Equicontinuity Classification of Hausdorff Locally Compact Topologies

# Equicontinuity Classification of Hausdorff Locally Compact Topologies

Theorem 1: Let $E$ be a Hausdorff locally convex topological vector space, let $\tau$ denote its Hausdorff and locally convex topology and let $\mathcal A$ be the collection of all $\tau$-equicontinuous sets of linear forms on $E$. Then $\tau$ is the topology of $\mathcal A$-convergence. |

**Proof:**Let $\mathcal A$ be the set of all $\tau$-equicontinuous sets of linear forms on $E$. We first need to show that $\mathcal A$ is a collection of $\sigma (E', E)$-weakly bounded sets to even consider the topology $\mathcal A$-convergence.

- Since $\tau$ is a locally convex topology, it has a base of closed and absolutely convex neighbourhoods and absorbent of the origin as proven on the Every LCTVS Has a Base of Closed Absolutely Convex Absorbent Neighbourhoods of the Origin page. Let $\mathcal U$ denote this base.

- Let $V \in \mathcal A$. Since $\tau$ is Hausdorff and since $V$ is $\tau$-equicontinuous, by The Polar Criterion for Equicontinuity of a Set of Linear Forms there exists a $U \in \mathcal U$ such that $V \subseteq U^{\circ}$. Taking polars again and using one of the properties on The Polar of a Set page, and we see that $U^{\circ \circ} \subseteq V^{\circ}$.

- Since $\tau$ is Hausdorff and since each $U \in \mathcal U$ is closed and absolutely convex, by considering the pair of dual pairs $(E, E^{\tau'})$ and $(E^{\tau'}, E)$, we have by the theorem on the If E is a Hausdorff LCTVS then A°° (in E) is the Closed Absolutely Convex Hull of A page that for each $U \in \mathcal U$:

\begin{align} U^{\circ \circ} = \overline{\mathrm{abs \: conv} (U)} = U \end{align}

- Therefore $U \subseteq V^{\circ}$. But since $U$ is absorbent, so is $V^{\circ}$. Indeed, if $x \in E$ then by the absorbency of $U$ there exists a $\lambda > 0$ such that if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda$ then $x \in \mu U$. Since $U \subseteq V^{\circ}$ we see that if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda$ then $x \in \mu V^{\circ}$. So $V^{\circ}$ is absorbent (in $E$). So by the Criteria for a Subset A to be σ(E, F)-Weakly Bounded, each $V \in \mathcal A$ is a $\sigma(E', E)$-weakly bounded set.

- All that remains to show is that $\tau$ is the topology of $\mathcal A$-convergence.

- First note that since $U = U^{\circ \circ}$ for all $U \in \mathcal U$, we have that $\tau$ is the topology of uniform convergence of the sets $\{ U^{\circ} : U \in \mathcal U \}$ and that $\{ U^{\circ} : U \in \mathcal U \}$ is a base for $\tau$. If $V \in \mathcal A$ then there exists a $U \in \mathcal U$ such that:

\begin{align} V \subseteq U^{\circ} \end{align}

- However, $\{ V^{\circ} : V \in \mathcal A \}$ is a base of neighbourhoods for the topology of $\mathcal A$-convergence, while $\{ U^{\circ \circ} : U \in \mathcal U \} = \mathcal U$ is a base of neighbourhoods for $\tau$. Therefore the topology of $\mathcal A$-convergence is coarser than $\tau$.

- On the otherhand, by taking polars again we see that:

\begin{align} U \subseteq V^{\circ} \end{align}

- So every neighbourhood in $\mathcal U$ is contained in a neighbourhood of the topology of $\mathcal A$-convergence. Therefore $\tau$ is coarser than the topology of $\mathcal A$-convergence.

- Thus $\tau$ is the topology of $\mathcal A$-convergence. $\blacksquare$