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Math Help - Open sets

  1. #1
    Junior Member
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    Open sets

    I'm quite new to topology. Just wondering if anyone can help me out a bit.

    Question
    Let \Omega = \{(x,y) \in \mathbb {R}^{2}: x + y \neq 0 \} Show \Omega is an open subset of \mathbb {R}^{2}

    My Attempt
    Let \bold {x}_{o} = (x_1,x_2) \in \Omega.

    We want to show that there exists an \varepsilon > 0 such that B(\bold {x}_{o}, \varepsilon ) \subset \Omega.

    Suppose that \bold {a} = (a_1,a_2) \in B(\bold {x}_{o}, \varepsilon ).
    Then, d_\infty(\bold {a},\bold {x}_{0})< \varepsilon .
    So, max \{ |a_1 - x_1|, |a_2 - x_2| \} <\varepsilon .
    \Rightarrow |a_1 - x_1| < \varepsilon  \ \ or \ \ |a_2 - x_2|< \varepsilon
    \Rightarrow -\varepsilon + x_1 < a_1 < \varepsilon + x_1  \ \ or \ \ -\varepsilon + x_2 < a_2 < \varepsilon + x_2
    Thus, if we add those inequalities, we have -2\varepsilon +x_1 + x_2 <a_1+a_2 \ \ or \ \ a_1+a_2 < 2\varepsilon +x_1 + x_2

    Then, choose \varepsilon = min \{\frac{1}{2}(x_1 + x_2),-\frac{1}{2}(x_1 + x_2)\} for a_1 + a_2 < 0 \ \ or \ \ a_1 + a_2 > 0.

    Thus, \bold {a} \in \Omega and d_\infty(\bold {a},\bold {x}_{0}) \subset \Omega

    Hence, \Omega must be open.

    Comment
    Is the above working correct?

    Is using the infinity distance function d_\infty(\bold {a},\bold {x}_{0}) instead of the euclidean distance function, mathematically correct? Could I have used something like d_1(\bold {a},\bold {x}_{0}) instead?

    Is there anyway to simplify the working out? What is the best way choosing an epsilon?

    Thanks in advance.
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  2. #2
    MHF Contributor

    Joined
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    Quote Originally Posted by shinn View Post
    I'm quite new to topology. Just wondering if anyone can help me out a bit.

    Question
    Let \Omega = \{(x,y) \in \mathbb {R}^{2}: x + y \neq 0 \} Show \Omega is an open subset of \mathbb {R}^{2}

    My Attempt
    Let \bold {x}_{o} = (x_1,x_2) \in \Omega.

    We want to show that there exists an \varepsilon > 0 such that B(\bold {x}_{o}, \varepsilon ) \subset \Omega.

    Suppose that \bold {a} = (a_1,a_2) \in B(\bold {x}_{o}, \varepsilon ).
    No! you can't talk about such a B until you have shown that it exists!

    Better: if (x_1, y_1)\in \Omega then it is NOT on the line x+ y= 0. Therefore, the distance from (x_1, y_1) to that line is not 0. Let \epsilon be that distance and show that no point in the neighborhood of (x_1, y_1) with radius \epsilon is on that line.

    Then, d_\infty(\bold {a},\bold {x}_{0})< \varepsilon .
    So, max \{ |a_1 - x_1|, |a_2 - x_2| \} <\varepsilon .
    \Rightarrow |a_1 - x_1| < \varepsilon  \ \ or \ \ |a_2 - x_2|< \varepsilon
    \Rightarrow -\varepsilon + x_1 < a_1 < \varepsilon + x_1  \ \ or \ \ -\varepsilon + x_2 < a_2 < \varepsilon + x_2
    Thus, if we add those inequalities, we have -2\varepsilon +x_1 + x_2 <a_1+a_2 \ \ or \ \ a_1+a_2 < 2\varepsilon +x_1 + x_2

    Then, choose \varepsilon = min \{\frac{1}{2}(x_1 + x_2),-\frac{1}{2}(x_1 + x_2)\} for a_1 + a_2 < 0 \ \ or \ \ a_1 + a_2 > 0.

    Thus, \bold {a} \in \Omega and d_\infty(\bold {a},\bold {x}_{0}) \subset \Omega

    Hence, \Omega must be open.

    Comment
    Is the above working correct?

    Is using the infinity distance function d_\infty(\bold {a},\bold {x}_{0}) instead of the euclidean distance function, mathematically correct? Could I have used something like d_1(\bold {a},\bold {x}_{0}) instead?

    Is there anyway to simplify the working out? What is the best way choosing an epsilon?

    Thanks in advance.
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