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I understand that the axiom of choice, given the axioms of ZF set theory, is equivalent to the statement that "the Cartesian product of any family of nonempty sets is nonempty." I've been unable to find this proof. Could someone sketch it for me? Or provide me with a source at least?

Asaf Karagila
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mathNotebook
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1 Answers1

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Suppose $X=\{X_i\mid i\in I\}$ is a family of nonempty sets.

If there exists a choice function, then $\langle f(i)\mid i\in I\rangle$ is an element of the product $\prod_{i\in I}X_i$.

If $\prod_{i\in I}X_i$ is nonempty then there is $f=\langle x_i\mid i\in I\rangle$ in this product, which is a sequence of $x_i$ such that $x_i\in X_i$. The function $f(i)=x_i$ is a choice function.

Indeed as Nate comments, it is most common to define the product $\prod_{i\in I}X_i$ as the set of functions $f:I\to\bigcup\{X_i\mid i\in I\}$ such that $f(i)\in X_i$ for all $i\in I$.

One can easily observe that under this definition the product is exactly the set of choice functions, therefore the product is nonempty if and only if there exists a choice function.

Asaf Karagila
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  • Thank you. I'm very impressed. This was very helpful. – mathNotebook Jan 21 '12 at 14:52
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    Indeed, $\prod_{i \in I} X_i$ is precisely the set of choice functions for the family ${X_i : i \in I}$. – Nate Eldredge Jan 21 '12 at 14:55
  • How about proving the equivalent statement "the Cartesian Product of any family is empty only if a member of the family is empty"? – mathNotebook Jan 21 '12 at 15:00
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    @mathNotebook: Once one of the members of the family is empty the product is obviously empty; on the other hand if they are all nonempty then we return to this formulation as above. – Asaf Karagila Jan 21 '12 at 15:04
  • Is it obvious because that is the definition of the Cartesian Product? – mathNotebook Jan 21 '12 at 15:07
  • @mathNotebook: Yes. $x\in A\times B$ if and only if $x=\langle a,b\rangle$ and $a\in A, b\in B$. If $B=\varnothing$ then there is no such $b$, therefore the product is empty. – Asaf Karagila Jan 21 '12 at 15:09
  • This formulation of the Cartesian Product is new to me. I'll need to study it some more. Thanks again. – mathNotebook Jan 21 '12 at 15:11
  • @mathNotebook: I'm confused... how do you define products if not using ordered pairs (or tuples)? – Asaf Karagila Jan 21 '12 at 15:20
  • I define them using ordered pairs. I'm just not familiar with this: – mathNotebook Jan 21 '12 at 15:23
  • "Indeed as Nate comments, it is most common to define the product ∏i∈IXi as the set of functions f:I→⋃{Xi∣i∈I} such that f(i)∈Xi for all i∈I." – mathNotebook Jan 21 '12 at 15:23
  • I've been asking these questions about set theory because I've come across someone online who believes that by restricting multiplication by 0 to the finite cardinals he can prove the continuum hypothesis. I think he is completely wrong about this. But it got me curious about what the consequences of such a restriction would be. Based on the last few threads, I believe that, in such a system, it is impossible to formulate the Axiom of Choice. (Among other issues) – mathNotebook Jan 21 '12 at 15:31
  • But, apart from that, these proofs are interesting in their own right. I was an applied mathematician and this is the first time I've really looked a infinite cardinals. – mathNotebook Jan 21 '12 at 15:35
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    @mathNotebook: I see. The reason products are defined in such way is that you have ordered pairs, and triplets, but what happens when your index set is infinite? It's impossible to write a formula saying that you have a $\kappa$-tuple, instead we just write it as a function from $\kappa$ into the sets. As for the crank you ran into... I wouldn't worry too much about it, people who misunderstand set theory and infinite sets are as common as rhinovirus, and twice as annoying. – Asaf Karagila Jan 21 '12 at 15:42
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    Thanks. Your replies have been incredibly helpful. – mathNotebook Jan 21 '12 at 16:09
  • A choice function is defined as a function on a collection of sets, i.e. a set of (distinct) sets. However, in the case of Cartesian product, each time we need to choose from the same set $\bigcup{X_i∣i\in I}$. How do we reconcile? – Elucidase May 24 '21 at 06:28
  • @Elucidase: A choice function is not defined only for pairwise disjoint families of sets. Where did you hear that? – Asaf Karagila May 24 '21 at 07:26
  • @Asaf Karagila♦: No, I mean different, not pairwise disjoint, sets. Each set appear only once in a family (a family is a set after all). I'm thinking that for a function $x: \alpha \mapsto A_\alpha$, where $\alpha \in J$ and $J$ is an index set, we can define $A'\alpha = {(\alpha, y): y \in A\alpha }$, so that we can guarentee $(A'\alpha){\alpha \in J}$ has distinct element for each $\alpha$. Is this implied when we mention choice functions? – Elucidase May 24 '21 at 09:14
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    @Elucidase: Yes, that's one way of doing that. Another, perhaps more "cumbersome" is to remove the redundancies, and then simply declare that once you've chosen an element from each set, you can choose the same one in all the other copies of that set. This is just to prove that the product is non-empty, after which we can do whatever we want. But the general approach you suggest is the right way to solve this. – Asaf Karagila May 24 '21 at 09:50