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Since each open set $\Omega \subset \mathbb{R}^n$ admits a increasing exhaustion of compact set $K \subset \Omega$, then the space of test functions are rewritten as a countable union $\mathcal{D}(\Omega):=\bigcup_{j \in \mathbb{N}} \mathcal{D}_{K_j}(\Omega)$, and $\mathcal{D}(\Omega)$ is not metrizable, because by contradiction if it is metrizable then it is a complete metric space and by theorem/corollary of Baire $\mathcal{D}_{K_j}(\Omega)$ has interior non empty for some $j \in \mathbb{N}$ and this is a contradiction because $\mathcal{D}_{K_j}(\Omega)$ must necessarily have interior set empty. I justified this because $\mathcal{D}_{K_j}(\Omega)$ is a vectorial topological subspace cloed of $\mathcal{D}(\Omega)$.

Can you have a better formulation of this fact?

The space of distributions $\mathcal{D}'(\Omega)$ is a locally convex space with seminorms family $\mathcal{F}=\lbrace u_{\varphi}(u)=|u(\varphi)| : \varphi \in \mathcal{D}(\Omega) \rbrace$ and $u \in \mathcal{D}'(\Omega)$ which defines the weak* topology $\sigma(\mathcal{D}'(\Omega), \mathcal{D}(\Omega))$ on $\mathcal{D}'(\Omega)$.

How do I prove that the distribution spaces is not metrizable?

Thanks for any help

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  • $\begingroup$ You mean $D(\Omega)$ is the space of test functions and $D'(\Omega)$ the space of distributions. And $D(\Omega)$ should be the same as $D''(\Omega)$ so it reduces to $D(\Omega)$ metrizable ? $\endgroup$ – reuns Sep 28 '16 at 21:41
  • $\begingroup$ I'm not assuming anything, I "know" that the space of test functions is not metrizable by Baire theorem, so I ask for a better explanation than the one I gave, if possible. On the space of distributions I know that is a locally convex space but do not know prove that it is not metrizable. Your vision is perhaps true for normed spaces, since $E \subset E''$, but in this case I do not know. $\endgroup$ – user288972 Sep 28 '16 at 21:47
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    $\begingroup$ I'm not sure if what they say here mathoverflow.net/a/87822/84768 means the non-metrizability can be transferred or not $\endgroup$ – reuns Sep 28 '16 at 22:23
  • $\begingroup$ Thanks for the link, basically it depends on which topology I consider on the space of distributions. I was thinking that the distribution space is not metrizable for this result: math.stackexchange.com/questions/1817305/… at least with respect to the weak* topology $\endgroup$ – user288972 Sep 28 '16 at 22:41

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