Explicit construction of a nested local base in the weak*-topology Consider the $C^*$-algebra $C([a,b])$ of complex continuous functions in a closed interval of $\mathbb{R}$. I want to construct a countable nested local base in the state space of the algebra with respect to the $weak*$-topology. Such a base exists since the algebra is separable. How can we construct it explicitly?
 A: Theorem.  Let $X$ be a normed space.

*

*If the algebraic dimension of $X$ is  uncountable then the origin of the dual space $X'$ does not admit a countable
neighborhood base relative to the weak$^*$ topology.


*If $X$ is separable then every bounded subset of $X'$ is metrizable relative to the weak$^*$ topology.
Proof.  (1) Suppose by contradiction that $\{V_n\}_n$ is a countable neighborhood base of the origin.  By definition of
the weak$^*$ topology,  for each $n$  there is a finite set $F_n\subseteq X$, and a positive number $\varepsilon _n$,  such that the set
$$
  U(F_n, \varepsilon _n):= \{f\in  X': |f(x)|<\varepsilon _n,  \forall  x\in  F_n\}
  $$
is contained in $V_n$.
By hypothesis we have that
$$
  \text{span}\left(\bigcup_nF_n\right) \subsetneq X,
  $$
so we may pick some vector $y$ in $X$
not lying in the linear span of  $\bigcup_nF_n$.  Observing that
$U(\{y\},1)$ is a neighborhood of the origin in $X'$, we will reach a contradiction by proving that
$$
  V_n\not\subseteq U(G,1),
  $$
for any given $n$.  In fact, since $y\notin \text{span}(F_n)$, we may use Hahn-Banach to produce a continuous linear
functional $f$ on $X$ vanishing on $F_n$, and such that $f(y)=1$.    It then follows that any multiple of $f$ lies in
$U(F_n, \varepsilon _n)$, and hence also in
$V_n$, but $\lambda f$ is not in $U(G,1)$ if $|\lambda |\geq 1$.
(2)  Assuming that $X$ is separable,
choose a dense subset  $\{x_n\}_n$ of the unit ball of $X$.  Given any $f$ and $g$ in $X'$, define
$$
  d(f,g) = \sum_n 2^{-n}\min\{1,  |f(x_n)-g(x_n)|\}.
  $$
It is easy to see that $d$ is a metric on $X'$ and moreover that a net $\{f_i\}_i$ converges to some $f$ according to
this metric iff
$$
  f_i(x_n)\ {\buildrel i\to\infty \over \longrightarrow}\ f(x_n),\quad\forall n.
  \tag 1
  $$
On the other hand,  it is well known that a bounded net $\{f_i\}_i$ converges to  $f$ in the weak$^*$ topology iff (1)
holds.
So if $S$ is a bounded subset of $X'$, the nets within  $S$ converge weak$^*$ iff they converge according to $d$,
thus proving that the weak$^*$ topology on $S$ is metrizable by $d$.
QED

It is well known that the algebraic dimension of any infinite dimension Banach space  is uncountable,  so the Theorem
above applies to $C([0, 1])$ and hence the origin in $C([0, 1])'$ does not admit a countable neighborhood base relative
to the weak$^*$ topology.
On the other hand, the state space of  $C([0, 1])$ is a bounded set, so part (2) of the above Theorem applies.
