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.