Why do we give $C_c^\infty(\mathbb{R}^d)$ the topology induced by all good seminorms? Briefly, my question boils down to the following: What benefits do we gain from considering the space of test functions in the topology induced by all "good" seminorms, as opposed to other topologies which are more common or more easily described?
More precisely:
So far as I understand, the topology on the space $C_c^\infty(\mathbb{R}^d)$ of test functions is usually constructed as follows:


*

*For each compact $K \subset \mathbb{R}^d$ we define the "smooth topology" on $C_c^\infty(K)$ to be the topology generated by the family of seminorms $\{\|\cdot\|_{C^k(K)}\}_{k \in \mathbb{N}}$, where $\|\cdot\|_{C^k(\mathbb{R}^d)}$ is defined by


$$
\| f \|_{C^k(K)} = \sup_{x \in \mathbb{R}^d} \left\{ \sum_{j=0}^k \left|\nabla^j f(x)\right| \right\}
$$


*

*We then define a seminorm $\| \cdot \|$ on $C_c^\infty(\mathbb{R}^d)$ to be "good" if, for each compact $K \subset \mathbb{R}^d$, it is continuous function on $C_c^\infty(K)$ with respect to the smooth topology.

*Finally, we topologise $C_c^\infty(\mathbb{R}^d)$ by giving it the initial topology induced by the family $\mathscr{G}$ of all good seminorms.  
(The construction above is the one given by Prof. Tao here on his blog. Any mistakes made in the reproduction above are mine.) 
What I would like to know is:


*

*How is the topology induced by the "good" seminorms different from the one induced by the family $\mathscr{F} = \left\{ \|\cdot\|_{C_c^k(\mathbb{R}^d)} \right\}_{k \in \mathbb{N}}$. I believe we have the inclusion $\mathscr{F} \subset \mathscr{G}$, and I can't really tell what we gain by considering the other good seminorms. 


In fact, I'm not really sure what the other good seminorms look like. It appears we can obtain some other good seminorms by considering functionals of the form
$$ 
\eta(f) = \int_{\mathbb{R}^d} |f(x)g(x)| ~dm(x)
$$
for suitable functions $g$. But again, I'm not sure what these contribute.


*If the topology induced by $\mathscr{F}$ is different from the one induced by $\mathscr{G}$, why do we prefer the latter?


Any insight will be much appreciated. 
 A: There are many equivalent ways to construct the topology on $C^\infty_c(\Bbb{R}^d)$ that makes it an LF-space. I'm not familiar with Tao's method, but I've looked at a different construction that should be equivalent, and is in my opinion very intuitive.
First, we set our goal to be to find a topology on $C^\infty_c(\Bbb{R^d})$ that makes it a locally convex complete topological vector space.
Now, how do we make sure that sequences of test functions converge to test functions? We have to make sure that the limit is smooth and compactly supported. Let's look at the subspace $C^\infty_c(K)$ first. In this space, the only problem with convergence really is just the smoothness of the limit function.
How do we topologise $C^\infty_c(K)$ ? First naïve idea could be to use the sup-norm, but we quickly see that test functions can uniformly converge to non-smooth function. But if $(\varphi)_i$ is a sequence of test functions and for convergence we require that $D^\alpha \varphi_i \to g_\alpha$ uniformly for all multiindices $\alpha$, we can show that $(\varphi_i)$ then converges to a test function. What is this topology? Well, if the function $p_\alpha: C^\infty_c(K) \to \Bbb{R}$, 
$$
p_\alpha(\varphi) = \sup_{x \in K} \{ |D^\alpha \varphi(x)|\}\,,
$$
is continuous for all multiindices $\alpha$, our topology has the wanted convergence property. So, for $C^\infty_c(K)$ we choose the topology induced by the seminorms $(p_\alpha)_{\alpha \in \Bbb{N}_0^d}$.
Now back to $C^\infty_c(\Bbb{R^d})$. We could try to use the topology induced by the extensions of $p_\alpha$, namely the seminorms $p_\alpha: C^\infty_c(\Bbb{R}^d) \to \Bbb{R}$, but with this topology the "limit function" is not necessarily compactly supported. Take any $\varphi \in C^\infty_c(\Bbb{R}^d)$ that is not the zero function and consider the sequence that consists of the functions
$$
\psi_n(x) = \sum_{k=1}^n \frac{1}{k^2} \varphi(x-k)\,.
$$
The sequence converges uniformly to the smooth function $\sum_{k=1}^\infty k^{-2} \varphi(x-k)$, but this is not compactly supported.
This curious phenomenon is important: consider now the sequence
$$
\xi_n(x) = \frac{1}{n^2} \varphi(x-n) = \psi_n(x) - \psi_{n-1}(x)\,.
$$
We see that $\xi_n \to 0$ $C^\infty$-uniformly, but if our topology lets this happen, we can construct a sequence of smooth functions that converges to non-compactly supported function: the sequence $(\psi_n)$. The special thing about $(\xi_n)$ is that it converges to a test function $C^\infty$-uniformly, but the supports of $\xi_n$ "escape to infinity". So we don't want our topology to have this property. How?
We could formulate this problem as: if $(\xi_n)$ is a Cauchy sequence, we don't want the supports of $\xi_n$ to escape to infinity. So we want the union of all the supports of $\xi_n$ to belong to some compact set $K$. To achieve this we just "push" the Cauchy sequences from $C^\infty_c(K)$ to $C^\infty_c(\Bbb{R}^d)$ for each $K$. Of course, we need our topology to have the property that if $(\xi_n)$ is Cauchy in $C^\infty_c(K)$, it is also Cauchy in $C^\infty_c(\Bbb{R}^d)$. This requires that the inclusion maps $\iota_K: C^\infty_c(K) \to C^\infty_c(\Bbb{R}^d)$ have to be continuous.
But the topology coinduced (final topology/strong topology) by some family $(\iota_{K_n})_{n \in \Bbb{N}}$ of inclusions, such that $\Bbb{R^d} = \bigcup_n K_n$, doesn't quite do the job yet. But if we take the finest locally convex topology in which the inclusions are continuous, we get a topology with the wanted properties (see Rudin's functional analysis text for more details).
