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


$F_\sigma$-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 $n \in \mathbb{N}$?

  • $\begingroup$ Isn't this only true when the space is Hausdorff? $\endgroup$
    – Steve D
    Sep 21, 2018 at 3:50
  • $\begingroup$ I'm not sure, in my text book Hausdorff is not mentioned $\endgroup$
    – Andreo
    Sep 21, 2018 at 4:46
  • 1
    $\begingroup$ Can you name your textbook, and what problem this is? $\endgroup$
    – Steve D
    Sep 21, 2018 at 4:58
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 21, 2018 at 6:02
  • 1
    $\begingroup$ @HennoBrandsma: I think it makes sense to turn this into an answer, since I (at least) thought it required normality. $\endgroup$
    – Steve D
    Sep 21, 2018 at 8:19

1 Answer 1


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.


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