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
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.
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.