1. The problem statement, all variables and given/known data Let [tex]x,y\in R[/tex] such that [tex]x\leq y+\epsilon[/tex] for all [tex]\epsilon>0[/tex]. Prove that [tex]x\leq y[/tex]. 2. Relevant equations None other than what's given in the question. 3. The attempt at a solution It's obvious to me that it's true because I can pick some numbers [tex]x>y[/tex] so that the original inequality is not true for all [tex]\epsilon[/tex]. I've tried proof by contradiction, but it doesn't leave to any contradictions. For example, assuming that [tex]x>y[/tex], then I've come up with the following things: [tex]y<x\leq y+\epsilon[/tex] or [tex]\epsilon>0[/tex] (this is okay from the assumptions in the problem) [tex]x+\epsilon>y+\epsilon\geq x[/tex] since [tex]x>y[/tex] and [tex]x\leq y+\epsilon[/tex] (this is okay since [tex]\epsilon>0[/tex]) [tex]y-\epsilon<x-\epsilon\leq y[/tex] or [tex]y-\epsilon<y[/tex] (again okay from the assumptions of the problem). I've also tried to write it in terms of quantifiers, negate it and prove the contrapositive, but I'm not at all sure what I'm doing since I've never formally covered it. I've come up with the following as an equivalent statement for the problem. [tex]\forall x,y,\epsilon\in R, \epsilon>0 : x\leq y+\epsilon \Rightarrow x\leq y[/tex] My attempt to negate it (which is probably wrong) is [tex]\exists x,y,\epsilon\in R, \epsilon>0 : x\geq y+\epsilon\Rightarrow x>y[/tex]. Can anybody give me any help and/or let me know if I'm doing anything right? Thanks in advance!