Why are the elements of this basis that generates the product topology open

general-topology

In class (and in parallel with Topology by Munkres), we proved the following theorem:

Let $\mathcal{B}$ be a topological basis on $X$, $\mathscr{C}$ be a topological basis on $Y$. Then, $\mathscr{D} = \{B \times C : B \in \mathcal{B}, C \in \mathscr{C} \}$ is a topological basis for the product topology on $X \times Y$.

I can mostly follow the proofs from class and the book, and the crux is to use the lemma that states:

Let $(X,\mathcal{T})$ be a topological space, and $\mathscr{C}$ be a collection of open sets in X, that satisfies: $\forall U \in \mathcal{T}.\forall x \in U.\exists C \in \mathscr{C}. x \in C \subset U$. Then $\mathscr{C}$ is a topological basis generating $\mathcal{T}$.

From the definition and the lemma, I (think I) follow the proof in Munkres and in my notes that shows that the elements of $\mathscr{D}$ satisfy the condition of lemma 2. However one part of the proof I am simply failing to see is how do we know $\mathscr{D}$ is a collection of open sets in $X\times Y$?

The proof in Munkres doesn't address this because I think it's meant to be an obvious conclusion. In my notes I have written "First, by the definition of the product topology of $X \times Y$, all elements of $\mathscr{D}$ are open."

I understand why the definition implies the basis elements are open (or rather I think it's because the elements of any basis are open), but I don't see how this helps since we haven't yet proved that $\mathscr{D}$ is a basis, let alone the basis generating the product topology? Why are we using the definition to justify assumptions on the set for which we are trying to prove is a bases?

My apologies if I am missing something obvious (I suspect I am…)

Best Answer

Because the product topology is defined as a ( actually the smallest) topology that makes the projections continuous, we know $\mathcal{D}$ is a collection of open sets: $B \in \mathcal{B}$ implies $B$ is open in $X$ and $C \in \mathcal{C}$ implies $C$ is open in $Y$ (the bases on $X$ and $Y$ also consist of open sets of their resp. space!).

Then $$B \times C= \pi_X^{-1}[B] \cap \pi_Y^{-1}[C]$$

is the (finite) intersection of two open subsets of $X \times Y$ in the product topology (using that projections are continuous), and so is open.