- #1

FallenApple

- 566

- 61

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|<eBasically, 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

So basically its an application of this: for every A there exists a B <=> for every B there exists an A.

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|<eBasically, 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.

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