# Point on the proof that $(C([0,1]),d_u)$ is complete.

The following is an excerpt from my notes on the example that $(C([0,1]),d_u)$ is a complete metric space.

I understand that to prove that $(C([0,1]),d_u)$ is complete, we need to show two things, namely that,

1. Every Cauchy sequence of elements in $C([0,1])$ converges to its limit in $C([0,1])$.

2. That these limits are also in $C([0,1])$, that is to say, that we need to show that the functions to which our sequence of functions converges are also continuous functions.

In moving to show the first of these points, I have the following in my notes:

"Let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in our space $C([0,1])$, so that for every $\epsilon\gt0$ there is a $N$ so that,

$$d_u(f_n,f_m)=\sup_{x\in[0,1]}|f_n(x)-f_m(x)|\lt\epsilon$$

if $m,n\ge N$.

Here now is the point that I am not too sure I understand. In my notes it then goes on to say that,

"But in that case, for every single $x\in[0,1]$, $|f_n(x)-f_m(x)|\lt\epsilon$, which means that the sequence $(f_n(x))_{n=1}^\infty$ is a Cauchy sequence in $\mathbb R$ for each $x$."

How is it that we know that this is the case for every single $x\in [0,1]$? How is it that what is written above shows this fact?

• If it's true in the worst case (supremum) it's true everywhere. Sep 26 '15 at 14:43

Well, for every $x\in [0, 1]$ you have
$$\left|f_n(x)-f_m(x)\right|\leq\sup\limits_{x\in [0, 1]}\left|f_n(x)-f_m(x)\right|$$
and that's just because the $\sup$ is the least upper bound for the $\left|f_n(x)-f_m(x)\right|$ ranging over all $x$. So if the $\sup$ is less than $\epsilon$, then so are all of the $\left|f_n(x)-f_m(x)\right|$.
It might be also helpful to note that, since $\left|f_n(x)-f_m(x)\right|$ is continuous, the $\sup$ is actually attained at some $x_0\in [0, 1]$.