Precompact Sets in a LCTVS

# Precompact Sets in a LCTVS

## Sets that are Small of Order U

 Definition: Let $E$ be a locally convex topological vector space, let $A \subseteq E$, and let $U$ be a neighbourhood of the origin. Then $A$ is Small of Order $U$ if for all $x, y \in A$ we have that $x - y \in U$.

## Precompact Sets

 Definition: Let $E$ be a locally convex topological vector spaces and let $A \subseteq E$. Then $A$ is said to be Precompact if for every absolutely convex neighbourhood $U$ of the origin, there exists a finite collection of sets $A_1, A_2, ..., A_n$ that are all small of order $U$, and such that $\displaystyle{A \subseteq \bigcup_{i=1}^{n} A_i}$.

That is, $A$ is precompact if for every absolutely convex neighbourhood $U$ of the origin, $A$ can be covered by a finite collection of subsets of $E$ that are all small of order $U$.

We have an alternative classification for precompact sets:

 Proposition 1: Let $E$ be a locally convex topological vector space and let $A \subseteq E$. Then $A$ is precompact if and only if for every absolutely convex neighbourhood $U$ of the origin there exists finitely many points $a_1, a_2, ..., a_n \in E$ such that $\displaystyle{A \subseteq \bigcup_{i=1}^{n} (a_i + U)}$.
• Proof: $\Rightarrow$ Suppose that $A$ is precompact. Then there exists 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}
• For each $1 \leq i \leq n$, take any point $a_i \in A_i$. Then, for each $a \in A$, since $A \subseteq \bigcup_{i=1}^{n} A_i$ there exists a $1 \leq j \leq n$ such that $a \in A_j$. Since $A_j$ is small of order $U$ and since $a, a_j \in A_j$, we have that $a - a_j \in U$. Thus $a \in a_j + U$. So every $a \in A$ is contained in $a_i + U$ for some $1 \leq i \leq n$, and thus:
(2)
\begin{align} \quad A \subseteq \bigcup_{i=1}^{n} (a_i + U) \end{align}
• $\Leftarrow$ Suppose that for every absolutely convex neighbourhood $U$ here exists $a_1, a_2, ..., a_n \in E$ such that $\displaystyle{A \subseteq \bigcup_{i=1}^{n} (a_i + U)}$. For each $1 \leq i \leq n$, let $A_i := a_i + U$. Let $x, y \in a_i + U$. Then $x = a_i + u_1$ and $y = a_i + u_2$ for some $u_1, u_2 \in U$. Then $x - y = (a_i - u_1) - (a_i - u_2) = u_2 - u_1 \in U - U = U$ (note that $U - U = U$ since $U$ is absolutely convex).
• Thus $A_i$ is small of order $U$, and clearly $\displaystyle{A \subseteq \bigcup_{i=1}^{n} (a_i + U) = \bigcup_{i=1}^{n} A_i}$. So $A$ is precompact. $\blacksquare$