Space of continuous functions with compact support dense in space of continuous functions vanishing at infinity How can we prove that the space of continuous functions with compact support is dense in the space of continuous functions that vanish at infinity?
Presumably your norm is the maximum absolute distance at any point. If I give you a continuous $f$ that vanishes at $\infty$ and an $\epsilon \gt 0$, you need to find a continuous $g$ with compact support that is everywhere within $\epsilon$ of $f$. Since $f$ vanishes at $\infty$, there is some $N$ such that $f(x) \lt \epsilon/2$ whenever $ x \gt N$. So if $g=f$ out to $N$, drops quickly to $0$ and stays there...

Added:  We are trying to show that any continuous function that vanishes at infinity can be approximated closely (within $\epsilon$) by a function with compact support.  The range where $f$ is large (greater that $\epsilon/2$) is some interval $(M,N)$  Our approximating function will agree with $f$ over that range.  We then know that $|f(x)| \lt \epsilon/2$ on $(\infty,M)$ and $(N,\infty)$.  If we let the approximating function go to zero on $(M',M)$ and $(N,N')$ and be zero on $(-infty,M')$ and $(N',\infty)$ it will have compact support and be within $\epsilon$ of $f$ over the whole line.
In other words: Let me denote by $C_c$ the space of continuous function with compact support and $C_0$ the one of continuous function vanishing at infinity. 


  $C_c$ dense in $C_0$ if and only if $\forall g \in C_0$ it exists a sequence $g_n\in C_c$ s.t. $g_n\to g$ w.r.t. the supremum-norm


So we can define 
\begin{equation}
g_N(x)=g(x) \text{ for } \vert x\vert\leq N \text{ and } g_N(x)=0 \text{ if } \vert x\vert&gt;N
\end{equation}
So $g_N$ is compactly supported. 
Thus $\lim_{N\to \infty} \Vert g_N-g\Vert_{\infty}=0$