How do we understand the axiom of Choice?

Question

I was pondering about the infamous statement given by Russell to understand choice, "To choose one sock from an infinitely many pairs of socks one requires axiom of choice, for shoes the axiom is not needed." But my question is, do such 'socks' exist in the Universe of Set Theory? Axiom of Extensionality says that each set should be identified with the sets it constitutes, so it means each sock from any particular pair must have something to distinguish them, so if we have a 'property' to distinguish them, then why do we need Choice?

Answer 1

We can prove an arbitrary product $\prod_{i\in I} G_i$ of groups forms a group without axiom of choice, as we can pick identity $e_i\in G_i$ uniformly. And for each element $(a_i)\in \prod_{i\in I} G_i$, we can form $(a_i^{-1})\in\prod_{i\in I}G_i$ without axiom of choice, etc. This corresponds to the intuitive idea that the product of shoes is not empty.

Now given an arbitrary family of sets $S_i$ for $i\in I$, is there an element in $\prod_{i\in I} S_i$? For any specific example you can describe, $S_i$'s almost certainly have some structure to exploit like being shoes rather than socks, so you can prove $\prod_{i\in I} S_i$ is nonempty without axiom of choice. The point of AC is that you don't need to exploit any specific feature of the sets, and you are allowed to assume the product is nonempty. It's unconstructive, but also consistent with ZF.

Similarly, for each vector space you can describe, there is probably some canonical basis based on the context, but we can assume any vector space has a basis without knowning the specifc structure or being able to explicitly exhibit such a basis (because of AC).

As proving such general theorems about specific structures is universal in modern math, one can argue that sets of socks are everywhere. It's not that "socks" exist in the Universe of Sets, but rather in our description of the universe of sets whatever it means.