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Hausdorff spaces

  1. Nov 18, 2005 #1
    Hi all, here's the problem I'm working on.

    Suppose (X,T) is a Hausdorff space and that f:(X,T) -> (Y,T') is a bijection such that f inverse is continuous.
    1) Prove that (Y,T') is Hausdorff.
    2) Suppose that (Y,T') is a Hausdorff space instead of (X,T). With the rest of the statement as given above, can we then prove that (X,T) is Hausdorff?

    For 1), I could just show two distinct points in (X,T) such that there exists open sets U and V where x1 is in U, x2 is in V, where U intersect V = 0. From there, I map those distinct points onto Y, which will give unique f(x1) and f(x2) due to the bijection condition. Hence, I then map the U and V in a similar fashion which will lead to f(U) intersect f(V) = 0. It seems correct but I'm still not sure with this proof.

    For 2), I don't think there's any difference in the statement if (Y,T') is Hausdorff instead of (X,T) but constructing a proof for the statement seems difficult...
     
  2. jcsd
  3. Nov 18, 2005 #2

    matt grime

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    f inverse is continuous thus f sends open sets to open sets, but f is not continuous so the situation is definitely not symmetric in the arguments so there IS a difference between 1 and 2.

    your proof of 1 is wrong. Indeed you suppose that the hausdorff condition is true in X to produce U and V. You can't do that.
     
  4. Nov 18, 2005 #3
    I believe that 2) is wrong. Hint: Think of the discrete topology on X to be (Y, T') with f=id.
     
  5. Nov 18, 2005 #4

    StatusX

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    To prove 1 you need to show that given any two points in X, there exist open sets U and V each containing one of these points and disjoint from each other. That is, pick two arbitrary points, and then construct the corresponding sets.
     
  6. Nov 18, 2005 #5
    You mean in Y.
     
  7. Nov 18, 2005 #6
    Thanks for the help.

    Yeah, I pretty much got the proof down but I was confused about the second part. I can't assume f is continuous judging from f inverse being continuous, so there's no way to prove that (X,T) is Hausdorff.
     
  8. Nov 18, 2005 #7

    matt grime

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    You must give an example to demonstrate that it fails (it might n ot fail, there might be another way to prove it.. there isn't, as it is false, and a counter example exists, but just because one proof fails doesn't mean all proofs fail).

    If by 'got a proof down' for the first you meant the one you gave then it is wrong, as has been explained.
     
  9. Nov 18, 2005 #8

    StatusX

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    Yea, sorry about that. I'm used to the more reasonable way of doing this, by showing that if there's a continuous function from X to Y, then X is Hausdorff if Y is.
     
  10. Nov 18, 2005 #9
    So am I, to be honest. For some reason this exercise seems to be 'backwards'.:smile:

    Edit: That's as long as the function is 1:1, of course.
     
    Last edited: Nov 18, 2005
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