I realize that the point of the Schroder-Bernstein theorem is to not use the axiom of choice (AC), but I keep reading that the theorem is trivial if one assumes AC. And since much of set theory is performed with AC, I would like to prove the theorem with AC just for the fun of it. So, is the following proof valid?
Assume AC, and $\vert x \vert \le \vert y \vert$ and $\vert y \vert \le \vert x \vert$. Then $\vert x \vert = \vert y \vert$
By hypothesis we are given one-to-one functions $f : x \to y$ and $g : y \to x$. By AC there are bijections $s : \alpha \to x$ and $t : \beta \to y$ for ordinals $\alpha$ and $\beta$. We can therefore write $x = \{x_\delta \vert \delta < \alpha \}$ where $x_\delta = s(\delta)$ and $y = \{y_\gamma \vert \gamma < \beta \}$ where $y_\gamma = t(\gamma)$. If we can show $\vert \alpha \vert = \vert \beta \vert$, then the map $x_\delta \mapsto y_\delta$ is a bijection from $x$ to $y$. Since $f$ is one-to-one and since $f[x] \subseteq y$, we have $\vert \alpha \vert \le \vert \beta \vert$. This conclusion doesn't follow unless $f$ is one-to-one. Similarly for $g$, we can show that $\vert \beta \vert \le \vert \alpha \vert$. Therefore $\vert \beta \vert = \vert \alpha \vert$.
