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Sequence of Compact Spaces

  1. Aug 20, 2012 #1
    I was just googling around and I came across this problem.

    Let (X,d) be a metric space.

    Let (An)n [itex]\in[/itex] N be a sequence of closed subsets of X with the property An [itex]\supseteq[/itex] An+1 for all n [itex]\in[/itex] N. Suppose it exists an m [itex]\in[/itex] N such that Am is compact. Prove that [itex]\bigcap[/itex]n[itex]\in N[/itex]An is not empty.



    I'm wondering if there is a typo here. Take some metric space. The we can set Am = ∅, and this is compact and closed, so it satisfies the conditions but the intersection is empty. What am I missing, thanks.
     
  2. jcsd
  3. Aug 20, 2012 #2

    jgens

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    The result is true with the added restriction that An ≠ ∅ for each n in N.
     
  4. Aug 20, 2012 #3
    If this were the case, then why would we need compactness?
     
  5. Aug 20, 2012 #4

    micromass

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    Just because each [itex]A_n[/itex] is nonempty, doesn't mean that the intersection is.

    For example, take [itex]A_n=[n,+\infty[[/itex], then the intersection [itex]\bigcap A_n = \emptyset
    [/itex]. We need a compactness hypothesis somewhere.
     
  6. Aug 20, 2012 #5
    Oh I see, that was stupid. Thanks!
     
  7. Aug 21, 2012 #6

    Bacle2

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    Notice that since Am is assumed compact, X is metric ( so Hausdorff ), and

    Am [itex]\supseteq[/itex] Am+1 , all of which are closed, then

    the Am+i ;i=1,2,... , can be seen as compact subspaces of

    Am. This is a standard theorem in Analysis/Topology.
     
  8. Aug 21, 2012 #7
    Yeah, I just had a momentarily lapse in brain function where I didn't realize closed and nested doesn't imply a non-empty intersection.
     
  9. Aug 22, 2012 #8

    Bacle2

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    Aah,.., welcome to the club :) .
     
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