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Define $2^\mathbb{N}$ as the set $$ 2^\mathbb{N} = \prod_{i \in \mathbb{N}} \{0,1\} = \{ f: \mathbb{N} \to \{0,1\} \}.$$ It seems we do not need the axiom of choice to show that this set is non-empty; one can just point to the function $f(i) = 0\ \forall i \in \mathbb{N}$. Is this correct?

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    Yes, it is correct. In general, you can prove that $X^A\ne\emptyset$ whenever $X\ne\emptyset$. – Sassatelli Giulio Jan 15 '24 at 13:44
  • So it seems the axiom of choice is only necessary in the case of an infinite cartesian product of an infinite variety of sets? – G. Bellaard Jan 15 '24 at 13:51
  • @G.Bellaard It is indeed true that ZF proves that if there are a finite set $S$ such that $\emptyset\notin S$ and a function $g:A\to S$, then $\prod_{x\in A} g(x)\ne\emptyset$. – Sassatelli Giulio Jan 15 '24 at 14:10

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This exactly goes back to Russell's explanation of AoC:

For any (even infinite) collection of pairs of shoes, one can pick out the left shoe from each pair to obtain an appropriate collection (i.e. set) of shoes; this makes it possible to define a choice function directly. For an infinite collection of pairs of socks (assumed to have no distinguishing features), there is no obvious way to make a function that forms a set out of selecting one sock from each pair, without invoking the axiom of choice.

$\{0, 1\}$ is analogous to a pair of shoes instead of socks.

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