A question about Cantor's Power Set theorem without the Axiom of Choice 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: 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.)
