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Topology question; examples of non-homeomorphic metric spaces

  1. Oct 8, 2006 #1
    Hello,

    Here's a problem that I'm having trouble with:

    Give an example of metric spaces X and Y and continuous maps f: X->Y and g: Y->X such that f and g are both bijective but X and Y are not homeomorphic.

    I can find plenty of examples where I can find one such function, but finding the second is always a problem.
     
  2. jcsd
  3. Oct 8, 2006 #2

    HallsofIvy

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    Let X be the real numbers with the "usual" topology: d(x,y)= |x-y|.

    Let Y be the real numbers with the "discrete" topology: d(x,y)= 0 if x= y, 1 otherwise.

    Let f(x)= x.
     
  4. Oct 8, 2006 #3
    HallsofIvy,

    Thanks for the reply, but that won't work with this definition of continuity:

    f:A->B is continuous if
    f^(-1)(U) is open in A for every open set U of B.
     
  5. Oct 8, 2006 #4

    matt grime

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    I suggest you re-examine the definition of homeomorphic. Given any space X and two topologies S and T, (X,S) and (X,T) are never homeomorphic.
     
  6. Oct 8, 2006 #5
    Matt,

    I understand that. I was not refuting the fact that those two topoligies are not homeomorphic. Finding such examples of non-homeomorphic spaces is easy, but finding the functions satisfying the conditions is the hard part.
     
  7. Oct 8, 2006 #6

    StatusX

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    You can form a topology on R from the basis of half open intervals whose endpoints are integers. Can you think of anything to do with this? EDIT: I don't know if this will work- in the example I had in mind it turns out the spaces are homeomorphic.
     
    Last edited: Oct 8, 2006
  8. Oct 8, 2006 #7

    matt grime

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    Did you stop to think about which topology we take as that on the domain and the range? The map f(x)=x is a bijection from R to R, with obvious bijective inverse. Give one copy of R the norm topology, the other the discrete. Done.

    If you've found one bijective function, then it is invertible, as a function. That is either the definition of bijective, or a trivial consequence of the definition of bijective.
     
    Last edited: Oct 8, 2006
  9. Oct 8, 2006 #8

    AKG

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    I don't think HallsofIvy's example will work. For every y in Y, {y} is open. For any bijection from X to Y, the inverse image of {y} is a one-point set, which is not open in X, so there are no continuous bijections from X to Y.
     
  10. Oct 8, 2006 #9

    matt grime

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    Ah, see the question properly now.

    Remember that any continuous bijection from a compact space to a hausdorff space is a homeomorphism (or perhaps I mean from a hausdorff space to a compact space), thus one will need to find and use a non-hausdorff space, or a non-compact space.
     
  11. Oct 8, 2006 #10
    Matt,

    Sorry for the confusion. I was a little confused by your response so I didn't reply, I was trying to see if I was understanding you correctly. AKG expressed what I meant to say.

    You're right about the Hausdorff/compactness conditions... we are stuck with finding both X and Y being noncompact spaces, since they must also be metric spaces (thus Hausdorff). I still have problems finding such an example.
     
  12. Oct 8, 2006 #11
    Let X = [0,1) with the normal metric on R

    Let Y be the uinit square, ie: [itex]Y = \{(x_1,x_2): x_1^2 + x_2^2 = 1\}[/itex], with the normal metric on R^2

    Let [itex] f:X \rightarrow Y, \ \ \ \theta \mapsto (cos2 \pi \theta, sin2 \pi \theta) [/itex].

    Take U = [0, 1/2). U is open in X but f(U) is not open in Y.
     
  13. Oct 8, 2006 #12

    AKG

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    nocturnal, do you understand the stated problem? You've found a continuous bijection from X to Y and shown that it isn't a homeomorphism. This is almost entirely pointless. You don't need to show that one particular continuous bijection from X to Y isn't a homeomorphism, you have to show that no continuous bijection from X to Y is a homeomorphism. Although you haven't done this, it can be done. More importantly, you have to find a continuous bijection from Y to X as well. This cannot be done.
     
  14. Oct 9, 2006 #13
    oops, I misread the question.
     
  15. May 25, 2007 #14
    do you really think that problem is true,
    I thnk it is wrong, it is a different defination of homemorphizm
     
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