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

Question

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.

Answer 1

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.

Answer 2

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.