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Topology proof

  1. Feb 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Conjecture: If K=a union of subsets of G with K open then each subset in the union is open






    3. The attempt at a solution
    Can't really see the proof. In fact it's false as any non discrete topology have open sets which are a union of subsets whch may not be open.
     
    Last edited: Feb 18, 2008
  2. jcsd
  3. Feb 18, 2008 #2

    morphism

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    How do you expect to see the proof if you already know that the statement is false?!
     
  4. Feb 18, 2008 #3

    quasar987

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    Or consider the classic example where one takes the reunion of the non-opens sets [1/n,+infty) and get the open sets (0,+infty)
     
  5. Feb 18, 2008 #4

    JasonRox

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    You can practically create a counterexample for any topology except that of the discrete topology.
     
  6. Feb 18, 2008 #5

    JasonRox

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    Why not just take the union of 0 and (-1,1). We get the open set (-1,1) but the point 0 is closed.
     
  7. Feb 18, 2008 #6
    After I created this thread, I realised the conjecture was false.
     
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