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Let $C[0,1]$ denote the metric space of all continuous functions $f:[0,1] \rightarrow \mathbb{R}$, with the metric

$d(f,g)=\sup |f(x) - g(x)|$ for $0\leq x \leq 1$

Show that $C[0,1]$ is a complete metric space

So far I have started with this:

Since $f$ is given to be continuous then it follows that for every sequence $(x_n)$ in $[0,1]$ which converges to every $x \in [0,1]$, then the sequence $(f(x_n))$in $\mathbb{R}$ converges to $f(x)$ for every $x\in [0,1]$ thus the sequence $(f(x_n))$ is Cauchy for every sequence $(x_n)$ in $[0,1]$ which converge to all points $x\in [0,1]$. Overall, by the definition of continuity the sequence $(f(x_n))$ is Cauchy for every $(x_n)\in [0,1]$ that converges to every $x\in[0,1]$.

And since every $(f(x_n))$ is Cauchy then $f$ is complete in $C$

I'm just wondering what I am doing wrong and what I can do to steer the proof in the right direction.

Thanks!

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  • $\begingroup$ You "have shown" that $C[0,1]$ is complete under any norm (you havent even used the fact that your given metric is very specific), which is clearly not true. $\endgroup$
    – luka5z
    Oct 6 '15 at 20:36
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Judging from your proof, you've completely misunderstood the problem. You're supposed to show that every Cauchy sequence $(f_n)_{n = 1}^\infty$ in $C[0,1]$ converges to an element $f$ of $C[0,1]$. Given a Cauchy sequence $(f_n)$ in $C[0,1]$, we have for each $x\in [0,1]$, $(f_n(x))_{n = 1}^\infty$ is Cauchy in $\Bbb R$. Since $\Bbb R$ is complete, there exists a function $f : [0,1] \to \Bbb R$ such that $\lim\limits_{n\to \infty} f_n(x) = f(x)$ for all $x\in [0,1]$. Now you need to show that $f \in C[0,1]$.

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    $\begingroup$ +1 -- "the uniform limit of a sequence of continuous functions on a compact interval is continuous". Cauchy himself didn't quite get that right :) $\endgroup$
    – BrianO
    Oct 6 '15 at 20:42
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If $\lbrace{f_n}\rbrace $is a Cauchy sequence of functions over $[0,1]$ then it converges uniformly. In this metric a sequence converges if and only if it converges uniformly (in canonical way for a sequence of functions).

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Firstly, let's write what does it mean for sequence of functions being Cauchy:

$$ \forall \varepsilon > 0 \space \exists N: \forall n, p > N \implies \sup|f_n-f_p|<\varepsilon $$

From here it follows that $(f_n)$ converges uniformly: $$ \forall x \in [0, 1] \implies |f_n(x)-f_p(x)| \le \sup |f_n-f_p| < \varepsilon $$

Now we can conclude that the limit of our Cauchy sequence $f(x)=\lim\limits_{n \to \infty} f_n(x)$ is continous as a uniform limit of continuous functions.

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