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.
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