Do you need axiom of choice to proof $2^\mathbb{N}$ is nonempty?

Question

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?

Answer 1

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.