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Consider in $\newcommand{\ZF}{{\sf (ZF)}}\ZF$ the following statement:

Weak Power Hypothesis (WPH): if $X,Y$ are sets and there is a bijection between $\newcommand{\P}{{\cal P}}\P(X)$ and $\P(Y)$, then there is a bijection between $X$ and $Y$.

The Axiom of Dependent Choice states that

(DC): if $X$ is a set and $R\subseteq X\times X$ such that for all $a\in X$ there is $b\in X$ such that $(a,b)\in R$, then there is a function $s:\omega\to X$ such that $\big(s(n),s(n+1)\big) \in R$ for all $n\in \omega$.

Does (WPH) imply (DC)?

Note: It appears to be open whether (WPH) implies the much stronger Axiom of Choice (AC).

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1 Answer 1

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I've reduced the problem to "every infinite set is Dedekind infinite", which is a consequence of WPH if my memory serves me right (this was answered before on the site).

We'll show that well-ordered choice holds, which is known to imply Dependent Choice. In fact, we will argue the equivalent, $\forall A(\aleph(A) = \aleph^*(A))$.

Suppose that $A$ is an infinite set, then the following holds: $$\aleph(A) = \aleph(A\times\omega),$$ since $A$ is Dedekind infinite. And of course, $$|A\times\omega|=|A\times\omega|\cdot 2.$$

If $\kappa<\aleph^*(A\times\omega)$, then:

$$2^{A\times\omega}\leq 2^{A\times\omega}\cdot 2^\kappa = 2^{A\times\omega + \kappa}\leq 2^{A\times\omega}\cdot 2^{A\times\omega} = 2^{A\times\omega}.$$

By WPH, $|A\cdot\omega|=|A\cdot\omega+\kappa|$, so in particular, $\kappa<\aleph(A\cdot\omega)=\aleph(A)$.

Therefore, $\aleph(A) = \aleph^*(A\cdot\omega)$, and so we get $\aleph(A) =\aleph^*(A)$, so Dependent Choice holds.

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