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I'm trying to find an example of a topological space that has countably (but not finitely) many elements, is regular, but not 2nd countable. It seems to me that if a space has countably many elements, it should have a countable basis, right? My reasoning is that every topology is a subset of the discrete topology, and the discrete topology on a countable set has a countable basis, so every topology should have a countable basis? This is less of an argument and more of an intuition.

A4F
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    It's a reasonable intuition to have but sadly it is wrong. A coarser topology can require a larger basis. – Eric Wofsey Nov 03 '21 at 04:01
  • Adding to @EricWofsey's comment, we get lucky in the case of the discrete topology. But as he said, for other topologies this need not be the case. In general, the basis for a topology is a subset of the power set, which need not be countable. – Aniruddha Deshmukh Nov 03 '21 at 04:20
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    To push back against the intuition you mention, consider the topology $\alpha$ on $\mathbb{N}$ whose (nonempty) open sets are exactly those sets which have asymptotic density $1$; that is, $U\in\alpha$ iff ($U=\emptyset$ or) we have $$\lim\inf_{n\rightarrow\infty}{\vert U\cap{1,...,n}\vert\over n}=1.$$ This is indeed a topology (exercise) and given any countable $\mathfrak{X}\subseteq\alpha$ we can find a $U\in\alpha$ which doesn't contain any element of $\mathfrak{X}$ (exercise); so we have a countable non-second-countable topology! (Unfortunately this $\alpha$ is not regular ...) – Noah Schweber Nov 03 '21 at 04:24

2 Answers2

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Define a topology $\tau$ on the set $\mathbb N$ of positive integers so that a set $U\subseteq\mathbb N$ is open iff $$\text{either }1\notin U\text{ or else }\sum_{n\in\mathbb N\setminus U}\frac1n\lt\infty.$$

The space $(\mathbb N,\tau)$ is a countably infinite $T_4$ space but is not first countable.

bof
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A countable dense subset of $[0,1]^{\Bbb R}$ (which exists) is

  • countable, and hereditarily normal Tychonoff.
  • has no isolated points.
  • the minimal size for a base for its topology is continuum, i.e. $|\Bbb R|$.

This is a strong counterexample. For a proof of the last fact and the existence (the others are obvious) see here.

Henno Brandsma
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