For any sets $X$ and $Y$, does $|\mathcal{P}(X)| = |\mathcal{P}(Y)|$ imply $|X| = |Y|$?
It is well-known that the converse true; for a bijection $f: X \to Y$, we can define the bijection $g: \mathcal{P}(X) \to \mathcal{P}(Y)$ with $g(A) = f[A]$ for all $A \subseteq X$.
If $X$ and $Y$ are finite, it is obviously true; there is $m, n \in \mathbb{N}$ such that $|X| = m$ and $|Y| = n$. Because $|\mathcal{P}(X)| = 2^m = |\mathcal{P}(Y)| = 2^n$, we have $m = n$.
But is it also true for infinite sets? How can I construct a bijection $g:X \to Y$ from the bijection $f$ from $\mathcal{P}(X)$ onto $\mathcal{P}(Y)$?
