The goal is to prove that the axiom of choice is implied by the statement "for every infinite set A: |A| = |A \times A|". It is known that the
well-ordering theorem is equivalent to the axiom of choice; thus it is enough to show that the statement implies that for every set B there exists a
well-order. Since the collection of all
ordinals such that there exists a
surjective function from B to the ordinal is a set, there exists an infinite ordinal, \beta, such that there is no
surjective function from B to \beta. We assume
without loss of generality that the sets B and \beta are
disjoint. By the initial assumption, |B \cup \beta| = |(B \cup \beta) \times (B \cup \beta)|, thus there exists a
bijection f : B \cup \beta \to (B \cup \beta) \times (B \cup \beta). For every x \in B, it is impossible that \beta \times \{x\} \subseteq f[B], because otherwise we could define a surjective function from B to \beta. Therefore, there exists at least one ordinal \gamma \in \beta such that f(\gamma) \in \beta \times \{x\}, so the set S_x = \{\gamma : f(\gamma) \in \beta \times \{x\}\} is not empty. We can define a new function: g(x) = \min S_x. This function is well defined since S_x is a non-empty set of ordinals, and so has a minimum. For every x, y \in B, x \neq y the sets S_x and S_y are disjoint. Therefore, we can define a well order on B, for every x, y \in B we define x \leq y \iff g(x) \leq g(y), since the image of g, that is, g[B] is a set of ordinals and therefore well ordered. ==References==