# I Epsilon Confusion

1. Nov 22, 2016

### FallenApple

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. Nov 22, 2016

### FactChecker

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?

3. Nov 22, 2016

### FallenApple

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.

4. Nov 22, 2016

### FactChecker

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.

5. Nov 22, 2016

### FallenApple

Ah got it. Thanks.