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Connectedness math problem

  1. May 8, 2008 #1
    "Let S be a subset of R^n.
    If S is connected, then the interior of S is connected."

    Is this true or not?

    I can't think of a counterexample, but I don't know how to prove it either...
     
  2. jcsd
  3. May 8, 2008 #2

    Hurkyl

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    It might help to think "backwards" -- rather than thinking of starting with a set S, and then working with its interior, why not start with an open set, and then consider its closure (possibly excluding part of the boundary)?
     
  4. May 8, 2008 #3
    But that "backward" one is not equivalent to the original one.

    How can I find a counterexample?
     
  5. May 8, 2008 #4

    Hurkyl

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    True, the backward case is just a subset of the possibilities -- but I assert that it covers enough of the possibilities that it should suggest a proof or yield a counterexample.
     
  6. May 8, 2008 #5
    I think you can take something like
    an interval (a,b] or [a,b) (the same should work also for R^n).
    obviously if [a,b) is connected, then (a,b) is'nt necessarily.
     
  7. May 8, 2008 #6
    Is the claim true or false?
    I can't figure it out and I don't know which direction to push my proof towards...
     
  8. May 8, 2008 #7
    It seems true with simple examples, but we can't generalize from specifics. How can I start the proof in the general situation?
     
  9. May 8, 2008 #8

    HallsofIvy

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    ?? All of [a,b], [a,b), (a,b], (a,b) are connected!
     
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