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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?)

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  • $\begingroup$ The second part follows from theorem 1.10 p. 10 of Rudin's book. For a topological vector space to be Hausdorff, it suffices that $\forall\, x\neq 0,\ \exists\, V$ neighborhood of $0$ such that $x\notin V$ $\endgroup$ – Noix07 Mar 18 at 16:39
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As for the first question, the singleton $\{\varphi_1\}$ is closed because you have just shown that the complement $\mathcal{D}(\Omega)\setminus\{\varphi_1\}$ is open. I suspect that the Hausdorffness argument is not correct, as the equivalent of this in $\mathbb{R}$ would be that $\varphi_1$ and $\varphi_2$ are points, and the set $\varphi_2+U$ would be the open interval centred at $\varphi_2$ with one of the endpoints coinciding with $\varphi_1$.

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  • $\begingroup$ Could you explain the first point? note that here $\varphi_1 \notin \varphi_2 + U$, and in general, if $E$ is a topological vector space, we have that $A \subset E$ is open if and only if each point of $a \in A$ is internal to $A$, i.e. if $\mathcal{U}$ is a local basis of $0 \in E$ we have $a+U \subset A$ for some $U \in \mathcal{U}$. About the second conclusion, I think because $\varphi_1 \neq \varphi_2$, I can choose a small neighborhood of the origin $U'$ such that $\varphi_1 + U' \cap \varphi_2 + U = \emptyset$. Probably I am confused with the definition of $U$. $\endgroup$ – user288972 Jun 19 '16 at 18:33
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    $\begingroup$ To show that the complement $\mathcal{D}(\Omega)\setminus\{\varphi_1\}$ is open, you need to show that for any $\varphi_2\in\mathcal{D}(\Omega)\setminus\{\varphi_1\}$, there is $U\in\mathcal{U}_0$ such that $\varphi_2+U\subset\mathcal{D}(\Omega)\setminus\{\varphi_1\}$. This is exactly what you did. $\endgroup$ – timur Jun 19 '16 at 18:39
  • $\begingroup$ Right! and for the second question? the only possibility I see is what I wrote in my comment. $\endgroup$ – user288972 Jun 19 '16 at 18:41
  • $\begingroup$ For the second point, you have already chosen $U$, which restricts your freedom in choosing $U'$. In fact, I think a cleaner way would be to show that $\mathcal{D}(\Omega)$ is continuously embedded into $C(\Omega)$, and then use Hausdorffness of $C(\Omega)$. $\endgroup$ – timur Jun 19 '16 at 18:43
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    $\begingroup$ My suggestion would be to choose $U$ and $U'$ at the same time, similarly to how you chose $U$ the first time, but now with smaller "radius". $\endgroup$ – timur Jun 19 '16 at 19:44
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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.

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  • $\begingroup$ The point is to follow the proof, this must be a consequence of the fact that singletons are closed. $\endgroup$ – user288972 Jun 20 '16 at 14:43

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