3.9.
2 Reducing to Cartesian Products of Initial Segments (rewrite)
Suppose that S is an infinite set that has been well ordered in such a way that for every member x of S for which the
initial segment PS, x is infinite, we have
PS, x PS, x PS, x.
Then we have S S S.
Proof
Suppose that S is an infinite set that is well ordered in such a way that for every member x ∈ S for which the initial
segment PS, x is infinite, we have PS, x PS, x PS, x.
Case 0: There is an intial segment of S that is equivalent to S.
Choose x such that PS, x S. Thus we have that S S PS, x PS, x PS, x S.
Case 1: There is no intial segment of S that is equivalent to S.
1. Suppose for a given x ∈ S, the initial segment PS, x is finite. PS, x PS, x x is the union of two finite
sets, and thus is finite.
Thus PS, x PS, x is finite by 3.6.3 since it is the cartesian product of two finite sets.
Since PS, x PS, x is finite, it is strictly subequivalent to the infinite set S.
2. Suppose for a given x ∈ S, the initial segment PS, x is infinite. PS, x PS, x by 3.8.4.
Thus by the given proposition PS, x PS, x PS, x PS, x PS, x which cannot be equivalent to S and
thus PS, x PS, x must be strictly subequivalent to S.
Thus whether PS, x is finite or infinite, PS, x PS, x must be strictly subequivalent to S.
3. Apply the box order to S S. Thus when x, y ∈ S S, we can define the larger of x and y to be c, which shows
us that
PS S, x. y ⊆ PS S, c, c ⊆ PS, c PS, c
Since PS, x PS, x is strictly subequivalent to S, then PS S, x. y clearly must be strictly subequivalent to S
By Theorem 3.1.14, since both S S and S are well ordered, either S ⊆ S S or S S ⊆ S.
However, because PS S, x. y is strictly subequivalent to S then we know that S cannot be order isomorphic to
an initial segment of S S and so S ⊆ S S is only true when S S S.
Thus we must have either S S S or S S ⊆ S which can simply be written as S S ⊆ S.
If we choose any element x ∈ S then we can see that S S x ⊆ S S and thus we have that S ⊆ S S.
Thus we have that S S ⊆ S and also that S ⊆ S S therefore by the Bernstein theorem we have S S ⊆ S.
3.9.3 The Set Equivalence S S S (rewrite)
If S is an infinite set that can be well ordered, then S S S.
Proof
Suppose that S is an infinite set that can be well ordered and choose to be a well order of S.
Consider all x ∈ S for which PS, x is infinite.
For sake of contradiction, suppose that there exist some x such that PS, x PS, x PS, x does not hold.
Define z to be the least such x - in other words, PS, z PS, z PS, z does not hold.
Define T PS, z. Since z is the least such x where PS, x PS, x PS, x does not hold, then for all x z
where PT, x is infinite, we have that PT, x PT, x PT, x.
However this means that by 3. 9. 2 we have that T T T which means that PS, z PS, z PS, z which
contradicts the definition of z.
1
Thus the supposition fails and we have no x ∈ S for which PS, x is infinite and for which
PS, x PS, x PS, x does not hold.
