Cartesian product of 0 sets Since the notions of products (resp. sums) of numbers include a convention for the product (resp. sum) of 0 terms,
${\prod_{i \in \emptyset} i = 1}$ and ${\sum_{i \in \emptyset} i = 0}$,
I was wondering if there was such a convention for the cartesian product of 0 sets:
${\prod_{i \in \emptyset} A_i = ?}$, where the terms $A_i$ are sets.
Intuitively I would imagine that such a product is equal to the singelton containing the empty tuple {()} since an n-fold cartesian products produces a set of n-tuples and the empty tuple is the only 0-tuple. I'm not sure, however, since I wasn't able to find an answer by searching on the internet or the math stack exchange.
 A: Yup, it's $\{\emptyset\}$! And this isn't by convention, it follows directly from the definition. If I'm using $\emptyset$ as my index set, then - for any $\emptyset$-indexed family of sets $(A_i)_{i\in\emptyset}$ - we have that $\prod_{i\in\emptyset}A_i$ is the set of functions $f$ from $\emptyset $ to $\bigcup_{i\in\emptyset}A_i$ satisfying $f(i)\in A_i$ for each $i\in\emptyset$. Now we don't even need to observe that the only $\emptyset$-indexed family of sets is $\emptyset$ itself and so $\bigcup_{i\in\emptyset}A_i=\emptyset$ as well: for any set $X$ there is exactly one function $\emptyset\rightarrow X$, namely the empty function. So we get $\{\emptyset\}$ as expected.

Here's a way to phrase this which, while more technical, might sound less strange (since it avoids using the language of indexed sets as such):

Suppose $F$ is a function with domain $\emptyset$ ($F$ here is our $\emptyset$-indexed family of sets). Then there is exactly one function $p$ with domain $\emptyset$ satisfying $p(i)\in F(i)$ for all $i\in\emptyset$, namely $p=\emptyset$ (this is the unique element of our Cartesian product).

