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My question is about finding a bijection between $\mathcal{P}(\aleph_\alpha)$ and $\mathcal{P}(\aleph_\alpha \times \aleph_\alpha)$ without the axiom of choice.

This is an unproved statement in an answer to a question that I can't find at the moment. With Choice, this is trivial. Without Choice, I can only think of an injection $f: \mathcal{P}(\aleph_\alpha) \to \mathcal{P}(\aleph_\alpha \times \aleph_\alpha)$, with $f(X) = Y$ iff $Y$ is the usual well-order on $X$. But I'm lost as to how to find an injection in the other direction. Any help/hint is appreciated!

Jason Zesheng Chen
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Yes because there is one $\aleph_\alpha\to \aleph_\alpha\times\aleph_\alpha$ without the axiom of choice : the proof of this fact does not use it.

Moreover the proof of "if there is a bijection $X\to Y$ then there is one $\mathcal{P}(Y)\to\mathcal{P}(X)$" does not use the axiom of choice either.

Maxime Ramzi
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  • Without AC, does it hold that there exists a bijection between $\aleph_\alpha$ and $\aleph_\alpha \times \aleph_\alpha$, for arbitrary $\aleph_\alpha$? Isn't that equivalent to AC? – Jason Zesheng Chen Nov 08 '18 at 08:39
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    No, what's equivalent to AC is to have such a bijection for any infinite $X$, but you don't have in general that an arbitrary infinite $X$ is in bijection with an $\aleph_\alpha$ without AC. There are many proofs of this fact on this site, see e.g. my answer here https://math.stackexchange.com/questions/2153102/cardinals-arithmetic/2153155#2153155 – Maxime Ramzi Nov 08 '18 at 09:30