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Homework Help: Topology: Nested, Compact, Connected Sets

  1. Mar 1, 2008 #1
    [SOLVED] Topology: Nested, Compact, Connected Sets

    1. Assumptions: X is a Hausdorff space. {K_n} is a family of nested, compact, nonempty, connected sets. Two parts: Show the intersection of all K_n is nonempty and connected.

    That the intersection is nonempty: I modeled my proof after the widely known analysis proof. I took a sequence (x_n) such that [tex]x_n\in K_n[/tex] for all n. Assuming x_n has a limit point x (AM I ALLOWED TO ASSUME THE SEQUENCE HAS A LIMIT POINT?), then x is in the sequential closure of K_n, which is contained in the closure of K_n, which is equal to K_n: [tex]x \in SCl(K_n) \subset Cl(K_n) = K_n[/tex] (since X is Hausdorff, all compact sets are closed). Thus [tex]x\in K_n[/tex] for all n, so it is in the intersection. Therefore the intersection is non-empty. This all hinges on the fact that I assumed there was a limit point ... am I talking in circles, or is this okay?

    That the intersection is connected: I'm guessing I should be using contradiction. So, suppose the intersection [tex]K=\bigcap^{\infty}K_n[/tex] is not connected, then there exists open sets U, V such that [tex]U\cap V=\emptyset[/tex], [tex]U\cap K\neq\emptyset[/tex], [tex]V\cap K\neq\emptyset[/tex], and [tex]K\subset U\cup V[/tex]. I also know then that [tex]U\cap K_n\neq\emptyset[/tex] for any n and likewise for V. But I don't know that there is any n for which [tex]K_n\subset U\cup V[/tex] - which would be the contradiction I am looking for, since every K_n is connected. Or is this not the right method at all?
     
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 1, 2008 #2

    morphism

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    What do you mean by "limit point"?

    For connectedness, we require our set to be in the union of U and V, not their intersection (which is empty!).
     
  4. Mar 2, 2008 #3
    Eek, you're right. That was a horrible typo.

    By "limit point," I mean any point such that any neighborhood of that point contains points of the sequence... I think.
     
  5. Mar 2, 2008 #4
    For showing that the intersection is connected, two ideas - can somebody check them?

    METHOD 1
    [tex]K_1 \cap K_2 = K_2[/tex] is connected. Then, for any [tex]n\in\mathbb{N}[/tex], we have [tex]\bigcap_{i=1}^{n} K_i = K_n[/tex] is connected. Then let [tex]n\rightarrow\infty[/tex]..? Or is that oversimplifying the problem?

    METHOD 2
    Suppose not connected. There exists open sets U, V such that (all assumptions from above). Then consider [tex]U\cup V[/tex]. There must exist n such that [tex]K_n\subset U\cup V[/tex] since {K_n} is a decreasing sequence of nested subsets. In other words, I can view the intersection as the "limit" of the sequence of intersections [tex]I_n=\bigcap_{i=1}^n K_i[/tex]. Thus, any neighborhood containing K must also contain an element of the sequence. So I take [tex]U\cup V[/tex] as my neighborhood containing K and then get my contradiction..?

    Please check these for me. Still having trouble with showing K is nonempty. Can someone please offer a hint? Thank you! :)
     
  6. Mar 2, 2008 #5
    I got that K is nonempty using the finite intersection property, so part 1 is done.

    Still wondering about part 2 (connectedness). Can someone please check the ideas I posted previously?
     
  7. Mar 2, 2008 #6
    Nevermind, got it. :)
     
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