**Proof plan.** We verify the three claims in order. The separating property follows because a smooth function vanishing on every compact subset of $\Omega$ must vanish identically. Independence of the exhaustion is a standard comparison argument. Completeness is the heart of the proof: a [Cauchy sequence](/page/Cauchy%20Sequence) in every seminorm $p_{m,\alpha}$ produces, for each multi-index $\alpha$, a [continuous](/page/Continuity) function $g_\alpha$ that is the locally uniform limit of $\partial^\alpha f_k$. The key step is showing that $g_\alpha = \partial^\alpha g_0$, which reduces to the classical fact that the locally uniform limit of a [sequence](/page/Sequence) of [functions](/page/Function) whose derivatives also converge locally uniformly is [differentiable](/page/Derivative) with derivative equal to the limit of the derivatives.
**Step 1 (Separating family).** Suppose $p_{m,\alpha}(f) = 0$ for every $m$ and every $\alpha$. Taking $\alpha = 0$: $\sup_{K_m} |f| = 0$ for every $m$, so $f$ vanishes on every $K_m$. Since $\Omega = \bigcup_m K_m$, $f \equiv 0$.
The family is countable because $\mathbb{N} \times \mathbb{N}_0^n$ is countable (as a countable product of [countable sets](/page/Countable%20Set)).
**Step 2 (Independence of the exhaustion).** Let $\{K_m'\}$ be another compact exhaustion of $\Omega$. For every $m$ and $\alpha$, the [set](/page/Set) $K_m$ is compact and contained in $\Omega = \bigcup_j \mathrm{int}(K_j')$, so by compactness $K_m \subseteq K_{j_0}'$ for some $j_0$. Therefore $p_{m,\alpha}(f) \le p_{j_0,\alpha}'(f)$, which shows that every seminorm in the first family is bounded by a seminorm in the second family. By symmetry, the two families generate the same [topology](/page/Topology).
**Step 3 (Completeness).** Let $\{f_k\}_{k=1}^\infty \subseteq C^\infty(\Omega)$ be Cauchy in every $p_{m,\alpha}$. Fix $\alpha \in \mathbb{N}_0^n$ and $m \in \mathbb{N}$. The Cauchy condition says: for every $\varepsilon > 0$, there exists $K_0$ such that $\sup_{K_m} |\partial^\alpha f_j - \partial^\alpha f_k| < \varepsilon$ for all $j, k \ge K_0$. In particular, $\{\partial^\alpha f_k|_{K_m}\}$ is a uniform Cauchy sequence of continuous functions on $K_m$.
[claim:Locally Uniform [Limits](/page/Limit) Exist]
For each $\alpha \in \mathbb{N}_0^n$, there exists a continuous function $g_\alpha: \Omega \to \mathbb{R}$ such that $\partial^\alpha f_k \to g_\alpha$ uniformly on every compact subset of $\Omega$.
[/claim]
[proof]
By the [uniform limit theorem](/theorems/258), the uniform Cauchy sequence $\{\partial^\alpha f_k|_{K_m}\}$ [converges uniformly](/page/Uniform%20Convergence) to a continuous function $h_{m,\alpha}: K_m \to \mathbb{R}$. Since $K_m \subseteq K_{m+1}$, the restriction of $h_{m+1,\alpha}$ to $K_m$ equals $h_{m,\alpha}$ (both are the uniform limit of $\partial^\alpha f_k|_{K_m}$). Defining $g_\alpha(x) = h_{m,\alpha}(x)$ for any $m$ with $x \in K_m$ gives a well-defined continuous function on $\Omega$, and $\partial^\alpha f_k \to g_\alpha$ uniformly on each $K_m$.
[/proof]
[claim:The Limit Is Smooth And Its Derivatives Are The Expected Ones]
$g_0 \in C^\infty(\Omega)$ and $\partial^\alpha g_0 = g_\alpha$ for every $\alpha \in \mathbb{N}_0^n$.
[/claim]
[proof]
We prove $\partial_i g_0 = g_{e_i}$ for each standard basis vector $e_i$; the general case follows by induction on $|\alpha|$.
Fix $x_0 \in \Omega$ and choose $m$ with $x_0 \in \mathrm{int}(K_m)$. On $K_m$, $f_k \to g_0$ uniformly and $\partial_i f_k \to g_{e_i}$ uniformly. The [fundamental theorem of calculus](/theorems/632) gives, for $x_0$ and $x_0 + t e_i$ both in $\mathrm{int}(K_m)$:
\begin{align*}
f_k(x_0 + te_i) - f_k(x_0) &= \int_0^t \partial_i f_k(x_0 + se_i) \, ds.
\end{align*}
Since $\partial_i f_k \to g_{e_i}$ uniformly on $K_m$, we may pass to the limit $k \to \infty$ under the [integral](/page/Integral):
\begin{align*}
g_0(x_0 + te_i) - g_0(x_0) &= \int_0^t g_{e_i}(x_0 + se_i) \, ds.
\end{align*}
Since $g_{e_i}$ is continuous, differentiating in $t$ at $t = 0$ gives $\partial_i g_0(x_0) = g_{e_i}(x_0)$. As $x_0$ was arbitrary, $\partial_i g_0 = g_{e_i}$ on $\Omega$.
By induction: assume $\partial^\beta g_0 = g_\beta$ for all $|\beta| \le k$. For $|\alpha| = k+1$, write $\alpha = \beta + e_i$ with $|\beta| = k$. Then $\partial^\beta g_0 = g_\beta$, and $\partial^\beta f_k \to g_\beta$ uniformly on compacta, while $\partial^{\beta + e_i} f_k = \partial_i (\partial^\beta f_k) \to g_{\beta + e_i}$ uniformly on compacta. The same fundamental-theorem-of-calculus argument shows $\partial_i g_\beta = g_{\beta + e_i}$, i.e. $\partial^\alpha g_0 = g_\alpha$.
[/proof]
**Step 4 (Convergence in the Fréchet topology).** By Claim 1, $\partial^\alpha f_k \to g_\alpha = \partial^\alpha g_0$ uniformly on $K_m$ for every $m$ and $\alpha$, i.e. $p_{m,\alpha}(f_k - g_0) \to 0$ for every $m, \alpha$. This is exactly convergence in the Fréchet topology. $\blacksquare$
