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Proof using interiors

  1. Jul 28, 2013 #1
    A friend gave me this to prove as part of an ongoing "game." I'm having a serious amount of difficulty with it, and I don't know what I need to do.
    1. The problem statement, all variables and given/known data
    "Prove the following:
    If ##U\subset\mathbb{R}^n## is open, ##A\subset U## is homeomorphic to ##S^{n-1}##, and ##\varphi:U\to\mathbb{R}^n## is a continuous bijection from ##U## to ##\mathbb{R}^n##, then ##\varphi(\operatorname{int} A)=\operatorname{int} \varphi(A)##."


    2. Relevant [strike]equations[/strike] lemma
    My friend said that I "might" need to know that "if ##B\subset\mathbb{R}^n## is homeomorphic to ##D^n=\left\{x\in\mathbb{R}^n : |x|\leq 1\right\}##, then ##\mathbb{R}^n\setminus B## is connected."

    Here, the ##|x|## means ##\displaystyle \sqrt{\sum_{i=1}^n x_i^2}##.
    3. The attempt at a solution
    I'm ashamed to say that I don't know where to start on this one. From the lemma that I "might" need, I know that ##\mathbb{R}^n\setminus \operatorname{int} A## is connected, but I don't know what this implies or how this gets me anywhere.

    I'm really confused. I'd really appreciate any help anyone can give me. :confused:
     
  2. jcsd
  3. Jul 28, 2013 #2

    Office_Shredder

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    I'm a bit confused, A is a codimension 1 submanifold of Rn so has no interior.
     
  4. Jul 28, 2013 #3

    micromass

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    Use invariance of domain.
     
  5. Jul 29, 2013 #4
    So, by invariance of domain, we know that, since ##\operatorname{int} A## is open, then ##\varphi(\operatorname{int} A)## is also open, right? I also know that ##\operatorname{int} \varphi(A)## is open. This is at least something, but I don't know where this goes. Could you please give another hint?
     
  6. Jul 29, 2013 #5

    micromass

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    Prove that if ##\varphi## is a homeomorphism, then for each set ##B## holds that ##int \varphi(B) = \varphi(int(B))##.
     
  7. Jul 29, 2013 #6
    I'm sorry. I knew that comes next because ##\varphi## being a homeomorphism follows from invariance of domain. I should have said that I don't know what property of a homeomorphism implies that equality. I think what I'm looking for can be stated in the form "[property] is conserved under homeomorphism," but I don't know. :confused:

    Edit: Never mind. I figured it out. Invariance of domain implies homeomorphisms between subsets of ##\mathbb{R}^n## map interior points of one subset to interior points of the other. The result follows from this. Thank you for your patience. :biggrin:
     
    Last edited: Jul 29, 2013
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