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Math Help - Proving espilon > x in S

  1. #1
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    Proving espilon > x in S

    Hello,
    Would someone please give me some clue as how I start to solve this problem: Let S = (0,1), Show that for each (epsilon) e > 0 there exist an x in S such that x < e (epsilon)
    Thanks
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  2. #2
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    Re: Proving espilon > x in S

    Quote Originally Posted by Mathlv View Post
    Would someone please give me some clue as how I start to solve this problem: Let S = (0,1), Show that for each (epsilon) e > 0 there exist an x in S such that x < e (epsilon)
    If \epsilon>0 then \frac{1}{\epsilon}>0. Because the positive integers are not bounded above, \exists N\in\mathbb{Z}^+\left[N>\frac{1}{\epsilon}\right].

    Can you finish?
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  3. #3
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    Re: Proving espilon > x in S

    I think it is sufficient to note that, for every \epsilon> 0, 0< \dfrac{\epsilon}{2}< \epsilon.

    (If \dfrac{\epsilon}{2}> 1 take x= 1/2.)
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  4. #4
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    Re: Proving espilon > x in S

    Plato,
    Thank you for helping me. But I think I will try to continue from the clue of HallsofIvy below.
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  5. #5
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    Re: Proving espilon > x in S

    HallsofIvy,
    Thank you for the help,
    Here is what I have as a solution for the problem using your helping clue.
    Please let me know if I did not answer it correctly.

    The Problem: Let S = (0,1)

    Prove for each \varepsilon > 0 there exist an x  \in S such that x < \varepsilon .

    Solution:

    For every \varepsilon > 0, 0 < \frac{\varepsilon }{2} < \varepsilon

    Consider \frac{\varepsilon }{2} and x = \frac{1}{2}
    then

    x < \varepsilon

     \Leftrightarrow \frac{1}{2} < \frac{\varepsilon }{2}

     \Leftrightarrow 1 < \varepsilon

    Thus

    \varepsilon > 0 and \varepsilon = 1 > x = \frac{1}{2}

    Therefore x < \varepsilon

    Q.E.D
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  6. #6
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    Re: Proving espilon > x in S

    Quote Originally Posted by Mathlv View Post
    HallsofIvy,
    Thank you for the help,
    Here is what I have as a solution for the problem using your helping clue.
    Please let me know if I did not answer it correctly.

    The Problem: Let S = (0,1)

    Prove for each \varepsilon > 0 there exist an x  \in S such that x < \varepsilon .

    Solution:

    For every \varepsilon > 0, 0 < \frac{\varepsilon }{2} < \varepsilon

    Consider \frac{\varepsilon }{2} and x = \frac{1}{2}
    then

    x < \varepsilon

     \Leftrightarrow \frac{1}{2} < \frac{\varepsilon }{2}

     \Leftrightarrow 1 < \varepsilon

    Thus

    \varepsilon > 0 and \varepsilon = 1 > x = \frac{1}{2}

    Therefore x < \varepsilon

    Q.E.D
    Suppose \varepsilon<\dfrac{1}{2}...

    What sort of consideration should one do with \dfrac{\varepsilon}{2} and x = \dfrac{1}{2}? Then you state that x<\varepsilon if and only if \dfrac{1}{2}<\dfrac{\varepsilon}{2} (which is false... suppose \varepsilon = \dfrac{3}{4}, then \varepsilon > x = \dfrac{1}{2} > \dfrac{3}{8} = \dfrac{\varepsilon}{2}) then you say that is true if and only if 1<\varepsilon (this if and only if may be true, but is not pertinent to what you are trying to prove).

    Here is a way to think about what you are trying to prove (hopefully it might even be a useful way of thinking about it):

    You are given a number \varepsilon. All you know about it is that it is greater than zero. It could be 10 million. It could be 0.0000000000001. You have no clue. All you know about it is that it is greater than zero. Prove that there is some number (between zero and one) that is smaller than \varepsilon, but still bigger than zero.
    Last edited by SlipEternal; October 9th 2013 at 06:44 PM.
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  7. #7
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    Re: Proving espilon > x in S

    SlipEternal,

    Thank you for the help, but still scratching my head. I will continue my thinking process. I clearly understand what the problem is asking me to do, except that I still confuse or unknown as how I come up with a logical reasoning.
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  8. #8
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    Re: Proving espilon > x in S

    Try this: Let x = \min \left\{\dfrac{\varepsilon}{2},\dfrac{1}{2}\right\}. Now show  x \in S and x < \varepsilon.
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  9. #9
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    Re: Proving espilon > x in S

    Thank you for the help SlipEternal,
    I think I will get it.
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