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, [itex]x > y[/itex]. Then, take [itex]\epsilon = 2 (x - y)[/itex]. Is the first inequality satisfied?

oleador

Quote by Dickfore

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

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

*DonAntonio*

Quote by oleador

The contrapositive is [itex]x>y \Rightarrow x>y+ε[/itex]

*** No, it is not. The contrapositive is [itex]x>y\Longrightarrow x\nleq y+\epsilon[/itex] , for some [itex]\epsilon > 0[/itex]

DonAntonio

[itex]\epsilon = 2 (x - y)[/itex] would not work:
[itex]x>y+ε \Rightarrow x>y+2 (x - y) \Rightarrow -x>-y[/itex], a contradiction unless [itex]x=y[/itex].
[itex]\epsilon = (x - y)/2[/itex] would work though.

....

oleador

True. Confused [itex]\forallε>0[x≤y+ε]\Rightarrow x≤y[/itex] with [itex]\forallε>0[x≤y+ε\Rightarrow x≤y][/itex].
The former is true.

This, however, does not change my conclusion. [itex]ε=2(x−y)[/itex] doesn't work, while [itex]ε=(x−y)/2[/itex] does.

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samsun2024	proof: if x≤y+ε for every ε>0 then x≤y Tweet 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, [itex]x > y[/itex]. Then, take [itex]\epsilon = 2 (x - y)[/itex]. Is the first inequality satisfied?

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