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

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.

• 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 $A:=\{x_3=0\}$). Commented Feb 28, 2022 at 1:54
• I'm not sure why you think you need to prove $\mathscr{V}$ is nonempty. Commented Feb 28, 2022 at 1:55
• 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. Commented Feb 28, 2022 at 2:00
• @DogeChan Thank you very much. I have now edited the beginning of my attempt. It is right? Commented Feb 28, 2022 at 2:42
• 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. Commented Feb 28, 2022 at 17:15

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.

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.

• 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? Commented Feb 28, 2022 at 5:10
• 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 $X$ Commented Feb 28, 2022 at 16:48