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I saw the settings in the title above in a book of probability. The author then goes on and states Urysohn's lemma: every two closed sets can be separated by a continuous functions. But he only gives a proof for metric spaces and doesn't give any proof for this case.

If this is true, it also implies that any locally compact Hausdorff and $\sigma$-compact ($X$ is a union of countable compact subsets) is normal. Is this correct? I'm a student in probability (mostly) so I'm not so good with topology in general and I don't know how to prove this. Any guidance would be highly appreciated.

P/s: I also searched for similar threads on this site, but I didn't find the answer.

Jeffrey Jao
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    Yes. A locally compact Hausdorff space is regular; a $\sigma$-compact space is Lindelöf; and a regular Lindelöf space is normal. – bof Nov 19 '23 at 00:50
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    https://topology.pi-base.org/spaces?q=Locally%20Compact%2Bt_2%2B%24%5Csigma%24-compact%2B~Normal provides some details – Steven Clontz Nov 19 '23 at 03:03

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Every locally compact and $\sigma$-compact space is covered by compacts $\{K_n:n<\omega\}$ (where $n<\omega$ is standard notation for $n\in\{0,1,2,\dots\}$) such that $K_n\subseteq int(K_{n+1})$, see A question about local compactness and $\sigma$-compactness This property is known as exhaustible by compacts. We will assume $K_{-1}=\emptyset$.

Each exhaustible by compacts Hausdorff space $X$ is normal. This is asserted in Steen/Seebach's Counterexamples but I cannot find a convenient proof. So let's write one.

First, we just assume $X$ is just an exhaustible by compacts space for which compacts are closed (a weakening of Hausdorff). Then this space is paracompact: given an open cover $\mathcal U$ of $X$, let $\mathcal F_n$ be a finite subcover for $K_n$. Then $\mathcal R_n=\{F\cap int(K_{n+1})\setminus K_{n-1}:F\in\mathcal F_n\}$ is a finite open refinement of $\mathcal U$ covering $K_n\setminus K_{n-1}$. Finally, $\mathcal R=\bigcup_{n<\omega}\mathcal R_n$ is a refinement of $\mathcal U$ covering $X$, and it is locally finite: given $x\in K_n$, $x\in int(K_{n+1})\subseteq K_{n+1}$, so $int(K_{n+1})$ is a neighborhood of $x$ disjoint from every member of $\mathcal R_N$ where $N\geq n+2$. Thus $int(K_{n+1})$ only intersects members of $\mathcal R$ belonging to the finite subset $\bigcup_{m\leq n+2} \mathcal R_m$.

Then to conclude, we note that Hausdorff paracompact spaces are normal: Hausdorf paracompact space is normal.

Steven Clontz
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  • There are a few confusing things in your answer. Do you mean that $U_n,V_n$ open in $K_{n+1}$ and closures are also closures in $K_{n+1}$? If so $U_n$ may not be open in our whole space $X$. Do you mean $\omega = \infty?$. After the part $c \in C\cap K_n$ for some $n < \omega$ why $c\in ...$ for all $m \leq n$, do you mean $m\geq n$? – Jeffrey Jao Nov 19 '23 at 18:34
  • All uses of "open", "closure", etc are to the entire space. I'll fix the other mistake, thanks. – Steven Clontz Nov 19 '23 at 19:34
  • Actually I decided to show a simpler intermediate result that can be used with https://math.stackexchange.com/questions/3473719/every-paracompact-hausdorff-is-normal to answer your question. – Steven Clontz Nov 19 '23 at 21:14