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Consider the space $\mathcal D(\mathbb{R}^n)$ of smooth functions (in the sense of having continuous derivatives of all orders) which are compactly supported. Endow it with its usual topology, i.e., the topology such that the dual space is the space of distributions.

Question: Is $\mathcal D(\mathbb{R}^n)$ separable?

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closed as off-topic by Benoît Kloeckner, Andrey Rekalo, Ricardo Andrade, Nate Eldredge, Daniel Moskovich Jan 15 '14 at 15:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Nate Eldredge
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up vote 4 down vote accepted

I suppose, you mean the usual topology on $D({\mathbb R}^n)$ defined for example in Rudin's book. Take $D_N=\{\varphi\in D({\mathbb R}^n):\ {\rm supp}\varphi\subseteq\{x\in{\mathbb R}^n:\ |x|\le N\} \}$, where $N\in{\mathbb N}$. Each $D_N$ can be considered as a subspace of the space $C^\infty({\mathbb R}^n)$ with the induced topology. Since this topology on $C^\infty({\mathbb R}^n)$ has a countable base, $D_N$ also has a countable base. As a corollary, $D_N$ is separable. Let $S_N$ be a countable dense subset in $D_N$. Then $S=\bigcup_{N\in{\mathbb N}}S_N$ is countable and dense in $D({\mathbb R}^n)=\bigcup_{N\in{\mathbb N}}D_N$.

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The above is an impeccable answer to the question posed but it is perhaps worth mentioning that many of the spaces which arise in distribution theory are nuclear F or LF spaces, resp., the duals thereof, and these are automatically separable. – 7891user Jan 15 '14 at 12:07
Thank you very much. – user45560 Jan 15 '14 at 12:47

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