This is a follow-up to this question.

Let $M$ be manifold (Hausdorff, countable base, finite dimensional, if it simplifies anything embedded in $\mathbb{R}^n$).

I'm interested in the topological vector space $\mathcal{C}_c^{\infty}(M)$ of smooth functions with compact support equipped with the Whitney-$\mathcal{C}^\infty$-topology (The strong one induced by the strong topologies on $\mathcal{C}^{k}(M)$ for finite $k$. Another description for this topology can be found in 4. at the linked question.)

If $M$ is noncompact $\mathcal{C}^{\infty}(M)$ is in general not first countable and restricting to $\mathcal{C}_c^{\infty}(M)$ doesn't make anything better.

Therefore, since I want $\mathcal{C}_c^{\infty}(M)$ to be first countable, I have to make it even smaller. The next attempt is the following: Let $K\subseteq M$ be a compact subset and consider the subspace of functions which vanish in $K$ $$V\colon=\{f\in\mathcal{C}_c^{\infty}(M)|f|_K=0\}.$$ Since $K$ is compact, this should be the "bigger part" of the whole $\mathcal{C}_c^{\infty}(M)$. Now take the quotient vectorspace $\mathcal{C}_c^{\infty}(M)/V$ with the quotient topology.

I reckon that $\mathcal{C}_c^{\infty}(M)/V$ is now first countable, but I did not succeed to prove it. Since it's a topological vector space, it suffices to construct a countable neighbourhood basis of $0$. Can someone point me in the right direction? Thanks.


Here is a proof:

Let $K\subset U\subset L\subset M$ where $U$ is open in $M$ and $L$ is compact. Then $C^\infty_c(M) = C^\infty_c(M\setminus K) + C^\infty_c(U)$ by a partition of unit element. Moreover $C^\infty_c(U)\subset C^\infty_L(M):=\{f\in C^\infty_c(M): \text{supp}(f)\subseteq L\}$, and the latter space is a Frechet space and is closed in $C^\infty_c(M)$. Thus $C^\infty_c(M) = C^\infty_c(M\setminus K) + C^\infty_L(M)$, and $C^\infty_c(M\setminus K)\subseteq V$. So finally $$ C^\infty_C(M)/V = C^\infty_L(M)/(V\cap C^\infty_L(M)) $$ is a quotient of the Frechet space $C^\infty_L(M)$ modulo a closed subspace, and is thus Frechet itself.


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