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The implication $|A| = |B|$ is already proven is this post.

How about the other way? It looks similar to $A = B$ iff $\mathcal{P}(A) = \mathcal{P}(B)$, but what about their cardinalities in general? That is, I want to prove that if $|\mathcal{P}(A)| = |\mathcal{P}(B)|$, then $|A| = |B|$.

I tried the following: Let $\hat{f}: \mathcal{P}(A) \to \mathcal{P}(B)$ be any bijection and $f: A \to B$ defined as $f(a) = b$ iff $\hat{f}(\{a\}) = \{b\}$.

Is this ok?

Yoyos Tutoring
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    Your solution doesn't work. Let $A={a}$ and $B={b}$. Then $P(A)={\varnothing,A}$ and $P(B)={\varnothing,B}$. If you're given the bijection $\varnothing \mapsto B$, $A \mapsto \varnothing$, your procedure doesn't give you a function $A \rightarrow B$. – Sambo Aug 06 '24 at 02:41
  • If $0<|A| \leq |B| < \infty$ and $\hat{f}: \mathcal{P}(B) - {\emptyset} \to \mathcal{P}(A) - {\emptyset}$ is a injection, then is $|A| = |B|$? – Yoyos Tutoring Aug 10 '24 at 22:41

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