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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- Oct 8th 2013, 08:53 PMMathlvProving 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 - Oct 9th 2013, 03:00 AMPlatoRe: Proving espilon > x in S
- Oct 9th 2013, 04:39 AMHallsofIvyRe: 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.) - Oct 9th 2013, 04:09 PMMathlvRe: 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. - Oct 9th 2013, 04:14 PMMathlvRe: 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 - Oct 9th 2013, 05:39 PMSlipEternalRe: Proving espilon > x in S
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. - Oct 9th 2013, 07:27 PMMathlvRe: 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. - Oct 10th 2013, 06:33 AMSlipEternalRe: 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 $\displaystyle x < \varepsilon$.

- Oct 10th 2013, 07:02 PMMathlvRe: Proving espilon > x in S
Thank you for the help SlipEternal,

I think I will get it.