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Thread: 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 $\displaystyle \epsilon>0$ then$\displaystyle \frac{1}{\epsilon}>0$. Because the positive integers are not bounded above, $\displaystyle \exists N\in\mathbb{Z}^+\left[N>\frac{1}{\epsilon}\right].$

    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 $\displaystyle \epsilon> 0$, $\displaystyle 0< \dfrac{\epsilon}{2}< \epsilon$.

    (If $\displaystyle \dfrac{\epsilon}{2}> 1$ 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 $\displaystyle \varepsilon $ > 0 there exist an x $\displaystyle \in $ S such that x < $\displaystyle \varepsilon $.

    Solution:

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

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

    x < $\displaystyle \varepsilon $

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

    $\displaystyle \Leftrightarrow $ 1 < $\displaystyle \varepsilon $

    Thus

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

    Therefore x < $\displaystyle \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 $\displaystyle \varepsilon $ > 0 there exist an x $\displaystyle \in $ S such that x < $\displaystyle \varepsilon $.

    Solution:

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

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

    x < $\displaystyle \varepsilon $

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

    $\displaystyle \Leftrightarrow $ 1 < $\displaystyle \varepsilon $

    Thus

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

    Therefore x < $\displaystyle \varepsilon $

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

    What sort of consideration should one do with $\displaystyle \dfrac{\varepsilon}{2}$ and $\displaystyle x = \dfrac{1}{2}$? Then you state that $\displaystyle x<\varepsilon$ if and only if $\displaystyle \dfrac{1}{2}<\dfrac{\varepsilon}{2}$ (which is false... suppose $\displaystyle \varepsilon = \dfrac{3}{4}$, then $\displaystyle \varepsilon > x = \dfrac{1}{2} > \dfrac{3}{8} = \dfrac{\varepsilon}{2}$) then you say that is true if and only if $\displaystyle 1<\varepsilon$ (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 $\displaystyle \varepsilon$. All you know about it is that it is greater than zero. It could be 10 million. It could be $\displaystyle 0.0000000000001$. 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 $\displaystyle \varepsilon$, but still bigger than zero.
    Last edited by SlipEternal; Oct 9th 2013 at 05:44 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 $\displaystyle x = \min \left\{\dfrac{\varepsilon}{2},\dfrac{1}{2}\right\}$. Now show $\displaystyle x \in S$ and $\displaystyle x < \varepsilon$.
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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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