Assume that we are working in ZF set theory without the Axiom of Choice. If S is an infinite set, let $S(f)$ denote the set of all finite subsets of $S$, let $S(I)$ denote the set of all infinite subsets of $S$ and let $\operatorname{Card}(S)$ denote the cardinal number of $S$.


Even though we can prove Cantor's theorem which states that the Power Set of $S$ always has a greater cardinal number than $\operatorname{Card}(S)$, could there exist an uncountable set $X$ such that we could not disprove the statement $\operatorname{Card}(X)=\operatorname{Card}(X(f))=\operatorname{Card}(X(I))$?


The answer would, of course, be negative if-without the Axiom of Choice-one could prove in ZF that, given any infinite set $S$, $\operatorname{Card}(S(I))$ is always greater than $\operatorname{Card}(S)$. Is this possible?



A question about Cantor's Power Set theorem without the Axiom of Choice
For the first equality, the answer is true.


It is quite easy to construct examples where the set of finite subsets is strictly larger. For example if $X$ is an infinite Dedekind-finite set which is the countable union of finite sets (e.g. Russell socks sets), then the set $X(f)$ is not Dedekind-finite anymore, since there is a countably infinite subset to it.


Finally, $\operatorname{Card}(X)=\operatorname{Card}(X(I))$ is provably false, unless $X=\varnothing$, in which case both cardinals are $0$. And if $X$ is non-empty finite, then $X(I)$ is empty while $X$ is not, so there is no equality.


If $\operatorname{Card}(X)=\operatorname{Card}(X(I))$, then $\operatorname{Card}(\mathcal P(X))=\operatorname{Card}(X(I))+\operatorname{Card}(X(f))=\operatorname{Card}(X)+\operatorname{Card}(X(f))$.


But since if $X$ is infinite, then there is at least an injection from the finite subsets to the infinite subsets: $A\mapsto X\setminus A$. So it follows that $\operatorname{Card}(X(f))=\operatorname{Card}(X(I))$ so we have: 


$$\operatorname{Card}(\mathcal P(X))=2\cdot\operatorname{Card}(X)$$


But this is a contradiction, since if $\operatorname{Card}(X)>4$, then $$2\cdot\operatorname{Card}(X)<2^{\operatorname{Card}(X)}=\operatorname{Card}(\mathcal P(X)).$$ ([See here for a sketch of the proof of that last inequality](https://math.stackexchange.com/a/1266932/622).)


