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

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...)

• What is your definition of the product topology? For a finite product, it's usually the topology given by $B \times C$ where $B$ is open in $X$ and $C$ is open in $Y$ right? So since $\mathcal{B}$ and $\mathcal{C}$ are bases for their respective topologies, in particular their elements are open, and so $\mathcal{D}$ like definitially consists of opens on $X\times Y$. – user113102 Oct 21 '19 at 19:22
• What is the definition of the product topology you are using. One standard definition is the topology generated by $\{U \times V \mid U \subseteq X \text{ open}, V \subseteq Y \text{ open}\}$. If this is the definition, the openness of the elements of $\mathscr{D}$ is trivial since elements of $\mathscr{B,C}$ are open as they are basis elements. – Keefer Rowan Oct 21 '19 at 19:25
• Yes that's exactly the definition I am using! I'll add that into the question when I am off mobile. However I am still struggling to see how it is trivially open. I see how B and C are trivially open, but I don't see why their cartesian product is necessarily open? – masiewpao Oct 21 '19 at 19:37
• I thought about the comment and I completely realise why now. The answer also helped me identify it precisely, hank you! – masiewpao Oct 22 '19 at 1:00

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.