# Proof that a locally convex space $E$ is regular

Where a topological space $$E$$ is a regular space if, given any closed set $$F$$ and any point $$x$$ that does not belong to $$F$$, there exists a neighbourhood $$U$$ of $$x$$ and a neighbourhood $$V$$ of $$F$$ that are disjoint. Concisely put, it must be possible to separate $$x$$ and $$F$$ with disjoint neighborhoods.

Remember that the topology of a locally convex space $$E$$, is defined by a separable family $$\lbrace p_j \rbrace_{j \in J}$$ of seminorms and it is a vector topology of Hausdorff, where the family of finite intersections \begin{align*} \displaystyle U_{p_1}(\epsilon) \cap \cdot \cdot \cdot \cap U_{p_N}(\epsilon) = \lbrace x\in E: \max \lbrace p_1(x),...,p_N(x)\rbrace < \epsilon \rbrace \end{align*} is a local basis $$\mathcal{U}$$ for this topology $$\mathcal{T}_P$$.

Any suggestions?

• I do not understand what you would like to ask. – gerw Sep 23 '16 at 13:50
• @gerw how do I proof that the locally convex spaces are regular? – user288972 Sep 23 '16 at 13:55

I have changed your notation $U_{p_i}(\epsilon)$ into $B_i(x;\epsilon)$, which I believe to be nicer.
Let $x\in V-F$, then there exists a finite amount balls $B_{i}(x;\epsilon), i \in I$, so that the intersection of these balls lies in $V-F$. If $y$ is any point of $F$, since $y\notin V-F$ you have that $p_i(x-y)≥\epsilon$ for at least one $i\in I$ as otherwise $y$ would lie in $\bigcap_{i\in I}B_{i}(x;\epsilon)\subseteq V-F$.
Since semi-norms obey the triangle inequality you have for any $y\in F$ that if $x\notin B_{i}(y;\epsilon)$ then $B_i(x;\epsilon/2)\cap B_i(y;\epsilon/2)=\emptyset$, as otherwise $p_i(x-y)<\epsilon$.
It follows that $$\bigcup_{y\in F}\left(\bigcap_{i\in I} B_i(y;\epsilon/2)\right)$$ has empty intersection with $\bigcap_i B_i(x;\epsilon/2)$. But the first set is a neighbourhood of $F$ and the second set a neighbourhood of $x$.