Problem Set 1: A Set-Theory diagnostic
Solution 1.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
A B and B C
A B and B C
A B or B C
A B or B C
A \ (A \ B)
A \ (B \ A)
A \ (B \ C)
A [ (B \ C)
A C and B D
(A B) [ (C D)
(A B) \ (C D)
A (B \ C)
(A B) \ (C D)
)
)
None
=
)
=
=
A (B [ C)
A (B \ C)
A (B [ C)
A (B \ C)
B
A\B
(A \ B) \ (A \ C)
(A [ B) \ (A [ C)
(A B) (C D)
(A [ C) (B [ D)
(A \ C) (B \ D)
(A B) \ (A C)
(A \ C) (B \ D)
)
)
)
)
(
(
(
(
,
,
,
,
=
=
=
=
( ,
=
=
=
=
None
None
None
None
None
None
None
None
None
None
None
None
None
Solution 2.
(n)
(o)
(p)
(q)
S
x 2 A2A A
S
x 2 A2A A
T
x 2 A2A A
T
x 2 A2A A
,
(
)
,
x 2 A for at least one A 2 A
x 2 A for every A 2 A
x 2 A for at least one A 2 A
x 2 A for every A 2 A
)
)
)
)
(
(
(
(
,
,
,
,
None
None
None
None
Solution 3.
(r)
(s)
(t)
(u)
(v)
(w)
g f is injective, then f is
g f is injective, then g is
g f is surjective, then f is
g f is surjective, then g is
Let A0 A. If A0 = f 1 (B0 )
for some B0 B, then f 1 f (A0 )
Let B0 B. If B0 f (A)
then f f 1 (B0 )
Solution 4.
inj.
inj.
inj.
inj.
inj.
None
None
surj.
surj.
surj.
surj.
surj.
bij.
bij,
bij.
bij.
none.
none.
none.
none.
A0
None
B0
None
a. Define the inverse map
: AC B C ! (A B)C
by sending (g, h) to the unique function that maps c 2 C to (g(c), h(c) 2
A B. Let
denote the given map (A B)C ! AC B C . Given
C
f 2 (A B) , ( (f )) = (p1 f, p2 f ) sends c to (p1 f (c), p2 f (c)) =
f (c), for all c 2 C, hence it is equal to f . On the other hand given
(g, h) 2 AC B C , ( (g, h)) = (p1
(g, h), p2
(g, h)). But (g, h)
sends c 2 C to (g(c), h(c); hence, p1
(g, h) = g and p2
(g, h) = h.
Thus ( (g, h)) = (g, h). Thus, and are inverse to each other and
is bijective.
b. Denote the given map by . We will define its inverse
: CA CB ! CA
by (g, h) 7! {(x, 0) 7! g(x), for x 2 A and (y, 1) 7! `
h(y), for y 2 B}. Then,
to show they are inverse to each other take f 2 C A B . Clearly, ((f )) =
(f i1 , f i2 ) sends (x, 0) 2 A {0} to f i1 (x) = f (x, 0) and (y, 1) 2
B {1} to f (y, 1), thus it is equal to f. On the other hand, given
(g, h) 2 C A C B , ((g, h)) = ((g, h) i1 , (g, h) i2 ). But (g, h) i1 (x) =
(g, h)(x, 0) = g(x) for x 2 A, by definition. Similarly for y 2 B,
(g, h) i2 (y) = h(y). Thus ((g, h)) = (g, h) and and are inverse to
each other.
S
S
S
Solution
a. By definition
S 5.
Sh(C) = h( n2NSCn ) = n2N h(Cn ) = n2N Cn+1 =
n2N 1 Cn . Hence, C =
n2N Cn = C0 [
n2N 1 Cn = C0 [ h(C).
b. A \ C A \ C0 = Im(g). So given x 2 A \ C there exist a unique y 2 B
such that x = g(y). If y 2 f (C), then x 2 g(f (C)) = h(C) C by part a.
But this cannot happen by assumption, so y 2 B \ f (C). Thus A \ C
g(B \ f (C)). Conversely, given y 2 B \ f (C), if g(y) 2 C = C0 [ h(C),
then it is in h(C) as C0 \ Im(g) = ;. Thus the iclusion holds the other
way as well, and we have A \ C = g(B \ f (C)).
c. A\C Im(g) and g is injective thus g 1 defines a bijection onto its image,
which is B \ f (C) by the above part. On the other hand, injectivity of
implies, its restriction to C defines a bijection onto f (C). This implies the
map k : A ! B defined by
(
f (x)
x2C
1
g (x) x 2 A \ C
is a bijection.
d. The function tan defines a bijection from ( /2, /2) to R. On the other
hand, there is a unique linear function sending a to /2 and b to /2
which is a bijection. The composition of these two functions give a bijection from (a, b) to R.
e. To use part c, we need to find injections both ways. The inclusion gives an
injection from U to R. On the other hand, the composition of a bijection
from R to the interval contained in U , which exists by part d, with the
inclusion map from the interval, gives as an injection from R to U . Hence
by part c, there is a bijection between U and R.
