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Nested sequence of closed sets and convergence in a topological space.

  1. Jun 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Let ##X## be a topological space. Let ##A_1 \supseteq A_2 \supseteq A_3...## be a sequence of closed subsets of ##X##. Suppose that ##a_i \in Ai## for all ##i## and that ##a_i \rightarrow b##. Prove that ##b \in \cap A_i##.



    2. Relevant equations



    3. The attempt at a solution

    Suppose, for contradiction, that ##b \not\in \cap A_i##. Then there exists ##n \in \Bbb{N}## with ##b \not\in A_n##. But then ##b \not\in A_i## for all ##i \geq n##.

    Since ##A_n## is closed, we can pick an open neighborhood ##V## of ##b \in A_1## with ##V \cap A_n = \emptyset##. By the definition of convergence, ##V## needs to contain all ##a_i## with ##i \geq N## for some ##N##. This is not possible because only a finite number of the ##a_i## can be contained ##V##, since ##b \not\in A_i## for all ##i \geq n##.

    Is my answer correct?

    Thanks in advance
     
  2. jcsd
  3. Jun 2, 2014 #2

    pasmith

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    This is false. The hypotheses allow successive subsets to be equal, so it may be that [itex]A_n = A_1[/itex] for some [itex]n > 1[/itex].

    The way to show that [itex]b \in \bigcap A_n[/itex] is to show that [itex]b \in A_n[/itex] for each [itex]n \in \mathbb{N}[/itex].

    You know that [itex]A_n[/itex] is closed, so it contains all its limit points. Does there exist a sequence lying within [itex]A_n[/itex] whose limit is [itex]b[/itex]?
     
  4. Jun 2, 2014 #3
    If ##A_n = A_1## for any ##n>1##, then we automatically get a contradiction, because ##a_i \rightarrow b## implies that ##b## is in the closure of ##A_1##. However, ##\bar{A_1} = A_1## and ##b \not\in A_1##. A contradiction, right?
     
  5. Jun 2, 2014 #4

    pasmith

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    In order for your proof to work, you need there to exist some [itex]A_n[/itex] which is a proper subset of [itex]A_1[/itex]. If, by assumption, [itex]b \notin A_n[/itex], then it can only be because [itex]A_n[/itex] is such a proper subset, and there then exists an open neighbourhood [itex]V[/itex] of [itex]b[/itex] such that [itex]V \cap A_n = \varnothing[/itex].

    The only circumstance in which you can't do this is when [itex]A_n = A_1[/itex] for all [itex]n \in \mathbb{N}[/itex], in which case trivially [itex]b \in A_1 = \cap A_n[/itex].

    Your proof just needs to be explicit about these points.

    My other point is that there is a simpler direct proof which doesn't have this difficulty: for each [itex]n \in \mathbb{N}[/itex], the subsequence [itex]\{a_n, a_{n+1}, \dots\}\subset A_n[/itex] has limit [itex]b[/itex].
     
  6. Jun 2, 2014 #5

    jbunniii

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    Two comments:
    1. What do you mean by "an open neighborhood ##V## of ##b \in A_1##"? Are you claiming that ##b \in A_1##? On what basis? And why does your proof need this to be true anyway?
    2. If ##b \not\in A_n##, then ##b \in A_n^c##. Since ##A_n## is closed, ##A_n^c## is an open neighborhood of ##b##. You don't need to introduce a new set ##V##. If ##i \geq n##, can ##a_i## be in ##A_n^c##?
     
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