Let $X$ and $Y$ be sets. It is undecidable in ZFC whether $2^{|X|} = 2^{|Y|}$ implies $|X| = |Y|$ (in Cohen's original model for ZFC + $\neg$CH, one has $2^{\aleph\_0} = 2^{\aleph\_1}$). What if we restrict our attention to the *finite* parts of $X$ and $Y$?


**Question.** Do $X$ and $Y$ have the same cardinality if the families of *finite* subsets of both sets do?


As noted by Nik Weaver [in a comment](https://mathoverflow.net/questions/450025/do-x-and-y-have-the-same-cardinality-if-their-families-of-finite-subsets-do#comment1162981_450025), the answer is yes when we assume in the background to be working with the axioms of ZFC. But, what about ZF?


I feel this must be either basic or very well known (to those who know it very well).



Do $X$ and $Y$ have the same cardinality if their families of finite subsets do?
It can be proved in $\mathsf{ZF}$ that ``for all cardianls $\mathfrak{a},\mathfrak{b}$, if $\mathrm{fin}(\mathfrak{a})=\mathrm{fin}(\mathfrak{b})$, then $\mathfrak{a}=\mathfrak{b}$'' implies $\mathsf{AC}$. Here $\mathrm{fin}(\mathfrak{a})$ denotes the cardinality of the set of all finite subsets of a set which is of cardinality $\mathfrak{a}$.


Assume that, for all cardianls $\mathfrak{a},\mathfrak{b}$, if $\mathrm{fin}(\mathfrak{a})=\mathrm{fin}(\mathfrak{b})$, then $\mathfrak{a}=\mathfrak{b}$. We prove $\mathsf{AC}$ by showing that, for an arbitrary limit ordinal $\lambda$, $\mathrm{V}\_\lambda$ is well-orderable. Let $\mathfrak{a}$ be the cardinality of $\mathrm{V}\_\lambda$ and let $\kappa$ be the Hartogs number of $\mathrm{V}\_\lambda$, i.e., the first ordinal that cannot be mapped injectively into $\mathrm{V}\_\lambda$. Note that every finite subset of $\mathrm{V}\_\lambda$ is an element of $\mathrm{V}\_\lambda$, so $\mathrm{fin}(\mathfrak{a})=\mathfrak{a}$.


It suffices to prove that $\mathrm{fin}(\mathfrak{a}+\kappa)=\mathrm{fin}(\mathfrak{a}\cdot\kappa)$, since then by our assumption we would have $\mathfrak{a}+\kappa=\mathfrak{a}\cdot\kappa$, and thus $\mathfrak{a}$ is well-ordered by a lemma of Tarski. Clearly, $\mathrm{fin}(\mathfrak{a}+\kappa)=\mathrm{fin}(\mathfrak{a})\cdot\mathrm{fin}(\kappa)=\mathfrak{a}\cdot\kappa$. It remains to show that $\mathrm{fin}(\mathfrak{a}\cdot\kappa)=\mathfrak{a}\cdot\kappa$. It is proved as follows.


Let $u$ be an arbitrary finite subset of $\mathrm{V}\_\lambda\times\kappa$.
Then $\mathrm{ran}(u)=\{\alpha\mid\exists x(\langle x,\alpha\rangle\in u)\}$ is a finite subset of $\kappa$, and thus there is a unique isomorphism $h\_u$ of $\mathrm{ran}(u)$ onto some natural number. Note that $u$ is uniquely determined by $\langle\mathrm{ran}(u),\{\langle x,h\_u(\alpha)\rangle\mid\langle x,\alpha\rangle\in u\}\rangle$, so
$$
\mathrm{fin}(\mathfrak{a}\cdot\kappa)\leqslant\mathrm{fin}(\kappa)\cdot\mathfrak{a}=\mathfrak{a}\cdot\kappa,
$$
which completes the proof.$\quad\square$


