Cartesian product indexed by a power set

I have been trying to solve a question related to set theory, specifically, related to the generalized cartesian product. So, the question is:

Let $$A$$ be a given set, find $$\prod_{X \in \mathcal{P}(A)} X$$, where $$\mathcal{P}(A)$$ is the power set of $$A$$. The question is: ¿Do we need Axiom of Choice to guarantee that $$\prod_{X \in \mathcal{P}(A)} X \neq \emptyset$$?

I know by definition that $$\prod_{X \in \mathcal{P}(A)} X := \left\lbrace f: \mathcal{P}(A) \longrightarrow \bigcup_{X \in \mathcal{P}(A)}X: \forall X \in \mathcal{P}(A), f(X) \in X\right\rbrace$$, but i still don't know what happen with $$\emptyset \in \mathcal{P}(A)$$. I think I'm not understanding such definition, so I'll be very grateful with some hint, idea or solution. Thanks in advance!

• Where did you find this question? Because as you already found out yourself: the empty set is definitely in the power set, so that product should be empty. Maybe they mean the power set, except for the empty set? Jun 1 '19 at 23:45

Since $$\varnothing\in\mathcal{P}(A)$$ for any set $$A$$, one of the sets in question is the empty set. That means that $$\prod_{X\in \mathcal{P}(A)} X = \varnothing$$ for every set $$A$$.

Now, if you want to look at $$\prod_{\varnothing\neq X\in\mathcal{P}(A)} X$$ then it is definitely the case that for $$A$$ infinite, you need the Axiom of Choice in order to show that this product is nonempty. In fact, the claim that this product is nonempty is equivalent to the Axiom of Choice.

With the Axiom of Choice you get that the product is nonempty.

To see that assuming this product is nonempty for any set $$A$$ yields the Axiom of Choice, let $$\{X_i\}_{i\in I}$$ be a nonempty family of nonempty sets. Let $$B=\{ X_i\times\{i\}\}_{i\in I}$$ (I do this just to ensure that the sets are disjoint). Now let $$A=\cup B = \cup_{i\in I}(X_i\times\{i\})$$. Note that $$X_i\times\{i\}\in \mathcal{P}(A)$$ for each $$i$$. That is, $$B\subseteq P(A)$$. Now let $$f\in \prod_{\varnothing\neq X\in P(A)} X$$ and let $$g=f|_B$$. Then for each $$i\in I$$, $$g(X_i\times\{i\})\in X_i\times\{i\}$$, so $$g$$ gives a choice function for $$B$$, and hence for $$A$$. Thus, $$\prod_{i\in I}X_i$$ is nonempty.

• One can say more in ZF. The product of the non-empty sets is non-empty if and only if the set can be well-ordered. Jun 2 '19 at 13:31
• @AsafKaragila: I’m not sure what you are trying to point out (which I’m sure I would want to know, given your expertise in this area). It sounds like you’re just saying “in ZF, AC is equivalent to the Well Ordering Theorem”, which, of course, is true; and you could also have said “if and only if given two sets $A$ and $B$, there is either an injection from $A$ to $B$ or an injection from $B$ to $A$”, or any number of other equivalents... so I’m sure I’m missing something. Jun 2 '19 at 13:47
• I mean that there is a choice function from a given set's power set (sans the empty set) if and only if it can be well-ordered. Without assuming AC. This can be "the hard lemma" in a simple proof that AC implies every set can be well-ordered. Jun 2 '19 at 13:53
• @AsafKaragila: Ah. Thanks for the clarification. Jun 2 '19 at 13:55
• ¡Thank you, Arturo! Thanks for the comments Asaf :) Jun 2 '19 at 14:59