= 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 $C=\bigcup_{n<\omega}A_n$. Define
$$
h(a)=\begin{cases}f(a),&a\in C,\\g^{-1}(a),&a\notin C.\end{cases}
$$
The 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 $z\mapsto(\operatorname{Re}z,\operatorname{Im}z)$ bijects $\mathbb C$ with $\mathbb R^2$. Therefore
$$
\boxed{|\mathbb C|=\mathfrak c^2=\mathfrak c.}
$$
Back to article page
