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
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.
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.