Cartesian Product of Infinite Set Equivalent to Itself implies Axiom of Choice

Theorem

Suppose that, for every Dedekind-infinite set $A$, it holds that:

$A \sim A \times A$

where $\sim$ denotes set equivalence, and $\times$ denotes the cartesian product.

Since every finite set is well-orderable, it suffices to consider only infinite sets.

Let $B$ be an arbitrary infinite set.

Without loss of generality, assume that $B \cap \On = \O$.

By Existence of Ordinal with no Surjection from Set, let $\beta \in \On \setminus \set \O$ such that there is no surjection from $B$ to $\beta$.

We have:

$\omega \subseteq \beta \subseteq B \cup \beta$

$B \cup \beta$ is Dedekind-infinite.

Therefore, by hypothesis, there exists a bijection:

$\phi : B \cup \beta \to \paren {B \cup \beta} \times \paren {B \cup \beta}$

For each $x \in B$, define:

$S_x = \set {\gamma \in \beta : \map \phi \gamma \in \beta \times \set x}$

Aiming for a contradiction, suppose for some $x_0 \in B, S_{x_0} = \O$.

Then, since $\phi$ is a bijection, it follows that $\phi^{-1} \sqbrk {\beta \times \set {x_0}} \subseteq B$.

Hence, the mapping $\psi : B \to \beta$ defined as:

$\map \psi y = \begin{cases} \gamma & : \map \phi y = \tuple {\gamma, x_0} \\ 0 & : \text{otherwise} \end{cases}$

Since the contradicts the definition of $\beta$, we conclude that each $S_x \ne \O$.

Finally, we define $\chi : B \to \beta$ as:

$\map \chi x = \min S_x$

Suppose that, for $x, y \in B$, we have:

$\map \chi x = \map \chi y$

Then, since:

$\map \phi {\map \chi x} \in \beta \times \set x$

$\map \phi {\map \chi y} \in \beta \times \set y$

it follows from the equality that $x = y$.

Therefore, $\chi$ is an injection.

Since $B$ was arbitrary, it follows that Zermelo's Well-Ordering Theorem holds.

$\blacksquare$