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Every closed subspace of a paracompact space $X$ is paracompact.

My attempt:

Let $A\subset X$ be closed and $\{U_{\alpha}\}_{\alpha \in I}$ an open cover of $A$. This means that $U_\alpha = A \cap U_{\alpha}^{'}$ for some open subset $U_{\alpha}^{'} \subset X. $Then the collection $\mathscr{U} = \{X \setminus A\} \cup \{U_{\alpha}^{'}\}_{\alpha \in I}$ is a cover of $X$. Since $X$ is paracompact, there is a locally finite open refinement $\{V_{\beta}\}_{\beta \in J}$ of $\mathscr{U}$ that covers $X$. Let $\mathscr{V} = \{A \cap V_{\beta}\}_{\beta \in J}$. So, I need to test three things:

(1) $\mathscr{V}$ is nonempty

(2) $\mathscr{V}$ is a locally finite refinement of $\{U_{\alpha}\}_{\alpha \in I}$

(3) $A \subset \bigcup_{\beta \in J}A \cap {V}_{\beta}$

I think an alternative way to write the set $\mathscr{V} $ is as follows: $\mathscr{V} = \left\{V_\beta \colon V_{\beta} \subset U'_{\alpha} \hspace{0.3cm}\text{for some} \hspace{0.3cm}\alpha \in I \right\}$. Let $W = \bigcup_{V_\beta \in \mathscr{V}}V_{\beta}$, then I must prove that $A \subset W$ and that $\mathscr{V}$ refines $\{U_{\alpha}\}_{\alpha \in I}$.

How do I deduce that from what I already have above? I need some help to do this.

Here are some definitions:

Definition. Let $X$ be a topological space. A collection of sets ${U_{\alpha} \subset X}$ (not necessarily open or closed) is said to be locally finite if to each ${x \in X}$, there is a neighborhood ${U}$ of ${x}$ that intersects only finitely many of the ${U_{\alpha}}$

Definition. Let ${\left\{U_{\alpha}\right\} }$ be a cover of a space ${X}$. Then a cover ${\left\{V_{\beta}\right\}}$ is called a refinement if each ${V_{\beta}}$ sits inside some ${U_{\alpha}}$

Definition. A Hausdorff space is paracompact if every open covering has a locally finite refinement.

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  • $\begingroup$ You have a mistake in your first paragraph. Just because $U_\alpha$ is an open subset of $A$, doesn't make $U_\alpha$ an open subset of $X$ (imagine $X:=\mathbb{R}^3$ and $A:=\{x_3=0\}$). $\endgroup$
    – 1Rock
    Commented Feb 28, 2022 at 1:54
  • $\begingroup$ I'm not sure why you think you need to prove $\mathscr{V}$ is nonempty. $\endgroup$ Commented Feb 28, 2022 at 1:55
  • $\begingroup$ Your original idea was good. $(1)$ holds of course and is an open cover by definition. Second, $A\cap V_{\beta} \subset V_{\beta} \subset U_\alpha$. Thirds, since $\{V_\beta\}_{\beta\in J}$ is a refinement, $x\in A$ implies $x\in V_\beta$ for some $\beta$ since the refinement is a cover. Hence $x\in V_\beta\cap A$. Just utilize the subspace definition a bit more. You need an open cover of $A$ in $A$'s subspace topology and then the corresponding open cover in $X$ in the beginning of the proof. $\endgroup$
    – Doge Chan
    Commented Feb 28, 2022 at 2:00
  • $\begingroup$ @DogeChan Thank you very much. I have now edited the beginning of my attempt. It is right? $\endgroup$
    – Inquirer
    Commented Feb 28, 2022 at 2:42
  • $\begingroup$ Now it's right. You just have to use a little bit of basic set theory after defining the refinement like $A\subset B$ implies $A\cap X \subset B\cap X$ to justify it. It's pretty much a simple one line proof at most for 1,2,3 though 1 need not be proven. $\endgroup$
    – Doge Chan
    Commented Feb 28, 2022 at 17:15

2 Answers 2

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Every $a \in A$ is in some $V_\beta$ (as these form a cover of $X$). But then $a \in V_\beta \cap A \in \mathcal{V}$. These trivial facts prove two things: $\mathcal{V}$ is non-empty if $A$ is (this is a necessary condition for (1) to hold of course), and $\mathcal{V}$ is a an open cover of $A$. (so 1 and 3 hold).

$\mathcal{V}$ refines $\{U_\alpha\}_{\alpha \in I}$: let $V_\beta \cap A$ be an arbitrary non-empty member of $\mathcal{V}$. Then $V_\beta \subseteq U'_\alpha$ for some $\alpha \in I$ (it cannot sit inside $X\setminus A$ because then its intersection with $A$ is empty). So $V_\beta \cap A \subseteq U'_\alpha \cap A=U_\alpha$, and we're done.

$\mathcal{V}$ is locally finite. Let $a \in A$. Then we can find a neighbourhood $N_a$ (in $X$) of $a$ so that $N_a$ intersects at most finitely many members of $\{V_\beta\}_{\beta \in J}$. But then $N_x \cap A$ also intersects at most finitely many sets (as many or fewer than before) of the form $V_\beta \cap A$ ($\beta \in J$) and this is a neighbourhood of $a$ in $A$. So $\mathcal{V}$ is locally finite.

There's no more to it than that. Besides remarking that a subspace of a Hausdorff space is also Hausdorff, but you rightly focused on the cover aspect of the definition.

Notice the similarity to the proof that a closed set of a compact space is again compact.

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I will summarize the other comments and try to give a final answer.

We will add the extra assumption that $A$ is non-empty so (1) does not fail.

(1) Let $a\in A\subseteq X$. Since $\{V_\beta\}_{\beta\in J}$ forms an open cover of $X$, there exists $b\in J$ such that $a\in V_b$. Hence, $A\cap V_b\neq \emptyset$ and $\mathscr{V}$ contains a non-empty set.

(2) Since $\{V_b\}_{b\in J}$ is locally finite, for each $a\in A$ there is some neighborhood $N$ such that $N$ intersects only finitely many elements of $\{V_b\}_{b\in J}$. The key property of the subspace topology here is that $A\cap N$ is a neighborhood of $a$ in the subspace topology of $A$. $A\cap N$ is contained in only finitely many elements of $\mathscr{V}$ because the elements of $\mathscr{V}$ are defined by intersections of elements of $\{V_b\}$ and $A$.

Furthermore, each $V_\beta$ is contained in some $U'_\beta$. So, $A\cap V_\beta \subseteq U_\beta$. Therefore, $\mathscr{V}$ is a refinement.

(3) Rearranging the given intersection, we have

$$\bigcup_{\beta\in J}A\cap V_\beta=A\cap\ \bigcup_{\beta\in J}V_\beta\supseteq A\cap X=A.$$ (Here is a post that discusses the set identity used here)

We have shown that each open cover of $A$ has a refinement $\mathscr{V}$ (1), (2) such that $\mathscr{V}$ covers $A$ (3) and for each $a\in A$ there is a neighborhood $N$ of $a$ such that $N$ intersects only finitely many elements of $\mathscr{V}$ non-trivially (2). $A$ is a subspace of a Hausdorff space and is therefore Hausdorff. By the given definition of paracompact, $A$ is a paracompact space.

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  • $\begingroup$ Your answer is very interesting. Thank you very much. In the last step you wrote $A= A\cap U$, which means that $A \subset U$. Who are you? $\endgroup$
    – Inquirer
    Commented Feb 28, 2022 at 5:10
  • $\begingroup$ Sorry about the typo in part (3). It should have been $A=A\cap X$ which is what it says now. This is because the various $V_\beta$ form an open cover of $X$ $\endgroup$
    – elevensor
    Commented Feb 28, 2022 at 16:48

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