TITLE: Axiom of Infinity needed in Cantor-Bernstein?
QUESTION [9 upvotes]: Can one prove the Cantor-Bernstein (or Schröder-Bernstein) theorem without using the Axiom of Infinity?

REPLY [10 votes]: Actually, the usual proof of the Cantor-Schröder-Bernstein Theorem does not use the Axiom of Infinity (nor the Axiom of Powersets). 
By the usual proof, I mean the one found on Wikipedia, for example. Using the notation from that proof, the main point of contention is whether we can form the sets $C_n$ and $C = \bigcup_{n=0}^\infty C_n$. These sets exist by comprehension:
$C_0 = \{x \in A: \forall y \in B\,(x \neq g(y))\}$
$C_n = \{x \in A: \exists s\,(s:\{0,\dots,n\}\to A \land s(0) \in C_0 \land (\forall i < n)(s(i+1) = g(f(s(i)))) \land s(n) = x\}$,
where abbreviations such as $s:\{0,\dots,n\}\to A$ should be replaced by the equivalent (bounded) formulas in the language of set theory. The definition of $C_n$ is uniform in $n$, and so
$C = \{x \in A: \exists n\,(\mathrm{FinOrd}(n) \land x \in C_n)\}$,
where $x \in C_n$ should be replaced by the above definition and $\mathrm{FinOrd}(n)$ is an abbreviation for "$n$ is zero or a successor ordinal and every element of $n$ is zero or a successor ordinal." An alternate definition of $C$ is
$C = \{x \in A: \forall D\,(C_0 \subseteq D \subseteq A \land g[f[D]] \subseteq D \to x \in D)\}$,
which shows that $C$ is $\Delta_1$-definable. The rest of the proof uses induction on finite ordinals, but since the definition of the sets $C_n$ is uniform these are special cases of transfinite induction. 
In conclusion, it looks like the proof could work in Kripke-Platek Set Theory — which has neither the Axiom of Infinity nor the Axiom of Powersets — provided that the two definitions of $C$ given above are provably equivalent. I haven't tried to check whether the two definitions are provably equivalent but, in any case, the proof can be carried out in Kripke-Platek Set Theory with $\Sigma_1$-Comprehension.