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

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

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

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

 Quote by oleador The contrapositive is $x>y \Rightarrow x>y+ε$ *** No, it is not. The contrapositive is $x>y\Longrightarrow x\nleq y+\epsilon$ , for some $\epsilon > 0$ DonAntonio $\epsilon = 2 (x - y)$ 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.
....
 True. Confused $\forallε>0[x≤y+ε]\Rightarrow x≤y$ with $\forallε>0[x≤y+ε\Rightarrow x≤y]$. The former is true. This, however, does not change my conclusion. $ε=2(x−y)$ doesn't work, while $ε=(x−y)/2$ does.