There's this proof that $\kappa \cdot \kappa = \kappa$ for $\kappa\in\text{Card}$ infinite by induction and well-ordering $\kappa \times \kappa$ with:
$(\alpha,\beta)<^*(\alpha',\beta')\iff\max(\alpha,\beta)<\max(\alpha',\beta')\vee[\max(\alpha,\beta)=\max(\alpha',\beta')\wedge(\alpha,\beta)<_{Lex}(\alpha',\beta')]$
and showing that every initial segment of $\kappa \cdot \kappa$ has cardinality $<\kappa$. However, I don't think I understand why this proof uses AC (or its equivalents), also because I understood that the same proof for $\aleph_\alpha$ (for $\alpha\in\text{Ord}$) instead of $\kappa$ does not use AC.
I'd like if someone can explain where AC comes in, and why it's not necessary when proving $\aleph_\alpha\cdot\aleph_\alpha=\aleph_\alpha$.
Thanks in advance
Best Answer
Some people use "cardinal" to mean an ordinal that is not in bijection with any smaller ordinal. With this definition, one can prove in ZF (without choice) that all infinite cardinals satisfy $\kappa\cdot\kappa=\kappa$. But one cannot prove that every set has a cardinality, i.e., that every set is in bijection with some cardinal. The latter statement is equivalent to the axiom of choice.
Other people use a different notion of "cardinal", designed so that every set $X$ has a cardinality $|X|$ and so that $|X|=|Y|$ if and only if there is a bijection between $X$ and $Y$. These people define the product of two cardinals by $|X|\cdot|Y|=|X\times Y|$ (where $\times$ refers to the cartesian product, the set of ordered pairs). With this notion of cardinal, the assertion that $\kappa\cdot\kappa=\kappa$ for all infinite cardinals $\kappa$ needs the axiom of choice; in fact, it's equivalent to the axiom of choice.
Regardless of what "cardinal" means, the assertion that every infinite set $X$ is in bijection with $X\times X$ is equivalent (in ZF) to the axiom of choice.
One method (the most common method) used by people in the second group to define cardinals is called "Scott's trick", named after Dana Scott. It defines $|X|$ to be the set of all those sets $Y$ such that (1) $Y$ is in bijection with $X$ and (2) the rank of $Y$ is as small as possible subject to (1). Unlike the first group, this second group cannot in general take $|X|$ to be a particular set in bijection with $X$; they have to take a whole family of such $Y$'s. The source for information about this last point is a paper "Cardinal representatives" by David Pincus.