This might be a dumb question, but I'll ask anyway. Without assuming the axiom of choice, are there necessarily cardinalities $\kappa$ not of the form $2^{...^{\aleph_0}}$ satisfying:
- $\kappa+\kappa=\kappa$
- $\kappa \times \kappa=\kappa$
I know that given the axiom of choice, if $\kappa\geq \aleph_0$ then this is true. But are there necessarily cardinalities other than $\aleph_0$ , $2^{\aleph_0}$ , $2^{2^{\aleph_0}}$ and so on, satisfying this?
Assuming that the axiom of choice is false, then for some infinite ordinal $\alpha$, $\mathcal P(\alpha)$ cannot be well-ordered (recall that in $\sf ZF$, "the power set of an ordinal can be well-ordered" is equivalent to $\sf AC$). This means that except possibly a set of ordinals, no $\aleph_\alpha$ is of the form $2^\kappa$.
But we can prove that $\aleph_\alpha\times\aleph_\alpha=\aleph_\alpha+\aleph_\alpha=\aleph_\alpha$. This answers your question.
But more to the point, given any set $X$, the set of finite sequences, $Y=X^{<\omega}$ will satisfy $Y\times Y\sim Y$. And those are not usually iterated power sets of an ordinal, especially if $X$ was "particularly non-well-orderable".