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Thread: Interior, closure, open set in topological space

  1. #1
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    Interior, closure, open set in topological space

    $\displaystyle T=\mathbb R^2 \: A=\mathbb Q^+ \times \mathbb R^+$

    $\displaystyle \overline{A}=?$ /closure of A/
    $\displaystyle Int A=?$
    In $\displaystyle (A, \tau_A)$ is $\displaystyle A$ open? $\displaystyle (A, \tau_A)$ /topological space/

    Thank you in advance!
    Last edited by zadir; Mar 2nd 2011 at 07:22 AM.
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    Super Member girdav's Avatar
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    You can show that if you have two topological spaces $\displaystyle (X,\mathcal T_1)$ and $\displaystyle (Y,\mathcal T_2)$ and if you consider the product space $\displaystyle (X\times Y,\mathcal T\otimes \mathcal U)$ you will have for any $\displaystyle A\subset X$ and $\displaystyle B\subset Y$ that $\displaystyle \overline{A\times B}=\overline{A}\times \overline B$ and $\displaystyle \mathrm{Int}(A\times B)=\mathrm{Int}(A)\times \mathrm{Int}(B)$.
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  3. #3
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    Quote Originally Posted by girdav View Post
    You can show that if you have two topological spaces $\displaystyle (X,\mathcal T_1)$ and $\displaystyle (Y,\mathcal T_2)$ and if you consider the product space $\displaystyle (X\times Y,\mathcal T\otimes \mathcal U)$ you will have for any $\displaystyle A\subset X$ and $\displaystyle B\subset Y$ that $\displaystyle \overline{A\times B}=\overline{A}\times \overline B$ and $\displaystyle \mathrm{Int}(A\times B)=\mathrm{Int}(A)\times \mathrm{Int}(B)$.
    Thank you for your help.
    To tell the truth I don't know how to continue, how to use that in the problem.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zadir View Post
    Thank you for your help.
    To tell the truth I don't know how to continue, how to use that in the problem.
    He's saying that if you consider giving $\displaystyle \mathbb{R}^2$ the usual topology induced by the usual norm then the induced topology is the same as if $\displaystyle \mathbb{R}^2$ the product topology (when both copies of the reals are endowed with the usual topology). That said, for the product topology it's easy to prove things like $\displaystyle \text{cl}_{\mathbb{R}\times\mathbb{R}}\left(A\time s B\right)=\text{cl}_{\mathbb{R}}(A)\times\text{cl}_ {\mathbb{R}}(B)$. So now since $\displaystyle \mathbb{Q},\mathbb{R}$ are both dense in $\displaystyle \mathbb{R}$ you can conclude that $\displaystyle \mathbb{Q}\times\mathbb{R}$ is dense in $\displaystyle \mathbb{R}^2$. etc.
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