Proving $C_c^\infty(\Omega)$ is not complete Let $\Omega \subseteq \mathbb{R}^n$ be a domain. 

How can I prove that $C_c^\infty(\Omega)$ (with the usual topology) is not sequentially complete?

I don't think I have ever seen a proof of this claim. I guess the idea is to construct a sequence of functions with increasing supports and small differences, but I don't know how to do that.       
 A: Let us do it for $C^\infty_c(\mathbb{R})$ first. With the help of $e^{-\frac{1}{x^2}}$, we can construct, for every $n$, a $C^\infty_c$ function $f_n$ such that $f_n(x)=0$ for $|x|\geq n+1$, and $f_n(x)=1$ for $|x|\leq n$.
Now for every compact $K$ in $\mathbb{R}$, $K$ is bounded so there exists $N$ such that $K\subseteq [-N,N]$. So for all $n,m\geq N$ and all $x\in K$, we have $f_n(x)=f_m(x)=1$. Hence $\sup_K|f_n^{(j)}(x)-f_m^{(j)}(x)|=0$ for all $j$ and all $n,m\geq N$.
So this a Cauchy sequence for the topology of compact convergence. But the pointwise limit is the constant function equal to $1$, which is not compactly supported. So $C^\infty_c(\mathbb{R})$ is not sequentially complete.
Working with $e^{-\frac{1}{\|x\|^2}}$ in $\mathbb{R}^n$ and with balls instead of intervals yields a very similat proof for $C^\infty_c(\mathbb{R}^n)$.
To do the case $C^\infty_c(\Omega)$, we need a compact exhaustion of $\Omega=\bigcup_{l\geq 1}K_l$ such that $K_l$ is contained in the interior of $K_{l+1}$ for every $l$. Then we construct for each $l$ a $C^\infty_c$ function $f_l$ such that $f_l=1$ on $K_l$ and $0$ on $K_{l+1}^c$. This is more difficult. It can be done by convolution. Still with the help of the exponential function. This sequence is Cauchy again, but converges pointwise to the constant function equal to $1$.
A: The answer depends on what you mean by ''the usual topology''. Dieudonne and Schwartz invented LF-spaces to have a complete locally convex toplogy on 
$\mathscr D(\Omega)=C^\infty_c(\Omega)$. 
Anyway, $C^\infty_c(\Omega)$ endowed with the topology of uniform convergence on compact sets of all derivatives is not complete because it is dense in $C^\infty(\Omega)$ (with the same topology) but clearly different.
A: Convolve a continuous non-differentiable function with a $C^{\infty}$ family of approximations to the identity. Each convolution will be $C^{\infty}$ but the limit won't be. 
A: We illustrate it in the case $\Omega=\mathbb{R}$.
Take a nonzero function $\varphi\in C^\infty_c(\mathbb{R})$ whose support is small and concentrated near $0$,
and consider the sequence
\begin{equation}
\varphi_k(x) = \varphi(x) + 2^{-1}\varphi(x-1) + \ldots + 2^{-k}\varphi(x-k),
\qquad k=1,2,\ldots.
\end{equation}
Obviously, this sequence is Cauchy with respect to the seminorms $\|\cdot\|_{C^m(\mathbb{R})}$,
but the support of the limit function is not compact.
