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Math Help - Proof-- Limits (Missed Class, please help)

  1. #1
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    Proof-- Limits (Missed Class, please help)

    Let
    x, y be elements ofR such that x not= y. Prove that there exists epsom > 0 such that y not element of[x − epsom, x + epsom]

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  2. #2
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    Quote Originally Posted by amm345 View Post
    Let
    x, y be elements ofR such that x not= y. Prove that there exists epsom > 0 such that y not element of[x − epsom, x + epsom]

    Since x\not = y, distance between x and y, denoted d(x,y)\not = 0

    Define \epsilon=d(x,y)/2

    The distance function is never negative so epsilon is greater than 0

    Then....

    can you finish it from here?
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  3. #3
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    I'm not sure, I had kinda had that based on the lecture notes he posts online, but I'm still not sure what to do next.
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    Quote Originally Posted by artvandalay11 View Post
    Since x\not = y, distance between x and y, denoted d(x,y)\not = 0

    Define \epsilon=d(x,y)/2

    The distance function is never negative so epsilon is greater than 0

    Then....

    can you finish it from here?

    Assume y\in[x-\frac{d(x,y)}{2}, x+\frac{d(x,y)}{2}]

    Then x-\frac{d(x,y)}{2}\leq y\leq x+\frac{d(x,y)}{2}

    So -\frac{d(x,y)}{2}\leq y-x\leq\frac{d(x,y)}{2}

    On the real line, d(x,y)=|x-y| so

    -\frac{1}{2}|x-y|\leq y-x\leq \frac{1}{2}|x-y|

    So -\frac{1}{2}|x-y|\leq y-x and

    y-x\leq\frac{1}{2}|x-y|

    Multiplying by -2 and 2 respectively yields

    -2(y-x)\geq |x-y| and

    2(y-x)\geq |x-y|


    These two facts imply x=y, but that is a contradiction with the given information
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