Tarski cardinal-square theorem
= Tarski cardinal-square theorem
{c}
In ZF, if $X\times X\cong X$ for every infinite $X$, apply this to $X\sqcup h(X)$. The <product-sum comparison lemma> applied to $X$ and its Hartogs ordinal gives either an impossible injection $h(X)\to X$ or a surjection $h(X)\to X$. In the latter case, ordering each $x\in X$ by its least ordinal preimage well-orders $X$. The converse is the well-orderable-cardinal identity $\kappa^2=\kappa$.