# Test

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• Oct 9th 2013, 04:04 PM
Mathlv
Test
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
• Oct 9th 2013, 05:47 PM
SlipEternal
Re: Test
See my post in the other forum.