I was wondering if you could help me settle a discussion with a co-author. Neither of us is a hard-core topologist but our current paper forced us into this, so we figured this forum was a good place to settle the discussion!
Setting We are thinking of the space $\mathbb{R}^X$ where $X$ is an arbitrary set (so, all real valued functions from $X$ to $\mathbb{R}$). We endow this with the product topology (or pointwise convergence topology). The basis we are using for this topology is the standard one, composed of sets that take the form $\Pi_{x\in X}I(x)$ where each $I(x)$ is an open set and only finitely many of these $I(x)$ sets are different from $\mathbb{R}$. Also, without loss of generality, I guess we can take the $I(x)$ sets to be intervals but that is not fundamental to the problem.
Question If $D_1, \dots, D_n$ are basis elements, is their intersection. $D_1 \cap D_2 \cap \dots \cap D_n$ a basis element? (I know that in general intersection of basis elements is not a basis element, but is it for this special case?)
Co-author 1 This person claims YES. The idea is that intersection and product commute. Therefore $D_1\cap D_2 \cap \dots \cap D_n = \prod_{x\in X} \bigcap_{i} D_i(x)$ Since the intersection is finite then the set $\bigcap_{i} D_i(x)$ is open, so $\prod_{x\in X} \bigcap_{i} D_i(x)$ fits the definition of basis elements. It even fits it when the sets $I(x)$ are intervals since the intersection of open intervals is another open interval (taking the convention that empty sets are intervals)
Co-author 2 Claims no because $\bigcap_{i} D_i(x)$ might be empty for some $x$. While we know empty sets are open this makes their head hurt. How can one think of a function such that $f(x)\in\emptyset$?
Who is right (if any, won't be the first time we get into an argument and both of us are wrong!)
There is no contradiction here. A basis $\mathscr B$ for a topology $\mathscr T$ is a subset $\mathscr B \subset \mathscr T$ such that each $U \in \mathscr T$ is a union of elements of $\mathscr B$. Equivalently, one can require that for each $U \in \mathscr T$ and each $x \in U$ there exists $B \in \mathscr B$ such that $x \in B \subset U$.
There is no problem to allow that $\emptyset \in \mathscr B$. Okay, the empty set does not contribute to unions, thus pragmatically it does not make much sense to have $\emptyset \in \mathscr B$. But it is no formal issue.
Of course one can define a basis to be a collection of non-empty sets; it is a matter of taste.
In your example there are two approaches:
As a basis take all $\Pi_{x\in X}I(x)$ where each $I(x)$ is an open set and only finitely many of these $I(x)$ sets are different from $\mathbb{R}$. Then $\emptyset$ is a basis element and Co-author 1 is right.
As a basis take all $\Pi_{x\in X}I(x)$ where each $I(x)$ is an open non-empty set and only finitely many of these $I(x)$ sets are different from $\mathbb{R}$. Then $\emptyset$ is no basis element and Co-author 2 is right. However, we can say that each finite intersections of basis elements is either a basis element or empty.
In my opinion 1. is the better approach and it seems to be the "usual" one. Actually one introduces the product topology as the topology generated by sets of the form $p_x^{−1}(U_x)$, where $p_x : \mathbb R^X \to \mathbb R$ is the projection onto the coordinate $x \in X$ and $U_x$ is open in $\mathbb R$. More formally, these sets are taken as subbasis $\mathscr S$, and clearly $\emptyset \in \mathscr S$. The finite intersections of elements of $\mathscr S$ are then a basis, and obviously this basis contains $\emptyset$.