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

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

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

    Suppose that $\displaystyle \bold {a} = (a_1,a_2) \in B(\bold {x}_{o}, \varepsilon )$.
    Then, $\displaystyle d_\infty(\bold {a},\bold {x}_{0})< \varepsilon $.
    So, $\displaystyle max \{ |a_1 - x_1|, |a_2 - x_2| \} <\varepsilon $.
    $\displaystyle \Rightarrow |a_1 - x_1| < \varepsilon \ \ or \ \ |a_2 - x_2|< \varepsilon $
    $\displaystyle \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 $\displaystyle -2\varepsilon +x_1 + x_2 <a_1+a_2 \ \ or \ \ a_1+a_2 < 2\varepsilon +x_1 + x_2$

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

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

    Hence, $\displaystyle \Omega$ must be open.

    Comment
    Is the above working correct?

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

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

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

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

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

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

    Then, $\displaystyle d_\infty(\bold {a},\bold {x}_{0})< \varepsilon $.
    So, $\displaystyle max \{ |a_1 - x_1|, |a_2 - x_2| \} <\varepsilon $.
    $\displaystyle \Rightarrow |a_1 - x_1| < \varepsilon \ \ or \ \ |a_2 - x_2|< \varepsilon $
    $\displaystyle \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 $\displaystyle -2\varepsilon +x_1 + x_2 <a_1+a_2 \ \ or \ \ a_1+a_2 < 2\varepsilon +x_1 + x_2$

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

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

    Hence, $\displaystyle \Omega$ must be open.

    Comment
    Is the above working correct?

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

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

    Thanks in advance.
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