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Compact nested sequences

  1. Jul 14, 2008 #1
    Hi everyone. I feel like I'm really close to the answer on this one, but just out of reach :) I hope someone can give me some pointers

    1. The problem statement, all variables and given/known data

    Let [tex]A1 \supseteq A2 \supseteq A3 \supseteq \ldots [/tex] be a sequence of compact, nonempty subsets of a metric space [tex](X, d)[/tex]. Show that [tex] \bigcap A_n \neq \emptyset [/tex]. (Hint: Let [tex] U_n = X-A_n [/tex])

    3. The attempt at a solution
    I tried showing by contradiction.

    Suppose [tex] \bigcap A_n = \emptyset [/tex]
    Choose an open subcover [tex] U_n = X-A_n [/tex] (that's supposed to be set minus but I don't know how to do \ in tex). Then [tex] \bigcup U_n = (X-A_1) \cup (X - A_2) \ldots = X - (\bigcap A_n) = X [/tex]

    but where's the contradiction? So X is not compact, but that goes without saying and we can't infer much from that. What am I overlooking here? Or is this the wrong approach entirely?

    Thank you for your assistance.
  2. jcsd
  3. Jul 14, 2008 #2


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    X may not be compact but A1 is. The U_n are a cover of A1 if you assume the intersection of all the Ai is empty. Hence there is a finite subcover. That seems headed for a contradiction.
  4. Jul 15, 2008 #3
    Thanks for your help, Dick. I was able to get the solution.

    I have one more question on my current HW.
    Is a set [tex] A_n = [n, \infty) [/tex] open or closed in [tex] \mathbb{R} [/tex]? I would think so, but it's unbounded.
  5. Jul 15, 2008 #4


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    What do you think it would be?
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