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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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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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    $\begingroup$ 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. $\endgroup$ – 7891user Jan 15 '14 at 12:07
  • $\begingroup$ Hi, how does one see that $C^\infty(\Bbb R^n)$ (or presumably for a more general open set) is second countable? $\endgroup$ – Ryan Unger Aug 27 '17 at 2:21
  • $\begingroup$ Let for any multiindex $k\in {\mathbb N}^n$ the record $f^k(t)$ mean the $k$-th derivative of $f$ in the point $t\in{\mathbb R}^n$, and let $||f||_{k,m}=\max_{l\le k}\max_{|t|\le m}|f^l(t)|$ be the standard seminorm on $C^\infty({\mathbb R}^n)$. Then the sets $U_{p,k,m,r}=\{f\in C^\infty({\mathbb R}^n):\ ||f-p||_{k,m}<\frac{1}{r}\}$, where $k\in {\mathbb N}^n$, $m,r\in{\mathbb N}$ and $p$ are polynomials with rational coefficients, form a countable base in $C^\infty({\mathbb R}^n)$. $\endgroup$ – Sergei Akbarov Aug 27 '17 at 4:14

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