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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?

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    $\begingroup$ $0$ and $1$ are shoes, not socks :) $\endgroup$ Jun 28, 2017 at 10:32
  • $\begingroup$ this is the set of indicator functions for subsets of $\Bbb R$. $\endgroup$
    – Masacroso
    Jun 28, 2017 at 10:32
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    $\begingroup$ @egreg The existence of $\mathbb{R}$ (that is of a complete Archimedean ordered field) can be proved in $\mathrm{ZF}$ alone $\endgroup$
    – abc
    Jun 28, 2017 at 10:55
  • $\begingroup$ @Achilles Maybe I got confused with “constructibility” and things like that. But it was not the main point of my comment. Thanks for noting. $\endgroup$
    – egreg
    Jun 28, 2017 at 10:58

2 Answers 2

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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.

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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).

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