1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Epsilon Confusion