We present here a short proof of Cantor-Berstein-Schroeder theorem based on a fixed-point argument, probably not known enough; this proof is less explicit that the usual one, but it is worth noticing that it does not depend on the axiom of choice.

As a lemma, we first prove a particular case of Knaster-Tarski fixed point theorem:

**Theorem:** *(Tarski)* Let be a set and be a nondecreasing function (ie. such that for all ). Then has a fixed point.

**Proof.** Let and .

Notice that for all , , hence .

Moreover, implies ie. , so .

We deduce that is a fixed point of from and , and the theorem follows.

**Theorem:** *(Cantor-Bernstein-Schroeder)* Let be two sets. Suppose that there exist two injections and . Then there exists a bijection .

**Proof.** First, we define the map

.

Easily, we see that is nondecreasing, so has a fixed point according to Tarski’s theorem. Now notice that

,

and that induces a bijection and induces a bijection . Therefore, and are equipotent. By the way, an explicit bijection from to is given by

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*Related*

The well-known proof of Cantor-Bernstein theorem does not need Axiom of Choice either. One can check this easily from Wikipedia.