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Connected components not relatively open?

  1. Jul 14, 2007 #1

    quasar987

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    Connected components not relatively open?!?

    I've been struggling with this problem.

    In a. I showed that if A is connected and B contains A but is itself contained in the closure of A, then B is connected.

    I must now show that the connected components of a disconnected subset A of a metric space are closed relative to A (i.e. in the subspace topology of A).

    (Sorry for the misleading title!)
     
    Last edited: Jul 14, 2007
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  3. Jul 15, 2007 #2

    StatusX

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    The connected components of any set closed in that set (try looking at the closure of a connected component).
     
    Last edited: Jul 15, 2007
  4. Jul 15, 2007 #3

    HallsofIvy

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    So B contains all points of A but any points of B not in A are limit points of A? Looks like a proof by contradiction. Suppose B= X U Y where X and Y are separated. Can you show that the closures of X and Y are also in A?

    Actually connected components of a topological space are both open and closed in that space. Suppose A were not closed. Can you show that its closure is connected?
     
  5. Jul 15, 2007 #4

    quasar987

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    lol, I seriously don't know what you guys are hinting at, sorry.

    First, I typed the question in a rush and I should have mentioned that part b. begun with "Use part a. to deduce that the connected components of a disconnected subset A of a metric space are closed relative to A."

    And at the back of the book, the hint says "First establish the lemma, "If A_0 is a connected subset of A and U,V separate A, then A_0 is contained entirely in U or entirely in V."

    And actually Halls, there is a subquestion in parenthesis after that that says "Give an example of when the connected components are not open."
     
  6. Jul 15, 2007 #5

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    If the closure of a connected component is connected, then since the component is maximal, it must be equal to its closure, and so closed. And it can turn out that connected components are not open, though only when there are infinitely many of them (eg, a totally disconnected space is one in which the connected components are individual points, and there's one familiar example of such a space).
     
  7. Jul 16, 2007 #6

    quasar987

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    By "maximal" I assume you mean that there is no larger connected subset of A that contains A_0.... but who said that the closure of a connected subet of A is necessarily in A? Why couldn't there be a connected component of A such that the closure of the component is not entirely contained in A?
     
  8. Jul 16, 2007 #7

    HallsofIvy

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    Obviously, I'm missing something!:redface:
    I had thought that in a totally disconnected set, all subsets are both open and closed. Trying to think how I would prove that, it suddenly occurs to me that the union of an infinite number of closed sets is not necessarily closed!
     
    Last edited: Jul 16, 2007
  9. Jul 16, 2007 #8

    StatusX

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    I mean the closure in A. The thing you're meant to show isn't true if you replace "closed relative to A" with "closed".
     
  10. Jul 16, 2007 #9

    quasar987

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    It hit me at work today.

    "I assert that if A_0 is a connected component of A, then cl(A_0) n A = A_0 and hence A_0 is closed in A. Indeed, suppose there were accumulation points of A_0 that were in A but not in A_0. Then A_0 u {these accumulation points} would be a connected subset of A (according to part a.), and it would contain A_0, which contradicts the fact that A_0 is a component (and hence "maximal" in the sense defined in post #6)." (N.B. I never used their hint-lemma though, so there must be another way)

    As for the counter-example, consider the connected component {0} of Q. Every open set containing 0 also contains other rationals, so {0} is not open in Q.
     
    Last edited: Jul 16, 2007
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