About a Weak Topology on TVS(part 2)

Question

Let $(X,\tau)$ be a topological vector space and suppose $P$ is a separating family of seminorms on $X$. Denote by $\sigma(X,P)$, the weak topology on $X$, the smallest topology on $X$ that makes each $p\in P$ continuous with respect to $\sigma(X,P)$. In here, it was shown that $\sigma(X,P)$ is compatible with the algebraic structure of $X$. My question is this. Is $\sigma(X,P)$ a Hausdorff topology? By the way, I assumed (like some authors did) that the original topology $\tau$ of $X$ is Hausdorff.

Answer 1

This is the reason why a family of seminorms $P$ such that for every $x \in X$ there is a $p \in P$ with $p(x) \gt 0$ is called separating.

The point is that for any two distinct points $x \neq y$ there is $p \in P$ such that $\varepsilon = p(x-y) \gt 0$. Observe that $U = \{z \in X \mid p(x-y) \gt \varepsilon /2\}$ and $V = \{ z \in X \mid p(x-y) \lt \varepsilon /2\}$ are non-empty, open and disjoint. Clearly, $0 \in V$ and $x-y \in U$ so that $y \in (y + V)$ and $x \in (y + U)$. In other words, $y + V$ and $y + U$ are open and disjoint neighborhoods of $y$ and $x$, respectively, so $(X, \sigma(X,P))$ is Hausdorff.

Notice that the original topology on $X$ plays no role in this argument.