Boundedness of Precompact Sets in a LCTVS

# Boundedness of Precompact Sets in a LCTVS

Proposition 1: Let $E$ be a locally convex topological vector space and let $A \subseteq E$. If $A$ is precompact then $A$ is bounded. |

**Proof:**Let $A$ be precompact and let $U$ be an absolutely convex and absorbent neighbourhood of the origin. Then there exists $a_1, a_2, ..., a_n \in E$ such that:

\begin{align} \quad A \subseteq \bigcup_{i=1}^{n} (a_i + U) \end{align}

- Since $U$ is absorbent, for each $1 \leq i \leq n$ there exists $\lambda_i > 0$ such that $a_i \in \mu U$ if $\mu \in \mathbf{F}$ and $|\mu| \geq \lambda_i$. In particular, $a_i \in \lambda_i U$. Let:

\begin{align} \quad \lambda = \max \{ \lambda_1, \lambda_2, ..., \lambda_n \} \end{align}

- Then $\lambda_i \leq \lambda$ for all $1 \leq i \leq n$. Since $U$ is absolutely convex, this implies that $\lambda_i \subseteq \lambda U$. Thus $a_i \in \lambda U$ for all $1 \leq i \leq n$. Hence:

\begin{align} \quad A \subseteq \bigcup_{i=1}^{n} (a_i + U) \subseteq \bigcup_{i=1}^{n} (\lambda U + U) = (1 + \lambda) U \end{align}

- Therefore $A$ is bounded. $\blacksquare$