Proposition 18

In an ordered topological vector space E, let H be a set directed by the relation $\leq$. If the section filter of H has a limit in E, then this limit is the upper bound of H.

For, let $b = \lim_{x \in H} x$; for every $y \in H$, the set of $x \in H$ such that $x \geq y$ is a set of the section filter of H, therefore b is a cluster point of this set; but as the set $x \geq y$ is closed in E, we have $b \geq y$, thus b is an upper bound of H. On the other hand, if a is an upper bound of H, then H is contained in the closed set $x \leq a$; as b is a cluster point of H, we have $b \leq a$, which completes the proof (II, p. 72, exerc. 42).