The space $C^\infty(\Bbb R)$ is usually defined to be the vector space of smooth functions on $\Bbb R$. We might define $C_0^\infty(\Bbb R)$ as the space of smooth functions whose derivatives all vanish at infinity.
On $C_0^\infty(\Bbb R)$ one has a bunch of semi-norms, namely: $\|f\|_n:=\sup_{x\in\Bbb R}|f^{(n)}(x)|$. These induce a topology and also a uniform structure on $C_0^\infty(\Bbb R)$.
I have a sort of pre-question: This topology/uniform structure does not seem to be normable, is it the standard topology/uniform structure one assigns the this vector space?
My actual question is:
Is $C^\infty_0(\Bbb R)$ complete with this uniform structure?
I don't know for sure - I haven't seen $C_0^{\infty}(\mathbb{R})$ used much - but I would expect that this is the standard topology/uniform structure on that space. It's rather natural, and has good properties, in particular this family of (semi-)norms makes it a Fréchet space.
We know that $C_0(\mathbb{R})$ endowed with the maximum norm is a Banach space. Hence the product
$$X = C_0(\mathbb{R})^{\mathbb{N}}$$
is a Fréchet space, and we have an embedding
$$\iota \colon C_0^{\infty}(\mathbb{R}) \to X;\quad f \mapsto (f, f', f'', \dotsc).$$
Now one checks that $\iota\bigl(C_0^{\infty}(\mathbb{R})\bigr)$ is a closed subspace of $X$: For $x < y$ and $m \in \mathbb{N}$, the functional
$$\eta_{x,y,m} \colon (g_n) \mapsto g_m(y) - g_m(x) - \int_x^y g_{m+1}(t)\,dt$$
is continuous, hence its kernel is closed, and
$$\iota\bigl(C_0^{\infty}(\mathbb{R})\bigr) = \bigcap_{x,y,m} \ker \eta_{x,y,m}.$$