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Homework Help: Classical Analysis I, difficult inequality

  1. Jan 30, 2008 #1
    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:

    1. [tex]y<x\leq y+\epsilon[/tex] or [tex]\epsilon>0[/tex] (this is okay from the assumptions in the problem)
    2. [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])
    3. [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!
  2. jcsd
  3. Jan 30, 2008 #2
    If x > y then x - y > 0. You're quite free to choose epsilon in any manner you like.
  4. Jan 30, 2008 #3
    Okay then, I can choose [tex]0<\epsilon<x-y[/tex] which implies [tex]y+\epsilon<x[/tex], a contradiction. But why is it that I can choose epsilon that way? I mean, am I not restricting epsilon to those such that that inequality holds? This goes against the fact that it is supposed to hold for any epsilon>0, right?
  5. Jan 30, 2008 #4


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    If x<=y, then you are done. So assume x>y. Pick epsilon=(x-y)/2. Now what?
  6. Jan 31, 2008 #5

    D H

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    Which is exactly why it is a contradiction. You are misinterpreting all to mean any. All means all, not just a select few. If some constraint is supposed to hold for all epsilon>0 but you can find even just one epsilon>0 then the constraint does not hold for all values. Contradiction.
  7. Jan 31, 2008 #6


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    The contradiction of the statement "for all p, q is true" is "for some p, q is not true".
  8. Jan 31, 2008 #7
    Ah, okay! That makes more sense now. The difference between all, any, and some was a distinction that I hadn't quite made. Thanks for your help everybody!
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