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Math Help - homeomorphism proof

  1. #1
    Junior Member
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    homeomorphism proof

    I have:

    X= \{ (x,y) \in \Re^2 : 0 < x^2+y^2<1 \}

    Y= \{ (x,y) \in \Re^2 : 1 < x^2+y^2<2 \}

    both have their usual topologies as subsbaces of \Re^2

    I want to show that X \equiv Y


    My work:


    Let S= \{ x:2<x<4 \}

    Let f map \{x \in \Re^2 : 0 < ||x|| < 1 \} to \{x \in \Re^2: 1< ||x||<2 \} by:

    f(x) = (1 + \frac{1}{||x||})x

    Now, =(1 + \frac{1}{||y||})y

    so f is one-to-one and onto.

    f(f^{-1}(y))=(1+ \frac{1}{||f^{-1}(y)||})f^{-1}(y)

    =(1+\frac{1}{||f^{-1}(y)||})(1- \frac{1}{||y||})y

    = ( 1 + \frac{1}{||(1-\frac{1}{||y||})y||})(1-\frac{1}{||y||})y

    = ( 1 + \frac{1}{(1-\frac{1}{||y||})||y||})(1-\frac{1}{||y||})y

    = \frac{1}{1-\frac{1}{||y||}}( 1 + \frac{1}{||y||}-\frac{1}{||y||})(1-\frac{1}{||y||})y=y

    and


    f^{-1}(f(x))=(1- \frac{1}{||f(x)||})f^(x)

    =(1-\frac{1}{||f(x)||})(1+ \frac{1}{||x||})x

    = ( 1 - \frac{1}{||(1+\frac{1}{||x||})x||})(1+\frac{1}{||x  ||})x

    = ( 1 - \frac{1}{(1+\frac{1}{||x||})||x||})(1+\frac{1}{||x  ||})x

    = \frac{1}{1-\frac{1}{||x||}}( 1 + \frac{1}{||x||}-\frac{1}{||x||})(1+\frac{1}{||x||})x=x

    f and f^{-1} are continuous in the unduced topologies on both sets, so... that's it?
    Last edited by MathSucker; March 5th 2011 at 12:52 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MathSucker View Post
    I have:

    X= \{ (x,y) \in \Re^2 : 0 < x^2+y^2<1 \}

    Y= \{ (x,y) \in \Re^2 : 1 < x^2+y^2<2 \}

    both have their usual topologies as subsbaces of \Re^2

    I want to show that X \equiv Y


    My work:


    Let S= \{ x:2<x<4 \}

    Let f map \{x \in \Re^2 : 0 < ||x|| < 1 \} to \{x \in \Re^2: 1< ||x||<2 \} by:

    f(x) = (1 + \frac{1}{||x||})x

    Now, =(1 + \frac{1}{||y||})y

    so f is one-to-one and onto.

    f(f^{-1}(y))=(1+ \frac{1}{||f^{-1}(y)||})f^{-1}(y)

    =(1+\frac{1}{||f^{-1}(y)||})(1- \frac{1}{||y||})y

    = ( 1 + \frac{1}{||(1-\frac{1}{||y||})y||})(1-\frac{1}{||y||})y

    = ( 1 + \frac{1}{(1-\frac{1}{||y||})||y||})(1-\frac{1}{||y||})y

    = \frac{1}{1-\frac{1}{||y||}}( 1 + \frac{1}{||y||}-\frac{1}{||y||})(1-\frac{1}{||y||})y=y

    and


    f^{-1}(f(x))=(1- \frac{1}{||f(x)||})f^(x)

    =(1-\frac{1}{||f(x)||})(1+ \frac{1}{||x||})x

    = ( 1 - \frac{1}{||(1+\frac{1}{||x||})x||})(1+\frac{1}{||x  ||})x

    = ( 1 - \frac{1}{(1+\frac{1}{||x||})||x||})(1+\frac{1}{||x  ||})x

    = \frac{1}{1-\frac{1}{||x||}}( 1 + \frac{1}{||x||}-\frac{1}{||x||})(1+\frac{1}{||x||})x=x

    f and f^{-1} are continuous in the unduced topologies on both sets, so... that's it?
    I didn't check it 100 %, but I looked at most and it looked correct. Good job!
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