Locally Convex tvs closure of $\{0\}$

Question

Let $E$ be a topological vector space locally convex, defined by the family of seminorms $\mathcal{F}=(p_j)_{j\in J}$.

I can't prove that $\underset{j\in J}\bigcap Ker(p_j)=\overline{\{0\}}$

Answer 1

Suppose $x \in \overline{\{0\}}$.

We know that all $p_j$ are by definition continuous in the topology generated by the semininorms, so for any $j \in J$: $$p_j(x) \in p_j[\overline{\{0\}}] \subseteq \overline{p_j[\{0\}]} = \overline{\{0\}} = \{0\}$$ as the last closure is taken in the reals (or complex numbers) where singletons are closed, and so $x \in \operatorname{Ker}(p_j)$ for all $j$, also using that $p_j(0)=0$ of course.

This shows $$\overline{\{0\}} \subseteq \bigcap_{j \in J} \operatorname{Ker}(p_j)$$

To see the other inclusion, let $x$ be in $\bigcap_{j \in J} \operatorname{Ker}(p_j)$ and let $O$ be a basic neighbourhood of $x$. This means that there are finitely many $p_{j_1},\ldots,p_{j_n}$ and an $\epsilon >0$ such that $O = \bigcap_{i=1}^n B_{j_i}(x,\varepsilon)$, where $B_{j_i}(x,\varepsilon)= \{y \in X: p_{j_i}(x - y) < \varepsilon\}$ is the open ball around $x$ induced by the seminorm $p_{j_i}$.

The fact that $x \in \operatorname{Ker}(p_{j_i})$ implies that $0 \in B_{j_i}(x,\varepsilon)$, and so $0 \in O$. As $O$ is an arbitary base element containing $x$ we have shown that $x \in \overline{\{0\}}$ as required.