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Math Help - prove or disprove

  1. #1
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    prove or disprove

    For any topological space (X,\tau ) and for any group A\subseteq X

    int(cl(A))=cl(int(A))


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  2. #2
    Member Focus's Avatar
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    Quote Originally Posted by sinichko View Post
    For any topological space (X,\tau ) and for any group A\subseteq X

    int(cl(A))=cl(int(A))


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    Think about reals and the rationals. What is the closure? What is the interior?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sinichko View Post
    For any topological space (X,\tau ) and for any group A\subseteq X

    int(cl(A))=cl(int(A))


    Thanks
    Quote Originally Posted by Focus View Post
    Think about reals and the rationals. What is the closure? What is the interior?
    Alternatively:

    Think about it like this. If K=\left(\overline{A}\right)^{\circ}=\overline{A^{\  circ}} then we have that \overline{K}=\overline{\overline{A^{\circ}}}=\over  line{A^{\circ}}=K and so K is closed. But, K^{\circ}=\left(\left(\overline{A}\right)^{\circ}\  right)^{\circ}=\left(\overline{A}\right)^{\circ}=K and so K is open. So, in a connected space X, K=X. But, X=\left(\overline{A}\right)^{\circ}\subseteq\overl  ine{A}\subseteq X and so A must be dense in X. So more examples would be be \mathbb{S}^1\subseteq\mathbb{R}^2, \text{SO}(n)\subseteq\text{GL}^+(n,\mathbb{R}), etc.

    The above also shows (like Focus) pointed out that your subgroup cannot have empty interior.


    Also, note the above applies equally well to non-subgroup subsets.
    Last edited by Drexel28; April 19th 2010 at 02:32 PM.
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