For an open $U \subseteq \mathbb{R}^n$, we give $C^{\infty}_c(U)$ (smooth compactly supported complex-valued functions on $U$) the limit topology from the inclusions $C^{\infty}_K(U)$ over all compact sets $K \subset U$, where $C^{\infty}_K(U)$ is the space of smooth functions on $U$ with support contained in $K$ (given the topology of uniform convergence of the function and all its derivatives within $K$). So this means a subset $\Omega \subseteq C^{\infty}_c(U)$ is open iff $\Omega \cap C^{\infty}_K(U)$ is an open subset of $C^{\infty}_K(U)$ for all compact $K \subset U$.
It is a fact (stated e.g. in Mitrea's 'Distributions, Partial Differential Equations and Harmonic Analysis' page 550) that a sequence $f_j \in C^{\infty}_c(U)$ converges to some $f \in C^{\infty}_c(U)$ if and only if there exists some compact $K \subset U$ with all $f_j$'s and $f$ in $C^{\infty}_K(U)$, with $f_j \to f$ in $C^{\infty}_K(U)$. How does one prove this? Particularly I don't see why there should be a compact $K$ with all $f_j$'s contained in $C^{\infty}_K(U)$. I realize Rudin's Functional Analysis has a proof of this statement, but his definition of the topology is quite different (even though I assume it ends up the same topology) so it doesn't help me much.
Say that $U= \Bbb{R}^n$ because it works quite the same for any open set (replacing $|x_k|\to \infty$ by $d(x_k,\partial U)\to 0$)
If the $f_j$ are not supported on a common compact then there is a strictly increasing sequence $|x_k|\to \infty$ and $e_k$ such that $|f_{e_k}(x_k)|\ne 0$.
Take a strictly positive continuous function such that $$h(x_k)=\frac1{|f_{e_k}(x_k)|}$$ then $$V=\{ \phi \in C^\infty_c(U), \|h \phi\|_\infty < 1/2\}$$ is open and $(f_j-\psi)_{j\ge J}$ is not contained in $V$ for any $J\ge 1,\psi\in C^\infty(U)$ whence $f_j$ isn't Cauchy for your topology.