# Counterexample on completely distributive lattices

I would like to see an example of a complete lattice $C$ which is both a frame and a dual-frame, i.e. finite meets distribute over arbitrary joins and finite joins distribute over arbitrary meets (http://en.wikipedia.org/wiki/Complete_Heyting_algebra), but $C$ is not completely distributive (http://en.wikipedia.org/wiki/Completely_distributive_lattice), i.e. arbitrary meets do not distribute over arbitrary joins (complete distributivity is a self-dual property).

I was unable to find this example in Gierz et al. "Compendium of continuous lattices" or in other standard references.

## 2 Answers

A complete Boolean algebra is both a frame and a co-frame. It is completely distributive if and only if it is atomic (see for example here), so all you need is a non-atomic complete Boolean algebra. The lattice of regular open sets (an open set is regular if it is equal to the interior of its closure) in say $[0, 1]$ is an example; this is complete because meets are calculated just as they are in the frame of all open sets.

• In case it helps, I have written out a proof that completely distributive Boolean algebras are atomic, in the nLab here: ncatlab.org/nlab/show/… – Todd Trimble Apr 18 '15 at 2:10

Let me give a few techniques which one can use to construct complete lattices which are frames and coframes but not completely distributive and I will explain why there are only a few classes of lattices which are both frames and coframes.

$\textbf{Constructions from Heyting algebras}$

The Dedekind-MacNeille completion of a Heyting algebra is always a Heyting algebra. In particular, if $X$ is both a a Heyting algebra and co-Heyting algebra, then the Dedekind-MacNeille completion of $X$ is both a Heyting algebra and co-Heyting algebra. Since the complete Heyting algebras are precisely the frames and the complete co-Heyting algebras are precisely the coframes, we conclude that if $X$ is a Heyting algebra and a co-Heyting algebra, then the Dedekind-MacNeille completion of $X$ is both a frame and a coframe. However, the Dedekind-MacNeille completion of a set $X$ is not necessarily completely distributive and the following proposition characterizes the posets whose Dedekind-MacNeille completion is a completely distributive lattice.

If $X$ is a poset, then let $\preceq$ be the refinement relation on $P(X)$. In other words, $R\preceq S$ if and only if for each $r\in R$ there is some $s\in S$ with $r\leq s$.

$\mathbf{Proposition}$ Suppose that $X$ is a poset and the Dedekind-MacNeille completion $DM(X)$ of $X$ is a frame. Then the following are equivalent:

1. $DM(X)$ is a completely distributive lattice

2. whenever $I$ is a set, $x\in X$, $R_{i}\subseteq X$, and $\bigvee^{X}R_{i}=x$ for $i\in I$, then there is some $S$ with $S\preceq R_{i}$.

3. for each $x\in X$, there is a smallest downwards closed subset $L\subseteq X$ with $\bigvee L=x$.

$\textbf{Being both a frame and a coframe is unusual}$

I should however mention that being both a frame and a coframe is a very special property in point-free topology: A frame $L$ is also a coframe if and only if the least upper bound $\bigvee_{i\in I}C_{i}$ of a system of closed sublocales $\{C_{i}|i\in I\}$ is a closed sublocale (this least upper bound is taken in the coframe of sublocales of $L$). This property is a very unusual property: If $\kappa$ is a regular cardinal, then the property that the least upper bound of less than $\kappa$ closed sublocales is a closed sublocale corresponds somewhat to the notion of a $P_{\kappa}$-space since the least upper bound of sublocales corresponds to the union of closed subsets of a topological space.

A frame $L$ is said to be subfit if whenever $a\not\leq b$ there is some $c\in L$ where $a\vee c=1\neq b\vee c$. Equivalently, a frame $L$ is subfit if every open sublocale is the least upper bound of closed sublocales. The notion of a subfit frame is a point-free separation axiom that is slightly weaker than $T_{1}$ in the sense that every $T_{1}$-space is subfit. Furthermore, every regular frame is subfit. Therefore, most frames that one encounters in point-free topology are subfit frames. However, every subfit frame which is also a coframe is a complete Boolean algebra. Therefore, most frames that are coframes that one encounters in point-free topology are simply complete Boolean algebras.

$\textbf{When$DM(X)$is a frame}$

The following remarks give necessary and sufficient conditions for when $DM(X)$ is a frame. By duality one obtains necessary and sufficient conditions for when $DM(X)$ is a coframe.

A closure system is a pair $(X,C)$ where $X$ is a set and $C\subseteq P(X)$ is a collection of subsets of $X$ closed under arbitrary intersection. It is easy to show that $(X,C)$ is a closure system if and only if for each $R\subseteq X$ there is a least $S\subseteq X$ with $R\subseteq S$. If $(X,C)$ is a closure system, then define a mapping $C^{*}:P(X)\rightarrow P(X)$ by letting $C^{*}(R)$ be the smallest set in $C$ containing $R$ whenever $R\subseteq X$. If $(X,C)$ is a closure system, then let $\leq$ be the preordering where $x\leq y$ iff $x\in C^{*}(\{y\})$. Then $\leq$ is called the specialization ordering. A $T_{0}$-closure system is a closure system where the specialization ordering is a partial ordering. If $X$ is a poset and $R\subseteq X$, then let $\uparrow R$ denote the set of all upper bounds of $R$ and let $\downarrow R$ denote the set of all lower bounds of $R$.

The following fact was proven by Marcel Erne in one of his papers.

$\mathbf{Proposition}$ Suppose that $(X,C)$ is a $T_{0}$-closure system. Then $C$ is a frame if and only if whenever $L$ is downwards closed and $x\in C^{*}(L)$, we have $x\in C^{*}(L\cap\downarrow x)$.

$\mathbf{Proposition}$ Let $X$ be a poset. Then $DM(X)$ is a frame if and only if whenever $R\subseteq X,x\in X$ and every upper bound of $R$ is also an upper bound of $x$, then there is some $S\subseteq X$ with $S\preceq R$ and with $\bigvee S=x$.

$\mathbf{Proof}$

$\rightarrow$ Suppose that $DM(X)$ is a frame. Let $R\subseteq X,x\in X$ and suppose that every upper bound of $R$ is also an upper bound of $x$. Let $L$ be the smallest downwards closed set containing $R$. Then $\uparrow R$ is the set of all upper bounds of $R$, but since every upper bound of $R$ is also an upper bound of $x$, we have $x\in\downarrow\uparrow R=DM(X)^{*}(R)=DM(X)^{*}(L)$. Therefore $x\in DM(X)^{*}(L\cap\downarrow x)$. Clearly $L\cap\downarrow x\preceq R$. Furthermore, $x$ is an upper bound of $L\cap\downarrow x$. If $y$ is another upper bound of $L\cap\downarrow x$, then $L\cap\downarrow x\subseteq\downarrow y$, so $x\in DM(X)^{*}(L\cap\downarrow x)\subseteq\downarrow y$. Therefore $x\leq y$. We conclude that $x$ is the least upper bound of $L\cap\downarrow x$.

$\leftarrow$ Let $L$ be a downwards closed set and assume that $x\in DM^{*}(L)=\downarrow\uparrow L$. Then every upper bound of $L$ is also an upper bound of $x$, so there is an $S\subseteq L$ with $x=\bigvee S$. However, since $DM(X)^{*}(S)=\downarrow\uparrow S$, we have $x\in DM^{*}(S)\subseteq DM^{*}(\downarrow x\cap L)$. $\mathbf{QED}$.

One may ask if there are other completions of a certain poset which are both frames and coframes. However, I only focused on the Dedekind-MacNielle completion of a poset since

1. the frames $L$ that extend a poset $X$ such that $L=\{\bigvee^{L}R|R\subseteq X\}$ can be characterized in terms of certain Grothendieck topologies on $X$ and are the most closely related to $X$ itself, and

2. the only extension of a poset $X$ to a complete lattice $L$ such that $L=\{\bigvee^{L}R|R\subseteq X\},\{\bigwedge^{L}R|R\subseteq X\}$ is the Dedekind-MacNeille completion of $X$.