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As the title, and $C[a,b]$ is the linear space of continuous real-valued functions on $[a,b]$

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closed as off-topic by Najib Idrissi, Fly by Night, M Turgeon, Tomás, Brad Jul 30 at 15:39

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Again, you should post a little motivation, label it homework since it appears to be homework and write down what work you have done, if any. –  Holdsworth88 Dec 2 '12 at 9:11
    
Please also take Holdsworth88 other advice: Post a little motivation and writhe down the work you have done already. If you didn't do any, do so before asking here. –  Julian Kuelshammer Dec 2 '12 at 9:28

1 Answer 1

Let $f_n\in \mathscr{C}[a,b]$ be a Cauchy sequence and $\epsilon>0$. Then, \begin{equation}\exists N\in \mathbb{N}:\forall x\in [a,b]\; n,m\ge N\Rightarrow \left| f_n(x)-f_m(x) \right|<\frac{\epsilon}{2} \end{equation} Fix $x\in [a,b]$. Then, $(f_n(x))$ is a Cauchy sequence of real numbers and thus converges to a real number $f(x)$. We can thus create a function $f:[a,b]\to \mathbb{R}$ that is the pointwise limit of $f_n$. $f\in \mathscr{C}[a,b]$ (why?) and \begin{equation}\forall x\in X\; n,m\ge N\Rightarrow \left| f_n(x)-f_m(x) \right|<\frac{\epsilon}{2} \Rightarrow \left|f_n(x)-f(x)\right|=\lim_{m\to \infty}\left| f_n(x)-f_m(x) \right|\le\frac{\epsilon}{2}<\epsilon\end{equation} and thus, $f_n$ converges uniformly (why?). I advise the OP to try explain the 2 "why" in my answer

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