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Proving homeomorphisms?

  1. Mar 1, 2008 #1
    I'm having difficulty in proving that two spaces are homeomorphic; I understand the definition and such, but working out the details is not coming easy. For instance, our teacher asked us to prove that all convex open non-empty subsets of R^n are homeomorphic to R^n. How does one go about defining a both-ways continuous bijection in such broad terms? It just seems intractible to me, how can I make these types of problems more tractible?
  2. jcsd
  3. Mar 1, 2008 #2


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    I believe that, in general, showing two spaces are or are not homeomorphic is a very difficult problem. Of course, specific instances can be easy.

    Applying general problem solving skills -- for example, examine special cases to see if they inspire any ideas.
  4. Mar 1, 2008 #3
    Okay well, to start I figured I could just play around with the R^2 case.

    For an open disk of radius r in R^2 (translated so its center is at the origin) I found that a both-ways continuous bijection from the disk to R^2 is

    f(x,y) = (x,y) / (r-|(x,y)|)

    Now to generalize a bit, I could try doing the case where the convex region in R^2 is bounded but not necessarily a disk. First I could translate the convex region so that an interior point is at the origin; now for each ray extending outward from the origin, there is an "effective" radius of the region which is the distance from the origin to the boundary of the convex region along that ray.

    If I could prove that the "effective" radius is continuous as a function of the angle of the corresponding ray then I could get somewhere. Is it true that, for an open bounded convex surface, the "effective radius" as described before is continuous as a function of the angle to which it corresponds? My gut tells me yes but I'm having a hard time proving it.

    Anyway I really appreciate your help. Am I heading in the right direction on this problem? I
    may not be able to prove the result in its full generality but it would be nice to at least make some progress!
  5. Mar 2, 2008 #4


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    It's the same idea I had, so I hope so. :smile:
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