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Locally Finite Collection

  1. Jan 2, 2017 #1
    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
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  3. Jan 2, 2017 #2

    andrewkirk

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    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?
     
  4. Jan 4, 2017 #3
    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##?
     
  5. Jan 4, 2017 #4
    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?
     
  6. Jan 4, 2017 #5

    andrewkirk

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    @Bashyboy Yes, that's right. Well done.
     
  7. Jan 4, 2017 #6
    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?
     
  8. Jan 4, 2017 #7

    andrewkirk

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    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.
     
  9. Jan 4, 2017 #8
    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}##.
     
  10. Jan 4, 2017 #9

    andrewkirk

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    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.
     
  11. Jan 4, 2017 #10
    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?
     
  12. Jan 4, 2017 #11

    andrewkirk

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