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Math Help - R^2 and R^2 - {(0,0)} are not homeomorphic

  1. #1
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    R^2 and R^2 - {(0,0)} are not homeomorphic

    Let R denote the real line and R^2 the real plane.

    How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic? I don't even know where to start since I cannot think of a topological property that is in one but not in the other. Can anyone give me a clue on what that property is?

    A similar problem is this:
    Consider S = the union of the x and y- axes together with the subspace topology inherited from R^2. Show that S and R are not homeomorphic.

    Thanks in advance!
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  2. #2
    Super Member girdav's Avatar
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    For the second problem. Let f :S\rightarrow \mathbb R an homeomorphism. We denote by g :S\setminus \left\{(0,0)\right\}\rightarrow \mathbb R\setminus \left\{f(0,0)\right\}. g is a homeomorphism. The first set has four connected components whereas the second has two.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mysteriouspoet3000 View Post
    Let R denote the real line and R^2 the real plane. How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic?

    Hint:


    \mathbb{R}^2 is contractible and \mathbb{R}^2-\{(0,0)\} it is not.
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  4. #4
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    Thanks for your replies! It really helps. :-)

    Sir fernandorevilla, I haven't encountered the concept of contractible (contractibility?) yet. Do you have any other suggestions?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mysteriouspoet3000 View Post
    Do you have any other suggestions?

    Simply connected is a topological property, \mathbb{R}^2 is simply connected but \mathbb{R}^2 minus the origin is not.
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  6. #6
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    consider the path p(t) = (cos(2πt), sin(2πt)) and the path q(t) = (cos(2πt) - 4, sin(2πt)). in \mathbb{R}^2 you can deform p to q continuously without ever leaving \mathbb{R}^2.

    why can't you do this in \mathbb{R}^2 - \{(0,0)\}?
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  7. #7
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    Quote Originally Posted by girdav View Post
    For the second problem. Let f :S\rightarrow \mathbb R an homeomorphism. We denote by g :S\setminus \left\{(0,0)\right\}\rightarrow \mathbb R\setminus \left\{f(0,0)\right\}. g is a homeomorphism. The first set has four connected components whereas the second has two.
    g is clearly bijective. But why is g and g^-1 continuous? Am I missing something obvious here?
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