Concerning $\kappa+\kappa=\kappa$ for any infinite cardinal $\kappa$ In Introduction to Set Theory by Hrbacek and Jech, on page 94, the authors state that $\aleph_0+\aleph_0=\aleph_0$, and the Axiom of Choice implies that $\kappa+\kappa=\kappa$ for any infinite cardinal $\kappa$. To me, it is clear that $\aleph_0+\aleph_0=\aleph_0$, as the (disjoint) union of two countable sets is countable, but I am not sure about the latter statement. In particular, I'm not sure how the Axiom of Choice is used.
Can someone please explain this to me?
 A: The point here is what does cardinal mean.
In many contexts, and perhaps naively, one might consider cardinals to mean initial ordinals, so that a set which cannot be well-ordered does not have a cardinal. In this context it is not hard to prove that $\kappa+\kappa=\kappa$ for any infinite cardinal, simply because this easily holds for well-ordered cardinals.
In many contexts, which are different, and also perhaps naively, one considers a cardinal to be something representing cardinality of sets. This can be done in multiple ways, and it doesn't really matter, since cardinal arithmetic is really just the "quotient" of considering sets and injections, along with unions and products. In this case, you will need the axiom of choice. Suppose, for example, that $A$ is a set such that if $B\subseteq A$ then $B$ is finite or $A\setminus B$ is finite. These sort of sets are called amorphous and their existence is consistent with $\sf ZF$.
But now consider $|A|+|A|$, it's the cardinal of $A\times\{0,1\}$, which certainly can be split into two infinite subsets, $A\times\{0\}$ and $A\times\{1\}$. So, $|A|\neq|A|+|A|$ in that case. There are other examples that are consistent.
Sageev proved that if $\kappa+\kappa=\kappa$ for any infinite cardinal, in the latter sense, then we can deduce that every infinite set has a countably infinite subset (which, for example, prevents the existence of amorphous sets). But at the same time, we cannot even prove that every countable family of non-empty sets admits a choice function. So while the axiom of choice is needed, its "absolute necessity" here is very mild and hard to notice.
Finally, if we do assume the axiom of choice, then every set can be well-ordered, so the two contexts merge into one, and then we can prove the equality in a straightforward manner by replacing a set with whatever ordinal is in bijection with it.
