On Some Locally Convex Topologies of a Vector Space(Update)

On Some Locally Convex Topologies of a Vector Space(Update)

This is an update of my previous question in here. Suppose that $(X,\tau)$
is already a locally convex TVS. Let us denote by $X'$, the space of all
$\tau$-continuous linear functionals on $X$, the topological dual of $X$.
Then $$\Gamma=\{|f|: f\in X'\}$$ is a separating family of seminorms on
$X$. By a theorem of Rudin, there exists a locally convex topology on $X$,
commonly denoted by $\sigma:=\sigma(X,X')$, with the property that every
member of $\Gamma$ is $\sigma$-continuous. Specifically, $\sigma$ consists
of arbitrary unions of the translates of finite intersections of the sets
$\{x\in X: |f(x)|<\epsilon\}$, $f\in X'$ and $\epsilon>0$. Let us go back
to what Rudin claimed. Since $X$ is locally convex TVS, the family of
Minkowski's functionals $$P=\{\mu_A: A\mbox {is a convex balanced nbd of
}0\}$$ is a separating family of seminorms on $X$ and hence again induces
a locally convex topology $\tau_1$. He remarked further that
$\tau=\tau_1$. I got no problem with this. Norbert in here commented that
$\tau=\sigma$. This is what I am confused with. In the first place I don't
have any idea on how to relate the two generated families $\Gamma$ and
$P$. How I wish I could get a direct proof (if only possible) that
$\tau\subset \sigma$. I don't have any problem of showing that
$\sigma\subset \tau$. With this, I will be totally convinced that
$\tau=\sigma$.