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Compactness under homeomorphisms

  1. Oct 4, 2007 #1
    1. The problem statement, all variables and given/known data
    I want to show that homeomorphism preserve compactness on a set or a space. The definition of a homeomorphism is a continuous function with a continuous inverse.
    The definition of a continuous function is a function such that the pre-image of an open set is open.

    Let f: X to Y be continuous. Let X be compact
    So, if you have an open cover in X then you have a finite subcover. But if you have an open cover in Y, then you could map it to X but how would you know that it is still an open cover in X?



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 4, 2007 #2

    morphism

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    Suppose f:X->Y is a homeomorphism. If {U_i} is an open cover of Y, then {f^-1(U_i)} is an open cover of X.
     
  4. Oct 4, 2007 #3
    That is exactly the statement form of my question. Could you please explain why that is true?
     
  5. Oct 4, 2007 #4

    morphism

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    We know that f is continuous and that each U_i is open. It follows that each f^-1(U_i) is open.

    If Y = [itex]\cup[/itex] U_i, can you guess what [itex]\cup[/itex] f^-1(U_i) should be? Can you prove it?
     
  6. Oct 4, 2007 #5
    Say, that there an x in X s.t. x is not in [itex]\cup[/itex] f^-1(U_i). Then take a nbhd of x N_x. f(N_x) must be in [itex]\cup[/itex] U_i.

    OK. So, it comes down to whether the function is injective. Can we assumed it is? That is not part of the definition, is it?
     
  7. Oct 4, 2007 #6

    morphism

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    A homeomorphism is a bijection, so yes it's injective... But what does that have to do with anything?

    Look, Y = [itex]\cup[/itex] U_i. So X = f^-1(Y) = f^-1([itex]\cup[/itex] U_i) = ...
     
    Last edited: Oct 4, 2007
  8. Oct 4, 2007 #7
    OK. I guess the problem is more fundamental. I do not understand why a homeomorphism is a bijection. How would one show from the definition I wrote in the first post that it is surjective? That it is injective?
     
  9. Oct 4, 2007 #8

    morphism

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    If it has an inverse it must be a bijection. :smile:

    By the way, the f^-1's I used above were used to indicate taking the preimage.
     
  10. Oct 4, 2007 #9
    I see. Thanks.
     
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