Basic Properties of Precompact Sets in a LCTVS

Recall from the Precompact Sets in a LCTVS page that if $E$ is a locally convex topological vector space and if $A \subseteq E$ then $A$ is said to be precompact if for every absolutely convex neighbourhood $U$ of the origin there exists finitely many subsets $A_1, A_2, ..., A_n$ of $E$ that are small of order $U$ and such that:

(1)

\begin{align} \quad A \subseteq \bigcup_{i=1}^{n} A_i \end{align}

Equivalently, $A$ is precompact if for every absolutely convex neighbourhood $U$ of the origin there exists $a_1, a_2, ..., a_n \in E$ such that:

(2)

\begin{align} \quad A \subseteq \bigcup (a_i + U) \end{align}

We will now look at some properties of precompact sets.

Proposition 1: Let $E$ be a locally convex topological vector space. (1) If $A$ precompact then $\overline{A}$ is precompact. (2) If $A$ is precompact then $\lambda A$ is precompact for every $\lambda \in \mathbf{F}$. (3) If $B \subseteq A$ and $A$ is precompact, then $B$ is precompact. (4) If $A_1, A_2, ..., A_n$ is a finite collection of precompact sets, then $\displaystyle{\bigcup_{i=1}^{n} A_i}$ is precompact. (5) If $\{ A_{\alpha} : \alpha \}$ is an arbitrary collection of precompact sets, then $\displaystyle{\bigcap_{\alpha} A_i}$ is precompact. (6) If $A_1, A_2, ..., A_n$ is a finite collection of precompact sets, then $\displaystyle{\sum_{i=1}^{n} A_i}$ is precompact.

• Proof of (1): Let $A$ be precompact. Let $U$ be a closed absolutely convex neighbourhood of the origin. Then by the precompactness of $A$, there exists $a_1, a_2, ..., a_n \in E$ such that $\displaystyle{A \subseteq \bigcup_{i=1}^{n} (a_i + U)}$. Since $U$ is closed, $a_i + U$ is closed, and so $\displaystyle{\bigcup_{i=1}^{n} (a_i + U)}$ is closed as it is a finite union of closed sets. Therefore, taking the closure of the inclusion above yields that:

(3)

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

• Now, since $E$ is a locally convex topological vector space, it has a base of closed and absolutely convex neighbourhoods (see the Every LCTVS Has a Base of Closed Absolutely Convex Absorbent Neighbourhoods of the Origin page). Thus, given an arbitrary absolutely convex neighbourhood $V$ of the origin, there exists a closed absolutely convex neighbourhood $U$ of the origin for which $U \subseteq V$. Then by the precompactness of $A$, there is $a_1, a_2, ..., a_n \in E$ such that $\displaystyle{\overline{A} \subseteq \bigcup_{i=1}^{n} (a_i + U) \subseteq \bigcup_{i=1}^{n} (a_i + V)}$. Since such $a_1, a_2, ..., a_n$ exist for every absolutely convex neighbourhood $V$ of the origin, we have that $\overline{A}$ is precompact. $\blacksquare$

• Proof of (3): Let $A$ be precompact. Then for every absolutely convex neighbourhood $U$ of the origin there exists $A_1, A_2, ..., A_n$ which are small of order $U$ and such that $B \subseteq A \subseteq \displaystyle{A \subseteq \bigcup_{i=1}^{n} A_i}$. So $B$ is precompact. $\blacksquare$

• Proof of (4): Let $A_1$, $A_2$, …, $A_n$ be precompact. Let $U$ be an absolutely convex neighbourhood of the origin. For each $1 \leq i \leq n$ there exists $A_{i, 1}, A_{i, 2}, ..., A_{i, n_i}$ which are small of order $U$ and such that $\displaystyle{A_i \subseteq \bigcup_{k=1}^{n_i} A_{i, k}}$. Then $\displaystyle{\bigcup_{i=1}^{n} A_i \subseteq \bigcup_{i=1}^{n} \bigcup_{k=1}^{n_i} A_{i, k}}$, so $\bigcup_{i=1}^{n} A_i$ is precompact. $\blacksquare$

• Proof of (5): Let $\{ A_{\alpha} : \alpha \}$ be a collection of precompact sets. Then for each $\alpha^*$, $\displaystyle{\bigcap_{\alpha} A_{\alpha} \subseteq A_{\alpha^*}}$. Since $A_{\alpha^*}$ is precompact, apply (3) to see that $\displaystyle{\bigcap_{\alpha} A_{\alpha}}$ is precompact too. $\blacksquare$