# second countable and Hausdorff doesn't imply normal

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?

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 $$D_{q,r}=\{(q,0)\}\cup \{(x,y)\in \mathbb{R}\times \mathbb{R}_{>0}\mid (x-q)^2+y^2 Equip $$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.

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.

• Wow! This is a great database! Thank you so much for sharing Sep 26 at 22:32