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I think the definition should be as following: for example, $f_n \in \mathcal{C}\,[0,1]$ is said to be Cauchy if for every positive real number $\varepsilon \gt 0$ there is a positive integer $N$ such that for all positive integers $m,n \gt N$, the distance $\left|f_n(x) - f_m(x)\right|\lt\varepsilon$ while $x\in[0,1]$. Is that right?

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that's the gist of it. However you should be careful and check what metric you are using for your functional space. – mathemagician Jan 16 '14 at 2:18
@mathemagician: Yes. I've made a mistake that infinity norm should be invovled – Bear and bunny Jan 16 '14 at 2:23
@Frank_W : note: $C([0,1])$ is a metric space, and the usual metric used on it is $d(f,g)=\max\{|f(x)-g(x)| \mid x \in [0,1]\}$. – Stefan Smith Jan 16 '14 at 2:24
up vote 1 down vote accepted

Recall the topology on $C([0,1])$ is that of uniform convergence. Thus the condition you're looking for is that the sequence of functions be uniformly Cauchy: given $\epsilon >0,$ there is a positive integer $N$ such that for all positive integers $m,n > N,$ one has $\sup_{x \in [0,1]} |f_m(x) - f_n(x)| < \epsilon.$

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