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In my lecture, I need prove the existence of the anyone Cauchy sequence $(f_n)_{n\in \mathbb{N}}$ belonging $C(K)$ with the norm ${\Vert \cdot \Vert}_{\infty}=\sup_{x \in K}|f(x)|$ where $K$ is a compact space. How I can do it? I'm trying that below

Fixing $\varepsilon>0$, there exist any $x_0$ such that

${\Vert f_n - f_m \Vert}_{\infty} < |(f_n - f_m)x_0| + \varepsilon/2$. Then for $n,m$ such that

$|(f_n - f_m)x_0|<\varepsilon/2$. Then $(f_n)_{n\in \mathbb{N}}$ is a Cauchy sequence.

Is this correct?

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(I fixed a bit your English :-) ) –  Mariano Suárez-Alvarez Aug 17 '12 at 18:11
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I have no idea what's going on here... –  David Mitra Aug 17 '12 at 18:14
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well, it looks like you're trying to prove the completeness of $C(K)$. If that is so, observe first that for $(f_n)$ a Cauchy sequence in $C(K)$ you have for each $x \in K$ that $|f_n(x) - f_m(x)| \leq \|f_n - f_m\|_\infty$, so $(f_n(x))_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$. Put $f(x) = \lim_{n\to\infty} f_n(x)$. Prove first that $\sup_{x \in K} |f(x) - f_n(x)| \xrightarrow{n\to\infty} 0$ and deduce that $f$ is in $C(K)$. –  t.b. Aug 17 '12 at 18:20
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The question still is unclear to me. Are you trying to show the existence of some Cauchy sequence in that space? If your mother tongue is spanish you can use it here: I and several others have that language as mother tongue. Puedes escribir en español: varios aquí lo hablamos y podemos ayudar en la traducción. –  DonAntonio Aug 17 '12 at 18:38
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Related: Is the space $C\[0,1\]$ complete? –  Martin Sleziak Aug 17 '12 at 18:44

1 Answer 1

up vote 4 down vote accepted

To have an answer, I'll expand on the comments.

You claim that $C(K)$ is complete with respect to $\|\cdot\|_\infty$. What this means is that every Cauchy sequence $f_k$ has its limit in $C(K)$.

So let $f_k$ be a Cauchy sequence in $C(K)$, that is, for $\varepsilon > 0$ there is $N$ such that for $n,m > N$ you have $\|f_n - f_m \|_\infty < \varepsilon$. You want to show that there is an $f$ in $C(K)$ such that $\|f-f_k\|_\infty \to 0$. Your guess for $f$ is that $f$ is the pointwise limit of $f_k$. For this observe first that the pointwise limit exists: for every $x \in K$ fixed, $f_k (x)$ is a Cauchy sequence in $\mathbb R$, $\mathbb R $ is complete hence its limit exists.

Now that we have a candidate for the limit let's denote it by $f(x) = \lim_{n \to \infty} f_n (x)$.

To finish the proof you need to show that $f$ is in $C(K)$ that is, that $f$ is continuous, and also that $f_k$ converges to $f$ in norm.

If you show that $f_k$ converge to $f$ in norm you will get continuity for free by the uniform limit theorem. So let's show that $f_k \to f$ in $\|\cdot\|_\infty$:

Let $N$ be such that for $k \geq N$ you have $\|f_k - f_N \|_\infty < \varepsilon / 2$ (by Cauchyness of $f_k$). Then $$ |f(x) - f_k(x)| \leq |f(x) - f_N (x)| + |f_N (x) - f_k(x)|$$

The second term is $< \varepsilon / 2$. For the first term observe that $$ |f(x) - f_N (x)| = |\lim_{n \to \infty} f_n (x) - f_N(x)|$$

So $|f(x) - f_N (x)| \leq \varepsilon /2$, so that $$ |f(x) - f_k(x)| \leq |f(x) - f_N (x)| + |f_N (x) - f_k(x)| < \varepsilon$$

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Hope this helps. Perhaps one could explain the argument for the first term in a little less clumsy way. –  Matt N. Aug 17 '12 at 18:57
    
Why $|f(x) - f_k(x)| < \varepsilon$ imply that $f_k \to f$ in $\|\cdot\|_\infty$? –  juaninf Aug 17 '12 at 19:58
    
@Juan Because if you have $|f(x) - f_k(x)| \leq \varepsilon$ for all $x$ in $X$ then you also have $\sup_{x \in X} |f(x) - f_k(x)| \leq \varepsilon$. –  Matt N. Aug 17 '12 at 20:00
    
Why $|\lim_{n \to \infty} f_n (x) - f_N(x)| < \varepsilon/2$ –  juaninf Aug 17 '12 at 20:39
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@Juan Yes, that's the part I thought was clumsily written. We have picked $N$ such that for $n \geq N$ we get $|f_n (x) - f_N(x)| < \varepsilon / 2$. Then $\lim_{n \to \infty} f_n (x)$ is an index above $N$. I don't like it either how I explain it but I don't know how to be more explicit and rigorous. –  Matt N. Aug 17 '12 at 20:42

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