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A friend of mine told me that $X \cong Y \Rightarrow 2^X \cong 2^Y$ ($X$ and $Y$ being sets), which is very easy to prove, but he was wondering about the converse in ZF, i.e., can one take logarithms? Since the (apparently) simpler question of whether it is possible to divide by a natural number is not particularly trivial without assuming the axiom of choice (see Doyle, Conway: Division by three), I would imagine that this problem doesn't have an easy answer either.

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See also here. – Andrés E. Caicedo Jun 15 '13 at 3:21
up vote 7 down vote accepted

Actually this statement is not even provable within ZFC.

See this mathoverflow question

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In fact, not only is it not provable without the Axiom of Choice, it's not provable with the Axiom of Choice! For instance, it's consistent with AC that $2^{\aleph_0} = 2^{\aleph_1} = \aleph_2$. On the other hand, if $X$ and $Y$ are finite then certainly $2^X\cong 2^Y \implies X\cong Y$, and proving this doesn't require AC at all since all the quantities involved are finite. More broadly, Easton's Theorem says that aside from some mostly-trivial constraints (e.g., $A \gt B \implies 2^A \gt 2^B$), the cardinalities of power sets (of regular cardinals) can be entirely arbitrary.

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I hadn't even thought of the continuum hypothesis! It seems that now I should post a new question asking the same in ZFC+GCH. – Abel Oct 21 '11 at 3:54
@Abel: Well, under ZFC+GCH it is obviously true that $X\not\cong Y$ implies $2^X\not\cong 2^Y$. Not worth a separate question, I think – Henning Makholm Oct 21 '11 at 4:03
Assuming GCH, then the statement is true because the exponentiation operation corresponds to the ordinal operation 'add 1', and it's true under AC that for two ordinals $\alpha$ and $\beta$, if $\alpha+1=\beta+1$ then $\alpha=\beta$. With just ZFC+CH, the statement is still false; by Easton's theorem we can have $2^{\aleph_0}=\aleph_1$ but $2^{\aleph_1} = 2^{\aleph_2} = \aleph_3$. – Steven Stadnicki Oct 21 '11 at 4:06
@Steven, are you implying that $\alpha+1=\beta+1\Rightarrow \alpha=\beta$ requires AC? I'm quite sure it doesn't. A successor ordinal is an order type with a maximal element, and the order type of everything except that maximal element is clearly determined by the order type of the whole. – Henning Makholm Oct 21 '11 at 13:15
@HenningMakholm I didn't mean to imply that it required AC, but presumed it wasn't wholly trivial without AC (it's mostly-but-not completely trivial even with AC), and since we get AC 'for free' with GCH I just piled it on the hypotheses; mea culpa. – Steven Stadnicki Oct 21 '11 at 13:55

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