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Analysis Q: Convex subsets

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Let U be a non-empty, convex, open subset of R^2. Prove that U is homeomorphic to R^2.
    Hint: First prove that the intersection of a line in R^2 with U (if non-empty) is homeomorphic to an open interval in R^1. Then use radial projections.


    2. Relevant equations

    We just have the basic definition of homeomorphism and some standard results about open/closed sets to work with. And of course the completeness axiom for R.


    3. The attempt at a solution

    Ok well, I have proven that the intersection of a line in R^2 with U (if non-empty) is homeomorphic to an open interval of R^1. I see the idea of the proof is to translate U homeomorphically so that the origin is at an interior point of U, and then radiate lines outward from the origin in all directions (think: polar coordinates). Since U is convex, the lines from the origin to the "boundary" of U are fully contained in U and comprise all of U; Since U is open, such lines to the "boundary" can be mapped to the FULL lines extending forever outward in the corresponding direction in R^2.

    So I can make a bijection from such projected "lines" in U to all of R^2; in fact for the case where U is bounded I've made an explicit bijection which I suspect is both-ways continuous but I'm having trouble proving it.

    The notation I'm using is

    R(theta) = supremum of the distances of points of the "line" in U which points in the direction of theta. (ie, R(theta) is the distance from the origin to the "boundary" of U along the direction of theta)

    I believe the following bijection sends a "line" in U in the direction of theta to the full line in R^2 in the direction of theta:

    f(r) = r/ [R(Theta)- r]

    where r is interpreted as distances along the direction of theta in U.

    If I could prove that R(theta) is a continuous function of theta then I think I would know how to proceed. It seems obviously true because a jump discontinuity in the boundary of a convex space would seem to produce a contradiction to convexity. But that's just for the bounded case!

    In any case, I am stumped, and I'm thinking there must be a better way to do this. Is there a better approach? Thanks guys.
     
    Last edited: Feb 29, 2008
  2. jcsd
  3. Mar 1, 2008 #2
    any ideas?
     
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