Closure, Conv. Hull, and Abs. Conv. Hull of a Bounded Set in a LCTVS
The Closure, Convex Hull, and Absolutely Convex Hull of a Bounded Set is a Bounded Set in a LCTVS
Recall from the Bounded Sets in a LCTVS page that if $E$ is a locally convex topological vector space and $A \subseteq E$ then $A$ is said to be bounded if for every neighbourhood $U$ of the origin, there exists a $\lambda_U > 0$ such that if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda_U$ then:
(1)\begin{align} \quad A \subseteq \mu U \end{align}
Equivalently, if $\mathcal U$ is a base of absolutely convex neighbourhoods of the origin, then $A$ is bounded if and only if for every $U \in \mathcal U$ there exists a $\lambda_U > 0$ such that:
(2)\begin{align} \quad A \subseteq \lambda_U U \end{align}
Using this characterization, we see that the closure, convex hull, and absolutely convex hull of a bounded set is bounded in a locally convex topological vector space.
| Proposition 1: Let $E$ be a locally convex topological vector space and let $A \subseteq E$. If $A$ is bounded then the closure of $A$, the convex hull of $A$, and the absolutely convex hull of $A$ are all bounded too. |
- Proof: Suppose that $A$ is bounded. Since $E$ is a locally convex topological vector space, by the result on the Every LCTVS Has a Base of Closed Absolutely Convex Absorbent Neighbourhoods of the Origin page, $E$ has a base of closed, absolutely convex, and absorbent neighbourhoods of the origin. Let $\mathcal U$ denote such a base.
- Since $E$ is a locally convex topological vector space and $A$ is bounded, by one of the equivalent notions for boundedness on the Bounded Sets in a LCTVS page, we have that for each $U \in \mathcal U$ there exists a $\lambda_U > 0$ such that $A \subseteq \lambda_U U$. Taking closures yields:
\begin{align} \quad \overline{A} \subseteq \overline{\lambda_U U} = \lambda_U U \end{align}
- (Where the last equality above comes from the assumption that $U$ is closed, so that $\lambda_U U$ is closed, and hence $\overline{\lambda_U U} = \lambda_U U$). Thus we see that $\overline{A}$ is bounded too.
- Furthermore, $A \subseteq \lambda_U U$ tells us that $\mathrm{conv}(A) \subseteq \lambda_U U$ since by definition, the convex hull of $A$ is the smallest convex set containing $A$, and that $\mathrm{abs \: conv}(A) \subseteq \lambda_U U$ since by definition, the absolutely convex hull of $A$ is the smallest absolutely convex set set containing $A$.
- Hence $\mathrm{conv}(A)$ and $\mathrm{abs \: conv}(A)$ are bounded too. $\blacksquare$