∈ B.
By Exercise 21.12 choose d ⊆ ω such that ∀a ∈ A ∀b ∈ B[a ⊆∗ d ⊆∗ b]. Thus either
∀a ∈ A [a ⊂∗ d] or ∀b ∈ B[d ⊂∗ b].
Case 1. ∀a ∈ A [a ⊂∗ d]. By Exercise 21.13 choose e ⊆ ω such that ∀a ∈ A [a ⊆∗
e ⊂ d]. By Exercise 11.14 choose c ∈ [ω]ω such that e ⊂∗ c ⊂∗ d.
∗

Case 2. ∀b ∈ B[d ⊂∗ b]. Then ∀b ∈ B[(ω\b) ⊂∗ (ω\d)]. By exercise E21.13 choose

e ⊆ ω such that ∀b ∈ B[(ω\b) ⊆∗ e ⊂∗ (ω\d)]. By exercise E21.14 choose c ⊆ ω such that
e ⊂∗ c ⊂∗ (ω\d). Then ∀a ∈ A ∀b ∈ B[a ⊂∗ (ω\c) ⊂∗ b].
E21.16 Suppose that am ∈ [ω]ω for all m ∈ ω, am ⊂∗ an whenever m < n ∈ ω, b ∈ [ω]ω ,
and am ⊂∗ b for all m ∈ ω. Then there is a c ∈ [ω]ω such that ∀m ∈ ω[am ⊂∗ c ⊂∗ b] and
c is near to {an : n ∈ ω}.
∗ ∗
S 21.15 choose d ⊆ ω such that ∀m ∈
By Exercise S ω[an ⊂ d ⊂ S b]. Now for each
m ∈ ω, Si≤m (am ∩ am+1 ) is finite, and am+1 \ S a
i≤m i = a m+1 \ i≤m (am+1 ∩ ai ),
so am+1S\ i≤m ai is infinite. Choose em ⊆ am+1 \ i≤m ai such that |em | = m. Let
c = d\ m∈ω em . Thus c ⊆∗ d ⊂∗ b.
If n ∈ ω, then
[ [
an \c = (an \d) ∪ (an ∩ em ) = (an \d) ∪ (an ∩ em ),
m∈ω m<n

and this last set is finite. Hence an ⊆∗ c. Since n is arbitrary, it follows that an ⊂∗ c for
all n ∈ ω.
Also for any m ∈ ω we have am+1 \c ⊇ am+1 ∩ em = em , and so |am+1 \c| ≥ m. It
follows that for any n ∈ ω, {am : am \c ⊆ n} ⊆ {a0 , . . . , an }. So c is near to {am : m ∈ ω}.
E21.17 Suppose that A ⊆ [ω]ω , ∀x, y ∈ A [x ⊂∗ y or y ⊂∗ x], b ∈ [ω]ω , ∀x ∈ A [x ⊂∗ b],
and ∀a ∈ A [b is near to {d ∈ A : d ⊂∗ a}].
Then there is a c ∈ [ω]ω such that ∀a ∈ A [a ⊂∗ c ⊂∗ b] and c is near to A .
Proof. We consider several cases.
Case 1. ∃a ∈ A ∀d ∈ A [d ⊆∗ a]. By Exercise 21.14, choose c such that a ⊂∗ c ⊂∗ b.
Choose n ∈ ω such that c\b ⊆ n. Then for any m ∈ ω and any d ∈ A , if d\c ⊆ m then
d\b ⊆ (d\c) ∪ (c\b) ⊆ max(m, n). Hence

{d ∈ A : d\c ⊆ m} ⊆ {a} ∪ {d ∈ A : d ⊂∗ a and d\b ⊆ max(m, n)},

and the later set is finite, since b is near to {d ∈ A : d ⊂∗ a}. Thus c is as desired.
Case 2. ∀a ∈ A ∃d ∈ A [a ⊂∗ d] and b is near to A . By Exercise 21.15, choose c so
that ∀a ∈ A [a ⊂∗ c ⊂∗ b]. Choose n ∈ ω such that c\b ⊆ n. Then for any m ∈ ω and any
d ∈ A , if d\c ⊆ m then d\b ⊆ (d\c) ∪ (c\b) ⊆ max(m, n). Hence

{d ∈ A : d\c ⊆ m} ⊆ {a} ∪ {d ∈ A : d\b ⊆ max(m, n)},

95
and the later set is finite, since b is near to A . Thus c is as desired.
Case 3. ∀a ∈ A ∃d ∈ A [a ⊂∗ d] and b is not near to A . For each m ∈ ω let
Bm = {a ∈ A : a\b ⊆ m}. Since b is not near to A , choose m so that Bm is infinite.
Note that p < q → Bp ⊆ Bq . Hence Bn is infinite for every n ≥ m. Now we claim

(1) ∀n ≥ m∀a ∈ A ∃d ∈ Bn [a ⊆∗ d].

In fact, otherwise we get n ≥ m and a ∈ A such that ∀d ∈ Bn [d ⊂∗ a]. Now b is near to

{d ∈ A : d ⊂∗ a} by a hypothesis of the lemma, so {d ∈ A : d ⊂∗ a and d\b ⊆ n} is finite.
But Bn ⊆ {d ∈ A : d ⊂∗ a and d\b ⊆ n}, contradiction. So (1) holds.
Next we claim

(2) ∀n ≥ m∀d ∈ Bn [{e ∈ Bn : e ⊂∗ d} is finite].

In fact, suppose that n ≥ m, d ∈ Bn and {e ∈ Bn : e ⊂∗ d} is infinite. Since b is

near to {a ∈ A : a ⊂∗ d}, the set {a ∈ A : a ⊂∗ d and a\b ⊆ n} is finite. But
{e ∈ Bn : e ⊂∗ d} ⊆ {a ∈ A : a ⊂∗ d and a\b ⊆ n}, contradiction. So (2) holds.
∗
S (2) it follows that Bn has order type ω under ⊂ , for each n ≥ m. Now clearly
From
A = p∈ω Bp , so A is countable.
Now by Exercise 21.16, choose cm such that ∀d ∈ Bm [d ⊂∗ cm ⊂∗ b] and cm is near
to Bm . By (1), a ⊂∗ cm for each a ∈ A . Now suppose that n ≥ m and cn has been
defined so that a ⊂∗ cn for each a ∈ A . Again by Exercise 21.16 choose cn+1 such that
∀d ∈ Bn+1 [d ⊂∗ cn+1 ⊂∗ cn ] and cn+1 is near to Bn+1 . Thus we have

∀a ∈ A [a ⊂∗ · · · ⊂∗ cn+1 ⊂∗ cn ⊂∗ · · · ⊂∗ cm ⊂∗ b].

By Exercise 21.15, choose d so that ∀a ∈ A ∀n ≥ m[a ⊂∗ d ⊂∗ cn ]. We claim that d is

near to A , completing the proof. For, let n ∈ ω. Let p = max(m, n), and choose q ≥ p
such that d\cp ⊆ q. Then

{a ∈ A : a\d ⊆ n} ⊆ {a ∈ A : a\d ⊆ p}
= {a ∈ Bp : a\d ⊆ p}
⊆ {a ∈ Bp : a\cp ⊆ q},

where the last inclusion holds since a\cp = (a\d) ∪ (d\cp ). The last set is finite since cp is
near to Bp , as desired.
E21.18 (The Hausdorff gap) There exist sequences haα : α < ω1 i and hbα : α < ω1 i of
members of [ω]ω such that ∀α, β < ω1 [α < β → aα ⊂∗ aβ and bβ ⊂∗ bα ], ∀α, β < ω1 [aα ⊂∗
bβ ], and there does not exist a c ⊆ ω such that ∀α < ω1 [aα ⊂∗ c and c ⊂∗ bα ].
We construct by recursion aα , bα ⊆ ω for α < ω1 so that aα ⊂∗ bα , α < β → aα ⊂∗ aβ and
bβ ⊂∗ bα , and for all α < ω1 , bβ is near to {aα : α < β}.
Let a0 = ∅, b0 = ω. Suppose that aα and bα have been constructed for all α < β
so that aα ⊂∗ bα , α < γ < β → aα ⊂∗ aγ and bγ ⊂∗ bβ , and α < β → bα is near to
{aγ : γ < α}. By exercise 21.15 choose c such that ∀α < β[aα ⊂∗ c ⊂∗ bα ]. Suppose that

96
α < β. We claim that c is near to {aγ : γ < α}. In fact, suppose that m ∈ ω. Choose
n ≥ m such that c\bα ⊆ n. Now for any γ < α we have aγ \bα ⊆ (aγ \c) ∪ (c\bα ), so

{aγ : γ < α and aγ \c ⊆ m} ⊆ {aγ : γ < α and aγ \bα ⊆ n},

and the latter set is finite since bα is near to {aγ : γ < α}. Thus indeed c is near to
{aγ : γ < α}. Now by exercise E21.17 there is a bβ such that ∀α < β[aα ⊂∗ bβ ⊂∗ c] and
bβ is near to {aa : α < β}. By exercise E21.15 choose aβ so that ∀α < β[aα ⊂∗ aβ ⊂∗ bβ ].
This finishes the construction. S
Now suppose that d ⊆ ω and ∀α < ω1 [aα ⊂∗ d ⊂∗ bα ]. Now ω1 = m∈ω {α < ω1 :
aα \d ⊆ m}, so we can choose m ∈ ω such that |{α < ω1 : aα \d ⊆ m}| = ω1 . Hence
there is an α < ω1 such that {β < α : aβ \d ⊆ m} is infinite. Choose p ≥ m such that
d\bα ⊆ p. Now aβ \bα ⊆ (aβ \d) ∪ (d\bα ), so {β < α : aβ \d ⊆ m} ⊆ {β < α : aβ \bα ⊆ p},
contradicting bα near to {aβ : β < α}.
Solutions to exercises in Chapter 22
E22.1 Let κ be an uncountable regular cardinal. We define S < T iff S and T are
stationary subsets of κ and the following two conditions hold:
(1) {α ∈ T : cf(α) ≤ ω} is nonstationary in κ.
(2) {α ∈ T : S ∩ α is nonstationary in α)} is nonstationary in κ.
Prove that if ω < λ < µ < κ, all these cardinals regular, then Eλκ < Eµκ , where

Eλκ = {α < κ : cf(α) = λ},

and similarly for Eµκ .

First of all, {α ∈ Eµκ : cf(α) ≤ ω} is empty, so of course it is nonstationary in κ.
For (2), let C = (µ, κ). We claim that

{α ∈ Eµκ : Eλκ ∩ α is nonstationary in α} ∩ C = ∅;

this will prove (2). In fact, suppose that α is in the indicated intersection. Let D be club
in α such that Eλκ ∩ D = ∅. Now α ∈ Eµκ , so cf(α) = µ. Define αξ for all ξ < λ as follows.
Let α0 be the least member of D. If αξ ∈ D has been S defined, take any member αξ+1
of D greater than αξ . If ξ S is limit less than λ, let αξ = η<ξ αη . Then αξ ∈ D because
D is closed. Now let β = ξ<λ αξ . Then β ∈ D since D is closed, and cf(β) = λ. So
β ∈ Eλκ ∩ D, contradiction.
E22.2 Continuing exercise E22.1: Assume that κ is uncountable and regular. Show that
the relation < is transitive.
Suppose that A < B < C. Then by definition
(1) {α ∈ C : cf(α) ≤ ω} is nonstationary in κ.
(2) {α ∈ B : α ∩ A is nonstationary} is nonstationary in κ.
(3) {α ∈ C : α ∩ B is nonstationary} is nonstationary in κ.

97
We want to show

{α ∈ C : α ∩ A is nonstationary} is nonstationary.

Our assumptions give us clubs M, N in κ such that

{α ∈ B : α ∩ A is nonstationary} ∩ M = ∅ and
{α ∈ C : α ∩ B is nonstationary} ∩ N = ∅.

Let M ′ be the set of all limits of members of M ; so also M ′ is club in κ. Now it suffices
to show that
{α ∈ C : α ∩ A is nonstationary} ∩ M ′ ∩ N = ∅.
So, suppose that α ∈ C ∩ M ′ ∩ N ; we show that α ∩ A is stationary in α. To this end, let
P be club in α, and let P ′ be the set of all of its limit points. Now α ∈ C ∩ N , so α ∩ B is
stationary. Since α ∈ M ′ , it follows that α ∩ M is club in α. So M ∩ P ′ is club in α, and
so we can choose β ∈ α ∩ B ∩ M ∩ P ′ . Now β ∈ B ∩ M , so β ∩ A is stationary in β. Since
β ∈ P ′ , it follows that P ∩ β is club in β. So β ∩ A ∩ P 6= ∅, hence A ∩ P 6= ∅, as desired.
E22.3 If κ is an uncountable regular cardinal and S is a stationary subset of κ, we define

Tr(S) = {α < κ : cf(α) > ω and S ∩ α is stationary in α}.

Suppose that A, B are stationary subsets of an uncountable regular cardinal κ and A < B.
Show that Tr(A) is stationary.
Assume the conditions of the exercise. Thus by definition, {α ∈ B : cf(α) ≤ ω} is
nonstationary in κ, and also {α ∈ B : A ∩ α is non-stationary in α} is non-stationary
in κ. Hence there is a club C in κ such that C ∩ {α ∈ B : cf(α) ≤ ω} = ∅ and also
C ∩ {α ∈ B : A ∩ α is non-stationary in α} = ∅. Thus B ∩ C ⊆ Tr(A), and it follows that
Tr(A) is stationary in κ.
E22.4 (Real-valued measurable cardinals) We describe a special kind of measure. A mea-
sure on a set S is a function µ : P(S) → [0, ∞) satisfying the following conditions:
(1) µ(∅) = 0 and µ(S) = 1.
(2) If µ({s}) = 0 for all s ∈ S,
S
P If hXi : i ∈ ωi is a system of pairwise disjoint subsets of S, then µ(
(3) i∈ω Xi ) =
i∈ω µ(Xi ). (The Xi ’s are not necessarily nonempty.)

Let κ be an infinite cardinal. Then µ is κ-additive iff for every system hXα : α < γi of
nonempty pairwise disjoint sets, wich γ < κ, we have
!
[ X
µ Xα = µ(Xα ).
α<γ α<γ

98
Here this sum (where the index set γ might be uncountable), is understood to be
X
sup µ(Xα ).
F ⊆γ,
F finite α∈F

We say that an uncountable cardinal κ is real-valued measurable iff there is a κ-additive

measure on κ. Show that every measurable cardinal is real-valued measurable. Hint: let µ
take on only the values 0 and 1.
Suppose that κ is measurable. Thus κ is uncountable, and there is a κ-complete nonprin-
cipal ultrafilter U on κ. Now for any X ⊆ κ we define
n
1 if X ∈ U ,
µ(X) =
0 otherwise.

Conditions (1) and (2) in the definition of measure are clear. We can check (3) and κ-
additivity simultaneously, by assuming
S that hXα : α < βi is a system of pairwise disjoint
subsets of κ, with β < κ. If µ( α<β Xα ) = 0, clearly µ(Xα ) = 0 for all α < β, and so
!
[ X
µ Xα = µ(Xα ).
α<γ α<γ

S S
Suppose that µ( α<β Xα ) = 1. Thus α<β Xα ∈ U . If Xα ∈
/ U for all α < β, then
κ\Xα ∈ U for all α < β, and hence by κ-completeness,
 
[ \
κ\  Xα  = (κ\Xα ) ∈ U,
α<β α<β

contradiction. Hence Xα ∈ U for some α < β. There can be only one such α, since if
γ 6= α and Xγ ∈ U , then ∅ = Xα ∩ Xγ ∈ U , contradiction. Hence again
!
[ X
µ Xα = µ(Xα ).
α<γ α<γ

E22.5 Suppose that µ is a measure on a set S. A subset A of S is a µ-atom iff µ(A) > 0
and for every X ⊆ A, either µ(X) = 0 or µ(X) = µ(A). Show that if κ is a real-
valued measurable cardinal, µ is a κ-additive measure on κ, and A ⊆ κ is a µ-atom, then
{X ⊆ A : µ(X) = µ(A)} is a κ-complete nonprincipal ultrafilter on A. Conclude that κ is
a measurable cardinal if there exist such µ and A.
Let F be the indicated set. Obviously A ∈ F . Suppose that X ∈ F and X ⊆ Y ⊆ A.
Then

µ(A) = µ(X ∪ (Y \X) ∪ (A\Y )) = µ(X) + µ(Y \X) + µ(A\Y ) = µ(A) + µ(Y \X) + µ(A\Y ),

99
and so µ(A\Y ) = 0. Hence µ(A) = µ((A\Y ∪ Y ) = µ(A\Y ) + µ(Y ) = µ(Y ). So Y ∈ F .
Now suppose that Y, Z ∈ F . Then

µ(A) = µ(Y ) = µ(Y ∩ Z) + µ(Y \Z) and

µ(A) = µ(Z) = µ(Y ∩ Z) + µ(Z\Y ).

It follows that µ(Y \Z) = µ(Z\Y ). If µ(Y \Z) = µ(A), then also µ(Z\Y ) = µ(A), and
hence
2µ(A) = µ(Y \Z) + µ(Z\Y ) = µ((Y \Z) ∪ (Z\Y )) ≤ µ(A),
contradiction. So µ(Y \Z) = 0, and hence µ(A) = µ(Y ∩ Z). It follows that Y ∩ Z ∈ F .
So, F is a filter.
Clearly ∅ ∈/ F , so F is proper.
If X ⊆ A, then µ(A) = µ(X) + µ(A\X), and hence µ(X) = µ(A) or µ(A\X) = µ(A).
So X ∈ F or A\X ∈ F . Thus F is an ultrafilter. T
<κ
T Finally, for κ-completeness, suppose that A ∈ [F ] Suppose that A ∈
/ F . Then
A\
T A ∈ F . Let hXα : α < λi be an enumeration of A . For each α < λ let Yα =
β<α Xβ \Xα .

[ [
(1) Yα = (A\Xα )
α<λ α<λ

S
In fact, ⊆ is clear. Suppose that ξ ∈ α<λ (A\Xα ), and choose α < λ minimum such that
ξ ∈ (A\Xα ). Then ξ ∈ Yα . So (1) holds.
Clearly the Yα ’s are pairwise disjoint. So from (1) we get
 \ 
µ(A) = µ A\ A
!
[
=µ (A\Xα )
α<λ
!
[
=µ Yα
α<λ
X
= µ(Yα ),
α<λ

and hence there is a α < λ such that µ(Yα ) = 1. Hence µ(A\Xα ) = µ(A) also, contradic-
tion.
Hence F is κ-complete.
Since all members of F have size κ by κ-completeness and nonprincipality, it follows
that |A| = κ. So κ is a measurable cardinal.
E22.6 Prove that if κ is real-valued measurable then either κ is measurable or κ ≤ 2ω .
Hint: if there do not exist any µ-atoms, construct a binary tree of height at most ω1 .

100
Let µ be a κ-additive measure on κ. By exercise E22.5, if there is a µ-atom, then κ is
measurable. So, suppose that there does not exist any µ-atoms. We construct a tree
under ⊃ by constructing the levels Lα , as follows. L0 = {κ}. Suppose that Lα has been
constructed, and that it is a nonempty collection of subsets of κ each of positive measure.
For each X ∈ Lα let YX be a subset of X such that 0 < µ(YX ) < µ(X); such a set exists
since X is not a µ-atom. Then we define

Lα+1 = {YX , X\YX : X ∈ Lα }.

If α is a limit ordinal and Lβ has been constructed for every β < α, then we define
(   )
\ \
Lα = Zβ : Zβ ∈ Lβ for all β < α and µ  Zβ  > 0 ,
β<α β<α

except that if Lα = ∅ the construction stops.

Clearly this gives a tree. Let α be the least ordinal such that Lα is not defined. So α
is a limit ordinal.
(1) α ≤ ω1 , and in fact, if hZβ : α < γi is a branch of the tree, thus with Zβ ⊂ Zδ if
δ < β < γ, then γ is countable.
In fact, we have µ(Zβ \Zβ+1 ) > 0 for every β < γ, and the sets Zβ \Zβ+1 are pairwise
disjoint. If γ ≥ ω1 , then
[ 1

γ= β < γ : µ(Zβ \Zβ+1 ) > ,
n∈ω
n + 1

and hence there would be an n ∈ ω such that

 
1
β < γ : µ(Zβ \Zβ+1 ) >
n+1

is uncountable, which is not possible. So (1) holds.

Similarly each level of our tree is countable. It follows that the tree has at most 2ω
branches.
T Let B be the collection of all branches in this tree, and for each B ∈ B let WB =
ω
X∈B X. Let C = {WB : B ∈ B}\{∅}. Now clearly |C | ≤ 2 , and C consists of measure
0 sets.
S
(2) κ = C .
In fact, if α ∈ κ, then B = {X ∈ T : α ∈ X} is a branch, and so α ∈ WB .
From (2) it follows that κ ≤ 2ω , since the measure µ is κ-additive and µ(κ) = 1. In
fact, 2ω < κ would imply by (2) that µ(κ) = 0, contradiction.
E22.7 Let κ be a regular uncountable cardinal. Show that the diagonal intersection of the
system h(α + 1, κ) : α < κi is the set of all limit ordinals less than κ.

101
For any β ∈ κ,

β ∈ △α<κ (α + 1, κ) iff ∀α < β[β ∈ (α + 1, κ)]

iff ∀α < β[α + 1 < β]
iff β is a limit ordinal.

E22.8 Let F be a filter on a regular uncountable cardinal κ. We say that F is normal

iff it is closed under diagonal intersections. Suppose that F is normal, and (α, κ) ∈ F for
every α < κ. Show that every club of κ is in F . Hint: use exercise E22.7.
Let C be a club, and let hαξ : ξ < κi be the strictly increasing enumeration of C, and let
D be the set of all limit ordinals less than κ. By exercise E22.7 suffices to show that

D ∩ △ξ<κ (αξ , κ) ⊆ C.

So, take any β ∈ D ∩ △ξ<κ (αξ , κ). Thus β is a limit ordinal, and ∀ξ < β[β ∈ (αξ , κ)], i.e.,
∀ξ < β[αξ < β]. Now ξ ≤ αξ for all ξ, so C ∩ β is unbounded in β. Hence β ∈ C.
E22.9 Let F be a proper filter on a regular uncountable cardinal κ. Show that the following
conditions are equivalent.
(i) F is normal
(ii) For any S0 ⊆ κ, if κ\S0 ∈ / F and f is a regressive function defined on S0 , then
there is an S ⊆ S0 with κ\S ∈ / F and f is constant on S.
(i)⇒(ii): Assume (i), and suppose that S0 ⊆ κ, κ\S0 ∈ / F , and f is a regressive function
on S0 . Suppose that the conclusion fails. Then for every γ < κ we have κ\f −1 [{γ}] ∈ F ,
as otherwise we could take S = f −1 [{γ}]. By (i), take β ∈ △γ<κ (κ\f −1 [{γ}]). Then
∀γ < β[β ∈ κ\f −1 [{γ}]; in particular, β ∈ κ\f −1 [{f (β)}], contradiction.
(ii)⇒(i): Assume (ii), and suppose that haα : α < κi is a system of members of
F . Suppose that △α<κ aα ∈ / F . Now ∀α ∈ κ\△α<κ aα ∃β < α[α ∈ / aβ ]. This gives us a
regressive function f defined on κ\△α<κ aα such that for every α in that set, α ∈ / af (α) .
Hence by (ii) choose S ⊆ κ\△α<κ aα such that f is constant on S, say with value γ, with
κ\S ∈/ F . Since aγ ∈ F , we have aγ 6⊆ κ\S. Choose β ∈ aγ ∩ S. Then β ∈ / af (β) gives a
contradiction.
E22.10 A probability measure on a set S is a real-valued function µ with domain P(S)
having the following properties:
(i) µ(∅) = 0 and µ(S) = 1.
(ii) If X ⊆ Y , then µ(X) ≤ µ(Y ).
(iii) µ({a}) = 0 for all a ∈ S.
S
P (iv) If hXn : n ∈ ωi is a system of pairwise disjoint sets, then µ( n∈ω Xn ) =
n∈ω µ(Xn ). (Some of the sets Xn might be empty.)

Prove that there does not exist a probability measure on ω1 . Hint: consider an Ulam
matrix.

102
Suppose that µ is a probability measure on ω1 . Let f = hfρ : ρ < ω1 i be a family of
injections fρ : ρ → ω. Define the function A : ω × ω1 → P(ω1 ) by setting, for any ξ < ω
and α < ω1 ,
Aξα = {ρ ∈ ω1 \(α + 1) : fρ (α) = ξ}.
S
Take any α < ω1 . Since n∈ω Anα = ω1 \(α+1), Anα ∩Am α = ∅ for α 6= β, and µ(ω1 \(α+1)) =
n(α)
1, choose n(α) ∈ ω such that ϕ(Aα ) > 0. Then there exist M ∈ [ω1 ]ω1 and m ∈ ω such
that n(α) = m for every α ∈ M . Then hAm α : α ∈ M i is a system of pairwise disjoint sets
each of positive measure, contradiction.
E22.11 Show that if κ is a measurable cardinal, then there is a normal κ-complete non-
principal ultrafilter on κ. Hint: Let D be a κ-complete nonprincipal ultrafilter on κ. Define
f ≡ g iff f, g ∈ κ κ and {α < κ : f (α) = g(α)} ∈ D. Show that ≡ is an equivalence relation
on κ κ. Show that there is a relation ≺ on the collection of all ≡-classes such that for all
f, g ∈ κ κ, [f ] ≺ [g] iff {α < κ : f (α) < g(α)} ∈ D. Here for any function h ∈ κ κ we
use [h] for the equivalence class of h under ≡. Show that ≺ makes the collection of all
equivalence classes into a well-order. Show that there is a ≺ smallest equivalence class x
such that ∀f ∈ x∀γ < κ[{α < κ : γ < f (α)} ∈ D. Let E = {X ⊆ κ : f −1 [X] ∈ D}. Show
that E satisfies the requirements of the exercise.
≡ is reflexive: κ = {α < κ : f (α) = f (α)}, hence f ≡ f .
≡ is symmetric: Assume that f ≡ g. Thus {α < κ : f (α) = g(α)} ∈ D. Hence
{α < κ : g(α) = f (α)} = {α < κ : f (α) = g(α)} ∈ D. Hence g ≡ f .
≡ is transitive: Assume that f ≡ g ≡ h. Thus {α < κ : f (α) = g(α)} ∈ D and
{α < κ : g(α) = h(α)} ∈ D. Hence

{α < κ : f (α) = g(α)} ∩ {α < κ : g(α) = h(α)} ∈ D;

since

{α < κ : f (α) = g(α)} ∩ {α < κ : g(α) = h(α)} ⊆ {α < κ : f (α) = h(α)},

we get {α < κ : f (α) = h(α)} ∈ D, so f ≡ h.

Now define

x≺y iff ∃f, g[x = [f ] and y = [g] and {α < κ : f (α) < g(α)} ∈ D].

(1) ∀f, g ∈ κ κ[[f ] ≺ [g] iff {α < κ : f (ϕ) < g(α)} ∈ D].
In fact, ⇐ is immediate from the definition. Now suppose that [f ] ≺ [g]. Choose f ′ , g ′ ∈ κ κ
such that [f ] = [f ′ ], [g] = [g ′ ], and {α < κ : f ′ (α) < g ′ (α)} ∈ D. Then

{α < κ : f (α) = f ′ (α)}∩{α < κ : f ′ (α) < g ′ (α)} ∩ {α < κ : g(α) = g ′ (α)}
⊆ {α < κ : f (α) < g(α)};

the left side is in D, hence also the right side is in D, so {α < κ : f (α) < g(α)} ∈ D. Thus
(1) holds.

103
 ≺ is irreflexive: {α < κ : f (α) < f (α)} = ∅ ∈
/ D, so [f ] 6≺ [f ].
≺ is transitive: Assume that [f ] ≺ [g] ≺ [h]. Then

{α < κ : f (α) < g(α)} ∩ {α < κ : g(α) < h(α)} ⊆ {α < κ : f (α) < h(α)};

the left side is in D, hence also the right side is in D, so [f ] ≺ [h].

≺ is a linear order: Suppose that f, g ∈ κ κ are such that [f ] 6= [g] and [f ] 6≺ [g]. Now

κ = {α < κ : f (α) < g(α)} ∪ {α < κ : f (α) = g(α} ∪ {α < κ : g(α) < f (α)};

The first two sets are not in D, so the third one is in D, and hence [g] ≺ [f ].
≺ is a well-order: Suppose not. Then we get a sequence hf m : m ∈ ωi of members of
κ
κ such that [fm+1 ] ≺ [fm ] for all m ∈ ω. Thus {α < κ : fm+1 (α) < fm (α)} ∈ D for all
m ∈ ω. It follows that
\
{α < κ : fm+1 (α) < fm (α)} ∈ D;
m∈ω

taking any element α in this intersection, we get . . . fm+1 (α) < fm (α) . . ., contradiction.
Now let k(α) = α for all α < κ. Then for any γ < κ we have

{α < κ : γ < k(α)} = {α < κ : γ < α} = κ\(γ + 1) ∈ D.

It follows that we can take the smallest equivalence class [f ] such that for any γ < κ we
have {α < κ : γ < f (α)} ∈ D. Now we let E = {X ⊆ κ : f −1 [X] ∈ D}. We claim that E
is as desired in the exercise.
∅∈ / E: This is true since f −1 [∅] = ∅ ∈/ D.
If X ⊆ Y ⊆ κ and X ∈ E, then Y ∈ E: In fact, assume that X ⊆ Y ⊆ κ and X ∈ E.
Then f −1 [X] ⊆ f −1 [Y ] and f −1 [X] ∈ D, so f −1 [y] ∈ D, so that Y ∈ E.
If X, Y ∈ E, then X ∩ Y ∈ E: In fact, f −1 [X ∩ Y ] = f −1 [X] ∩ f −1 [y], so this is clear.
If X ⊆ κ, then X ∈ E or (κ\X) ∈ E: For, suppose that X ∈ / E. Then f −1 [X] ∈ / D,
−1 −1
so f [κ\X] = (κ\f [X]) ∈ D, and hence (κ\X) ∈ E.
E is nonprincipal: for any α < κ we have {β < κ : α < f (β)} ∈ D, and {β < κ : α <
f (β)} ⊆ {β < κ : α 6= f (β)}, so {β < κ : α 6= f (β)} ∈ D, hence {β < κ : α = f (β)} ∈ / D,
−1
hence f [{α}] ∈ / D and so {α} ∈ / E.
E is κ-complete: Suppose that hXα : α < βi is a system of subsets of κ, with
−1
β <h κ and with i [X α ] ∈ E for all α < β. Thus f [Xα ] ∈ D for all α < β. Since
−1
T T −1
T
f α<β Xα = α<β f [Xα ] ∈ D, it follows that α<β Xα ∈ E.
E is normal: We apply exercise 22.9. Suppose that S0 ∈ E and g is regressive on S0 .
Note that f −1 [S0 ] ∈ D. Let h = g ◦ f . Then for any α ∈ f −1 [S0 ] we have h(α) < f (α),
so that [h] ≺ [f ]. By the definition of f it then follows that there is a γ < κ such that
{α < κ : γ < h(α)} ∈ / D. Hence {α < κ : h(α) ≤ γ} ∈ D. Now
[
{α < κ : h(α) ≤ γ} = {α < κ : h(α) = δ},
δ≤γ

104
and so there is a δ ≤ γ such that {α < κ : h(α) = δ} ∈ D. Now {α < κ : h(α) = δ} =
h−1 [{δ}] = f −1 [g −1 [{δ}], so g −1 [{δ}] ∈ E. This checks the condition of exercise 22.9.
Solutions to exercises in Chapter 23
E23.1 In the ordering <L determine the first four sets and their order. Hint: use Lemma
23.2
Recall that L is well-ordered by ordering each set Lα , placing the elements of Lα+1 after
all of the elements of Lα . We have L0 = ∅. By 23.23(viii) we have L1 = V1 = {∅},
L2 = V2 = {∅, {∅}}, L3 = V3 = {∅, {∅}, {{∅}}, {∅, {∅}}}. Thus the first four elements of L
are ∅, {∅}, {{∅}}, {∅, {∅}}. We have ∅ <L {∅} and {∅} <L both {{∅}} and {∅, {∅}}. We
just need to determine the relative order of {{∅}} and {∅, {∅}}.
(1) Df(L2 , ∅, 1) = P(1 L2 ).
For, clearly ⊆ holds. We can see ⊇ by applying Lemma 23.2:
For ϕ equal to v0 = v0 we get 1 L2 ∈ Df(L2 , 1).
For ϕ equal to ¬(v0 = v0 ) we get ∅ ∈ Df(L2 , 1).
For ϕ equal to ∃v1 (v0 ∈ v1 ) we get {h∅i} ∈ Df(L2 , 1).
For ϕ equal to ∃v0 (v0 ∈ v1 ) we get {h{∅}i} ∈ Df(L2 , 1).
This proves (1).
Now L2 = {x ∈ L2 : ∅⌢ hxi ∈ 1 L2 } and {{∅}} = {x ∈ L2 : ∅⌢ hxi ∈ {h{∅}i}. It follows
that in determining the order of our two elements we have n(L2 ) = n({{∅}}) = 0. It
follows that s(L2 ) = s({{∅}}) = ∅. Now we need to determine R(L2 ) and R({{∅}}). Since
1
L2 = Diag= (∅, 1, 0, 0), we have 1 L2 ∈ Df′ (0, L2 , ∅, 1). Clearly {{∅}} ∈
/ Df′ (0, L2 , ∅, 1), so
it follows that 1 L2 <4L2 1 {h{{∅}}i}. Hence L2 preceeds {{∅}} in the order <L .
Thus the first four sets are, in order, ∅, {∅}, {∅, {∅}}, and {{∅}}.
E23.2 Suppose that M is a nonempty transitive class satisfying the comprehension ax-
ioms, and also ∀x ⊆ M∃y ∈ M[x ⊆ y]. Show that M is a model of ZF.
Extensionality holds by Theorem 10.11.
Foundation holds by Theorem 10.17
Comprehension is given.
Pairing: given x, y ∈ M, we have {x, y} ⊆ M, hence there is a z ∈ M such that
{x, y} ⊆ z. So pairing holds by Theorem 10.13
S S
Union: Suppose that x ∈ M. Then x ⊆ M, so choose z ∈ M such that x ⊆ z.
Hence union holds by Theorem 10.14
Power set: Clear by Theorem 10.15 and the hypothesis.
Replacement: We apply Theorem 10.16 Assume that A, w1 , . . . , wn ∈ M and

∀x ∈ A∃!y ∈ MϕM (x, y, A, w1, . . . , wn ).

Let
z = {y ∈ M : ∃x ∈ AϕM (x, y, A, w1, . . . , wn )}.

105
Then choose t ∈ M such that z ⊆ t. So replacement holds.
Infinity: By induction we show that m ∈ M for all m ∈ ω. To show that ∅ ∈ M, take
any a ∈ M. By comprehension in M choose z ∈ M such that ∀x[x ∈ z ↔ x ∈ a ∧ x 6= x].
So z = ∅, as desired. Suppose m ∈ M. By the pairing axiom in M applied to m, m,
choose a set a ∈ M such that m ∈ a. Then by the compehension axiom in M take b ∈ M
such that ∀x[x ∈ b ↔ x ∈ a ∧ x = m]. Thus b = {m}. By the pairing axiom in M
choose c ∈ M such that m, {m} ∈ c. By the union axiom in bf M choose d ∈ M such that
∀Y ∀x[x ∈ Y ∈ c → x ∈ d]. So m, {m} ⊆ d. By the comprehension axiom in M choose
e ∈ M such that ∀x[x ∈ e ↔ x ∈ d ∧ (x ∈ m ∨ x = m)]. Hence e = m ∪ {m}. This finishes
the induction. So ω ⊆ M. Hence by hypothesis there is a y ∈ M such that ω ⊆ y. Now
by comprehension in M let z ∈ M be such that ∀x[x ∈ z ↔ x ∈ y ∧ x is a finite ordinal].
Thus z = ω. Hence infinity holds in M by Theorem 10.27.
E23.3 Show that if M is a transitive proper class model of ZF, then ∀x ⊆ M∃y ∈ M[x ⊆
y].
Suppose that x ⊆ M. Choose α such that x ⊆ Vα . Thus x ⊆ Vα ∩ M.
(1) If α ∈ M, then Vα ∩ M ∈ M.
This holds by Theorem 10.31(ii). By (1), it suffices to show that Ord ⊆ M.
Let ϕ(x, y) be the formula “rank(x) = y”. So ϕ is absolute for M by Theorem
10.30(iv). Given an ordinal β, choose x ∈ M such that rank(x) > β; this is possible
because M is a proper class. Now M |= ∀u∃vϕ(u, v), so choose y ∈ M such that ϕM (x, y).
Thus ϕ(x, y), so y is an ordinal greater than β. Since M is transitive, β ∈ M.
E23.4 Show that for every ordinal α > ω, |Lα | = |Vα | iff α = iα .
Assume that α > ω; write α = ω + β. By Lemma 23.24, |Lα | = |α|. By Theorem 10.10(ii),
|Vα | = iβ . So |Lα | = |Vα | iff |α| = iβ .
Now suppose that α = iα . Now if γ < α, then ω, γ < α since α is an uncountable
cardinal. Hence ω + γ < α. Clearly then α = ω + α, and so α = β. It follows that

|Lα | = |α| = iα = iβ = |Vα |.

Now suppose that α 6= iα . If α is countable, then

|Lα | = |α| ≤ ω < iβ = |Vα |,

as desired. Suppose that α ≥ ω1 . Then β ≥ ω1 . Write β = ω1 + γ. Then

α = ω + β = ω + ω1 + γ = ω1 + γ = β.

Hence |Lα | = |α| ≤ α < iα = iβ = |Vα |, as desired.

E23.5 Assume V = L and α > ω. Then Lα = Vα iff α = iα .
⇒: by exercise E23.4
⇐: First we claim

106
(*) (V = L) Vβ ⊆ Liβ for every ordinal β.
We prove this by induction on β. It is obvious for β = 0, and the inductive step with β
limit is obvious. Now assume it for β. Then using Theorem 23.32

Vβ+1 = P(Vβ ) ⊆ P(Liβ ) ⊆ Li+ = Liβ+1 ,

β

finishing the inductive proof.

Now assume that α = iα . Then Vα ⊆ Liα = Lα ⊆ Vα using Theorem 23.23(vii).
E23.6 Assume V = L and prove that Lκ = H(κ) for every infinite cardinal κ.
First, Lω = Vω = H(ω). Now suppose that κ is uncountable and regular. Take any x ∈ Lκ .
Choose α < κ such that x ∈ Lα . We may assume that α is infinite. Since Lα is transitive,
we also have trcl(x) ⊆ Lα . Hence |trcl(x)| ≤ |Lα | = |α| < κ. So x ∈ H(κ). Thus we
have shown that Lκ ⊆ H(κ). Suppose that H(κ) 6⊆ Lκ . By the foundation axiom, choose
x ∈ H(κ)\Lκ such that x ⊆ Lκ . Since |x| < κ and κ is regular, choose β < κ such that
x ⊆ Lβ . Hence x ∈ Lβ + ⊆ Lκ by Theorem 23.32. This is a contradiction. So Lκ = H(κ).
This finishes the case when κ is regular.
Now suppose that κ is singular. Then
[ [
Lκ = Lα+ = H(α+ ) = H(κ).
α<κ α<κ

E23.7 Show that if ϕ(y1 , . . . , yn , x) is a formula with at most the indicated variables free,
then
∀α1 , . . . , αn ∀a[∀x[ϕ(α1 , . . . , αn , x) ↔ x = a] → a ∈ OD].
Also show that ∅ ∈ OD.
Fix α1 , . . . , αn and assume that ∀x[ϕ(α1 , . . . , αn , x) ↔ x = a]. By Corollary 10.39 fix
β > max(α1 , . . . , αn , rank(a)) so that ϕ is absolute for Vβ . Let

R = {z ∈ n+1 Vβ : ϕ(z0 , . . . , zn )}.

Let s = hα1 , . . . αn i ∈ n β. Then

∀x ∈ Vβ [s⌢ hxi ∈ R ↔ x = a].

By the absoluteness of ϕ,

R = {s ∈ n+1 Vβ : ϕ(s0 , . . . , sn )Vβ ,

so R ∈ Df(Vβ , ∅, n + 1) by Lemma 23.2. Hence a ∈ OD.

Now for the last part of the exercise, we apply the first part to the formula y = x and
the ordinal 0. This gives

∀a[∀x[0 = x ↔ x = a] → a ∈ OD],

107
which means that 0 ∈ OD.
E23.8 We define s ⊳ t iff s, t ∈ <ω ON and one of the following holds:
(i) s = ∅ and t 6= ∅;
(ii) s, t 6= ∅ and max(rng(s)) < max(rng(t));
(iii) s, t 6= ∅ and max(rng(s)) = max(rng(t)) and dmn(s) < dmn(t);
(iv) s, t 6= ∅ and max(rng(s)) = max(rng(t)) and dmn(s) = dmn(t) and ∃k ∈
dmn(s)[s ↾ k = t ↾ k and s(k) < t(k)].
Prove the following:
(v) ⊳ well-orders <ω ON.
(vi) ∀t ∈ <ω ON[{s : s ⊳ t} is a set].
(vii) For every infinite ordinal α we have |<ω α| = |α|.
(viii) For every uncountable cardinal κ, the set <ω κ is well-ordered by ⊳ in order type
κ and is an initial segment of <ω ON.
(ix) <ω ω is well-ordered by ⊳ in order type ω 2 .
(v): Clearly ⊳ is irreflexive, and ∀s, t ∈ <ω ON[s 6= t → s ⊳ t or t ⊳ s]. Now suppose that
s ⊳ t ⊳ w.
Case 1. s = ∅. Then t, w 6= ∅, and so s ⊳ w.
Case 2. s 6= ∅. Then clearly t, w 6= ∅ and max(rng(s)) ≤ max(rng(t)) ≤ max(rng(w)).
Subcase 2.1. max(rng(s)) < max(rng(t)). Then max(rng(s)) < max(rng(w)), and
hence s ⊳ w.
Subcase 2.2. max(rng(s)) = max(rng(t)). Thus dmn(s) ≤ dmn(t).
Subsubcase 2.2.1. max(rng(t)) < max(rng(w). max(rng(s)) < max(rng(w))
follows, and so s ⊳ w.
Subsubcase 2.2.2. max(rng(t)) = max(rng(w)). So dmn(t) ≤ dmn(w).
Subsubsubcase 2.2.2.1. dmn(s) < dmn(t) or dmn(t) < dmn(w). Then
dmn(s) < dmn(w) and hence s ⊳ w.
Subsubsubcase 2.2.2.2. dmn(s) = dmn(t) = dmn(w). Then there exist
k, l ∈ dmn(s) such that s ↾ k = t ↾ k, s(k) < t(k), t ↾ l = w ↾ l, and t(l) < w(l).
Subsubsubsubcase 2.2.2.2.1. k < l. Then s ↾ k = w ↾ k and s(k) < t(k) =
w(k), so s ⊳ w.
Subsubsubsubcase 2.2.2.2.2. k = l. Then s ↾ k = w ↾ k and s(k) < t(k) <
w(k), so s ⊳ w.
Subsubsubsubcase 2.2.2.2.3. l < k. Then s ↾ l = w ↾ l and s(l) = t(l) =
w(l), so s ⊳ w.
This completes the proof that ⊳ is a linear order. Now suppose that A is a nonempty
subset of <ω ON. If ∅ ∈ A, then it is the least element of A. So, suppose that ∅ ∈
/ A. Let
B = {max(rng(s)) : s ∈ A}, and let α be the least member of B. Let C = {dmn(s) : s ∈
A, max(rng(s)) = α}. Thus C 6= ∅, so let m be the least member of C. Let D = {s(0) :
s ∈ A, max(rng(s)) = α, dmn(s) = m}, and let γ0 be the least member of D. Suppose that
γj has been defined for all j ≤ i so that i + 1 < m and

def
E = {s ∈ A : max(rng(s)) = α, dmn(s) = β, s ↾ (i + 1) = hγ0 , . . . , γi i}

108
is nonempty. Let F = {s(i + 1) : s ∈ E}, and let γi+1 be the least element of F . By
construction, there is an s ∈ A such that max(rng(s)) = α, dmn(s) = m, and s(i) = γi
for all i < m. We claim that s is the least element of A. For, take any t ∈ A with s 6= t.
Then max(rng(s)) = α ≤ max(rng(t)); if < holds here, then s ⊳ t, as desired. So suppose
= holds. Then dmn(s) = m ≤ dmn(t). It < holds here, then s ⊳ t, as desired. So assume
that = holds. Let i ≤ m be maximum such that s ↾ i = t ↾ i. Then i < m since s 6= t.
Then by construction s(i) < t(i). Hence s ⊳ t, as desired. Therefore, ⊳ is a well-order.
(vi): If t = ∅, then {s ∈ <ω ON : s ⊳ t} = ∅. Now suppose that t 6= ∅. Clearly
{s ∈ <ω ON : s ⊳ t} ⊆ |<ω (α + 1), where α = max(rng(t)).
(vii): We have X X
<ω
α ≤ |k α| = |α|k = |α|.
k∈ω k∈ω

Also, ξ 7→ {(0, ξ)} is a one-one function from α into 1 (α + 1) ⊆ <ω (α + 1), so equality
holds.
(viii): Assume that κ is an uncountable cardinal. We verify the second statement
first. Suppose that s, t ∈ <ω ON and s ⊳ t ∈ <ω κ. If s = ∅, then s ∈ <ω κ. If s 6= ∅, then
max(rng(s)) ≤ max(rng(t)) ∈ κ; so max(rng(s)) ∈ κ and hence s ∈ <ω κ.
Now |<ω κ| = κ by (vii), so κ ≤ |<ω κ|. If s ∈ <ω κ, then {t ∈ <ω On : t ⊳ s} ⊆
<ω
(max(rng(s))), and so by (vii), |{t ∈ <ω On : t ⊳ s}| ≤ | max(ω, rng(s))| < κ. Hence by
the preceding paragraph, the order type of <ω κ is κ.
(ix): For each m ∈ ω let Am = {s ∈ <ω ω : s 6= ∅ and S max(rng(s)) = m}. Thus
<ω
∀m ∈ ω∀s ∈ Am ∀t ∈ Am+1 [s ⊳ t]. Moreover, ω = {∅} ∪ m∈ω Am . For m ∈ ω and n a
positive integer, let Bmn = {s ∈ Am : dmn(s) = n}. So ∀m ∈ ω∀n ∈ ω\1∀s ∈ Bmn ∀t ∈
Bm,n+1 [s ⊳ t]. Each Bmn is finite. Hence Am has order type ω and <ω ω has order type
ω2.
E23.9 Prove that for any m ∈ ω and any set A, Df(A, ∅, n) = {En(m, A, n) : m ∈ ω}.
We prove ⊇ by complete induction on m. Suppose that En(m′ , A, n) ∈ Df(A, ∅, n) when-
ever m′ < m. If m = 0, then En(m, A, n) = ∅ ∈ Df(A, ∅, n) using Lemma 23.2. Suppose
that m 6= 0, and write m = 2i · 3j · 5k · r with r not divisible by 2,3, or 5. If r = 1, k = 0,
and i, j < n, then En(m, A, n) = Diag∈ (A, n, i, j) ∈ Df ′ (0, A, ∅, n) ⊆ Df (A, ∅, n). If r = 1,
k = 1, and i, j < n, then En(m, A, n) = Diag= (A, n, i, j) ∈ Df ′ (0, A, ∅, n) ⊆ Df (A, ∅, n). If
r = 1 and k = 2, then i < m, hence by the inductive assumption En(i, A, n) ∈ Df(A, ∅, n),
and clearly then En(m, A, n) = n A\En(i, A, n) ∈ Df(A, ∅, n). If r = 1 and k = 3, then
i, j < m, hence by the inductive assumption En(i, A, n), En(i, A, n) ∈ Df(A, ∅, n), and
clearly then En(m, A, n) = En(i, A, n) ∩ En(j, A, n) ∈ Df(A, ∅, n). If r = 1 and k = 4, then
i < m, hence by the inductive assumption En(i, A, n + 1) ∈ Df(A, ∅, n), and clearly then
En(m, A, n) = Proj(A, En(i, A, n + 1), n) ∈ Df(A, ∅, n). Finally, in any other case, again
En(m, i, n) = ∅ ∈ Df(A, ∅, n) by Lemma 23.2. This finishes the proof of ⊇.
For ⊆, we prove by induction on k that Df ′ (k, A, ∅, n) ⊆ {En(m, A, n) : m ∈ ω}. For
k = 0, take any x ∈ Df ′ (k, A, ∅, n). We have three possibilities.
Case 1. x = Rel(A, ∅, n, i). Then x = ∅ = En(0, A, n).
Case 2. x = Diag∈ (A, n, i, j). If not(i, j < n), then x = ∅ = En(0, A, n). If i, j < n,
then x = En(2i · 3j , A, n).

109
 Case 3. x = Diag= (A, n, i, j). If not(i, j < n), then x = ∅ = En(0, A, n). If i, j < n,
then x = En(2i · 3j · 5, A, n).
Now suppose that Df ′ (k, A, ∅, n) ⊆ {En(m, A, n) : m ∈ ω}. Take any x ∈ Df ′ (k +
1, A, ∅, n). If x ∈ Df ′ (k, A, ∅, n), then the inductive hypothesis gives the desired conclusion.
Suppose that x ∈ / Df ′ (k, A, ∅, n). Then we have three possibilities.
Case 1. There is an R ∈ Df′ (k, A, ∅, n) such that x = n A\R. By the inductive
hypothesis choose i so that R = En(i, A, n). Then x = En(2i · 52 , A, n).
Case 2. There are R, S ∈ Df′ (k, A, ∅, n) such that x = R ∩ S. By the inductive
hypothesis choose i, j so that R = En(i, A, n) and S = En(j, A, n). Then x = En(2i · 3j ·
53 , A, n).
Case 3. There is an R ∈ Df′ (k, A, ∅, n + 1) such that x = Proj(A, R, n). By the
inductive hypothesis choose i so that R = En(i, A, n + 1). Then x = En(2i · 54 , A, n).
This completes the inductive proof.
E23.10 Prove that if ϕ(x0 , . . . , xn−1 ) is a formula with free variables among x0 , . . . , xn−1 ,
then there is an m ∈ ω such that for every set A,

{s ∈ n A : ϕA (s(0), . . . , s(n − 1))} = En(m, A, n).

We proceed by induction on the number of quantifiers in ϕ, and within that, by induction

on formulas in the usual sense. For brevity let S(ϕ) be the set {s ∈ n A : ϕA (s(0), . . . , s(n−
1))}. Then

S(xi ∈ xj ) = Diag∈ (A, n, i, j) = En(2i · 3j , A, n);

S(xi = xj ) = Diag= (A, n, i, j) = En(2i · 3j · 5, A, n);

if S(ψ) = En(i, A, n), then

S(¬ψ) = n A\S(ψ) = En(2i · 52 , A, n);

if S(ψ) = En(i, A, n) and S(χ) = En(j, A, n), then

S(ψ ∧ χ) = S(ψ) ∩ S(χ) = En(2l · 3j · 53 , A, n).

The inductive step from ψ to ∃yψ requires more care. By a change of bound variable we
obtain a formula χ such that ∀yψ is logically equivalent to ∀xn χ, with the free variables
of χ among x0 , . . . , xn . Hence with S(χ) = En(i, A, n + 1),

S(∃yψ) = {s ∈ n A : ∃xn χ(s(0), . . . , s(n − 1), xn )}

= Proj(A, S(χ), n + 1) = En(2i · 54 , A, n).

E23.11 By exercise E23.8, for each uncountable cardinal κ there is an isomorphism fκ

from (κ, ∈) onto (<ω κ, ⊳). Then fκ ⊆ fλ for κ < λ. It follows that there is a function
Enon mapping ON onto <ω ON such that α < β iff Enon(α) ⊳ Enon(β).

110
 Now we define a class function Enod with domain ON, as follows. For any ordinal γ,

 a if there exist s, β, m, n such that Enon(γ) = s⌢ hβ, n, mi

Enod(γ) = with m, n ∈ ω, β ∈ ON, s ∈ <ω β, dmn(s) = n, and

 ∀x ∈ Vβ [s⌢ x ∈ En(m, Vβ , n + 1) ↔ x = a],
0 otherwise.

Prove that OD = {Enod(γ) : γ ∈ ON}.

First suppose that γ ∈ ON. If Enod(γ) = 0, then Enod(γ) ∈ OD by the second part
of exercise E23.7. Suppose that Enod(γ) = a by the first part of the definition of Enod.
Choose s, β, m, n accordingly. Now En(m, Vβ , n + 1) ∈ Df(A, ∅, n) by Exercise 23.9. So
a ∈ OD by definition.
Conversely, suppose that a ∈ OD. Choose β, n, s, R accordingly. By Exercise 23.9
choose m so that R = En(m, Vβ , n+1). Let γ be such that Enon(γ) = s⌢ hβ, m, ni. Clearly
then Enod(γ) = a.
E23.12 Now we define HOD = {x ∈ OD : trcl(x) ⊆ OD}.
Prove that ON ⊆ HOD and HOD is transitive.
Let α ∈ ON. Now α is transitive, so trcl(α) = α. So it suffices to show that α ∈ OD. We
apply exercise E23.7 to the formula y = x to obtain ∀x[α = x ↔ x = α] → α ∈ OD. So
α ∈ OD.
If y ∈ x ∈ HOD, then trcl(x) ⊆ OD. Since y ∈ trcl(x), it follows that y ∈ OD, and
also trcl(y) ⊆ trcl(x), so trcl(y) ⊆ OD. Hence y ∈ HOD.
E23.13 Show that (Vα ∩ HOD) ∈ HOD for every ordinal α.
Take any ordinal α. It suffices to show that (Vα ∩ HOD) ∈ OD. Applying exercise E23.7
to the formula x = Vy ∩ HOD, with n = 1, α in place of α1 , and Vα ∩ HOD in place of
a, we have

∀x[x = Vα ∩ HOD ↔ x = Vα ∩ HOD] → (Vα ∩ HOD) ∈ OD,

and the desired result follows.

E23.14 Prove without using the axiom of choice that HOD is a model of ZFC.
Extensionality: holds since HOD is transitive.
Comprehension axioms: by Theorem 10.12 it suffices to take a formula ϕ with free
variables among x, z, w1 , . . . , wn , assume that z, w1 , . . . , wn ∈ HOD, and prove that

def
a = {x ∈ z : ϕHOD ϕ(x, z, w1 , . . . , wn )} ∈ HOD.

By exercise E23.11, choose ordinals α, β1 , . . . , βn such that z = Enod(α) and wi = Enod(βi )

for i = 1, . . . , n. Let ψ(y0 , . . . , yn , v) be the formula

v = {x ∈ Enod(y0 ) : ϕHOD (x, Enod(y1 ), . . . , Enod(yn ))}.

111
Then we apply exercise E23.7, with ψ(y1 , . . . , yn , v) in place of ϕ(y1 , . . . , yn , x) to obtain

∀v[ψ(α, β1 , . . . , βn , v) ↔ v = a] → a ∈ OD.

Now ψ(α, β1 , . . . , βn , v) is the formula

v = {x ∈ Enod(α) : ϕHOD (x, Enod(β1 ), . . . , Enod(βn ))},

which is the same as

v = {x ∈ z : ϕHOD (x, w1 , . . . , wn )},
or simply v = a. So a ∈ OD. Since a ⊆ HOD, we get a ∈ HOD.
Pairing: Given x, y ∈ HOD, take α > max(rank(x), rank(y)). Then x, y ∈ Vα ∩
HOD ∈ HOD by exercise E23.13. So Theorem 10.13 applies.
S
Union: Given x ∈ HOD, take α > rank(x). Then x ∈ Vα ∩ HOD ∈ HOD; apply
Theorem 10.14
Power set: Given x ∈ HOD, write x = Enod(α). Apply exercise E23.7 to the formula
x = P(Enod(y)); we get

∀w[w = P(Enod(α)) ↔ w = P(Enod(α))] → P(Enod(α)) ∈ OD,

and hence P(x) = P(Enod(α)) ∈ OD. Now P(x) ⊆ HOD, so P(x) ∈ HOD. Hence
power set holds by Theorem 10.15
Replacement: to apply Theorem 10.16, let ϕ be a formula with free variables among
x, y, A, w1, . . . , wn , take any A, w1 , . . . , wn ∈ HOD, and assume that

∀x ∈ A∃!y[y ∈ HOD ∧ ϕHOD (x, y, A, w1, . . . , wn )].

Choose α, β1 , . . . , βn such that A = Enod(α) and wi = Enod(βi ) for i = 1, . . . , n. By

the replacement axiom there is a set Z such that for all x ∈ A there is a y ∈ Z such
that y ∈ HOD ∧ ϕHOD (x, y, A, w1, . . . , wn ). Let a = {y ∈ Z : y ∈ HOD ∧ ∃x ∈
AϕHOD (x, y, A, w1, . . . , wn )}. Then

a = {y ∈ HOD : ∃x ∈ A[ϕHOD (x, y, A, w1, . . . , wn )]}.

In fact, ⊆ is clear. Conversely, suppose y ∈ HOD ∧ ∃x ∈ A[ϕHOD (x, y, A, w1, . . . , wn )]}.

Take such an x. Choose y ′ ∈ Z such that y ′ ∈ HOD ∧ ϕHOD (x, y ′ , A, w1 , . . . , wn ). By the
uniqueness condition, y = y ′ . Hence y ∈ a, as desired.
Since a ⊆ HOD and HOD is transitive, it follows that trcl(a) ⊆ HOD, and so
trcl(a) ⊆ OD. Hence it suffices to show that a ∈ OD. Now let ψ(v, s1, . . . , sn , z) be the
formula

z = {y ∈ HOD : ∃x ∈ Enod(v)[ϕHOD (x, y, Enod(v), Enod(s1 ), . . . , Enod(sn ))]}.

Now we apply exercise E23.7 to the formula ψ(v, s1, . . . , sn , z) in place of ϕ, and take the
ordinals α, β1 , . . . , βn ; this gives

∀z[ψ(α, β1 , . . . , βn , z) ↔ z = a] → a ∈ OD;

112
so

∀z[z = {y ∈ HOD : ∃x ∈ Enod(α)

[ϕHOD (x, y, Enod(α), Enod(β1 ), . . . , Enod(βn ) ↔ z = a]
→ a ∈ OD,

giving

∀z[z = {y ∈ HOD : ∃x ∈ A[ϕHOD (x, y, A, w1, . . . , wn )} ↔ z = a] → a ∈ OD.

Recalling the definition of a, it follows that a ∈ OD.

Foundation: holds since HOD is transitive.
Infinity: holds since ω ∈ HOD by exercise E23.12; see Theorem 10.27
Axiom of Choice: given A ∈ HOD, we want to find a relation R ∈ HOD which
well-orders A. By absoluteness (Proposition 10.32), this suffices. Let

R = {(x, y) ∈ A × A : ∃ξ[x = Enod(ξ) ∧ ∀η ≤ ξ[y 6= Enod(η)]].

Clearly R well-orders A. To see that R ∈ HOD we will apply exercise E23.7 again. Choose
α so that A = Enod(α). Let ϕ(y, x) be the formula

x = {(u, v) ∈ Enod(y) × Enod(y) : ∃ξ[u = Enod(ξ) ∧ ∀η ≤ ξ[v 6= Enod(η)]}.

Applying exercise E23.7, we get

∀x[ϕ(α, x) ↔ x = R] → R ∈ OD.

Now ψ(α, x) is

x = {(u, v) ∈ A × A : ∃ξ[u = Enod(ξ) ∧ ∀η ≤ ξ[v 6= Enod(η)]},

i.e., it is x = R. So we conclude that R ∈ OD. It remains only to check that R ⊆ HOD.

If (u, v) ∈ R, then clearly u, v ∈ HOD. Hence by Theorem 10.28, (u, v) ∈ HOD.
E23.15 Show that the axiom of choice holds in L(B) iff trcl({A}) can be well-ordered in
L(B).
The direction ⇐ is obvious. For ⇒, we just need to modify a usual proof of AC in L. Note
that since C = ∅ in the definition of L(B), we have Rel(A, ∅, n, i) = ∅. By the proof of
Lemma 23.1, we can simply ignore Rel(A, ∅, n, i).
Let ≺ be a well-order of trcl({B}). Let A be any set, and n any natural number. For
each R ∈ {Diag∈ (A, n, i, j) : i, j < n}, let (Ch(0, A, n, R), Ch(1, A, n, R)) be the smallest
pair (i, j), in the lexicographic order of ω × ω such that i, j < n and R = Diag∈ (A, n, i, j).
Now we define

R <0An S iff (Ch(0, A, n, R), Ch(1, A, n, R)) <lex Ch(0, A, n, S), Ch(1, A, n, S)).

113
Clearly this is a well-order of {Diag∈ (A, n, i, j) : i, j < n}.
In a very analogous way we can define a well-order <1An of {Diag= (A, n, i, j) : i, j <
n}.
Now we can define a well-order <2An of

{Diag∈ (A, n, i, j) : i, j < n} ∪ {Diag= (A, n, i, j) : i, j < n}

as follows. For any R, S in this union,

R <2An S iff R, S ∈ {Diag∈ (A, n, i, j) : i, j < n} and R <0An S

or R ∈ {Diag∈ (A, n, i, j) : i, j < n}, S ∈
/ {Diag∈ (A, n, i, j) : i, j < n}
or R, S ∈
/ {Diag∈ (A, n, i, j) : i, j < n} and R <1An S.

For the next few constructions, suppose that X and A are sets, n ∈ ω, and we are given a
well-ordering < of X. Then we well-order {n A\R : R ∈ X} by setting

S ≺0,A,n,<,X T iff ∃S ′ , T ′ ∈ X[S ′ < T ′ and S = n A\S ′ , T = n A\T ′ ].

We well-order {R ∪ S : R, S ∈ X} as follows. Suppose that U, V ∈ {R ∪ S : R, S ∈ X}.

Let (R, S) be lexicographically smallest in X × X (using <) such that U = R ∪ S, and let
(R′ , S ′ ) be lexicographically smallest in X × X (using <) such that V = R′ ∪ S ′ . Then
U <1,A,n,<,X V iff (R, S) <lex (R′ , S ′ ).
We well-order {proj(A, R, n) : R ∈ X} as follows. Suppose that U, V ∈ {proj(A, R, n) :
R ∈ X}. Let R be <-minimum in X such that U = proj(A, R, n), and let S be <-minimum
in X such that V = proj(A, S, n). Then U <2,A,n,<,X V iff R < S.
Next, for any set A and any k, n ∈ ω we define a well-order <3kAn of Df ′ (k, A, n) by
induction on k. Let <30An be <2An . Assume that <3kAn has been defined for all n ∈ ω,
and let R, S ∈ Df ′ (k + 1, A, n). Then we define R <3(k+1)An S iff one of the following
conditions holds:
(1) R, S ∈ Df ′ (k, A, n) and R <3kAn S
(2) R ∈ Df ′ (k, A, n) and S ∈
/ Df ′ (k, A, n).
/ Df ′ (k, A, n), R, S ∈ {n A\T : T ∈ Df ′ (k, A, n)}, and
(3) R, S ∈

R ≺0,A,n,<3kAn ,Df ′ (k,A,n) S.

(4) R, S ∈ / Df ′ (k, A, n), R ∈ {n A\T : T ∈ Df ′ (k, A, n)}, and S ∈

/ {n A\T : T ∈
′
Df (k, A, n)}.
(5) R, S ∈/ Df ′ (k, A, n), R, S ∈
/ {n A\T : T ∈ Df ′ (k, A, n)}, R, S ∈ {T ∪ U : T, U ∈
′
Df (k, A, n)}, and
R ≺1,A,n,<3kAn ,Df ′ (k,A,n) S.
(6) R, S ∈/ Df ′ (k, A, n), R, S ∈
/ {n A\T : T ∈ Df ′ (k, A, n)}, R ∈ {T ∪ U : T, U ∈
′
Df (k, A, n)}, and S ∈/ {T ∪ U : T, U ∈ Df ′ (k, A, n)}.

114
(7) R, S ∈/ Df ′ (k, A, n), R, S ∈
/ {n A\T : T ∈ Df ′ (k, A, n)}, R, S ∈
/ {T ∪ U : T, U ∈
′
Df (k, A, n)}, and
R ≺2,A,n,<3kA(n+1),Df ′ (k,A,n) S.
Finally, for any set A and any natural number n, we well-order Df(A, n) as follows. Let
R, S ∈ Df(A, n). Let k be minimum such that R ∈ Df ′ (k, A, n), and let l be minimum
such that S ∈ Df ′ (l, A, n). Then we define

R <4An S iff k < l, or k = l and R <3kAn S.

{B}∅
Now we define a well-ordering <B5α of Lα by recursion. First of all, <B50 =≺, the given
well-ordering of trcl({B}). If α is a limit ordinal, then for any x, y ∈ L{B}∅ α we define

x <B5α y iff ρ(x) < ρ(y) ∨ [ρ(x) = ρ(y) and x <B5ρ(x) y].

{B}∅
Clearly this is a well-order of Lα .
{B}∅
Now suppose that a well-order <B5α of Lα has been defined. Then for each n ∈ ω
n {B}∅ {B}∅
we define the lexicographic order <B6nα on Lα : for any x, y ∈ n Lα ,

x <B6nα y iff ∃k < n[x ↾ k = y ↾ k and x(k) <B5α y(k)].

{B}∅ {B}∅
Clearly this is a well-order of n Lα . Now for any X ∈ L{B}∅ α + 1 = D(Lα ), let n(X)
be the least natural number n such that
{B}∅ {B}∅ {B}∅
∃s ∈ n Lα ∃R ∈ Df(Lα , n + 1)[X = {x ∈ Lα : s⌢ hxi ∈ R}].

{B}∅
Then let s(X) be the least member of n(X) Lα (under the well-order <B6n(X)α ) such
that
{B}∅ {B}∅
∃R ∈ Df(Lα , n(X) + 1)[X = {x ∈ Lα : s(X)⌢ hxi ∈ R}].
{B}∅
Then let R(X) be the least member of Df(Lα , n + 1) (under <B4L{B}∅ (n+1) ) such that
α

{B}∅
X = {x ∈ Lα : s(X)⌢ hxi ∈ R(X)}].

Finally, for any X, Y ∈ L{B}∅ α + 1 we define X <B5(α+1) Y iff one of the following
conditions holds:
{B}∅
(i) X, Y ∈ Lα and X <B5α Y .
{B}∅ {B}∅
(ii) X ∈ Lα and Y ∈
/ Lα .
{B}∅
(iii) X, Y ∈
/ Lα and one of the following conditions holds:
(a) n(X) < n(Y ).
(b) n(X) = n(Y ) and s(X) <B6n(X)α s(Y ).
(c) n(X) = n(Y ) and s(X) = s(Y ) and R(X) <4L{B}∅ (n+1) R(Y ).
α

{B}∅
Clearly this gives a well-order of Lα+1 .

115
We denote the union of all the well-orders <B5α for α ∈ On by <BL . Under V = L(B) it
is a well-ordering of the universe.
E23.16 Recall from elementary set theory the following definition of the standard well-
ordering of On × On:

(α, β) ≺ (γ, δ) iff (α ∪ β < γ ∪ δ)

or (α ∪ β = γ ∪ δ and α < γ)
or (α ∪ β = γ ∪ δ and α = γ and β < δ).

Prove that ≺ is absolute for transitive class models of ZF.

Now define ∆ : On → On × On by recursion as follows:

∆(0) = (0, 0);


 (β, γ + 1) if ∆(α) = (β, γ) and γ < β,

(0, β + 1) if ∆(α) = (β, γ) and γ = β,
∆(α + 1) =
 (β + 1, γ)
 if ∆(α) = (β, γ) and β + 1 < γ,
(γ, 0) if ∆(α) = (β, γ) and β + 1 = γ;
∆(α) =≺ -least(β, γ) such that ∀δ < α[∆(δ) ≺ (β, γ)] if α is limit.

Prove:
(1) If α < β, then ∆(α) ≺ ∆(β).
(2) ∆ maps onto On × On.
(3) ∆ is absolute for transitive class models of ZF.
(4) ∆−1 is absolute for transitive class models of ZF.
Clearly ≺ is absolute, by the basic absoluteness results.
For (1), fix α; we go by induction on β. Assume that the implication “α < β implies
that ∆(α) ≺ ∆(β)” holds for all β < γ; we prove it for γ. If γ ≤ α, the implication
vacuously holds. Now suppose that α < γ. If γ = β + 1, then α ≤ β, and hence
∆(α)  ∆(β) either trivially if α = β, or by the inductive hypothesis if α < β. Clearly
∆(β) < ∆(β + 1), so ∆(α) < ∆(γ). The case of γ limit is clear.
For (2), suppose to the contrary that rng(∆) ⊂ On × On, and let (β, γ) be the ≺-
least pair of ordinals not in rng(∆). Then the following list of all possiblities gives a
contradiction in each case:
(1) β = γ = 0.
(2) β = δ + 1, γ = 0; then (β, γ) is the immediate successor of (δ, δ + 1).
(3) β limit, γ = 0; then (β, γ) is the least upper bound of {(δ, β) : δ < β}.
(4) β = 0, γ = δ + 1; then (β, γ) is the immediate successor of (δ, δ).
(5) β = ε + 1 < δ + 1 = γ; then (β, γ) is the immediate successor of (ε, γ).
(6) β = δ + 1 = γ; then (β, γ) is the immediate successor of (δ + 1, δ).
(7) β = ε + 1 > δ + 1 = γ; then (β, γ) is the immediate successor of (β, δ).
(8) β limit, β < γ = δ + 1; then (β, γ) is the lub of {(δ, γ) : δ < β}.
(9) β limit, β > δ + 1 = γ; then (β, γ) is the immediate successor of (β, δ).
(10) β = 0, γ limit; then (β, γ) is the lub of {(δ, 0) : δ < γ}.

116
 (11) β = δ + 1 < γ with γ limit; then (β, γ) is the immediate successor of (δ, γ).
(12) β = δ + 1 > γ with γ limit; then (β, γ) is the lub of {(β, ε) : ε < γ}.
(13) β and γ are limit, with β < γ. Then (β, γ) is the lub of {(δ, γ) : δ < β}.
(14) β = γ limit; then (β, γ) is the lub of {(δ, β) : δ < β}.
(15) β and γ are limit, γ < β; then (β, γ) is the lub of {(β, δ) : δ < γ}.
(3) is clear by the theorem about absoluteness of recursive definitions.
(4) holds on general grounds; the inverse of a one-one absolute function is clearly
absolute.
E23.17 Suppose that M is a transitive class model of ZFC, and every set of ordinals is
in M . Show that M = V . Hint: take any set X. Let κ = |trcl({X})|, and let f : κ →
trcl({X}) be a bijection. Define αEβ iff α, β < κ and f (α) ∈ f (β). Use exercise E23.16
to show that E ∈ M . Take the Mostowski collapse of (κ, E) in M , and infer that X ∈ M .
We follow the hint. Let ∆ be as in exercise E23.16. Then ∆−1 [E] is a set of ordinals, so
by supposition it is in M . It follows that E itself is in M . Now we work in M . Clearly E
is well-founded, since ∈ is, and obviously E is set-like. Hence we can apply the Mostowski
collapse; let G be the Mostowski collapsing function.
Now working outside M , we claim that G = f (hence f ∈ M ). Since E is clearly
well-founded (outside M too), we can suppose that G 6= f and take the E-least α ∈ κ such
that G(α) 6= f (α). Then for any x ∈ M ,

x ∈ G(α) iff ∃β ∈ κ[x = G(β) and βEα (definition of G)

iff ∃β ∈ κ[x = f (β) and βEα (minimality of α)
iff ∃β ∈ κ[x = f (β) and f (β) ∈ f (α) (definition of E)
iff x ∈ f (α) (transitivity of rng(f )));

Thus G(α) = f (α), contradiction.

Hence our claim holds: G = f . So f ∈ M . Since X ∈ trcl({X}) = rng(f ), and
rng(f ) ∈ M by absoluteness, if follows that X ∈ M .
E23.18 Show that if X ⊆ ω1 then CH holds in L(X). Hint: show that if A ∈ P(ω) in
L(X), then there are α, β < ω1 such that X ∈ Lα (X ∩ β).
We apply a reflection principle, Theorem 10.41. Let ϕ1 be enough of ZFC to prove that
“x ∈ Ly (z)” is absolute for transitive models of ϕ1 . Let ϕ2 be the extensionality axiom,
let ϕ3 be the formula “x ∈ Ly (z)”, and let ϕ4 be the formula “x is an ordinal”. Now
A ∈ Lα (X) for some ordinal α. By that reflection principle, let B be a countable set
such that {A, α, X, ω1} ⊆ B ⊆ L(X) and ϕ1 , ϕ2 , ϕ3 ϕ4 are absolute for B, V . Let C be
the Mostowski collapse of B, with collapsing function G. Then (A ∈ Lα (X))B , since ϕ3
is absolute for B, V . Hence also (G(A) ∈ LG(α) (G(X))C , since G is an isomorphism. So
G(A) ∈ LG(α) )G(X) since C, being isomorphic to B, is a model of ϕ1 . Now A ⊆ ω, so
clearly G(A) = A. Let γ be the least ordinal not in C. An easy argument using the
transitivity of C shows that C ∩ Ord = γ.
(1) If δ ∈ γ, then G(δ) = δ.

117
In fact, suppose that δ ∈ γ is ∈-minimal such that G(δ) 6= δ. If ε ∈ G(δ), choose β ∈ δ such
that ε = G(β). Since β ∈ δ, we have G(β) = β; so ε = β ∈ δ. This shows that G(δ) ⊆ δ.
If β ∈ δ, then G(β) = β and G(β) ∈ G(δ); so β ∈ G(δ). This shows that δ ⊆ G(δ). so
G(δ) = δ, contradiction. Hence (1) holds.
Now α is an ordinal, so (α is an ordinal)B since ϕ4 is absolute for B, V . Hence (G(α) is
an ordinal)C since G is an isomorphism. Hence G(α) is an ordinal, by absoluteness since
C is transitive. Morover, C is countable and G(α) ∈ C, so G(α) is countable. This is part
of the desired conclusion. Here is the other part:
G(X) = X ∩ γ.
def
For, suppose that y ∈ G(X). Choose β ∈ X such that y = G(β). Now δ = G(ω1 ) is an
ordinal since B is a model of ϕ4 . Since G(X) ⊆ G(ω1 ), it follows that β < δ < γ. So by
(1) we have y = G(β) = β ∈ X ∩ γ.
On the other hand, if β ∈ X ∩ γ, then by (1), β = G(β) ∈ G(X). This finishes the
proof of the statement of the hint. S
It follows that in L(X) we have P(ω) ⊆ α,β<ω1 Lα (X ∩ β). Now it is clear (by
induction) that |Lα (X ∩ β)| ≤ max(ω, |α|, |β|), so CH follows.
E23.19 Show that if X ⊆ ω1 then GCH holds in L(X).
By induction it is clear that |Lα (X)| ≤ max(α, ω, |X|) for any ordinal α. Next, we claim
(*) There is a sentence ϕ which is a finite conjunction of members of ZF + V = L(X) such
that
ZF C ⊢ ∀M [M transitive ∧ ϕM → M = Lo(M ) (X)].
Proof. Let ϕ be a conjunction of V = L(X) together with enough of ZF to prove that
hLα (X) : α ∈ Oni is absolute, and also enough to prove that there is no largest ordinal.
Then for any transitive set M , if ϕM , then o(M ) is a limit ordinal, (∀x(x ∈ L(X)))M and
hence M = L(X)M , and
[
M = L(X)M = {x ∈ M : (∃α(x ∈ L(α)(X)))M } = Lα (X) = Lo(M ) (X),
α∈M

as desired in (*).
(**) If V = L(X), then for every infinite ordinal α we have P(Lα (X)) ⊆ Lω1 ∪α+ .
For, let ϕ be as in (*). Assume that V = L(X) and α is an infinite ordinal. Take any A ∈
P(Lα (X)). Let Y = Lα (X) ∪ {A}. Clearly X is transitive. Now |Y | ≤ |ω1 ∪ α|. Now by a
reflection theorem, let M be a transitive set such that X ⊆ M , |M | = |α|, and ϕM ↔ ϕV .
But ϕV actually holds, so ϕM holds. Hence M = Lo(M ) by (**). Now o(M ) = M ∩ On,
and |M | = |ω1 ∪ α|, so o(M ) < ω1 ∪ α+ . Hence A ∈ X ⊆ M = Lo(M ) (X) ⊆ Lω1 ∪α+ (X).
Clearly (**) and exercise E23.18 imply GCH.
Solutions to exercises in Chapter 24
E24.1 Show that fin(ω, ω1 ) collapses ω1 to ω, but preserves cardinals ≥ ω2 .

118
For collapsing, see the beginning of Chapter 12.
Clearly fin(ω, ω1 ) has size ω1 , and hence satisfies ω2 -cc. So the second assertion follows
from Proposition 12.5.
E24.2 Suppose that κ is an uncountable regular cardinal of M , and P ∈ M is a κ-cc
quasi-order. Assume that C is club in κ, with C ∈ M [G]. Show that there is a C ′ ⊆ C
such that C ′ ∈ M and C ′ is club in κ. Hint: in M [G] let f : κ → κ be such that
∀α < κ[α < f (α) ∈ C]. Apply Theorem 12.4.
By Theorem 12.4 let F ∈ M be such that F : κ → P(κ), f (α) ∈ F (α) for all α < κ, and
(|F (α)| < κ)M for all α < κ. We now define g : κ → κ by recursion. Let g(0) = sup(F (0)).
If g(α) Shas been defined, let g(α + 1) = max(g(α) + 1, sup(F (g(α)))). For β limit, let
g(β) = α<β g(α).
(1) If β is limit, then g(β) ∈ C.
For, it suffices to show that g(β) ∩ C is unbounded in g(β), since g(β) is clearly a limit
ordinal. Suppose that γ < g(β). Choose α < β such that γ < g(α). Then γ < g(α) <
f (g(α)) ∈ F (g(α)). Now f (g(α)) ≤ sup(F (g(α))) ≤ g(α + 1) < g(β). So we have
γ < f (g(α)) < g(β) with f (g(α)) ∈ C, as desired.
def
It follows that C ′ = {g(β) : β limit} is as desired.
E24.3 Suppose that κ is an uncountable regular cardinal of M , and P ∈ M is a κ-cc
quasi-order. Assume that S ∈ M is stationary in κ, in the sense of M . Show that it
remains stationary in M [G].
Let C ∈ M [G] be club in κ. By exercise E24.2, let C ′ ∈ M be club in κ with C ′ ⊆ C.
6 C ′ ∩ S ⊆ C ∩ S.
Then ∅ =
E24.4 Suppose that κ is an uncountable regular cardinal of M , and P ∈ M is a κ-closed
quasi-order. Assume that S ∈ M is stationary in κ, in the sense of M . Show that it
remains stationary in M [G].
Suppose that P is κ-closed, C is club in κ, C ∈ M [G], and S ∩ C = ∅. Let f ∈ M [G] be
the strictly increasing enumeration of C. Say τG = f . Choose q ∈ P such that

q τ : κ → κ is strictly increasing, continuous, and such that ∀α ∈ κ̌[τ (α) ∈

/ Š]

Now by induction define in M sequences hpα : α < κi, hzα : α < κi such that each pα ∈ P ,
each zα ∈ κ\S, p0 = q, pα τ (α̌) = žα , and pβ ≤ pα for α < β < κ; this is possible by
κ-closure. We claim that rng(z) is club in κ with rng(z) ∩ S = ∅ (contradiction). Clearly
rng(z) ∩ S = ∅.
(1) If α < β < κ, then zα < zβ .
In fact,
q ∀γ < κ̌∀δ < κ̌[γ < δ → τ (γ) < τ (δ)],
so
pβ τ (α̌) < τ (β̌) ∧ τ (α̌) = žα ∧ τ (β̌) = žβ ;

119
hence pβ žα < žβ , so zα < zβ , as desired in (1).
S
(2) If α < κ is limit, then zα = β<α zβ .
In fact,  
[
q ∀γ < κ γ limit → τ (γ) = τ (δ) ;
δ<γ

S M with pSα ∈ H, and let g = τH . Now

Hence pα forces this too. Let H be P-generic over
pα τ (δ̌) = žδ for each δ ≤ α, so zα = g(α) = δ<α g(δ) = δ<α zδ , as desired in (2).
Now it follows that rng(z) is club in κ.
E24.5 Prove that if ZFC is consistent, then so is ZFC + GCH + ¬(V = L).
Let M be a c.t.m. of ZFC + GCH; it exists by the theory of constuctibility. Let P =
fin(ω1 , 2), and let G be P-generic over M . By Lemma 11.2, G ∈/ M . By Lemma 11.13, M
and M [G] have the same ordinals, so, since L is absolute, LM = LM [G] ⊆ M , and so M [G]
is not a model of V = L.
For GCH, let λ be any cardinal of M . Note that our quasi-order is ccc, so that
cardinals are preserved. Then by theorem 24.4,

λ+ ≤ (2λ )M [G] ≤ (ω1λ )M = (2λ )M = (λ+ )M = (λ+ )M [G] .

Solutions to exercises for Chapter 25

E25.1 Show that for any infinite cardinal κ, the partial order fn(κ, 2, ω) is isomorphic to
fn(κ × ω, 2, ω).
Let f : κ × ω → κ be a bijection. Now we define F : fn(κ, 2, ω) → fn(κ × ω, 2, ω) as follows.
Let a ∈ Fn(κ, 2, ω). So a is a finite function contained in κ × 2. Let F (a) = a ◦ f −1 . Thus
F (a) is a function whose domain is

{x ∈ dmn(f −1 ) : f −1 (x) ∈ dmn(a)} = f [dmn(a)],

and this is a finite subset of κ × ω. Obviously F (a) is a function which is a subset of

(κ × ω) × 2. So F (a) ∈ Fn(κ × ω, 2, ω).
If a, b ∈ fin(κ, 2, ω), clearly a ⊆ b iff F (a) ⊆ F (b). In particular, F is one-one. Clearly
F (∅) = ∅.
So it remains only to show that F maps onto fin(κ × ω, 2, ω). Let b ∈ fin(κ × ω, 2, ω).
Then clearly b ◦ f ∈ fin(κ, 2) and F (b ◦ f ) = b ◦ f ◦ f −1 = b, as desired.
E25.2 Prove that if P and Q are isomorphic forcing orders, then RO(P) and RO(Q) are
isomorphic Boolean algebras.
This is abstract generalized nonsense.
Let f be an isomorphism from P onto Q, and let eP and eQ be the embeddings of P
into RO(P) and of Q into RO(Q) given in Chapter 9. Then the following conditions are
clear:

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(1) (eQ ◦ f )[P ] is dense in RO(P).
(2) For all p, q ∈ P , if p ≤ q then eQ (f (p))) ≤ eQ (f (q)).
(3) For all p, q ∈ P , p ⊥ q iff eP (f (p)) · eP (f (q)) = 0.
Hence it follows from Theorem 9.22 that there is an isomorphism g from RO(P) into RO(Q)
such that g ◦ eP = eQ ◦ f .
By symmetry we get an isomorphism h from RO(Q) into RO(P) such that h ◦ eQ =
eP ◦ f −1 .
Hence h ◦ g ◦ eP = h ◦ eQ ◦ f = eP ◦ f −1 ◦ f = eP . It follows from Theorem 9.22 that
h ◦ g is the identity on RO(P).
Similarly, g ◦ h is the identity on RO(Q). So g is the desired isomorphism.
E25.3 Give an example of non-isomorphic forcing orders P and Q such that RO(P) and
RO(Q) are isomorphic Boolean algebras.
Let P be fn(ω, 2, ω), let A = RO(P), and let Q = (A\{0}, ≥, 1). For each a ∈ Q let
j(a) = a. Then the conditions of Theorem 9.22 are clear, and so RO(Q) is isomorphic to
A. So we just need to show that P and Q are not isomorphic. Suppose to the contrary
that f is an isomorphism from P onto Q. Let p = {(0, 0)}. Now p 6= ∅, so f (p) 6= 1, and
hence −f (p) 6= 0, so that −f (p) ∈ Q. Choose q ∈ P such that f (q) = −f (p). Let r be
any member of fin(ω, 2) such that q ⊂ r. Hence r < q, and so f (r) < f (q) = −f (p), and
so f (p) < −f (r). Since f (r) 6= 1, we have −f (r) 6= 0, so we can choose s ∈ P such that
f (s) = −f (r). Thus f (p) < f (s), so p < s. This means that s ⊂ p. Since p is a singleton,
it follows that s = ∅, hence 1 = f (s) = −f (r), so f (r) = 0, contradiction.
Q
E25.4 For any systemQhPi : i ∈ Ii, we define the weak product w i∈I Pi as follows: the
underlying set is {f ∈ i∈I Pi : {i ∈ I : f (i) 6= 1} is finite}, with f ≤ g iff f (i) ≤Pi g(i)
forQall i ∈ I. Prove that for any infinite cardinal κ, the forcing order fin(κ, 2) is isomorphic
w
to α<κ Pα , where each Pα is equal to fin(ω, 2).

Q F be a bijection from κ × ω onto κ. Now for each f ∈ fin(κ, 2) we define G(f ) ∈

Let
α<κ fin(ω, 2) as follows. For any α < κ, (G(f ))α has domain {i ∈ ω : F (α, i) ∈ dmn(f )}.
Clearly this is a finite set. For any i ∈ dmn((G(f ))α we Qset ((G(f ))α )(i) = f (F (α, i)).
w
Clearly {α < κ : dmn(G(f ))α 6= ∅} is finite, so G(f ) ∈ α<κ fin(ω, 2). Clearly f ⊆ g
implies that G(f ) ≥ G(g). Q
w
Conversely, for each x ∈ α<κ fin(ω, 2) we define

H(x) = {(α, i) : ∃β < κ∃j < ω[α = F (β, j), j ∈ dmn(xβ ), and xβ (j) = i]}.

Now H(x) is a function. For, suppose that (α, i), (α, k) ∈ H(x) Then with α = F (b, j) we
must have j ∈ dmn(xβ ) and xβ (j) = i and xβ (j) = k, so that i = k. The domain of H(x)
is
{F (β, j) : j ∈ dmn(xβ )};
there are only finitely many β such that xβ 6= ∅, and for each β the set dmn(xβ ) is finite,
so H(x) is finite. So H(x) ∈ fin(κ, 2). Clearly x ≤ y implies that H(x) ⊇ H(y).

121
 Now suppose that f ∈ fin(κ, 2). we claim that H(G(f )) = f . For, if β < κ and j < ω,
then

F (β, j) ∈ dmn(H(G(f )) iff j ∈ dmn((G(f ))β

iff F (β, j) ∈ dmn(f ),

and for any F (β, j) ∈ dmn(f ),

(H(G(f )))(F (β, j)) = (G(f ))β (j) = f (F (β, j)).

Hence H(G(f )) = f , asQwclaimed.

Finally, let x ∈ α<κ fin(ω, 2); we claim that G(H(x)) = x. (This completes the
solution.) For, if β < κ and j ∈ ω, then

j ∈ dmn((G(H(x))β ) iff F (β, j) ∈ dmn(H(x))

iff j ∈ dmn(xβ ),

and for any j ∈ dmn(xβ ) we have

(G(H(x))β (j) = (H(x))(F (β, j)) = xβ (j).

Thus G(H(x)) = x.
E25.5 We expand the language of set theory by adding an individual constant ∅. An
Urelement is an object a such that a 6= ∅ but a does not have any elements. (Plural is
Urelemente.) A set is an object x which is either ∅ or has an element. Both of these are
just definitions, formally like this:

U r(a) ↔ a 6= ∅ ∧ ∀x(x ∈
/ a);
Set(x) ↔ x = ∅ ∨ ∃y(y ∈ x).

Now we let ZFU be the following set of axioms in this language:

All the axioms of ZF except extensionality and foundation.
∀x[¬(x ∈ ∅)].
∀x, y[Set(x) ∧ Set(y) ∧ ∀z(z ∈ x ↔ z ∈ y) → x = y].
∀x[Set(x) ∧ x 6= ∅ → ∃y ∈ x∀z(z ∈ x → z ∈ / y)].
We also reformulate the axiom of choice for ZFU; it is the following statement:

∀A {Set(A ) ∧ ∀x ∈ A [Set(x) ∧ x 6= ∅]
∧ ∀x ∈ A ∀y ∈ A [x 6= y → ∀z[¬(z ∈ x ∧ z ∈ y)]]
→ ∃B∀x ∈ A ∃!y(y ∈ x ∧ y ∈ B)}.

We let ZF CU be all of these axioms.

122
One can adapt most of elementary set theory to use these axioms; browsing through the
first few chapters should convince one of this.
In this exercise, give a new definition of ordinal.
Also, show that if we add the axiom ¬∃a[U r(a)] we get a theory equivalent to ZF.
We define x is an ordinal to mean that x is a transitive set of transitive sets, just as before.
But note that “set” is taken in the new sense; so an ordinal, by definition, does not have
any Urelemente as elements.
Suppose that we add ¬∃a[U r(a)] to these new axioms. Since Set(a) is just the negation
of this, we are now assuming that everything is a set. So the new extensionality axiom
reduces to the old one, and the new foundation axiom reduces to the old one. The first new
axiom implies that ∅ is the same as the empty set given by the comprehension axiom. Thus
all of the new axioms are derivable in ZF (including ¬∃a[U r(a)]), and ZF is a consequence
of ¬∃a[U r(a)] plus the new axioms plus the part of ZF allowed in the new framework.
E25.10
(1) If γ < β, then Wγ ⊆ Wβ .
An obvious induction on β, with γ fixed, proves (1).
(2) If x ∈ y ∈ Wβ \U , then x ∈ Wβ .
We prove this by induction on β. It vacuously holds for β = 0. Now assume that it holds
for β, and x ∈ y ∈ Wβ+1 \U . If y ∈ Wβ , then x ∈ Wβ ⊆ Wβ+1 by the induction hypothesis.
Otherwise, y ∈ P(Wβ ), hence y ⊆ Wβ , and obviously x ∈ Wβ . Finally, the case β limit is
clear.
(3) Wβ ∩ Vα = ∅ for all β.
We prove this by induction on β. It is clear for β = 0. Assume it for β, and suppose that
x ∈ Wβ+1 . If x ∈ Wβ , then x ∈ / Vα by the inductive hypothesis. Otherwise, ∅ = 6 x ∈
P(Wα ). Choose y ∈ x. Now x ⊆ Wβ , so y ∈ Wβ . By the inductive hypothesis, y ∈ / Vα , so
also x ∈
/ Vα . This takes care of the successor case, and the limit case in the induction is
clear.
(4) Wβ = {x ∈ W : rankW (x) < β}.
In fact, first suppose that x ∈ Wβ . Let γ = rankW (x). If γ = −1, obviously rankW (x) < β.
Suppose that γ ≥ 0. Then x ∈ / Wγ , so γ < β. This proves ⊆.
def
Now suppose that x ∈ W and γ = rankW (x) < β. Then x ∈ Wγ+1 ⊆ Wβ , as desired.
(5) If x, y ∈ W and x ∈ y, then rankW (x) < rankW (y).
In fact, assume the hypotheses. Then y ∈
/ U , as otherwise x ∈ Vα , contradicting (3). Hence
def
β = rankW (y) > −1. So y ∈ Wβ+1 \Wβ , and hence y ∈ P(Wβ ). So y ⊆ Wβ and hence
x ∈ Wβ . Therefore rankW (x) < β, as desired.
(6) If x ∈ W \U , then rankW (x) = supy∈x (rankW (y) + 1).
For, suppose that x ∈ W \U . Then x 6= ∅ by (3). If y ∈ x, then y ∈ W by (2), and hence
rankW (y) < rankW (x) by (5). Hence ≥ holds. Let γ be the right side of (6). If y ∈ x, then

123
rankW (y) < γ, and so y ∈ WrankW (y)+1 ⊆ Wγ . This shows that x ⊆ Wγ . So x ∈ P(Wγ ).
Since also x 6= ∅, as noted above, it follows that x ∈ Wγ+1 , and hence rankW (x) ≤ γ,
giving ≤.
(7) If x ∈ W , then rank(x) = α + 1 + rankW (x).
We prove (7) by ∈-induction. So, suppose that the implication is true for all members of
x, with x ∈ W . If x ∈ U the conclusion is clear. Suppose that x ∈
/ U . Then

α + 1 + rankW (x) = α + 1 + sup(rankW (y) + 1) by (6)

y∈x
= sup(α + 1 + rankW (y) + 1)
y∈x
= sup(rank(y) + 1) by the inductive hypothesis
y∈x
= rank(x).

Here the last step uses a property of rank found in the M6730 notes.
(8) If a ∈ W \U , then W ∩ a 6= ∅.
In fact, by induction if a ∈ Wβ \U then W ∩ a 6= ∅, so (8) holds.
(9) For any a ∈ W we have U r W (a) iff a ∈ U \{Z}.
In fact, suppose that U r W (a). Thus a 6= Z and x ∈
/ a for all x ∈ W . Then a ∈ U by (8).
Thus a ∈ U \{Z}. Conversely, if a ∈ U \{Z}, then there is no x ∈ W such that x ∈ a, for
every member of a is in Vα , and (3) applies. Hence U r W (a).
(10) For any a ∈ W we have SetW (a) iff a ∈
/ U \{Z}.
This is clear from (9).
E25.11 The first axiom obviously holds, since every member of Z is in Vα ; so (3) applies.
(11) New extensionality holds.
For, suppose that x, y ∈ W , SetM (x), SetM (y), and for all z ∈ W , z ∈ x iff z ∈ y. Then
by (10), x, y ∈
/ U . Suppose that z ∈ x. Then by (2), z ∈ W , and hence z ∈ y. Similarly
y ⊆ x. So x = y. This proves (11).
(12) New foundation holds.
For, suppose that x ∈ W , SetW (x), and x 6= Z. Then there is a y ∈ W such that y ∈ x.
Choose such a y of smallest W -rank. Suppose that z ∈ W and z ∈ x. Then z ∈ / y by (5)
and the definition of y.
(13) Comprehension holds.
For, let ϕ be a formula with free variables among x, z, w1 , . . . , wn , and suppose that
z, w1 , . . . , wn ∈ W . If z ∈ U , let y = Z. Then for any x ∈ W , x ∈ / y, and also it is
W
not true that x ∈ z ∧ ϕ , so the desired equivalence holds. Now suppose that z ∈ / U.
Let y ′ = {x ∈ z : ϕW }. If y ′ = ∅, then clearly again y = Z works for our equivalence. If

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y ′ 6= ∅, let y = y ′ . Say z ∈ Wβ . If x ∈ y, then x ∈ z ∈ Wβ \U , and so x ∈ Wβ by (2). So
y ∈ P(Wβ )\{∅}, and hence y ∈ Wβ+1 . Clearly
∀x ∈ W [x ∈ y ↔ x ∈ z ∧ ϕW ]
as desired. So (13) holds.
(14) Pairing holds.
For, given x, y ∈ W , choose β so that x, y ∈ Wβ . Then {x, y} ∈ P(Wβ )\{∅}, so {x, y} ∈
Wβ+1 and the desired conclusion follows.
(15) Union holds.
For, let A ∈ W be given. We define
n S S
A= ( A ) ∩ W if ( A ) ∩ W 6= ∅,
Z otherwise.
S S
We claim that A ∈ W . This is clear if ( A ) ∩ W = ∅, so suppose that ( A ) ∩ W 6= ∅.
For each a ∈ A choose βa such that a ∈ Wβa . Let γ = supa∈A βa . Then A ⊆ Wγ and
A 6= ∅, so A ∈ Wγ+1 . Thus, indeed, A ∈ W . S
Now suppose that x, Y ∈ W and x ∈ Y ∈ A . So x ∈ A ∩ W , so x ∈ A, as desired.
(16) Power set holds.
To prove this, note that this axiom involves the defined notion ⊆. We claim:
(17) For any x, y ∈ W , (x ⊆ y)W iff x ∈ U or x ⊆ y.
In fact, assume that x, y ∈ W . First suppose that (x ⊆ y)W and x ∈ / U . Suppose that
z ∈ x. Then z ∈ W by (2), and so z ∈ y since (x ⊆ y)W . Thus x ⊆ y.
Conversely, if x ∈ U , then (x ⊆ y)W by (3). Suppose that x ⊆ y. Then obviously
(x ⊆ y)W .
Thus (17) holds.
Now for the power set axiom, let x be given. Define y = U ∪ (P(x) ∩ W ) Now suppose
that z ∈ W and (z ⊆ x)W . By (17) there are two possibilities. If z ∈ U , obviously z ∈ y.
If z ⊆ x, again obviously z ∈ y. So (16) is proved.
(18) Infinity holds.
To prove (18), we first define by recursion a sequence hun : n ∈ ωi of members of W . Let
u0 = Z and un+1 = (un ∪ {un })W . Let Ω = rngu. Now each un is in W , so there is a βn
such that un ∈ Wβn . Let γ = supn∈ω βn . Thus Ω ⊆ Wγ . Also, obviously Ω 6= ∅. Hence
Ω ∈ Wγ+1 . Clearly Ω is what is needed in the infinity axiom.
(19) Replacement holds.
For, suppose that ϕ is a formula with free variables among x, y, A, w1, . . . , wn , assume that
A, w1 , . . . , wn ∈ W , and suppose that (∀x ∈ A∃!yϕ)W . Written out more fully, this last
supposition is:
∀x ∈ W [x ∈ A → ∃y ∈ W [ϕ(x, y, A, w1, . . . , wn )W ∧
∀z ∈ W [ϕ(x, z, A, w1 , . . . , wn )W → y = z]]].

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Thus ∀x ∈ A ∩ W ∃!y[y ∈ W ∧ ϕW ]. Hence by the replacement axiom we obtain a set Y ′
such that ∀x ∈ A ∩ W ∃y ∈ Y ′ [y ∈ W ∧ ϕW ]. Let

u = {(x, y) : x ∈ A ∩ W and y ∈ Y ′ ∩ W and ϕW }.

Thus u is a function with domain A ∩ W . Define

n
Y = rng(u) if rng(u) 6= ∅,
Z otherwise.

We claim that Y ∈ W . This is clear if rng(u) = ∅, so suppose S that rng(u) 6= ∅. For

each y ∈ rng(u) choose βu such that y ∈ Wβu . Let γ = y∈rng(u) βu . Clearly then
rng(u) ∈ P(W + γ)\{∅}, so rng(u) ∈ Wγ+1 . This proves that Y ∈ W .
If x ∈ A ∩ W , then u(x) ∈ Y and ϕ(x, u(x), A, w1, . . . , wn )W , as desired.
E25.12 We extend f by recursion, denoting extension by f + . For any a ∈ W ,

 f (a) if a ∈ U \{Z},
+
f (a) = Z if a = Z,
 +
{x : ∃b ∈ a[x = f (b)]} if a ∈
/ U.

Note that if a ∈ W \U , then a ⊆ W , by (2). To show that f is one-one and onto, it

suffices by symmetry to show that f + ◦ (f −1 )+ is the identity on W . We prove that
f + ((f −1 )+ (a)) = a for all a ∈ W , by induction on a. This is clear if a ∈ U . Now suppose
that a ∈ W \U . Then

f + ((f −1 )+ (a)) = f + ({x : ∃b ∈ a[x = (f −1 )+ (b)]})

= {y : ∃z ∈ {x : ∃b ∈ a[x = (f −1 )+ (b)]}[y = f + (z)]}
= {y : ∃b ∈ a[y = f + ((f −1 )+ (b))]}
= {y : ∃b ∈ a[y = b]} (induction hypothesis)
= a.

So f : W → W is a bijection, and it takes Z to Z. If a, b ∈ W and a ∈ b, obviously

f + (a) ∈ f + (b). Conversely, suppose that f + (a) ∈ f + (b). Then a = (f −1 )+ (f + (a)) ∈
(f −1 )+ (f + (b)) = b. Thus f is the desired isomorphism.
E25.13 We define by recursion

D0 = a;
Dn+1 = Dn ∪ {b ∈ W : b ∈ c ∈ d for some c ∈ W ∩ Dn };
[
E= Dn .
n∈ω

By induction, Dn ⊆ W for each n ∈ ω, and hence E ⊆ W . Clearly a ⊆ E. Hence E 6= ∅,

and it follows easily, by an argument used several times above, that E ∈ W . Now suppose

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that b, c ∈ W and b ∈ c ∈ E. Choose n such that c ∈ Dn . It follows that b ∈ Dn+1 ⊆ E.
Thus E is W -transitive.
For the second part of the problem, it suffices to prove that if T is the W -transitive
closure of a, then f + [T ] is the W -transitive closure of f + (a). First, f + (a) ⊆ f + [T ]. For,
let x ∈ f + (a). Then we can write x = f + (b) with b ∈ a ∩ W . Hence b ∈ T , and so
x = f + (b) ∈ f + [T ]. So f + (a) ⊆ f + [T ]. Now suppose that S is any W -transitive set such
that f + (a) ⊆ S. Let R = {x : x ∈ W and f + (x) ∈ S}. Now a ⊆ R. For, if y ∈ a, then
f + (y) ∈ f + (a) ⊆ S, and hence y ∈ R.
Suppose that F ∈ W is W -transitive and a ⊆ F . By an easy induction, Dn ⊆ F for
every n ∈ ω, and hence E ⊆ F .

E25.14 (i): Let a ∈ U . Then the W -transitive closure of {a} is clearly just {a} itself. If
a = Z, clearly a is symmetric. If a 6= Z, then letting F = {a} in the definition shows that
a is symmetric.
(ii): Since a is symmetric, let F be a finite subset of U \{Z} such that g + (a) = a for
every permutation g of U \{Z} which is the identity on F . Let G = f [F ]. So G is a finite
subset of U \{Z}. If g is a permutation of U \{Z} which is the identity on G, Then for
any u ∈ F we have (f −1 ◦ g ◦ f )(u) = f −1 (g(f (u)) = f −1 (f (u)) = u. So f −1 ◦ g ◦ f is the
identity on F , and so (f −1 )+ (g + (f + (a))) = a = f −1 (f (a)), and hence g + (f + (a)) = f + (a).
This shows that f + (a) is symmetric.
(iii): By E25.13 and (ii), f + maps H into H. By considering f −1 , it is clear that
f + ↾ H is a permutation of H. By E25.12 it is an automorphism.
(iv): We prove this by induction on ϕ. It is obvious for atomic formulas, and the in-
duction hypothesis for ∨ is clear. Suppose that (¬ϕ)H holds. If ϕH (f + (v0 ), . . . , f + (vn−1 ))
holds, applying the inductive hypothesis to f −1 gives a contradiction. Finally, suppose
that (∃xϕ(x, v0 , . . . , vn−1 ))H holds. Choose x ∈ H such that ϕH (x, v0 , . . . , vn−1 ). Then
ϕH (f + (x), f +(v0 ), . . . , f + (vn−1 )), hence (∃xϕ(f + (x), f +(v0 ), . . . , f + (vn−1 ))H .
(v): We go through all of the axioms. First note that if a is hereditarily symmetric
and not in U , then its elements are also hereditarily symmetric. Hence U r and Set are
absolute for H, W .

• ∀x[¬(x ∈ Z)]. This holds since H ⊆ W .

• New extensionality. This holds by the initial remark in (ii).

• New foundation. This also holds by that initial remark.

• Comprehension. Let ϕ be a formula with free variables among x, z, w1 , . . . , wn , and

suppose that z, w1 , . . . , wn ∈ H. Let y = {x ∈ z : ϕH }. Note that every member of z is in
H, so ϕH makes sense here. Clearly it suffices to show that y ∈ H. Since each member of y
is in H, it suffices to show that y is symmetric. Let F, G1 , . . . , Gn be finite subsets of U \{Z}
such that for every permutation f of U \{Z}, if f ↾ F is the identity then f + (z) = z, and
for each i = 1, . . . , n, if f ↾ Gi is the identity then f + (wi ) = wi . Now take any permutation
f of U \{Z} which is the identity on F ∪ G1 ∪ . . . ∪ Gn . We claim that f + (y) = y, which
will show that y is symmetric. Take any u ∈ f + (y). Then we can write u = f + (x) with
x ∈ y. So x ∈ z and ϕH holds. Hence by (iv), also ϕH (f + (x), f + (z), f + (w1 ), . . . , f + (wn ))

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holds. Since f + (z) = z and f + (wi ) = wi for each i, it follows that u = f + (x) ∈ y. Thus
f + (y) ⊆ y. Hence y = (f −1 )+ (f + (y)) ⊆ f + (y) and so y = f + (y), as desired.
• Pairing. If x, y ∈ H, clearly {x, y} ∈ H.
• Union. Given A ∈ H, let A = {x ∈ W : ∃Y ∈ A [x ∈ Y ]}. Clearly A is as desired.
• Power set. Given x ∈ H, let

y = {z : ∀w ∈ W [w ∈ z → w ∈ x}.

Clearly y is as desired.
• Infinity. By induction it is clear that each un , defined in the proof of infinity in W given
above, is in H. Hence so is Ω defined there, and so infinity holds.
• Replacement. This is clear by the proof of replacement in W .
E25.15 (a): Clear, by induction on m.
(b): Again clear, by induction on n, with m fixed.
(c): We follow the hint. Note that certain simple notions like unordered and ordered
pairs, relations, and functions, are absolute in the usual sense. Also note, by induction,
that g + (i) = i for every permutation g of U \{Z} and every i ∈ ω.
Choose a finite subset F of U \{Z} such that g + (f ) = f for any permutation g of
U \{Z} which is the identity on F . Clearly hf (i) : i ∈ ωi is a one-one function, by (b)
and absoluteness. Choose u ∈ {f (i) : i ∈ ω}\F , and choose v ∈ U \F with u 6= v. Say
u = f (i). Let g be a permutation of U \{Z} which is the identity on F and takes u to v.
Now (i, u) ∈ f , so

(i, v) = (i, g(u)) = (g + (i), g +(f (i))) ∈ g + (f ) = f.

Hence f (i) = v, contradiction.

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