Cartesian Product of an indexed family This is an exercise I'm trying to do but I don't know if $A=\emptyset$ makes sense:
                  $$ A=\emptyset\ and\ \{X_\alpha:\alpha\in A\} $$ Calculate $ \prod \limits_{\alpha \in A}X\alpha$ Does it makes sense? Any help would be great!
 A: It makes perfect sense, and its value falls out from the definition.
The Cartesian product $\Pi_{a\in A} X_a$ is the set of all functions $f$ such that


*

*$f\colon A\to \bigcup_{a\in A} X_a$, and

*for all $a \in A$, $f(a)\in X_a$.


If $A = \emptyset$, this definition can be simplified. The second requirement, "for all $a\in \emptyset$, $f(a)\in X_a$", is vacuously true (just as "for all $a\in \emptyset, a \neq a$" is true), so only the first requirement matters in this case. 
Now note that $\bigcup_{a\in \emptyset} X_a = \emptyset$. Thus, $\Pi_{a\in \emptyset} X_a$ is equal to the set of all functions $f\colon\emptyset\to\emptyset$. There is exactly one such function — namely, $\emptyset$ itself. (Exercise: confirm this). It follows that
$$
\Pi_{a\in \emptyset} X_a = \{\emptyset\}.
$$

Note on "vacuous truth": "for all $a \in A, P(a)$" means "for all $a$, if $a\in A$ then $P(a)$". This is often written symbolically as $(\forall a\in A) P(a)$, but remember that that's shorthand for $(\forall a)(a\in A \implies P(a))$. When $A = \emptyset$, the antecedent "$a\in A$" is  false, therefore the entire conditional is true.
A: EDIT: As discussed in the comments, calling this a matter of convention isn't truly correct. The rest of the content still holds though.
This is an edge case and likely a matter of convention, but let me try to convince you that that product should be a singleton. If I ignored that $A = \emptyset$ and just saw $\prod_{\alpha \in A} X_\alpha$ I would say that this is the set of all tuples indexed by $A$ whose $\alpha$-th element lies in $X_\alpha$. In other words, it's the set of all maps $A \longrightarrow \coprod_{\alpha \in A} X_\alpha$ such that $\alpha$ maps to the $X_\alpha$ component. Now, if $A = \emptyset$ what is the codomain? Well it's a disjoint union indexed by the empty set, so it's a disjoint union of 0 sets. Thus, it's empty. We have now that this product is the set of maps $\emptyset \longrightarrow \emptyset$. Whether such maps exist is a matter of convention as well, but the typical definition of a function $f: X \longrightarrow Y$ is a subset of $X \times Y$ satisfying $\forall x \in X \exists! y \in Y$ such that $(x, y) \in f$, which we denote as $y = f(x)$. Well $\emptyset \subseteq \emptyset \times \emptyset$ and the empty set vacuously satisfies the condition I wrote down, so why not accept this as a function? I think it's the right convention to do so, so my answer is that the product indexed by the empty set is a singleton consisting of the empty map, alternatively written as the empty tuple $\{()\}$.
This convention about the existence of a map $\emptyset \longrightarrow \emptyset$ brings me to my next argument about this product: it aligns with category theory. Indeed, in category theory there is always a map $id_X: X \longrightarrow X$ for any object $X$. Then for the category of sets to make sense, we need a map $\emptyset \longrightarrow \emptyset$. There is also a category theoretic notion of a product. Given objects $X_\alpha$ indexed by $\alpha \in A$, a product of these consists of an object $\prod_{\alpha \in A} X_\alpha$ along with maps $p_\alpha: \prod_{\alpha \in A} X_\alpha \longrightarrow X_\alpha$. These must satisfy the following "universal property". For any object $Y$ with maps $f_\alpha: Y \longrightarrow X_\alpha$, there exists a unique map $f: Y \longrightarrow \prod_{\alpha \in A} X_\alpha$ such that $p_\alpha \circ f = f_\alpha$. This is a mouthful, but if you apply this definition to sets it becomes clear what I mean. Take the $p_\alpha$ to be the usual projection maps. Then $p_\alpha \circ f$ is just the $\alpha$-th component of $f$. Hence, all I'm saying is that a map to $\prod_{\alpha \in A} X_\alpha$ is determined wholly and uniquely by its components - which should not be surprising.
Anyway, the point of this definition is that it gives a description of what the product should mean in generality. The product of objects like groups, topological spaces, affine varieties, and sets should all be defined according to this definition. As such, we can look at it in order to answer your question about the empty set. What does the universal property say in that case? Well take an arbitrary set $Y$. Vacuously, for all $\alpha \in A = \emptyset$, there is a map $f_\alpha: Y \longrightarrow X_\alpha$. Then by universal property, there is a unique map $Y \longrightarrow \prod_{\alpha \in A} X_\alpha$ such that for all $\alpha \in A$, $p_\alpha \circ f = f_\alpha$. But there are no actual such $\alpha$! Any function $Y \longrightarrow \prod_{\alpha \in A} X_\alpha$ will satisfy this condition. Thus, given only a set $Y$, the universal property tells us that there is a unique map $Y \longrightarrow \prod_{\alpha \in A} X_\alpha$. In category theory, such objects are called terminal. Point being that singleton sets are precisely those which satisfy this property. There is exactly one map $Y \longrightarrow \{*\}$ for any set $Y$. Thus, to align with the category theoretic definition of the product, I think the right convention is that $\prod_{\alpha \in A} X_\alpha$ is a singleton for $A = \emptyset$.
