elementary set theory - Cardinality of the Cartesian Product of Two Equinumerous Infinite Sets

Is the cardinality of the Cartesian product of two equinumerous infinite sets the same as the cardinality of any one of the sets? I couldn't find this explicitly stated in any handout or text.

This certainly seems to be true from the examples I have seen:
- The Cartesian product of two infinitely countable sets is again infinitely countable.
- The Cartesian product of two sets with cardinality of continuum again has cardinality of continuum.

I found a question here, but it is with regard to finite sets only.

Answer

This depends on whether we assume the axiom of choice.

In the presence of choice, then yes, $\vert X^2\vert=\vert X\vert$ for all infinite $X$. This was proved by Zermelo.

If choice fails, however, this may no longer be the case: e.g. it is consistent with ZF that there is a set $X$ which is infinite but cannot be partitioned into two infinite sets. Since (exercise) if $X$ is infinite then $X^2$ can be partitioned into two infinite sets, this means that such an $X$ (called amorphous) is a counterexample to the rule.

In fact, this will happen whenever choice fails: the principle "$\vert X^2\vert=\vert X\vert$ for all infinite $X$" is exactly equivalent to the axiom of choice! See For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice.