# Every $F_\sigma$-set in a paracompact space is paracompact.

Every $$F_\sigma$$-set in a paracompact space is paracompact.

Definitions:

$$F_\sigma$$-set is a countable union of closed sets

paracompact: if every open cover has an open refinement that is locally finite

Let $$(X,\tau)$$ be a paracompact topological space and $$Y = \cup_{n=1}^{\infty}F_n$$ be an $$F_\sigma$$ subset of $$X$$. Then for any open cover $$\mathcal{A}$$ of $$Y$$, there exists a refinement $$\mathcal{W}_n$$ of $$\mathcal{A}$$, that is locally finite on $$F_n$$. Then the union $$\mathcal{W} = \cup_{n=1}^{\infty}\mathcal{W}_n$$ will be a refinement of $$\mathcal{A}$$ and a cover of $$Y$$, but will it be locally finite? If there is a point $$x \in Y$$, s.t. $$x \in F_n$$, for each $$n \in \mathbb{N}$$?

• Isn't this only true when the space is Hausdorff? Sep 21, 2018 at 3:50
• I'm not sure, in my text book Hausdorff is not mentioned Sep 21, 2018 at 4:46
• Can you name your textbook, and what problem this is? Sep 21, 2018 at 4:58
• In my note, in the proof of the first lemma (the $\sigma$-locally finite to locally finite step(2) to (3), bottom page 1) you can find the key ideas to prove this. The $F_\sigma$ part gives you a $\sigma$-locally finite refinement which you have to modify. Sep 21, 2018 at 6:02
• @HennoBrandsma: I think it makes sense to turn this into an answer, since I (at least) thought it required normality. Sep 21, 2018 at 8:19

A classic theorem in General Topology is the following characterisation of paracompactness in regular spaces:

Let $$X$$ be a regular space. Then TFAE (note: a refinement must itself also be a cover):

1. Every open cover of $$X$$ has a locally finite open refinement. (i.e. $$X$$ is paracompact)
2. Every open cover of $$X$$ has a $$\sigma$$-locally finite open refinement (where a family of sets is $$\sigma$$-locally finite iff it is a countable union of locally finite families)
3. Every open cover of $$X$$ has a locally finite refinement (of any kind).
4. Every open cover of $$X$$ has a locally finite closed refinement.

See Engelking, "General Topology" (2nd ed. thm 5.1.11); Munkres, "Topology" (first ed. chapter 6, lemma 4.4), (2nd edition: lemma 41.3 page 254), Kelley (chapter 5, Thm 28) etc. I also wrote its proof (somewhat sloppily) in my note here.

Then suppose that $$X$$ is paracompact and Hausdorff, and let $$S = \bigcup_n F_n$$ (all $$F_n$$ closed in $$X$$ of course) be an $$F_\sigma$$ subset of $$X$$. Then $$S$$ is also paracompact (in the subspace topology).

Proof: let $$\mathcal{U} = \{U_i : i \in I\}$$ be an open cover of $$X$$ by relatively open subsets. So we have for each $$i$$ an open subset $$V_i$$ of $$X$$ such that $$V_i \cap S = U_i$$. Now fix $$n$$ and consider the open cover $$\{V_i: i \in I\} \cup \{X\setminus F_n\}$$ of $$X$$. By paracompactness this has a locally finite refinement $$\mathcal{W}_n$$. Define $$\mathcal{W'}_n = \{S \cap W: W \in \mathcal{W}_n, W \cap F_n \neq \emptyset\}$$
and note that $$\mathcal{W}' = \bigcup_n \mathcal{W'}_n$$ is a $$\sigma$$-locally finite open (in $$S$$) refinement of $$\mathcal{U}$$ and as a paracompact Hausdorff space is regular and regularity is hereditary, we conclude by the lemma that $$S$$ is paracompact.

Note however that by using the fundamental lemma I committed myself to work in regular (and thus even normal) spaces (as paracompact plus Hausdorff implies normal). Also note that we could have started with a regular paracompact (not necessarily $$T_1$$) space $$X$$ as well.

This however seems essential: this post by Brian M. Scott gives an example of a compact (non-Hausdorff and non-regular), so paracompact, space with an $$F_\sigma$$ subspace $$Y$$ that is not paracompact.