# An LCH space with a non $\sigma$-compact open set

for every second-countable LCH space $$X$$, any open set of it is $$\sigma$$-compact, and I think it's not true for a separable but not second-countable LCH. I tried to start with an uncountable set $$X$$, and require closed sets to be countable, and thus compact sets are countable, ensuring that $$X$$ is not $$\sigma$$-compact first, but in this setup, every open set is huge and $$X$$ is unlikely to be an LCH space. I saw some examples of separable but not second-countable spaces here.

I tried the following concrete example. Consider the real line $$\mathbb{R}$$ and topology:

$$\tau = \{U| x \in U, \textbf{U is an interval with arbitrary endpoints}\} \cup \varnothing$$ for some fixed $$x \in \mathbb{R}$$.

This is not a Hausdorff space, but it satisfies all other conditions and requirements, further, I consider throwing in all $$U^c$$ to $$\tau$$ in order to make it Hausdorff, but then it's not even a topology, does anyone have a different example or ways to modify my example? Also, I was thinking maybe I should devise some topologies on $$\mathbb{R}$$ such that any compact set can only contain finitely many irrationals thus ruling out $$\sigma$$-compact.

• What is LCH ? Locally compact something... ?
– Ulli
Commented Apr 20, 2023 at 6:55
• @Ulli I guess it's locally compact Hausdorff. Commented Apr 20, 2023 at 7:07
• This is just a shot in the dark but what about $\beta\mathbb N$? I'm thinking that the complement of a non-isolated point will not be $\sigma$-compact. But it's been a long, long time since I took a course in topology, so I could be wrong. Commented Apr 20, 2023 at 7:12
• @user14111: yes, this is exactly, what I had in mind. But as far as the OP doesn't confirm your guess about LCH it is not clear, what he means.
– Ulli
Commented Apr 20, 2023 at 7:19
• By LCH I mean locally compact Hausdorff, sorry for the confusion Commented Apr 20, 2023 at 19:52

Ok, Wikipedia also says that LCH means locally compact Hausdorff. So, let's take this as granted.

Then the example mentioned by user14111 does the job:

$$\beta \mathbb N$$, the Stone-Cech compactification of the integers, is compact and separable. Let $$x \in \beta \mathbb N \setminus \mathbb N$$, $$X := \beta \mathbb N \setminus\{x\}$$ is open in $$\beta \mathbb N$$.

$$X$$ is countably compact. [Assume not, then there exists a countable, infinite, closed discrete set $$D \subset X$$. Then $$D \cup \{x\}$$ is closed in $$\beta \mathbb N$$. But each closed, infinite subset of $$\beta \mathbb N$$ has size $$2^c$$ (see Engelking, General topology, 3.6.14).]

Of course, $$X$$ is not closed in $$\beta \mathbb N$$, hence not compact, hence not Lindelöf, hence not $$\sigma$$-compact.

Therefore, either $$X$$ open in $$\beta \mathbb N$$ is the required example, or $$X$$ itself, since $$X$$ is locally compact and separable.

I ran a search on pi-Base: Separable+T_2+Locally Compact+~Second Countable+~$$\sigma$$-compact.

It returned the rational sequence topology on $$\mathbb R$$: it's separable because the rationals are dense, but it's neither second countable nor sigma-compact because the irrationals form an uncountable discrete closed subset. It's locally compact because each point has a compact neighborhood: either a singleton or a converging sequence.

• It's very concise, I got a similar idea myself but yours is clearer and a complete example. Commented Apr 26, 2023 at 22:28