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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?

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

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