A doubt about the vectorial topology on $\mathcal{D}(\Omega)$ We denote with $\mathcal{U}_0$ the family of all subsets $U \in \mathcal{D}(\Omega)$ convex and balanced such that $U \cap \mathcal{D}_K(\Omega) \in \mathcal{T}_K$, where $\mathcal{T}_K$ is the topology on $\mathcal{D}_K(\Omega):=\lbrace \varphi \in C^{\infty}(\Omega) : \mathrm{supp}(\varphi) \subset K \rbrace$, defined by seminorm $p_{K_N}(\varphi):=\sup_{|\alpha| \leq N ; x \in K} |D^\alpha \varphi(x)|$. It can be shown that $\mathcal{U}:=\lbrace \varphi + U : \varphi \in \mathcal{D}(\Omega), U \in \mathcal{U}_0 \rbrace$ is a base for the vectorial topology on $\mathcal{D}(\Omega)$. As in the book Functional Analysis by Rudin, page 152-153. In particular the topology $\mathcal{T}$ on $\mathcal{D}(\Omega)$ it is Hausdorff topology, and the question I have is on this step:
If $\varphi_1 \neq \varphi_2$ are function test, we define:
$\displaystyle U:= \lbrace \varphi \in \mathcal{D}(\Omega) : \sup_{x \in \Omega} |\varphi(x)| < \sup_{x \in \Omega} |\varphi_1(x) - \varphi_2(x)| \rbrace$
we have $U \in \mathcal{U}_0$ and $\varphi_1 \notin \varphi_2 + U$. It follows that the singleton $\lbrace \varphi_1 \rbrace$ is a closed for topology $\mathcal{T}$ (why?). Then, since $\varphi_1 \neq \varphi_2$, there is $U' \in \mathcal{U}_0$ such that $\varphi_1 + U' \cap \varphi_2 + U = \emptyset$, so $\mathcal{T}$ is Hausdorff topology. (Is it correct this conclusion?)

The functorial property of the inductive limit makes life much easier: The inclusion $\mathscr D(\Omega) \hookrightarrow C(\Omega)$ where the latter space is endowed with the topology of convergence on all compact sets is continuous because so are all $\mathscr D(K)\hookrightarrow C(\Omega)$. Since $C(\Omega)$ is Hausdorff (trivially: if $f\neq g$ there is $x\in\Omega$ with $f(x)\neq g(x)$ hence $p(h)=|h(x)|$ defines a seminorms separating $f$ and $g$) this implies that $\mathscr D(\Omega)$ is Hausdorff.
