Given an infinite set $A$ - does the cardinality of $A$ equal to the cardinality of $A^2$?
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$\begingroup$Zermelo proved that every well-ordered infinite set has this property, so if we assume the axiom of choice then the answer is yes. In fact the axiom of choice is equivalent to the assertion that for every infinite set $A$, $A$ and $A^2$ are equinumerous.
- The proof of Zermelo's theorem can be found here: About a paper of Zermelo
- The proof for the reverse implication can be found here: For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice
However if the axiom of choice is negated there are sets whose cardinality is strictly less of that of their square. We even know how to construct such set:
Suppose that $X$ cannot be well-ordered (such set exists, since we assume the axiom of choice is false, and therefore the well-ordering principle is false). Let $\kappa$ be an ordinal such that there is no $f\colon\kappa\to X$ which is injective.
In this case $A=X\cup\kappa$ has the property that $A^2$ is strictly larger than $A$.
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