Strict topology - what is the "importance" of the concept? From Conway, A course in functional analysis, page 105. Problem 21. 
Let $X$ be a locally compact space and for each $\phi\in C_{0}(X)$, define $p_{\phi}(f)=|\phi f|_{\infty}$ for $f\in C_{b}(X)$. Show that $p_{\phi}$ is seminorm on $C_{b}(X)$. Let $\beta\equiv$ the topology defined by these seminorms. Show that $(C_{b}(X),\beta)$ is a LCS that is complete, $\beta$ is called the strict topology.
I was able to solve this problem by following the definition and prove everything step by step. However I am wondering what is the importance of this concept? It seems to me the strict topology should be "finer" than the sup-norm topology inherited from $C_{b}(X)$, but it is not clear to me how exactly the difference lies. And on the practical side, what is the use of this concept?
Strict topology showed up again after a few pages; in page 107 I was asked to show if $X$ is locally compact, then a subset of $C_{b}(X)$ is $\beta$-bounded if and only if it is (sup-)norm bounded. I can prove this by following the definition. And $(C_{b}(X),\beta)$ is metrizable if and only if $X$ is compact. Since an LCS is metrizable if and only if it is (Hausdauff) and first countable, I have to prove that if $X$ is compact, then $(C_{b}(X),\beta)$ is first countable. This is the same as showing the zero map is a $G_{\delta}$ set in $(C_{b}(X),\beta)$ by problem 5. At this point I found I am stuck; I do not see the connection between $X$ being compact and $0$-map is a $G_{\delta}$ set right away. So I want to ask for a hint and importance of the concept, for clearly I do not really understand it very well.   
 A: 
Since an LCS is metrizable if and only if it is (Hausdorff and) first countable, I have to prove that if $X$ is compact, then $(C_b(X),\beta)$ is first countable.

Well, you can prove it indirectly. Assuming

$(C_b(X),\beta)$ is a LCS that is complete

is already proven, you have
$(C_b(X),\beta)$ is metrizable if and only if it is a Fréchet space. Since $\beta \subset \tau$, where $\tau$ is the topology of uniform convergence induced by the sup-norm $\lvert\: \cdot\,\rvert_\infty$, you have a continuous linear bijection
$$\operatorname{id} \colon (C_b(X),\tau) \to (C_b(X),\beta).$$
Now, $(C_b(X),\tau)$ is a Banach space, hence a Fréchet space, and for Fréchet spaces, we have the open mapping theorem, which says that a continuous linear surjection is an open mapping. That means $\beta$ is metrizable if and only if $\beta = \tau$.
For compact $X$, we have $1 \in C_0(X)$, and then, because $p_1(f) = \lvert f\rvert_\infty$, $\tau \subset \beta$, and we're done.
Conversely, if $\beta = \tau$, we must have a $\phi \in C_0(X)$ with $\lvert f\rvert_\infty \leqslant p_\phi(f)$ for all $f\in C_b(X)$.
($\tau \subset \beta$ means that there are $\phi_1,\,\dotsc,\, \phi_k \in C_0(X)$ such that $\lvert f\rvert_\infty \leqslant \sup\limits_{1 \leqslant i \leqslant k} p_{\phi_i}(f)$ for all $f \in C_b(X)$. Since $\sup\limits_{1 \leqslant i \leqslant k} \lvert\phi_i\rvert \in C_0(X)$, a single non-negative function that dominates the supremum norm exists.)
Now, $\bigl(\forall f\in C_b(X)\bigr)(\lvert f\rvert_\infty \leqslant p_\phi(f))$ implies that $\inf\limits_{x \in X} \lvert\phi(x)\rvert \geqslant 1$ (locally compact Hausdorff spaces are $T_{3\frac12}$).
But $\phi \in C_0(X) \land \inf\limits_{x \in X} \lvert\phi(x)\rvert \geqslant 1$ implies that $X$ is compact.

Without using the open mapping theorem, we have to put a little more work into the proof.
First, we note that $p_\phi = p_{\lvert\phi\rvert}$, hence we can restrict ourselves to functions taking only nonnegative values. Also $p_{r\cdot\phi} = r\cdot p_\phi$ for $r > 0$, hence - because $C_0(X) \subset C_b(X)$ - we can restrict to functions with values in $[0,\,1]$, and furthermore $\max \{p_\phi,\, p_\psi\} = p_{\max \{\phi,\,\psi\}}$, thus the family of $p_\phi$ balls is closed under finite intersections, and $\mathcal{F} = \{p_\phi : \phi \in C_0(X), \phi(X) \subset [0,\,1]\}$ closed under finite maxima.
Next, we note that in any case, $\beta \subset \tau$. That follows since $p(f) \leqslant \lvert f\rvert_\infty$ for all $p \in \mathcal{F}$.
Now, that $\beta$ is metrizable when $X$ is compact is still as easy to prove as before. For compact $X$, we have $C_0(X) = C(X)$, hence the constant $1 \in C_0(X)$, and therefore $\lvert\:\cdot\,\rvert_\infty = p_1 \in \mathcal{F}$, thus $\tau \subset \beta$. And then $\beta = \tau$ is generated by the single norm $p_1$.
On the other hand, if $X$ is not compact, we show that no countable family of $\beta$-neighbourhoods of $0$ is a neighbourhood basis.
Let $(U_n)_{n\in \mathbb{N}}$ be a countable family of $\beta$-neighbourhoods of $0$. By shrinking, if necessary, we can assume that there are sequences $0 \leqslant \phi_k \leqslant \phi_{k+1} \leqslant 1$ in $C_0(X)$ and $1 > \varepsilon_k > \varepsilon_{k+1}$ with $\varepsilon_k \to 0$ such that $U_n = \{f \in C_b(X) : p_{\phi_n}(f) < \varepsilon_n\}$.
Since $X$ is not compact, for all $n\in \mathbb{N}$, there is an $x_n \in X$ such that $\phi_n(x_n) < \dfrac{\varepsilon_n}{2^n}$. By continuity and local compactness, there is a compact neighbourhood $W_n$ of $x_n$, and a continuous $\chi_n \colon X \to [0,\,1]$ with $\phi_n(x) < \dfrac{\varepsilon_n}{2^n}$ for all $x \in W_n$, $\chi_n(x_n) = 1$, and $\operatorname{supp}\chi_n \subset W_n$.
Let $\psi = \sum\limits_{n\in \mathbb{N}} 2^{-(n+1)}\chi_n$. Then $\psi \in C_0(X)$, and $\psi(X) \subset [0,\,1]$, hence $p_\psi \in \mathcal{F}$. (The continuity of $\psi$ follows from uniform convergence, $\lim\limits_{x\to\infty} \psi(x) = 0$, because for every $\varepsilon > 0$ there is an $n_0$ with $2^{-n_0} < \varepsilon$, and outside the compact set $W_0 \cup \dotsb \cup W_{n_0}$, we have $\psi(x) < \varepsilon$.)
Now, for every $n \in \mathbb{N}$, we have $p_{\phi_n}(2^n\chi_n) < \varepsilon_n$, hence $2^n\chi_n \in U_n$, but $p_\psi(2^n\chi_n) \geqslant 2^n\psi(x_n)\chi_n(x_n) \geqslant 2^n2^{-(n+1)}\chi_n(x_n)^2 = \frac12$, hence $\{f \in C_b(X) : p_\psi(f) < \frac13\}$ is not contained in any $U_n$, therefore $(U_n)_{n\in\mathbb{N}}$ is not a local base at $0$.
Thus, for non-compact $X$, $\beta$ is not metrizable.


This is the same as showing the zero map is a $G_\delta$ set in $(C_b(X),\beta)$ by problem 5.

Hmm, no, not if I don't misunderstand what you're saying. $\{0\}$ can be a $G_\delta$ in a topological vector space $E$ without $E$ being metrizable. Consider, in our setting, $\beta$ on $C_b(\mathbb{R})$ for an example. Let $\phi$ be the characteristic function of $[-1,\,1]$, with linear cutoff for $1 \leqslant\lvert x\rvert \leqslant 2$. Let $\phi_n(x) = \phi(2^{-n}x)$. Let $U_n = p_{\phi_n}^{-1}([0,\,2^{-n}))$. Then each $U_n$ is an open neighbourhood of $0$, but $\bigcap\limits_{n\in\mathbb{N}} U_n = \{0\}$, although we just saw that $\beta$ is not metrizable.

Since it was mentioned, a couple of remarks about the topology $\kappa$ of compact convergence.
First, let us note that the construction $p_\phi(f) = \sup\limits_{x\in X} \lvert \phi(x)f(x)\rvert$ gives a seminorm on $C_b(X)$ for all bounded functions $\phi \colon X \to \mathbb{R}$ (or $\mathbb{C}$, if one wants, but one could also restrict to non-negative real-valued functions). These seminorms are monotonic and (positive) homogeneous in $\phi$, i.e. $\lvert \phi\rvert \leqslant \lvert\psi\rvert \Rightarrow p_\phi \leqslant p_\psi$, and $p_{r\phi} = rp_\phi$ for $r > 0$, and conserve maxima, $\max \{ p_\phi,\, p_\psi\} = p_{\max \{\lvert \phi\rvert,\, \lvert\psi\rvert\}}$.
Let us say a bounded function $\phi$ dominates another such function $\psi$ if there is a positive constant $c$ with $\lvert \psi\rvert \leqslant c\lvert\phi\rvert$. It is then clear that each $p_\psi$-neighbourhood contains a $p_\phi$-neighbourhood.
By definition, $\kappa$ is induced by the family $\{ p_{\chi_K} : K \subset X \text{ compact}\}$.
$\kappa$ is also induced by $\{p_\phi : \phi \in C_c(X)\}$, since $X$ is locally compact.
$\phi \in C_c(X)$ is dominated by $\chi_{\operatorname{supp} \phi}$, hence the topology induced by the $(p_\phi)_{\phi \in C_c(X)}$ is coarser than $\kappa$.
But, since $X$ is locally compact, each compact $K \subset X$ has a relatively compact open neighbourhood $W$. Then, since $\overline{W}$ is compact, hence $T_4$, we can, by Tietze's extension theorem, extend the function $\chi_K$ to a continuous function $\phi_K$ on $\overline{W}$ with values in $[0,\,1]$ that vanishes on $\partial W$. $\phi_K$ can be trivially (meaning, by $0$) extended to $\psi_K \in C_c(X)$. By construction, $\psi_K$ dominates $\chi_K$, hence the topology induced by the $(p_\phi)_{\phi \in C_c(X)}$ is also finer than $\kappa$.
(Remark: the topology $\tau$ of uniform convergence is induced by $(p_\phi)_{\phi \in C_b(X)}$.)
Since $C_c(X) \subset C_0(X) \subset C_b(X)$, we have the inclusions $\kappa \subset \beta \subset \tau$.
For compact $X$, all three trivially coincide. For non-compact $X$, we saw above that $\beta \neq \tau$. For $\sigma$-compact $X$ - hemicompact, since $X$ is locally compact -, it is easy to see that $\kappa$ is metrizable (the characteristic functions of the sets in a normal exhaustion of $X$ dominate all other $\chi_K$). Then - if $X$ is not compact - we also have $\kappa \neq \beta$. It follows from the open mapping theorem, that then $(C_b(X),\, \kappa)$ is not complete, its completion is $C(X)$ (That can also be seen directly: For a normal exhaustion $K_n \subset\mspace{-2mu}\subset \overset{\circ}{K}_{n+1}$, choose continuous cutoff functions $\phi_n$ with $\phi_n(K_n) \subset \{1\}$, $0 \leqslant \phi_n \leqslant 1$ and $\operatorname{supp} \phi_n \subset \overset{\circ}{K}_{n+1}$. For any $f \in C(X)$, $\phi_n\cdot f \to f$ compactly.).
More generally, whenever $X$ contains an infinite subset without accumulation point (that is the case if $X$ is $\sigma$-compact but not compact), we have $\kappa \neq \beta$. We can then construct a family $F \subset C_b(X)$ that is bounded in $\kappa$, but not in $\beta$.
