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I find the following two statements contradictory. Can anyone explain me why they are not?

  1. (from G. Gierz et al., Continuous Lattices and Domains, Proposition O-2.2) Let $L$ be a poset. For $L$ to be a complete lattice it is sufficient to assume the existence of arbitrary sups (or the existence of arbitrary infs).

  2. (from S. Willard, General Topology, 12B.1) The intersection of any number of filters on (a topological space) $X$ is a filter on X. But the set of all filters on $X$, ordered by $\mathcal{F}_1 \leq \mathcal{F}_2$ iff $\mathcal{F}_1 \subseteq \mathcal{F}_2$, is not a lattice because if $\mathcal{F}$ and $\mathcal{G}$ are different ultrafilters on $X$, then $\{\mathcal{F},\mathcal{G}\}$ has no supremum.

Giorgio Bacci
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1 Answers1

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Try to carry out the dual of the argument in Gierz et al. on the poset $\mathfrak{F}$ of filters on $X$ ordered by $\subseteq$. For any $\mathfrak{X}\subseteq\mathfrak{F}$ this requires you to form $\mathfrak{B}=\bigcap\{\uparrow\!\!\mathscr{F}:\mathscr{F}\in\mathfrak{X}\}$, setting $\mathfrak{B}=\mathfrak{F}$ if $\mathfrak{X}=\varnothing$. Then you must take $\inf\mathfrak{B}$. When $\mathfrak{X}=\{\mathscr{F}_1,\mathscr{F}_2\}$, where $\mathscr{F}_1$ and $\mathscr{F}_2$ are ultrafilters, this requires you to take

$$\inf(\uparrow\!\!\mathscr{F}_1\cap\uparrow\!\!\mathscr{F}_2)=\inf\big(\{\mathscr{F}_1\}\cap\{\mathscr{F}_2\}\big)=\inf\varnothing\;.$$

But in any poset $\inf\varnothing$ is the maximum element of the poset if it exists and is otherwise undefined. $\mathfrak{F}$ has no maximum element, so $\inf\varnothing$ is undefined. Thus, $\langle\mathfrak{F},\subseteq\rangle$ does not have arbitrary infs after all: it only has infs of non-empty subsets.

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Brian M. Scott
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