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If $\Omega$ is a bounded domain, and on $C(\bar{\Omega})$ we use the uniform distance $$d(f,g)=\max_{\bar{\Omega}} |f-g|,$$ a Cauchy sequence of functions (w.r.t. the distance $d$) converge and the convergence is the uniform convergence.

On some book I can read: "with the definition of a distance the convergence of functions in a metric space was reduced to convergence of sequences of real numbers; this leads to the Cauchy criterion for uniform convergence of sequences of functions", i.e. $\{f_n\}$ converges uniformly iff for every $\epsilon>0$ we can find an $N=N(\epsilon)$ such that for every $x\in\bar{\Omega}$ and every $m,n>N$ results $|f_m-f_m|<\epsilon$.

If $d(f_n, f_m)\to0$ when $m,n\to\infty$ how can I prove that $\{f_n\}$ converges? What is the "sequence of real numbers" I need to use?

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)You use the fact that if $(f_n)$ is Cauchy in the space $C (\bar{\Omega})$ then the sequence of real numbers $(f_n(x))$ is Cauchy for all $x \in \bar{\Omega}$, and hence converges to some limit which we denote by $f(x)$. Now you should be able to use a "$3\epsilon$ argument" to show that in the supremum norm (equivalently in your metric space $(C (\bar{\Omega}),d)$ , $f_n \rightarrow f$ as $n \rightarrow \infty$.

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  • $\begingroup$ I agree there is some $f$ to wich ${f_n}$ converges pointwise. But, the wording on the book, seems to point to build a numeric sequence like $a_n=d(f_n,...)$ and to use this to prove $f_n \to f$ not only pointwise but uniformly, i.e. to use $d(f_n, f) \to 0$. There is some path like this? $\endgroup$
    – unlikely
    Mar 19 '14 at 17:18
  • $\begingroup$ To complete the proof, forget about what the book says, and have a look here: math.stackexchange.com/questions/71121/… $\endgroup$
    – Frank
    Mar 19 '14 at 18:00
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    $\begingroup$ I am sure that the 'convergence of sequences of real numbers' just refers to using the completeness of $(\mathbb{R}, | \cdot |)$ to define the function $f$. $\endgroup$
    – Frank
    Mar 19 '14 at 18:00

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