 Solution

The Cantor-Schröder-Bernstein theorem states that injections $f:A\to B$ and $g:B\to A$ imply the existence of a bijection $A\to B$. Here is a direct proof, avoiding any appeal to a fixed-point theorem. Let $A_0=A\setminus g(B)$, $A_{n+1}=g(f(A_n))$, and $C={\bigcup }_{n<\omega }{A}_n$. DefineThe second case is defined because $A\setminus C\subseteq g(B)$. On $C$, $f$ is a bijection onto $f(C)$. For $b\in B$, injectivity of $g$ and the construction of $C$ give $g(b)\in C$ if and only if $b\in f(C)$: $g(b)$ cannot be in $A_0$, so membership in $C$ means membership in some $g(f(A_n))$. Hence $g^{-1}$ bijects $A\setminus C$ onto $B\setminus f(C)$. The two pieces have disjoint domains and images and together cover $A,B$, proving $h$ bijective. This proof itself uses no choice.

Write $\mathfrak{c}=|\mathbb{R}|=2^{{\aleph }_0}$. We use the standard infinite-cardinal identities $\kappa \cdot {\aleph }_0=\kappa$, ${\kappa }^n=\kappa$ for positive finite $n$ and infinite $\kappa$, and $({\kappa }^{\lambda }{)}^{\mu }={\kappa }^{\lambda \cdot \mu }$, under the permitted axiom of choice. The map $z\mapsto (\mathrm{Re}z,\mathrm{Im}z)$ bijects $\mathbb{C}$ with ${\mathbb{R}}^2$. Therefore