Criterion for the Coarsest Topology Determined by Q to be Hausdorff

# Criterion for the Coarsest Topology Determined by a Set of Seminorms to be Hausdorff

Recall from The Coarsest Topology Determined by a Set of Seminorms on a Vector Space page that if $E$ is a vector space and if $Q$ is a collection of seminorms on $E$, then there is a coarsest topology on $E$ such that every seminorm in $Q$ is continuous, and such that $E$ equipped with this topology forms a locally convex topological vector space. This topology is called the **Coarsest Topology Determined by $Q$**.

Proposition 1: Let $E$ be a topological space and let $Q$ be a collection of seminorms on $E$. Then $E$ equipped with the coarsest topology determined by $Q$ is Hausdorff if and only if for each nonzero $x \in E$ there exists a seminorm $p \in Q$ such that $p(x) > 0$. |

**Proof:**$\Rightarrow$ Suppose that $E$ equipped with the coarsest topology determined by $Q$ is Hausdorff. Let $x \in E$ with $x \neq o$. If $p(x) = 0$ for all $p \in Q$, then $x \in \{ y : p(y) \leq \epsilon \}$ for all $p \in Q$ and for all $\epsilon > 0$. But then $x \in \{ y : \sup_{1 \leq i \leq n} p_i(y) \leq \epsilon \}$ for all $p_1, p_2, ..., p_n \in Q$ and for all $\epsilon > 0$. So every neighbourhood of the origin contains $x$. So for $x \neq o$, there is no pair of neighbourhoods $U$ of $x$ and $V$ of $o$ such that $U \cap V \neq \emptyset$, contradicting the hypothesis that $E$ equipped with the coarsest topology by $Q$ being Hausdorff.

- Thus, there must exist some $p \in Q$ such that $p(x) > 0$.

- $\Leftarrow$ Now suppose that for each nonzero $x \in E$ there exists a seminorm $p_x \in Q$ with $p_x(x) > 0$. Then for each $x \in E$ we have that $x \not \in \left \{ y : p_x(y) \leq \frac{p_x(x)}{2} \right \}$. Thus:

\begin{align} \bigcap_{p_1, p_2, ..., p_n \in Q, \: \epsilon > 0} \{ y : \sup_{1 \leq i \leq n} p_i(y) \leq \epsilon \} \subseteq \bigcap_{x \in E} \left \{ y : p_x(y) \leq \frac{p_x(x)}{2} \right \} = \{ 0 \} \end{align}

- Since $E$ equipped with the coarsest topology determined by $Q$ is a (locally convex) topological vector space, by the proposition on the Criterion for a Topological Vector Space to be Hausdorff page, we have that $E$ equipped with the coarsest topology determined by $Q$ is Hausdorff. $\blacksquare$