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)