This is a statement from Rudin's Functional Analysis section 1.46:
Given an open set $\Omega \subset \mathbb{R}^n$, choose compact sets $K_i$ such that $\Omega = \bigcup K_i$ and $K_i \subset K_{i+1}^{\mathrm{o}}$. Then define seminorms for $\phi \in C^\infty(\Omega)$:
$$ p_N(\phi) = \max\{ |D^\alpha \phi(x)| : x \in K_N, |\alpha| \le N \} $$
The seminorms define a locally convex topology on $C^\infty(\Omega)$. My question is, how do we show that this topology is complete?
Rudin says after the definition that given a Cauchy sequence ${\phi_i}$ in this topology, we have $p_N(\phi_i - \phi_j) < \frac{1}{N}$ for $i, j$ sufficiently large, which means that $|D^\alpha\phi_i - D^\alpha\phi_j| < \frac{1}{N}$ on $K_N$. Then he says that this implies that $D^\alpha\phi_i \rightarrow g_\alpha$ for some function $g_\alpha$, and then we must have $\phi_i \rightarrow g_0$ and $g_\alpha = D^\alpha g_0$.
I'm not sure how the statements for the functions $g_\alpha$ follow here. How how would we show that $g_0 \in C^\infty(\Omega)$ and it's derivatives match the limits of the derivatives of $f_i$?
Let $\{\phi_j\}$ be a Cauchy sequence in $\mathcal{D}(\Omega)$. That is $\exists K \subseteq \Omega$ such that $\text{supp}(\phi_j) \subseteq K \ \forall j \in \mathbb{N}.$
Given $\epsilon > 0$ and $n \in \mathbb{N}_0$, there exists $N = N(\epsilon, n) > 0$ such that $\forall j, k > N$ we have $$ \|\phi_j - \phi_k\|_{C^n( \Omega)} = \left( \sum_{|\alpha| \leq n} \|D^{\alpha}(\phi_j - \phi_k)\|_{L^{\infty}(\Omega)} \right) < \epsilon $$ $$ = \sum_{|\alpha| \leq n} \sup_{x \in \Omega} |D^{\alpha} (\phi_j(x) - \phi_k(x))| < \epsilon. $$
Then, the sequence $\{\phi_j\}$ is Cauchy in $C^n_B = \{\phi \in C^n(\Omega), \|\phi\|_{C^n( \Omega)} < \infty\}.$
Since $C^n_B$ is a Banach space, then $\{\phi_j\}$ converges in $C^n_B(\Omega)$, say, to $\phi \in C^n_B(\Omega)$. Since this is true for each $n \in \mathbb{N}$, then $$ \lim_{j \to \infty} \|\phi - \phi_j\|_{C^n(\Omega)} = 0 \quad \forall n \in \mathbb{N}. $$ i.e., $\phi_j \to \phi$ in $\mathcal{D}(\Omega)$.
Hence, $\mathcal{D}(\Omega)$ is complete.