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Let $\Omega\subseteq\mathbb{R}^n$ be some nonempty open, and use the notation $U\Subset V$ to imply that $U$ is a compact subset of $V$. Then, for all $K\Subset\Omega$, we can define the space $$\mathscr{C}^\infty_K(\Omega)=\{\varphi\in \mathscr{C}^\infty(\Omega)\mid \operatorname{supp}\varphi\subseteq K\}$$ with the seminorms $\lVert f\rVert_{\alpha,K}=\lVert\partial^\alpha f\rVert_{L^\infty(\Omega)}$, so that $$\{\mathscr{C}^\infty_K(\Omega)\}_{K\Subset\Omega}$$ is an inductive system of topological spaces with maps given by inclusion as subspaces, so we can define the inductive limit of this system of $\mathscr{C}^\infty_0(\Omega)$ with the inductive limit topology, so that for any inclusion $K\Subset K'\Subset \Omega$ the inclusion $$\mathscr{C}^\infty_K(\Omega)\subset\mathscr{C}^\infty_0(\Omega)$$ is simply the chain of inclusions $$\mathscr{C}^\infty_K(\Omega)\subset\mathscr{C}^\infty_{K'}(\Omega)\subset\mathscr{C}^\infty_0(\Omega).$$

Now, I wish to show that this is equivalent to the usual definition of $\mathscr{C}^\infty_0(\Omega)$ using convergence, i.e. that for $\varphi_n,\varphi\in\mathscr{C}^\infty_0(\Omega)$, we have that $$\varphi_n\xrightarrow{n\to\infty}\varphi$$ in $\mathscr{C}^\infty_0(\Omega)$ if and only if $\varphi_n,\varphi$ all have support in some $K\Subset\Omega$, and $$\varphi_n\xrightarrow{n\to\infty}\varphi$$ in $\mathscr{C}^\infty_K(\Omega)$. Now, since $\mathscr{C}^\infty_K(\Omega)$ is a subspace of $\mathscr{C}^\infty_0(\Omega)$ it is trivial to see that convergence in the smaller space implies convergence in the larger one. However, I don't see how we can conclude that if $$\varphi_n\xrightarrow{n\to\infty}\varphi$$ in $\mathscr{C}^\infty_0(\Omega)$, we can restrict to some $K\Subset\Omega$. I see that $\mathscr{C}^\infty_K(\Omega)\subset\mathscr{C}^\infty_0(\Omega)$ is a closed subspace for all $K\Subset \Omega$, but I can't find any neighborhood of $\varphi$ that would help me restrict to some $\mathscr{C}^\infty_K(\Omega)$.

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(First, a quibble: it is probably better to denote compactly-supported smooth functions with a subscript "c" rather than "0", since $C^\infty_o$ often means smooth functions going to $0$ at infinity, and maybe their derivatives, etc.)

Anyway, yes, it is not obvious that the "inductive" or "strict colimit" characterization of the topology on test functions gives the sequential version that you quote. One way to approach this with dignity is to prove the lemma that any bounded (in the topological vector space sense) subset of a strict colimit must be contained in a limitand. In the present situation, this gives the constraint on supports. Of course, as one will see in the general case, it's not about "supports" per se, but about the limitands in such a colimit.

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  • $\begingroup$ While I see why any convergent sequence must be bounded in the t.v.s. sense, I don't see why boundedness implies containment in a limitand. $\endgroup$ – D. Wynter May 6 '17 at 1:30

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