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A Hausdorff space $(X, \tau)$ is said to be $k_w$-space if there is a countable collection $X_n$, $n \in \mathbb{N}$ of compact subsets of $X$, s.t.:

a) $X_n \subseteq X_{n+1}$, for all $n$

b) $X = \cup_{n \in \mathbb{N}}X_n$

c) any subset $A$ on $X$ is closed if and only if $A \cap X_n$ is compact for each $n \in \mathbb{N}$

Prove that every $k_w$ space is normal.

In order to show that space is normal - we need to show that for any closed disjoint subsets $A$ and $B$, there exist open sets $U$ and $V$, s.t. $A \subseteq U$ and $B \subseteq V$ and $U \cap V = \emptyset$. Let $A_i = A \cap X_i$, by property c) of $k_w$, $A_i$ is compact. Since $A_i$ and $B_i$ are disjoint compact sets, by Hausdorfness of $X$, there exist open $U_i$ and $V_i$, s.t. $A_i \subseteq U_i$ and $B_i \subseteq V_i$ and $U_i \cap V_i = \emptyset$.

But then how can I build $U$ and $V$? The problem is that $U_i$ can intersect $V_j$ for some $i \neq j$.

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Andreo
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  • Do you remember how to prove that two regular second-countable spaces are normal? There we perform a 'trick' of removing intersection of $V$'s from $U$'s and removing $U$'s from $V$'s. That trick seems to work here. – Prakhar Gupta Aug 20 '18 at 13:03
  • In light of the fact that there maybe multiple ways to prove that regular second-countable spaces are normal, in the previous comment I was talking about the trick given in Munkres book Topology-2nd edition in Theorem 32.1. – Prakhar Gupta Aug 20 '18 at 13:07
  • One interesting observation is that if $A\subset X$ is closed then it is clear that $A\cap X_{n}$ is compact. But we are given the extra information that converse is true. This seems to be in interesting property to use to prove that $X$ is normal. – Prakhar Gupta Aug 20 '18 at 13:16
  • @prakhar-gupta That 'trick' works when there are $U$ and $V$, s.t. $\overline{U} \cap B = \emptyset$ and $\overline{V} \cap A = \emptyset$. In my case $U_i$ can intersect $B$ at some $j$-th segment. – Andreo Aug 20 '18 at 18:02
  • In my case: $\overline{\cup_{i \in \mathbb{N}}U_i} \cap B \neq \emptyset$ and I can't use that trick. – Andreo Aug 20 '18 at 18:15
  • Yeah I tried and couldn't complete the proof. I thought that trick would work but maybe some other approach is needed. – Prakhar Gupta Aug 21 '18 at 10:42
  • If you want you can omit condition a). Given a family $X_n$ satisying b), setting $X'n = \bigcup{i=1}^n X_i$ yields a family of compact $X'_n$ satisfying a) and b). The $X'_n$ is satisfy c) if the $X_n$ do. A Hausdorff space satisyfing b) is called $\sigma$-compact. Perhaps it helps to observe that any regular $\sigma$-compact space is normal? See https://math.stackexchange.com/q/194301 . – Paul Frost Aug 29 '18 at 22:58

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The answer is yes as I found in

Franklin, S.P. and B.V. Smith Thomas, A survey of k$_\omega$-spaces, Topology Proceedings 2 (1978), 111–124.

See https://pdfs.semanticscholar.org/8b2a/1b2db52abdf330df52fc019468002ffe093e.pdf p.113.

Paul Frost
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  • @Andreo If the answer is okay, you should officially accept it. See https://math.stackexchange.com/help/someone-answers . – Paul Frost Sep 08 '18 at 10:32