Corollary

— Suppose in addition that $X$ is metrizable. Then the following properties are equivalent:

a) $a$ is an $\mathscr{A}$-extremal point of $X$.

b) There exists $u \in \mathscr{A}$ such that $|u(x)| < |u(a)|$ for all $x \neq a$.

c) Let $\mathfrak{M}$ be the set of subsets $M$ of $X$ such that for every function $f \in \mathscr{A}$, $|f|$ attains its supremum in $X$ at at least one point of $M$. Then $a$ belongs to all of the sets $M \in \mathfrak{M}$.

d) Let $\mathfrak{N}$ be the set of subsets $N$ of $X$ such that, for every function $f \in \mathscr{A}$, $\mathscr{R}(f)$ attains its supremum in $X$ at at least one point of $N$. Then $a$ belongs to all of the sets $N \in \mathfrak{N}$.

In other words,

Since, in a metrizable space, every point admits a countable fundamental system of neighborhoods, the equivalence of a) and b) follows from Th. 2. Let us show that b) implies c): indeed, $a$ is the unique point where $|u|$ attains its supremum; on the other hand, c) implies a) because, for every $f \in \mathscr{A}$, $\mathrm{Ch}_{\mathscr{A}}(X)$ intersects the set of points where $|f|$ attains its supremum (No. 4, Prop. 9). The same reasoning, using Prop. 5 of No. 3, shows that d) implies a). Finally, to see that b) implies d), we can restrict ourselves to the case that $X$ does not reduce to the single point $a$, therefore $u(a) \neq 0$; the function $v = u/u(a)$ then belongs to $\mathscr{A}$, and we have $v(a) = 1$ and $|v(x)| < 1$ for $x \neq a$, whence $\mathscr{R}(v(a)) = 1$ and $\mathscr{R}(v(x)) < 1$ for $x \neq a$. Since the function $\mathscr{R}(v)$ attains its supremum only at the point $a$, we have indeed $a \in N$ for every $N \in \mathfrak{N}$.