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?

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


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