Whitney C-infinity topology

Remarks

With the Whitney $C^\infty$ topology, $C^\infty(M, N)$ is a Baire space. If $M$ is compact, it coincides with the weak $C^\infty$ topology, the topology of uniform convergence of all derivatives, which is built from the compact-open topology on jets. If $M$ is not compact, it is much finer.

Example: for $M = N = \R$, let $\varepsilon \colon \R \to (0, \infty)$ be continuous and let $f \in C^\infty(\R, \R)$. The set $U = \set{ (x, y) \in \R \times \R : \abs{y - f(x)} < \varepsilon(x) }$ is open in $J^0(\R, \R) = \R \times \R$. So $\mathcal{M}(U) = \set{ g : \abs{g(x) - f(x)} < \varepsilon(x) \text{ for all } x }$ is an open neighbourhood of $f$.

Non-example: the constant maps $f_n = 1/n$ do not converge to $0$ in the Whitney $C^\infty$ topology on $C^\infty(\R, \R)$, although they converge uniformly with all derivatives. Take $f = 0$ and $\varepsilon(x) = 1/(1 + x^2)$ above. For each $n$, $1/n > \varepsilon(x)$ for large $\abs{x}$, so no $f_n$ lies in this neighbourhood of $0$.