Example of a topological space that has countably (but not finitely) many elements, is regular, but not 2nd countable

Question

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.

Answer 1

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.

Answer 2

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.