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

[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

 

[Dickfore]

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

 

[oleador]

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.

 

[DonAntonio]

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.

...

 

[oleador]

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.