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[Ex5] Let $X$ be paracompact. We proved a "shrinking lemma" for arbitrary indexed open coverings of $X$. Here is an "expansion lemma" for arbitrary locally finite indexed families in $X$.

Lemma. Let $\{ B_ \alpha \} _ { \alpha \in J}$ be a locally finite indexed family of subsets of the paracompact Hausdorff space $X$. Then there is a locally finite indexed family $\{ U_ \alpha \} _ { \alpha \in J}$ of open sets in $X$ such that $ B_ \alpha \subset U_ \alpha$ for each $\alpha$.

I read Brian M. Scott's answer in Expansion lemma for paracompact hausdorff space. and Showing a uniformity is complete. But I couldn't understand it because I haven't studied about barycentric refinement, star refinement, uniform spaces, completeness, and filters(also they haven't appeared in Munkres text yet). Are there any other ways to prove this?

Antonio Maria Di Mauro
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Sphere
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    You don’t need to know anything about uniform spaces, completeness, or filters to read the relevant parts of those answers, which, together with the first link in the second answer, contain all of the necessary information about barycentric and star refinements. – Brian M. Scott Nov 07 '20 at 01:47
  • I read again and understood it. Thanks. It was quite difficult though. – Sphere Nov 07 '20 at 09:17
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    I agree: it’s definitely not a simple argument. That one was handy, but there may be a simpler one. I just dug up a copy of Munkres and took a quick look to see what material he’d covered at that point. You might see whether you can prove the result by using the trick that he uses to prove $(3)\Rightarrow(4)$ in Lemma $\bf{41.3)$; I’ve not checked to be sure that it can be done this way, but it’s worth a try. – Brian M. Scott Nov 07 '20 at 18:21
  • @BrianM.Scott I've checked. It works. – Petra Axolotl Oct 08 '23 at 20:57

2 Answers2

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Here is an answer which uses only the basic definition of paracompactness and does not require a Hausdorff assumption. There are some gymnastic involved but it's not too messy.

Let $\mathcal{A}$ be a locally-finite family of closed subsets of a paracompact space $X$. Since the closures of the members of a locally-finite family themselves form a locally-finite family, there is no loss of generality in posing the additional assumption.

Notation: Write $Fin(\mathcal{A})$ for the set of finite nonempty subsets $\alpha=\{A_1,\dots,A_n\}\subset\mathcal{A}$. For a subset $U\subseteq X$ write $\mathcal{A}_U=\{A\in\mathcal{A}\mid A\cap U\neq\emptyset\}$. For $x\in X$ we write $\mathcal{A}_x=\mathcal{A}_{\{x\}}$. Since a locally-finite family is point finite, we have $\mathcal{A}_x\in Fin(\mathcal{A})$ for each $x\in X$.

Claim: There is a locally-finite family $\mathcal{U}=\{U_A\}_{A\in\mathcal{A}}$ of open subsets of $X$ such that $A\subset U_A$ for each $A\in\mathcal{A}$.

Proof: For $\alpha\in Fin(\mathcal{A})$ write $V(\alpha)=X\setminus\bigcup(\mathcal{A}\setminus\alpha)$. By the local-finiteness of $\mathcal{A}$, each $V(\alpha)$ is open. Moreover, if $x\in X$, then $x\in V(\mathcal{A}_x)$, and thus the family $\{V(\alpha)\}_{\alpha\in Fin(\mathcal{A})}$ is an open covering of $X$. Let $\mathcal{V}$ be a locally-finite open refinement of it.

Now for $A\in\mathcal{A}$ put $U_A=St(A;\mathcal{V})=\bigcup\{V\in\mathcal{V}\mid V\cap A\neq\emptyset\}$. Then $A\subseteq U_A$ and $U_A$ is open since the sets in $\mathcal{V}$ are. We claim that $\mathcal{U}=\{U_A\}_{A\in\mathcal{A}}$ is locally-finite. In showing this we will complete the proof.

So fix a point $x\in X$. Let $W\subseteq X$ be an open neighbouhood of $x$ which meets only finitely many sets in $\mathcal{V}$, say $V_1,\dots,V_n$. For each $i=1,\dots,n$ choose $\alpha_i\in Fin(\mathcal{A})$ such that $V_i\subseteq V(\alpha_i)$ and write $\alpha=\{A\in\mathcal{A}\mid W\cap U_A\neq\emptyset\}$. Then $\alpha\subset\bigcup_{i\leq n}\alpha_i$. In particular $\alpha\in Fin(\mathcal{A})$, which shows that $W$ meets only finitely many sets in $\mathcal{U}$. $\square$

Antonio Maria Di Mauro
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Tyrone
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  • Why is each $V(\alpha)$ open? For it to be open, $\bigcup(\mathcal{A}\setminus\alpha)$ should be closed, but it does not follow directly that arbitrary union of closed sets(each set in $\mathcal{A}\setminus\alpha$) is closed. – Sphere Nov 11 '20 at 06:09
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    @Sphere This is Lemma 39.1 in Munkres. This was why I assumed the sets of $\mathcal{A}$ are closed. If you prefer to work with an arbitrary collection, then replace $\bigcup(\mathcal{A}\setminus\alpha)$ with the union of the closures of these sets and the rest of proof is unchanged. – Tyrone Nov 11 '20 at 15:01
  • @Tyrone Hi, could I ask why the inclusion $$\alpha\subseteq\bigcup_{i\le n}\alpha_i$$ holds, please? Unfortunately I do not see it, sorry. – Antonio Maria Di Mauro Sep 25 '22 at 10:32
  • @AntonioMariaDiMauro can you show that if $V\cap A\neq\emptyset$ and $V_i\subseteq V(\alpha_i)$, then $A\in\alpha_i$? Think about how $V(\alpha_i)$ is defined. Let me know if that hint doesn't get you there. – Tyrone Sep 26 '22 at 16:33
  • @Tyrone So I think that $A\in\cal A$ is contained in $V(\alpha)$ for any $\alpha\in Fin(\cal A)$ if and only if $A\in\alpha$, right? Now if $A\in\alpha$ then $U_A$ is not disjoint from $W$ so that by definition of $U_A$ there must exists $i\le n$ such that $$V_i\subseteq U_A$$ and this (for the same definition of the set whose union is $U_A$) implies that $V_i$ and so $V(\alpha_i)$ is not disjoint from $A$ so that by what initially observed we conclude that $A\in\alpha_i$, right? – Antonio Maria Di Mauro Sep 27 '22 at 09:58
  • Does this argument work? – Antonio Maria Di Mauro Sep 27 '22 at 10:07
  • @AntonioMariaDiMauro Your first statement is not quite correct. However enough of it holds for the rest to go through. Observe that if $A\not\in\alpha_i$, then certainly $A\cap V(\alpha_i)=\emptyset$. As you have noted, $A\in\alpha\Leftrightarrow W\cap U_A\neq\emptyset\Leftrightarrow A\cap V_i\neq\emptyset$ for some $i=1,\dots,n$. Since $V_i\subseteq V(\alpha_i)$, if we assume that $A\cap V_i\neq\emptyset$, then we simply must have $A\in\alpha_i$. – Tyrone Sep 27 '22 at 11:53
  • @Tyrone Okay, however if the first statement is not correct I do not see why if $A$ is not disjoint from $V_i$ then $A\in\alpha_i$: so we know that $$ V(\alpha):=X\setminus\bigcup\cal (A\setminus α) $$ for any $\alpha\in Fin(\cal A)$ so that if $A\in\cal A$ is not disjoint from $V(α)$ then surely $A\in α$ because $V(α)$ cannot have elements of $\bigcup\cal A\setminus\alpha$ (right?) and so of any $A'\in\cal A\setminus\alpha$ so that in the matter of the above case we know that $A$ is not disjoint from $V(α_i)$ and thus $A\in α_i$ necessarly by the given argument. So am I wrong now? – Antonio Maria Di Mauro Sep 27 '22 at 13:49
  • More precisely we observe that $$ A'\subseteq\bigcup\mathcal A\setminus\alpha\Rightarrow X\setminus\Big(\bigcup\mathcal A'\setminus\alpha\Big)\subseteq X\setminus A'\Rightarrow A'\cap\Biggl(X\setminus\Big(\bigcup\mathcal A\setminus\alpha\Big)\Biggl)=\emptyset $$ for any $A'\in\cal A\setminus\alpha$ so that if $A$ is not disjoint from $V(\alpha)$ then necessarly $A\in\alpha$. – Antonio Maria Di Mauro Sep 27 '22 at 13:57
  • @AntonioMariaDiMauro Suppose $\mathcal{A}={A_1,A_2}$, where $A_1\subseteq A_2$. Then $V({A_1})=X\setminus A_2$ is disjoint from $A_1$. However the 'only if' part of your statement holds: if $A\cap V(\alpha)\neq\emptyset$, then $A\in\alpha$. This is all you need to complete. – Tyrone Sep 28 '22 at 14:03
  • @Tyrone Okay, so the correct statement would be $$\text{ }$$"if $A\in\cal A$ is not disjoint form $V(\alpha)$ then $A\in\alpha$ because $V(\alpha)$ is disjoint form $\bigcup(\cal A\setminus\alpha)$ so that the elements of $\alpha$ are the unique element of $\cal A$ from which $V(\alpha)$ could be not disjoint" $$\text{ }$$ right? – Antonio Maria Di Mauro Sep 28 '22 at 15:22
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As @Brian M. Scott noted in comment, Lemma 41.3 $(3)\implies(4)$ part used a specific trick, which can be applied directly here. The only differences, in Munkres' notation, are that:

  1. you need to take closure of elements in the refinement $\mathcal C$ to ensure closedness.
  2. you don't need $F(B)$ to make sure $\mathcal D$ is a refinement. The open sets $E(B_\alpha)$ will suffice.

Anything else will go through. Note that I just made a quick review, if I mistaken anything plz correct me.

Westlifer
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