Exercise 9, § 1

Let E be a fully lattice-ordered space. A filter $\mathscr{F}$ on E is said to be bounded above (resp. bounded below, bounded) if there is a set in $\mathscr{F}$ that is bounded above (resp. bounded below, bounded). For a filter $\mathscr{F}$ that is bounded above (resp. below), the limit superior (resp. limit inferior) of $\mathscr{F}$, denoted $\limsup \mathscr{F}$ (resp. $\liminf \mathscr{F}$) is defined to be the element $\inf_X (\sup X)$ (resp. $\sup_X (\inf X)$) of E, where X runs over the set of sets in $\mathscr{F}$ that are bounded above (resp. bounded below). If $\mathscr{F}$ is bounded, then $\liminf \mathscr{F} \leq \limsup \mathscr{F}$; $\mathscr{F}$ is said to have a limit, or to be convergent, for the order structure of E, if $\limsup \mathscr{F} = \liminf \mathscr{F}$; the common value of these two elements is then denoted $\lim \mathscr{F}$, and $\mathscr{F}$ is said to converge to this element of E. If a bounded filter has a limit, then every finer filter has the same limit (for the order structure).

a) In order that a bounded filter $\mathscr{F}$ have a limit for the order structure, it is necessary and sufficient that $\inf_X (\sup X - \inf X) = 0$ as X runs over the set of bounded elements of $\mathscr{F}$ (one can make use of the fact that if, in E, $(x_\alpha)$ and $(y_\alpha)$ are two decreasing families that are bounded below, having the same right-directed index set, then

to prove this formula, one observes that $\inf(x_\alpha + y_\alpha) \leq x_\beta + y_\gamma$ for every pair of indices $\beta, \gamma$.

b) Let A be a set filtered by a filter $\mathscr{G}$, $f$ a mapping of A into E; $f$ is said to have a limit with respect to $\mathscr{G}$ for the order structure of E if $f(\mathscr{G})$ is a base of a bounded filter having a limit in E (for the order structure). Show that if A is a directed ordered set and $f$ is an increasing mapping of A into E that is bounded above on A, then $f$ has a limit with respect to the section filter of A, equal to $\sup_{x \in A} f(x)$.

c) Let $(E_\ell)$ be a family of fully lattice-ordered spaces, E the product fully lattice-ordered space of the $E_\ell$. In order that a bounded filter $\mathscr{F}$ on E be convergent for the order structure, it is necessary and sufficient that, for every $\ell$, the filter with base $\mathrm{pr}_\ell(\mathscr{F})$ be convergent in $E_\ell$; if $a_\ell$ is its limit, then $a = (a_\ell)$ is the limit of $\mathscr{F}$.