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I am not an an expert in set theory, so this question could be trivial. I am sorry in that case.

Let $I$ be a set and $\{ X_i \}_{i \in I}$ be a collection of sets such that $X_i \neq \emptyset$ for all $i \in I$. The axiom of choice tells precisely that the set $$ \prod_{i \in I} X_i \neq \emptyset $$ is not empty. The use of the word "choice" here is clear. To produce an element $(x_i)\in \prod_{i \in I} X_i$ we need to choose an $x_i \in X_i$ for all $i \in I$.

Now I am wondering: what if there no choice at all? Namely, if $X_i = \{x_i\}$ has one element for each $i \in I$? Indeed, in this case $\prod_{i \in I} X_i$ has only one element, that is $(x_i)$, and there is no choice at all. So the question is the following:

Question: it the axiom of choice needed to prove that the product of a family of sets, each with exactly one element, is not empty?


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closed as too localized by Martin Brandenburg, Henry Cohn, Misha, Gerald Edgar, Asaf Karagila Apr 23 '13 at 13:18


Of course not. Please read the FAQ mathoverflow.net/faq –  Martin Brandenburg Apr 23 '13 at 12:58
No, obviously not. It is also not needed to prove that products of finite families of nonempty sets are nonempty, but this is a bit more subtle. In your case, you can just explicitly define the function, and the basic axioms of set theory suffice to verify that this definition gives you a set. –  Andres Caicedo Apr 23 '13 at 12:59
The questino is probably not appropriate for this site. No, AC is not needed to choose from a family of one-element sets. It may be needed, however, to choose from a family of two-element sets. –  Gerald Edgar Apr 23 '13 at 13:11
Note that if $X_i=\lbrace x_i\rbrace$, then $\bigcup_{i\in I}\lbrace\langle i,x_i\rangle\rbrace$ is a choice function. –  Asaf Karagila Apr 23 '13 at 13:20
Let $U$ be the union of the sets $X_i$. Let $Y=I\times U$. For $(i,x)\in Y$, let $\phi(i,x)$ be the property $x\in X_i$. Then the axiom schema of specification gives us a set $\{(i,x)|i\in I,x\in X_i\}$, which is the product set you're looking for. –  Steven Landsburg Apr 23 '13 at 13:20