Rudin's Construction of Inductive Limit Topology: unnecessarily abstruse? In Rudin's Functional Analysis Book, one of the examples in the first chapter is used later in the chapter on distributions. But when he gets to defining the inductive limit topology on a certain space, it seems unnecessarily abstruse and confusing to me, (probably because I am missing something essential). I want to check that I understand the ideas, and know if there is a particular reason for Rudin's definition. 
Here is the way he sets it up in Chapter one: take an open set $\Omega\subseteq \mathbb R^n$, a compact $K\subseteq \Omega$ and define $\mathcal D_K$ to be the collection of $C^{\infty}(\mathbb R^n)$ functions supported in $K$. Then, let $\{K_n\}$ be an exhaustion of $\Omega$, so that the norms $p_N(f) = \max \{D^{\alpha}f(x): x\in  K_N;\ |\alpha|<N\}$, (where $\alpha $ of course is a multi-index) induce a topology on $C^{\infty}(\Omega)$, and $\mathcal D_K$ is a closed subspace whenever $K\subset \Omega.$  Now, in the chapter on distributions, Rudin goes on to define $\mathcal D(\Omega)=\bigcup_{K\subset \Omega}\mathcal D_K$ and topologizes this by a collection of norms whose restriction to each $\mathcal D_K$ induces the same topology as that induced by the $p_N$. But $\mathcal D(\Omega)$ is not complete in this topology, so we look for a finer one that works. I see this as: to control what happens at the boundary of $\Omega$ we'd like to add seminorms to the ones we already have, until we get a complete space. 
The foregoing seems to be just the right setup for defining the topology we want on $\mathcal D(\Omega)$  to be the inductive limit topology, (even if we don't use the name) because clearly the inclusions $\mathcal D_{K_n}\to \mathcal D_{K_{n+1}}$ are continuous, so if we define $\tau_{D(\Omega)}$ to be the finest topology that makes the inclusions $\mathcal D_N\to \mathcal D(\Omega)$ continuous, then the results obtained in the rest of the chapter follow (more intuitively and clearly?) from this definition.
In fact, this definition implies that for $\textit{any}$ seminorm $p$ on $\mathcal D(\Omega)$,  we have that $p$ is continuous if and only if its restriction to $\mathcal D _K$ is continuous for each $K\subset \Omega.$ So we could also just have declared the desired topology to be that induced by the collection $\mathscr P$ of seminorms $p$ that satisfy: $p\in \mathscr P\Leftrightarrow p|_{\mathcal D_K}$ is continuous. In fact, using this, I was able to get all the proofs that Rudin obtained by his characterization of the topology:
$a).\ $ Let $\beta$ be the collection of all convex balanced sets $W\subseteq \mathcal D(\Omega)$ such that $\mathcal D_K\cap W\in \tau_K$ for every compact $K\subset \Omega.$ 
$b).\ $ the desired topology is then the collection of unions of the sets $\phi + W;\ \phi\in \mathcal D(\Omega)$.
In the first place, given the setup, why go to this more abstract approach? Why not do it the way the setup seems to lead naturally? I think part  $a).$ is a restatment of the above definition $p\in \mathscr P\Leftrightarrow p|_{\mathcal D_K}$ is continuous, in which case, all is well. 
In any case, wouldn't it just be cleaner to note that, as we already have topologies on the $\mathcal D_K$, why not just use the above definition in the first place? That is, topologize $\mathcal D(\Omega)$ by taking all seminorms on $\mathcal D(\Omega)$ such that their restrictions to each $\mathcal D_K$ are continuous.
 A: Before answering the question it might be useful to recall some generalities.
A common theme throughout mathematics is relating two types of descriptions for the same set. Take the unit circle $C$ around the origin.
One has a description by constraints (implicit equation):
$$
C=\{(x,y)\in\mathbb{R}^2\ |\ x^2+y^2=1\}\ .
$$
One also has a parametric description:
$$
C=\{(\cos t,\sin t)\ |\ t\in\mathbb{R}\}\ .
$$
Solving a linear system in linear algebra means going from a constraint description to a parametric one.
If one is asked to check if a given object is in the set, the constraint description is better. If one is asked to produce an element in the set, the parametric description is better.
Now about the question, let $\Omega$ be an open domain in $\mathbb{R}^n$ and consider the space $\mathscr{D}(\Omega)=C_{c}^{\infty}(\Omega)$. For each compact $K$ in $\Omega$ there is an obvious injection $\iota_{K}:\mathscr{D}_K\rightarrow \mathscr{D}(\Omega)$. Then let
$\mathscr{P}$ be the set of seminorms $p$ on $\mathscr{D}(\Omega)$ such that $\forall K$, $p\circ\iota_K$ is a continuous seminorm on $\mathscr{D}_{K}$. As the OP rightfully said the cleanest way of defining the topology of $\mathscr{D}(\Omega)$ is as the locally convex topology generated by the collection of seminorms $\mathscr{P}$. I agree with the OP that Rudin's presentation is unnecessarily abstruse.
There is however an issue with the $\mathscr{P}$ definition. It is a description by constraints. Given a seminorm on $\mathscr{D}(\Omega)$, this definition provides us with a way to check if it is continuous or not. When proving theorems about distributions one often needs to pull out of a hat some seminorms for certain estimates. Horváth gave a set of seminorms $\mathscr{H}\subset\mathscr{P}$ which generates the topology of $\mathscr{D}(\Omega)$. Basically it is a parametric description in terms of families of continuous functions.
Let $\mathbb{N}=\{0,1,\ldots\}$, and denote the set of multiindices by $\mathbb{N}^n$. A locally finite family $\theta=(\theta_{\alpha})_{\alpha\in\mathbb{N}^n}$ of continous functions $\mathbb{R}^n\rightarrow \mathbb{R}$ is one such that for all $x\in\mathbb{R}^n$ there is a neighborhood $V$ such that $V\cap {\rm Supp}\ \theta_{\alpha}=\varnothing$ for all but finitely many $\alpha$'s.
Let
$$
||f||_{\theta}=\sup_{\alpha\in\mathbb{N}^n}\sup_{x\in\mathbb{R}^n}
|\theta_{\alpha}(x)D^{\alpha}f(x)|\ ,
$$
The set $\mathscr{H}$ of seminorms $||\cdot||_{\theta}$ where $\theta$ runs over all such locally finite families defines the topology of $\mathscr{D}(\mathbb{R}^n)$.
For an example of use of these seminorms see:
https://mathoverflow.net/questions/234025/why-is-multiplication-on-the-space-of-smooth-functions-with-compact-support-cont
