$\{0,1\}^\mathbb{R}$ not empty

Question

Do we need the axiom of choice to show that $\{0,1\}^\mathbb{R}$ is not empty?

I do know that the axiom of choice is equivalent to the statement that the product of non-empty spaces is not empty. However, is there another way to show that $\{0,1\}^\mathbb{R}$ is not empty which doesn't use the axiom of choice?

Answer 1

It is true that

the product of any family of nonempty set is nonempty

is equivalent to the axiom of choice (under the standard ZF axioms).

However, this doesn't imply that you cannot decide for a specific product to be nonempty.

In particular, if $X$ is not empty, the set $X^Y$ of all maps $Y\to X$ is provably not empty, because if $x\in X$, the constant function mapping every element of $Y$ to $x$ is well defined and doesn't require choice.

The problem is instead in something like $$ \prod_{i\in I}X_i $$ where the fact that each $X_i$ is not empty doesn't allow for a single choice like before, but requires the “simultaneous choice” of an element from each $X_i$ (that is, a choice function).

Answer 2

The set in question corresponds to functions from $\mathbb R$ to $\{0,1\}$. You can select the constant function $0$ without the axiom of choice, thus showing that the set is non-empty.