I’m going through the proof by Asaf that $|S|=|S \times S|$ for all infinite $S$ implies the Axiom of Choice, here. I’m not understanding 100% of it and wondering if I could get some help. I'm self studying set theory and still learning...! Some of the confusion comes about because I am not that familiar with ZF and there is not a lot online about it.
- Although not in the proof, is it true that under ZF if $A,B$ are not well-orderable then $|A|=|B| ⇔A≈B$.
The forwards direction is obvious, but I can’t get the reverse (although I don’t think it is required in any event).
Whenever it is stated “let $κ$ be an infinite cardinal” does this mean that $κ$ is infinite and either it is the least ordinal equinumerous to some well-orderable set or it is the set of all sets equinumerous to some set that cannot be well-ordered and that are of minimal rank?
Looking at the main theorem part, it is shown that if $κ$ is an infinite cardinal, then it must be well orderable. The direct conclusion that if $A$ is an infinite set then it is well orderable is not given. But I guess if we assume that $A$ is an infinite set and $κ=|A|$ then because $κ$ is well orderable, $|κ|=κ$ and so $|A|=|κ|$ and so $A≈κ$ and $A$ is well orderable.
For the inequality part, I assume that we use the definition of cardinal arithmetic irrespective of if the cardinal is an initial ordinal or it is generated via Scott’s trick? In which case:
By our assumption, $|κ+ℵ(κ)|=|(κ+ℵ(κ))$ x $(κ+ℵ(κ))|$. And by definition $(κ+ℵ(κ)).(κ+ℵ(κ))=|(κ+ℵ(κ)$ x $(κ+ℵ(κ)|$. Expanding out $|κ+ℵ(κ)|≥κ.ℵ(κ)≥κ+ℵ(κ)$. But because $|κ|≠ κ$ for a cardinal $κ$ that cannot be well-ordered, how do we turn $|κ+ℵ(κ)|$ into $κ+ℵ(κ)$ and then conclude $κ+ℵ(κ)=κ.ℵ(κ)$ ?
The use of $\kappa$ in that answer is perhaps a bit misleading (and I have stopped using $\kappa$ for general cardinal notation, although some of my coauthors still do that).
The idea that a cardinal is inherently an ordinal is somewhat limiting when working in a choiceless context. We can instead of a cardinal as a representation of the equivalence class of all sets which are in bijection with $A$. This can be a member of this equivalence class, or it can just be some abstract representative (traditionally, Scott cardinals for sets which are not well-orderable).
But the important thing is that the concept is a well-defined one, and that we can use the term cardinal to talk about cardinality of any set. Not just well-ordered ones.
Indeed, in the case of a well-ordered set, we already have the Gödel pairing function which provides us a bijection with its square, and therefore we do not need the Axiom of Choice for proving that.