Question about Cauchy sequence in $C(K)$ 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?
 A: 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 
$$ |f(x) - f_k(x)| \leq |f(x) - f_N (x)| + |f_N (x) - f_k(x)| < \varepsilon$$
