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Consider $C[0,1]$ the space of continuous functions in $[0,1]$ and the norm $||f||_{\infty}=sup\{|f(t)|:0\leq t\leq 1\}$. How is proven that this space is complete?

Here is my attempt:

Let $(f_n)\subset C[0,1]$ Cauchy:

$\forall \varepsilon >0,\exists N\in \mathbb N:||f_n-f_m||_{\infty}\leq\varepsilon \quad \forall n,m\geq N$

Let $t_0\in [0,1]$:

$|f_n(t_0)-f_m(t_0)|\leq ||f_n-f_m||_{\infty}=sup\{|f_n(t)-f_m(t)|:0\leq t\leq 1\}\leq \varepsilon\quad \forall n,m\geq N$.

And since $\mathbb R$ is complete and $f_n(t_0)$ is Cauchy, $f_n(t_0)$ converges.

My doubt is that I don't see why the limit is also a continuous function.

Raphaelo
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    You have seen that your sequence $(f_n)$ converges pointwise to a function $f$. Now, you have to see that it also converges uniformly (because the convergence must be with the supremum norm). Then, you can use that the limit of a uniformly convergent sequence of continuous functions is continuous. – Eparoh Apr 04 '22 at 08:32
  • What has this question to do with [tag:hilbert-spaces]? – José Carlos Santos Apr 04 '22 at 08:34

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