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!
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.