# Homework Help: Locally Finite Collection

1. Jan 2, 2017

### Bashyboy

1. The problem statement, all variables and given/known data
I read somewhere that if $\{A_i\}$ is a collection of subsets in some topological space $X$ that is locally finite, then $\overline{\bigcup A_i} = \bigcup \overline{A}_i$, but I am having difficulty showing this.
2. Relevant equations

3. The attempt at a solution
I already know that $\bigcup \overline{A}_i \subseteq \overline{\bigcup A_i}$ holds independently of whether the collection is locally finite, so it suffices to prove the other inclusion. Let $a \in \overline{\bigcup A_i}$. Then for every open neighborhood $U_a$ containing $a$, it will intersect $\bigcup A_i$ which means that $A_i \cap U_a \neq \emptyset$ for some $i$. Since $\{A_i\}$ is locally finite, there exists an open neighborhood $U_a$ that intersects finitely many sets in the collection...

Clearly my goal is to show that $a \in \bigcup \overline{A}_i$, and this happens if and only if $a$ is the limit point of some $A_i$, or that every open of $a$ intersects "consistently" intersects some $A_i$. My problem is, I don't see how the hypothesis (that $\{A_i\}$ is a locally finite collection) is strong enough to show this happens for all open sets; I can only get that one open neighborhood intersects some of the sets in $\{A_i\}$, but that's far from showing all open neighborhoods intersect a given $A_i$. I have thought about this for some time; I could use a few hints.

Last edited: Jan 2, 2017
2. Jan 2, 2017

### andrewkirk

Proof by contradiction should do it. Let $a\in\overline{\bigcup A_i}\smallsetminus \bigcup\overline{A_i}$. What happens if you excise from an open neighbourhood $N_a$ of $a$ the closure of each set in the collection that intersects it?

3. Jan 4, 2017

### Bashyboy

Let me see if I understand what you are saying. let $S_a := \{i ~|~ A_i \cap N_a \neq \emptyset \}$. By excise, do you mean $N_a - \bigcup_{i \in S_a} \overline{A}_i$?

4. Jan 4, 2017

### Bashyboy

How is this for a proof. Suppose that $x \in \overline{\bigcup A_i}\smallsetminus \bigcup\overline{A_i}$. Then every open neighborhood of $x$ must intersect $\bigcup A_i$. Moreover, there exists an open neighborhood $N_x$ such that it intersects $A_i$ for finitely many $i$, whose indices we'll label $\alpha_1,...,\alpha_n$. Note that $x \notin A_i$ for every $i$, otherwise $x \in \bigcup \overline{A}_i$. Therefore $N_x - \bigcup_{k=1}^n A_{\alpha_k} = \bigcap_{k=1}^n (N_x - A_{\alpha_k})$ is an open neighborhood of $x$. However, this clearly does not intersect $\bigcup A_i$, which is a contradiction.

Does this sound right?

5. Jan 4, 2017

### andrewkirk

@Bashyboy Yes, that's right. Well done.

6. Jan 4, 2017

### Bashyboy

Okay, I have a related question. Let $X$ be some topological and $\{A_i\}$ a locally finite collection with each $A_i$ closed in $X$ such that $X = \bigcup A_i$. I am trying to prove that every open set has a finite cover. I believe this can be done using the above theorem. Here an attempt:

Let $U$ be open in $X$. Then $A_i \cap U \neq \emptyset$ for finitely many $i$, which we shall index as we did above. I claim that $U \subseteq \bigcup_{k =1}^n A_{\alpha_k}$. By way of contradiction, suppose that $x \in U - \bigcup_{k =1}^n A_{\alpha_k} = \cap_{k=1}^n (U-A_{\alpha_k})$, which is an open neighborhood of $x$. Since $X = \bigcup A_i = \bigcup \overline{A}_i = \overline{\bigcup A_i}$ and $x \in X$, this open neighborhood $\cap_{k=1}^n (U-A_{\alpha_k})$, must intersect $\bigcup A_i$, which is impossible since they are disjoint. Hence, $U \subseteq \bigcup_{k=1}^n A_{\alpha_k}$.

First, does this seem right? Second, is there a way of proving this without using what I proved in my previous post?

7. Jan 4, 2017

### andrewkirk

That doesn't follow. The locally finite property doesn't say that any open set has finitely many overlaps. It just says that for any point $x\in X$ there exists an open neighbourhood $U$ of $x$ with finitely many overlaps.

Also, this may be just me but I find the doubly-nested subscripts confusing and messy. It is cleaner and simpler to just let the collection be its own index set, rather than introducing a superfluous index set. For instance in the OP problem, this allows expressing it as follows.

Let $\mathscr C$ be a locally finite collection of subsets of $M$. Then
$$\overline{\bigcup_{A\in\mathscr C}A}=\bigcup_{A\in\mathscr C}\overline A$$
We can then refer to the finite set of overlapping sets as $A_1,...,A_n$ rather than having to use double subscripts.

8. Jan 4, 2017

### Bashyboy

Okay, so if we begin with the proof with an arbitrary point $x$, will the rest of the proof hold? That is, I begin by letting $x \in X$ be arbitrary, and by the local finiteness of $\mathcal{C}$, there exists an open neighborhood $U_x$ of $x$ that intersects finitely many sets in $\mathcal{C}$. Could I then use the rest of the proof to show that $U_x$ has a finite cover of sets in $\mathcal{C}$.

9. Jan 4, 2017

### andrewkirk

I'm afraid that won't work. Given $x\in X$, the locally finite property tells us that there exists an open nbd $N_x$ of $x$, which we can deduce to be covered by a finite subcollection $\mathscr C_x$ of $\mathscr C$. But, because $\mathscr C_x$ depends on $x$, that doesn't entail that any open set is covered by a finite subcollection.

In particular, for an open set $U$, the above allows us to deduce that the collection $\mathscr C_U=\bigcup_{x\in U}\mathscr C_x$ covers $U$. But unless $|U|$ is finite, we have no reason to expect that $|\mathscr C_U|$ is finite.

10. Jan 4, 2017

### Bashyboy

Ah, okay. This is actually all I need to prove a theorem I am working on. Just to be especially clear, though, we both agree that $N_x$ having a finite cover is provable, while an arbitrary open neighborhood may not necessarily have this property. Is this right?

11. Jan 4, 2017

### andrewkirk

Yes. For every point $x\in X$ there exists an open nbd $N_x$ that is covered by a finite subcollection of $\mathscr C$.