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Math Help - proof for open set

  1. #1
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    proof for open set

    Given the set S={(x,y): |x|<1 , |y|<1} prove that is open in R^2.

    Using the definition of the open set we have:

    S is open in R^2 iff for all (x_{1},y_{1})\in S ,there exists an ε>0 such that ,for all (x_{2},y_{2}):


    (x_{2},y_{2})\in B((x_{1},y_{1})\epsilon)\Longrightarrow(x_{2},y_{2  })\in S .


    .......................................OR......... ..................................................


    if |x_{1}|<1,|y_{1}|<1 ,there exists an ε>0 such that:


    \sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_{2})^2}<\epsilon\Longrightarrow |x_{2}|<1,|y_{2}|<1

    What is the epsilon ,so that the implication can be satisfied ??

    That is as far i can go.

    Any suggestions??
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  2. #2
    Senior Member Dinkydoe's Avatar
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    Choose a point x:=(x_0,y_0)\in S

    Then the maximum \epsilon > 0 such that B^o(x, \epsilon)\subset S is:

    \epsilon_{\max} = \min\left\{|1-x_0|, |1-y_0|\right\}.

    But, if you want to show the given square S is open, you need to show that \epsilon_{\max}> 0 for all (x_0,y_0)\in S
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  3. #3
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    Quote Originally Posted by Dinkydoe View Post
    Choose a point x:=(x_0,y_0)\in S

    Then the maximum \epsilon > 0 such that B^o(x, \epsilon)\subset S is:

    \epsilon_{\max} = \min\left\{|1-x_0|, |1-y_0|\right\}.

    But, if you want to show the given square S is open, you need to show that \epsilon_{\max}> 0 for all (x_0,y_0)\in S

    So we have:


    \sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_{2})^2}<\epsilon=min{ (1-|x_{1}|),(1-|y_{1}|)} \Longrightarrow |x_{2}|<1,|y_{2}|<1.

    But how do we get that : |x_{2}|<1,|y_{2}|<1 now??
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  4. #4
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    Quote Originally Posted by alexandros View Post
    So we have:


    \sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_{2})^2}<\epsilon=min{ (1-|x_{1}|),(1-|y_{1}|)} \Longrightarrow |x_{2}|<1,|y_{2}|<1.

    But how do we get that : |x_{2}|<1,|y_{2}|<1 now??
    If \left(x_1,y_1\right)\in \mathcal{B}\left(x_0,y_0\right); \varepsilon) then \left| {x_1 } \right| \leqslant \left| {x_1  - x_0 } \right| + \left| {x_0 } \right| < \left( {1 - \left| {x_0 } \right|} \right) + \left| {x_0 } \right| = 1.
    Likewise \left| {y_1 } \right| < 1

    That proves that  \mathcal{B}\left((x_0,y_0); \varepsilon\right) \subseteq \mathcal{S}.
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