# Is the divisor topology second countable?

Let $$X\neq\varnothing$$ and $$B\subseteq P(X)$$. We define the topology generated by $$B$$ as follows: $$T:=\bigcap\limits_{\tau\supseteq B}\tau$$ where $$\tau$$ is a topology containing $$B$$.

I’m trying to show that the topology on $$\mathbb{Z}_{\geq 2}=\{x\in \mathbb{Z},\,x\geq 2\}$$, generated by the sets $$U_n:=\{x\in \mathbb{Z}_{\geq 2},\,x|n\}$$ is second countable. My attempt was trying to give an explicit countable base but I know that the sets $$U_n$$ aren’t an option. Could anyone give me a hint?

• Any space with a countable generating set is second-countable (count the finite intersections). – Noah Schweber Apr 5 '20 at 4:54
• If I understood correctly, would the set of all finite intersections of the generating set produce an actual basis for the topological space? – pmorelli Apr 5 '20 at 4:58
• That's exactly right. – Noah Schweber Apr 5 '20 at 4:58
• I got it. Thanks a lot! – pmorelli Apr 5 '20 at 5:03

What you're describing is that $$B$$ is a subbase (i.e. generatin set) for $$\tau$$, and if we have a countable subbase, the set of all finite intersections from $$B$$ (including $$X$$ (the empty intersection) and all members of $$B$$ itself) is also countable (standard set theory) and forms a base for $$\tau$$, so yes, $$\tau$$ is then second countable.