Proposition 1

— Let E be a Hausdorff locally convex space, and A a subset of E. Let $E_\sigma$ denote the space E with the weakened topology. a) If every infinite sequence of points of A has a limit point in E, then A is precompact in E. b) In order that A be relatively compact in E, it is necessary and sufficient that it is precompact in E and relatively compact in $E_\sigma$.

We shall prove a) by reductio ad absurdum. If A is not precompact, then by th. 3 of GT, II, § 3, No. 7, it follows that there exists a symmetric convex neighbourhood V of 0 in E such that A cannot be covered by a finite number of translates of V.

In other words, if $x_0, x_1, ..., x_{n-1}$ are points of A, then $A \notin \bigcup_{0 \leq i < n} (x_i + V)$ and so there exists a point $x_n$ of A such that $x_n - x_i \notin V$ for $0 \leq i < n$. Then, by induction on the integer $n$, we can construct an infinite sequence $(x_n)_{n \in \mathbf{N}}$ of points of A such that $x_n - x_m \notin V$ whenever $n > m$; since V is symmetric, we also have $x_m - x_n \notin V$ for $m \neq n$ and the sets $x_n + \frac{1}{2} V$ are disjoint. For every point $x$ in E, there exists at most one integer $n \geq 0$ such that $x_n \in x + \frac{1}{2} V$, hence the sequence $(x_n)_{n \in \mathbf{N}}$ does not have any limit point. This proves a).

Now suppose that A is precompact in E and is contained in a compact subset B of $E_\sigma$. Then B is complete in $E_\sigma$, hence also in E (IV, p. 5, Remark 2). We have $\overline{A} \subset B$, hence A is relatively compact in E. The converse is evident and b) follows.