second countable and Hausdorff doesn't imply normal

Question

We know that if $X$ is a second countable and regular, then it is normal. Now assume that $X$ is second countable and Hausdorff. Are there any examples to show that $X$ need not necessarily be a normal space?

Answer 1

Consider the subset $X=\mathbb{Q}\times \{0\}\cup \mathbb{R}\times\mathbb{R}_{>0}$ of $\mathbb{R}\times \mathbb{R}$. For each $q\in \mathbb{Q}$ and $r\in \mathbb{Q}_{>0}$, we set $$ D_{q,r}=\{(q,0)\}\cup \{(x,y)\in \mathbb{R}\times \mathbb{R}_{>0}\mid (x-q)^2+y^2<r\}\subset X. $$ Equip $X$ with the weakest topology such that

• it is stronger than the topology induced from $\mathbb{R}\times \mathbb{R}$, and
• $D_{q,r}$ is open for each $q\in \mathbb{Q}$ and $r\in \mathbb{Q}_{>0}$.

Then $X$ is second countable and Hausdorff, but we cannot separate $(0,0)$ and the closed subset $(\mathbb{Q}\setminus\{0\})\times \{0\}$ by open neighborhoods.

Answer 2

Maybe the simplest example is Munkres' favourite example $\Bbb R_K$, see here, e.g.

It's Hausdorff, non-regular, separable and second countable.

Tip: to look for such examples in the future you can use $\pi$-base queries. Very useful, and highly recommended; the previous example is called Smirnov's deleted sequence topology there.