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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 :-) )2012-08-17
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    Do you have to show that any Cauchy Sequence converges in that space with that norm? Please correct the formulation of the question, since in it's current shape it is incomprehensible.2012-08-17
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    (Notice that all your questions have titles which begin with «Question abput...»: there is no need to include that, as it is assumed that what you are writing is a question!)2012-08-17
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    Are you trying to prove that there *is* some Cauchy sequence? A constant sequence is already Cauchy...2012-08-17
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    I have no idea what's going on here...2012-08-17
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    sorry I write again the question2012-08-17
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    Are you supposed to show that every Cauchy sequence in $C(K)$ is convergent?2012-08-17
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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 $C(K)$.2012-08-17
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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.2012-08-17
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    @t.b. but if $f_n$ is not a sequence Cauchy then $|f_n(x) - f_m(x)| \leq \|f_n - f_m\|_\infty$ yet is true, Why I have taking $f_n$ as Cauchy sequence?2012-08-17
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    Of course it is still true, but the conclusion "so $(f_n(x))_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$." no longer holds.2012-08-17
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    Related: [Is the space $C\[0,1\]$ complete?](http://math.stackexchange.com/questions/97356/is-the-space-c0-1-complete)2012-08-17
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    @t.b. Why ...?,2012-08-17
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    @t.b. is $\sup_{x \in K} |f(x) - f_n(x)| \xrightarrow{n\to\infty} 0$ or $\sup_{x \in K} |f - f_n| \xrightarrow{n\to\infty} 0$?2012-08-17
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    @Matt the norm is ${\Vert \cdot \Vert}_{\infty}=\sup_{x \in K}|f(x)|$, i don't understand why you write the norm $\|\cdot\|_\infty$ is independent of $x$.2012-08-17
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    @Juan I meant that $\|f\|_\infty$ is an expression independent of $x$ and meant to say that that is the reason why we write $\|f\|_\infty$ instead of $\|f(x)\|_\infty$. But then I thought my comment could be more confusing than helpful and also that I was interrupting your conversation with tb so I deleted my comment.2012-08-17

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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.2012-08-17
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    Why $|f(x) - f_k(x)| < \varepsilon$ imply that $f_k \to f$ in $\|\cdot\|_\infty$?2012-08-17
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    @Juan Because if you have $|f(x) - f_k(x)| \leq \varepsilon$ for all $x$ in $X$ then you also have $\sup_{x \in X} |f(x) - f_k(x)| \leq \varepsilon$.2012-08-17
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    Why $|\lim_{n \to \infty} f_n (x) - f_N(x)| < \varepsilon/2$2012-08-17
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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.2012-08-17
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    @Juan Or is your question why it's $\varepsilon/2$ and not $\varepsilon$?2012-08-17
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    @Juan Or perhaps my mistake was confusing you? I corrected it now.2012-08-17