Let $X$ be an infinite set.
Is it provable in ZF that there is a bijection between $X$ and the family of finite subsets of $X$?
If no, does the existence of such bijection imply (in ZF) that there is a bijection between $X$ and $X\times X$?
Let $X$ be an infinite set.
Is it provable in ZF that there is a bijection between $X$ and the family of finite subsets of $X$?
If no, does the existence of such bijection imply (in ZF) that there is a bijection between $X$ and $X\times X$?
Suppose that $X$ is an infinite, Dedekind-finite set. If there were a bijection $f:X\times X\to X$, we could fix $x_0\in X$ and define $g:X\to X:x\mapsto f(x,x_0)$ to get an injection of $X$ to a proper subset of $X$, contradicting the Dedekind-finiteness of $X$. Since the existence of such $X$ is consistent with ZF, we cannot prove in ZF that there is a bijection between an arbitrary infinite $X$ and $X\times X$.
Suppose now that we have a bijection $f:[X]^{<\omega}\to X$, where $[X]^{<\omega}$ is the set of finite subsets of $X$. Let $g:X\to[X]^{<\omega}:x\mapsto\{x\}$; then $g\circ f:X\to X$ is a bijection of $X$ with a proper subset of $X$, so $X$ must be Dedekind-infinite, and it follows as before that we cannot prove the existence of such an $f$ in ZF.
In 1904 Zermelo proves that for every $\alpha$ we have $\aleph_\alpha^2=\aleph_\alpha$. This implies two immediate things:
The reason for the first one to hold is that we can effectively enumerate subsets of ordinals, and we can well-order them. Even without the axiom of choice, if we are careful enough and our sets are such that can be well ordered, then we can prove about them some of the theorems we know from ZFC.
19 years later, 1923, Tarski publishes a paper in which he proves that if for every infinite set $|X|=|X\times X|$ then the axiom of choice holds. In particular this means that if we do not assume the axiom of choice we cannot assume for an arbitrary set that $|X|=|X\times X|$. As remarked above, we can still prove this for some sets.
Brian's example is very good, let me use that for a moment: $X$ is Dedekind-finite if $f\colon X\to X$ is injective if and only if it is a bijection.
Now let $X$ be an infinite Dedekind-finite set (this can even be a dense subset of $\mathbb R$ in some models), suppose $f\colon Fin(X)\to X$ was a bijection, where $Fin(X)$ denotes the finite subsets of $X$, let $g\colon X\to Fin(X)$ be the map $g(x)=\{x\}$.
Now consider the function $f\circ g\colon X\to X$. As a composition of two injective maps it is injective, and since $X$ is Dedekind-finite $f\circ g$ is actually a bijection, this means that $g$ was also a bijection.
However there are no sets of the form $\{x,y\}$ for $x\neq y$ in the image of $g$ so it cannot be surjective! Contradiction.
Some more to read on this site: