$C^{\infty}$ topology is finer than $C^0$ topology

Question

Let $C(\mathbb{R};\mathbb{R})$ be equipped with the topology of uniform convergence on compacts. This is clearly at-least as fine as the standard topology generated by the semi-norms on $C^{\infty}(\mathbb{R};\mathbb{R})$ (ie: uniform convergence of all derivatives on compacts)...

However, is the latter strictly finer?

Answer 1

The sequence $\frac1n\sin(nx)$ converges uniformly to $0$ on any compact, but its derivatives do not. So looking at neighborhoods of the zero function, the latter is strictly finer.