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Are countable spaces (i.e. $\mathbb{N}$ with any topology) second-countable? A countable space can have at most $2^\omega$ open subsets which suggests that a counterexample may exist. On the other hand both discrete and anti-discrete (or more generally with countable topology) spaces are second-countable. Also note that obviously a countable space is separable. So if it is additionally metrizable then it is second-countable.

But I couldn't prove that in general. Or is there a counterexample?

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  • $\begingroup$ Not in general. But the other way round is always correct. Every second-countable space will also be first countable. For a counterexample see: math.stackexchange.com/questions/2808369/… $\endgroup$
    – Alpha001
    Jun 24, 2019 at 10:52
  • $\begingroup$ @Alpha001 the counterexample in the linked question is based on $\mathbb{R}$ which is not countable. I'm not talking about first-countable at all. I'm refering to set theoretic countable space. Simplifying: an arbitrary topology on $\mathbb{N}$. $\endgroup$
    – freakish
    Jun 24, 2019 at 10:53
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    $\begingroup$ topology.jdabbs.com/spaces?q=countable%20%2B~second%20countable $\endgroup$ Jun 24, 2019 at 11:30

2 Answers 2

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Consider $ω$ many convergent sequences, and glue their limits. The resulting space won't have countable base at the common limit point.

Also note that a countable space is second-countable if and only if it is first-countable.

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    $\begingroup$ Isn't that space metrizable? E.g. I'll take some infinite dimensional normed vector space with countable basis $\{e_i\}$ and take sequences $\frac{1}{n}e_i$ together with $0$. Or maybe: why is there no countable basis around the common limit point? $\endgroup$
    – freakish
    Jun 24, 2019 at 11:07
  • $\begingroup$ That's a different space. By gluing the limits I meant: take a topological sum of $ω$-many copies of a space homeomorphic to a convergent sequence and take the quotient space obtained by identifying the limits. In this topology you may fix final segments of the sequences independently and their union will be a neighborhood of the common limit point. $\endgroup$
    – user87690
    Jun 24, 2019 at 11:11
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    $\begingroup$ I still think that your space is metrizable. The initial space is a disjoint union of sequences (with limits) and so it is metrizable. Thus the quotient space is a pseudometrizable space. But it is $T_0$ so it is metrizable. Am I missing something? $\endgroup$
    – freakish
    Jun 24, 2019 at 12:09
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    $\begingroup$ The quotient pseudometric does not induce the quotient topology in general. $\endgroup$
    – user87690
    Jun 24, 2019 at 12:09
  • $\begingroup$ Ah, yes, you are right. And your example seems to be correct. $\endgroup$
    – freakish
    Jun 24, 2019 at 12:17
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Take a free ultrafilter on $\mathbb{N}$ and add $\emptyset$ to it to make it a topology. Standard facts on ultrafilters tell us that this is not second countable.

Another advanced example :let $x \in \omega^\ast$ and let $X= \omega \cup \{x\}$ in the subspace topology.

Or let $D$ be a countable dense subset of $\{0,1\}^{\Bbb R}$ in the product topology. I wrote extensively on such spaces here.

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