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Proof: if x≤y+ε for every ε>0 then x≤y 
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#1
Apr812, 10:30 AM

P: 5

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 


#2
Apr812, 10:34 AM

P: 3,014

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



#3
Apr1912, 09:25 PM

P: 19

[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. 


#4
Apr1912, 10:51 PM

P: 606

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



#5
Apr1912, 11:10 PM

P: 19

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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