TITLE: Is the space $C[0,1]$ complete?
QUESTION [23 upvotes]: In order to prove $C[0,1]$ is complete, my functional analysis book says: 

It is only necessary to show that every Cauchy sequence in $C[0,1]$ has a limit.

It goes on by supposing $\{x_n\}$ is a Cauchy sequence in $C[0,1]$. Then for each fixed $t \in [0,1]$ $|x_n(t)-x_m(t)| \le \|x_n - x_m\| \to 0$, so $\{x_n(t)\}$ is a Cauchy sequence of real numbers. I guess this inequality makes sense, since difference of function sequences would be larger than pointwise difference (at a specific $t$) of those functions.. But I did not understand how the author could conclude the sequence  $\{x_n(t)\}$ is Cauchy. 
Second question: the explanation continues, 
"Since the set of real numbers is complete, there is a real number $x(t)$ to which the squence converges; $x_n(t) \to x(t)$". 
I know set of real numbers is complete, but how can $\{x_n(t)\}$ represent the entire set of real numbers?

REPLY [7 votes]: This is simpler to prove if we split the claim up in parts. I'm not giving proofs because others here have already done that.
Theorem 1a. A uniformly Cauchy sequence of $\mathbb K$-valued bounded functions is pointwise convergent to a bounded function.
Theorem 1b. A uniformly Cauchy sequence of $\mathbb K$-valued bounded functions is uniformly convergent to its pointwise limit.
Corollary 2. $B(X \to \mathbb K)$ is a Banach space when equipped with the uniform norm.
Theorem 3. Let $X$ be a Hausdorff space, $(Y,d)$ a metric space and $f$ and $f_n$ functions from $X$ to $Y$. If $(f_n)$ converges uniformly to $f$ on $X$ (*) and if each $f_n$ is continuous at $x \in X$, then $f$ is continuous at $x$.
Corollary 4. $C_b(X \to \mathbb K)$, the set of bounded continuous functions on a Hausdorff space $X$, is a closed subspace of $B(X \to \mathbb K)$, so it too is a Banach space.
(*): i.e. for all $\varepsilon > 0$ there exists an $N \in \mathbb N$ such that $x \in X \land \mathbb N \owns n \geq N \implies d(f_n(x),f(x)) < \varepsilon$.