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Thread: homeomorphism proof

  1. #1
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    homeomorphism proof

    I have:

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

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

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

    I want to show that $\displaystyle X \equiv Y$


    My work:


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

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

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

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

    so $\displaystyle f$ is one-to-one and onto.

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

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

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

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

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

    and


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

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

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

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

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

    $\displaystyle f$ and $\displaystyle f^{-1}$ are continuous in the unduced topologies on both sets, so... that's it?
    Last edited by MathSucker; Mar 5th 2011 at 11:52 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MathSucker View Post
    I have:

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

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

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

    I want to show that $\displaystyle X \equiv Y$


    My work:


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

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

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

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

    so $\displaystyle f$ is one-to-one and onto.

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

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

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

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

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

    and


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

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

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

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

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

    $\displaystyle f$ and $\displaystyle f^{-1}$ 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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