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If sequence of functions {$f_n$} converges uniformly, then {$f_n$} is a cauchy sequence.
That is, it satisfies $|f_n(x)-f_m(x)| \le \epsilon$.

Then if {$f_n$} is a cauchy sequence, $|f_n(x)-f_m(x)| \le \epsilon$,
this sequence of functions {$f_n$} converges uniformly?
Or only we can conclude is that it converges?

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It would be better to write a whole question.
In Rudin's book, theorem 7.8 says
The sequence of functions {$f_n$} defined on E converges uniformly on E if and only if for every $\epsilon>0$ there exists an integer N such that $N \le m,n$ , x belongs to E implies
$$|f_n(x)-f_m(x)| \le \epsilon$$.

My question is that: we have $|A_n-A_m| \le \epsilon$ for $N \le m,n$, so that {$A_n$} is a cauchy sequence converges to A. Therefore $|A_n-A| < 3/\epsilon$.
In here, that inequality is from convergence or uniform convergence?
I wonder whether I can use Theorem 7.8 in here or not. (because it says if and only if)

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    How do you define "Cauchy sequence" (of functions)? If you consider it pointwise (i.e. that for *each* $x$, $\{f_n(x)\}$ is Cauchy) I don't think the result will follow. A possible counterexample over some domain might be $f_n(x)=x^n$.2012-11-30

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It depends on what you mean by saying "Cauchy." There are uniformly Cauchy sequences (which will converge uniformly) and pointiwse Cauchy sequences (which are only guaranteed to converge pointwise). Both are described in the linked article.

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    In Rudin's book, theorem 7.8 says "{$f_n$} converges uniformly iff {$f_n$} is a cauchy sequence". Then, in here, I can say that it converges uniformly by this theorem?2012-11-30
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    Well, to be precise though, it says The sequence of functions $\{f_n\}$, defined on $E$ converges uniformly on $E$ if and only if for every $\epsilon>0$ there exists $N$ such that $m,n\geq N$ **and $x\in E$** implies $\vert f_n(x)-f_m(x)\vert\leq\epsilon$. This is a uniform statement - it is saying $\{f_n\}$ converges uniformly iff $\{f_n\}$ is *uniformly* Cauchy.2012-11-30
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    Oh, I see. I added my whole question, but I think your comment give me a clue. Maybe in here, I can't say "{$A_n$} converges uniformly" carelessly. So that inequality $|A_n-A|<3/\epsilon$ comes from convergence not uniformly convergence...2012-11-30
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    I'm not sure what your sequence $A_n$ is, but yes - you have to be very careful about claiming uniform convergence.2012-11-30
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    Actually, it is a part of proof theorem 7.11 in Rudin. If you have a book you can check.2012-11-30