Partitioning an infinite set into two equinumerous subsets

Question

Let $X$ be an infinite set. Then there exists a partition $\lbrace A, B \rbrace$ of $X$ such that $A, B, X$ are equinumerous.

Can you prove it in $\mathrm{ZFC}$?
Can you prove it in $\mathrm{ZF}$?

Answer 1

We can prove it in $\operatorname{ZFC}$, but not in $\operatorname{ZF}$:

In $\operatorname{ZFC}$ there is some initial ordinal $\kappa$ and a bijecion $f \colon \kappa \to X$ and also some bijection $g \colon \kappa \times 2 \to \kappa$ (the function $g$ exists also if we drop choice, but $f$ may not exist). Let $h = f \circ g$, $A = \{ h(\xi, 0) \mid \xi < \kappa \}$ and $B = \{ h(\xi, 1) \mid \xi < \kappa \}$. Clearly $A \cong B \cong X \cong \kappa$.

On the other hand, if we drop choice, there may be some infinite, but Dedekind finite set $X$ and those sets don't even allow for a proper subset $A$ that is in bijection with $X$.