# Proof: if x≤y+ε for every ε>0 then x≤y

1. Apr 8, 2012

### samsun2024

let x,y,ε in ℝ.
if x≤y+ε for every ε>0 then x≤y

hints: use proof by contrapositive .

i try to proof it, and end up showing that....
if x+ε≤y for every ε>0 then x≤y

2. Apr 8, 2012

### Dickfore

Suppose, $x > y$. Then, take $\epsilon = 2 (x - y)$. Is the first inequality satisfied?

3. Apr 19, 2012

The contrapositive is $x>y \Rightarrow x>y+ε$
$\epsilon = 2 (x - y)$ would not work:
$x>y+ε \Rightarrow x>y+2 (x - y) \Rightarrow -x>-y$, a contradiction unless $x=y$.
$\epsilon = (x - y)/2$ would work though.

4. Apr 19, 2012

### DonAntonio

....

5. Apr 19, 2012

True. Confused $\forallε>0[x≤y+ε]\Rightarrow x≤y$ with $\forallε>0[x≤y+ε\Rightarrow x≤y]$.
This, however, does not change my conclusion. $ε=2(x−y)$ doesn't work, while $ε=(x−y)/2$ does.