How to show that $C_0 (U) $ is a Banach space Let $ \emptyset \neq U \subset \mathbb{R} ^d $ be an open set.
Define $  C_0(U) :=  \left\{ f \in C(U) : \forall \epsilon > 0 \exists K \subset U , K \mbox{ compact and }\sup_{x \in U \setminus K} |f(x)| < \epsilon \right\} $.
To show is that $ (C_0 (U) , \left \lVert . \right \rVert_{\infty} ) $ is a Banach space.
I have huge struggles showing that $C_0 (U) $ is complete.
I know I need a Cauchy sequence which converges in that norm.
It would be great if someone can help me understand what to do with that set.
 A: If $f_n$ is a Cauchy sequence with respect to the uniform norm then you already know that $f_n\rightarrow f$ where $f$ is some continuous function on $U$ (because the continuous functions with the uniform norm is a Banach space). It is left to show that $f\in C_0(U)$.
Let $\varepsilon>0$, since $f_n\rightarrow f$ you have for $N$ sufficiently such that $\|f_n-f\|<\varepsilon/2$ whenever $n>N$. Fix such $n>N$, since $f_n\in C_0(U)$ there exists $K$ such that $\|f_n\|_{U\backslash K}<\varepsilon/2$
By the triangle inequality $\|f\|_{U \backslash K} \leq \|f_n-f\| +\|f_n\|_{U\backslash K} < \varepsilon $
*I use $\|f\|_{U \backslash K}$ to denote $\sup_{x\in U\backslash K} |f(x)|$.
A: It's a standard and not too difficult result that that if $f_n$ is a Cauchy sequence of continuous functions in the superemum norm, it converges with this norm to a continuous function.
So, it remains to show that the uniform limit of continuous functions small outside of sets $K_n$ is also small outside of a compact set. Indeed, let $f_n\to f$ uniformly. Then, for a given $\epsilon$, we can find an $N(\epsilon)$ large enough so that whenever $n\geq N(\epsilon)$, we have 
$$
|f_n(x)-f(x)|<\epsilon/2
$$
for any $x\in U$.
For this given epsilon, we can also find a $K_{N,\epsilon}$  compact in $U$ such that 
$$
|f_{N}(x)|<\epsilon/2
$$
for any $x\in U\setminus K_{N,\epsilon}$. So, by the triangle inequality, we show that $f$ is small outside of this set as well; let $x\in U\setminus K_{N,\epsilon}$, 
$$
||f(x)|-|f_n(x)||<\epsilon/2\implies |f(x)|<\epsilon/2+|f_n(x)|<\epsilon
$$
And the claim is shown.
