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I Epsilon Confusion

  1. Nov 22, 2016 #1
    Ok, I'm reviewing my real analysis and found a point of confusion.

    So for a sequence S(n) to converge to S, for every e>0 there exists N such that n>N implies |S(n)-S|<e.

    Ok that makes sense. However when I backsolve and find N as a function of epsilon, and then formally proving forwards,

    I'm supposed to use the statement, Given e>0, let N(e). Then for any n>N(e), we have |S(n)-S|<e

    Basically, I found N as a function of a specific epsilon, and that is somehow accounting for all of the epsilions? Because the original statement is for every epsilon>0. I can only show that it works for one of them.

    How to reconcile this?

    Is it because once N is a function of e, then for fixed e, n>N(e) wouldn't work anymore? I must say that for any n>N(e) since now n must vary based off of e?

    So basically its an application of this: for every A there exists a B <=> for every B there exists an A.
    Last edited: Nov 22, 2016
  2. jcsd
  3. Nov 22, 2016 #2


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    When you back-solved to find N as a function of ε, did you do it for a particular value of ε or for any ε > 0?
  4. Nov 22, 2016 #3
    Well it was for any I suppose. I mean if I want to find convergence of 1/n, setting N=1/e will solve the problem. And e is not a particular numeric value.

    But by back solving, you have to logically assume that e is fixed. Because it said for "any" e there exists an N such that n>N....

    So I pick epsilon= e. Then I can find an N(e). such that n>N(e).

    Going forward, I can only say for e>0, where N=N(e). So its for this particular e.
  5. Nov 22, 2016 #4


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    You can think of this as "for any fixed e, there exists ...". I say that, because in the rest of that section of the proof, you assume that e represents the same number everywhere it appears. The most important word is "any".
    The wording of the proof may be a little ambiguous, but the logic is sound and the wording is an accepted standard.
  6. Nov 22, 2016 #5
    Ah got it. Thanks.
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