4
$\begingroup$

Every complete Boolean algebra arises as the lattice of regular open sets in some topological space, namely given a complete Boolean algebra $B$, the corresponding Stone space $S(B)$ will be extremally disconnected, so in particular its regular open sets are precisely its clopen sets.

A semi-common generalization of topological spaces is the class of spaces $(X,\tau)$ such that $\varnothing,X\in\tau$ and such that $\tau$ is closed under arbitrary unions (no assumption about intersections). This is equivalent to a weakening of the Kuratowski closure axioms to the axioms of a general closure operator. (Wikipedia says that these are sometimes called 'closure spaces,' but I've also seen them called 'generalized topological spaces' in some papers. If anyone knows another name for these things I would really like to know it because searching for relevant literature has been difficult.)

A closure operator is equivalently specified by an interior operator, a function $\mathrm{int}:\mathcal{P}(X)\rightarrow \mathcal{P}(X)$ satisfying:

  • $\mathrm{int}(A)\subseteq A$
  • $A\subseteq B \Rightarrow \mathrm{int}(A)\subseteq \mathrm{int}(B)$
  • $\mathrm{int}(\mathrm{int}(A))=\mathrm{int}(A)$

(This definition makes sense in an arbitrary poset.) Under such an operator, the collection of open sets (sets such that $\mathrm{int}(U)=U$) forms a complete lattice with arbitrary meets given by set theoretic union. Unlike in a topological space finite joins are given by $U \wedge V = \mathrm{int}(U \wedge V)$. (This lattice is dual to the lattice of closed sets, but I'm phrasing this question in terms of open sets since I find it a little more familiar and because there's no standard name for the closure of complement operation.)

An interior operator gives rise to a natural exterior operator: $\mathrm{ext}(A)=\mathrm{int}(X \setminus A)$. Just like in a topological space one can show that for an open set $U$, $\mathrm{ext}(\mathrm{ext}(\mathrm{ext}(U)))=\mathrm{ext}(U)$, so call an open set $U$ 'regular' if it satisfies $\mathrm{ext}(\mathrm{ext}(U))=U$. (Note that unlike in topology it seems that the exterior operator is not definable from the structure of the lattice of opens alone.)

The map $U\mapsto \mathrm{ext}(\mathrm{ext}(U))$ gives a closure operator on the lattice of open sets, so we get that the collection of regular opens also forms a complete lattice whose join is the join of open sets (although again unlike in topology this is not set theoretic intersection) but whose meet is different, so in particular neither operation is set theoretic.

One can show that the collection of regular opens together with their meet, join, and the exterior operator form an orthocomplemented lattice, with the exterior playing the role of orthocomplementation, i.e. $ext$ is an order-reversing involution such that $U$ and $\mathrm{ext}(U)$ are always complements. This is a purely algebraic fact about Boolean algebras with interior operators satisfying the above axioms. With a finite counterexample search I found that in general the lattice does not need to be orthomodular.

I convinced myself that every complete lattice arises as the lattice of opens of a closure space, although it's easier to phrase in terms of closed sets and the dual lattice. Given a lattice $(L,0,1,\wedge,\vee)$ we can consider the collection of subsets of $L\setminus \{0\}$ and let a set be closed if it is of the form $(0,a]$ for some $a\in L\setminus \{0\}$ (so in particular the closure operator is $\mathrm{cl}(A)=(0,\bigvee A]$). By completeness of the lattice any intersection of sets of this form is of this form or empty and in particular the intersection's top is the greatest lower bound of the tops of the original sets, so the lattices agree (if we identify $\varnothing$ with $0$, this is why we need to remove $0$, otherwise we get an extra bottom element). To get a lattice of opens we can just start with the dual lattice.

That construction always produces trivial lattices of regular opens (no non-trivial open sets are disjoint, so the exterior of any non-empty set is empty), which shows that the exterior operator really does depend on the underlying set and not just the lattice of opens (since there are closure spaces with non-trivial lattices of regular opens). So finally we can come to the question:

Which complete orthocomplemented lattices arise as the lattice of regular opens in a closure space?

$\endgroup$
1
$\begingroup$

All orthocomplemented lattices may be so represented.

A complete ortholattice L is the system of regular closed sets in a closure space generated on the family F of all maximal increasing orthocomplement-free sets of L. The embedding of L in this space associates with each x in L a clopen set of F, whose clopen complement is the set associated with x', the orthocomplement of x. L is separated by F, so the lattice order corresponds isomorphically to the order on the clopen sets. Since these sets are clopen, they also form the system of regular open sets of L.

The work toward this result began in investigations into reconciliations of classical (Boolean) and quantum logics, and culminated with the independent and nearly simultaneous papers by Iturrioz, Katrnoska and Mayet in 1982. The relevant references are below.

L. Iturrioz, A simple proof of a characterization of complete orthocomplemented lattices, Bull. London Math. Soc. 14, no. 6 (1982), 542–544.

F. Katrnoˇska, On the representation of orthocomplemented posets, Comment. Math. Univ. Carolin. 23, no. 3 (1982), 489–498.

A.R. MARLOW: Quantum theory and Hilbert Space, Journal of Math. Phya. 19(1978), 1-15.

R. Mayet, Une dualit´e pour les ensembles ordonn´es orthocompl´ement´es, C. R. Acad. Sci. Paris S´er. I Math. 294, no. 2 (1982), 63–65.

Luigi Santocanale, Friedrich Wehrung. Lattices of regular closed subsets of closure spaces. International Journal of Algebra and Computation (IJAC), 2014, 24 (7), pp.969–1030.

M. Sekanina, On a characterisation of the system of all regularly closed sets in general closure spaces, Math. Nachr. 38 (1968), 61–66.

N. ZIERLER - M. SCHLESINGER: Boolean Embeddings of orthomodular sets and Quantum logics, Duke Math. Journal 32(1965), 251.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ How do you know that every regular open set is clopen? $\endgroup$ – James Hanson Oct 10 '19 at 0:53
  • 1
    $\begingroup$ Also can you give a precise definition of 'maximal increasing orthocomplement-free set'? $\endgroup$ – James Hanson Oct 10 '19 at 1:36
  • $\begingroup$ A subset X of L is increasing if $a \in X$ and $a \leq b \implies b \in X$. It is orthocomplement-free if $a \in X \implies a' \notin X$. It is maximal if it is orthocomplement-free and not contained in any other increasing set. Zorn's Lemma is invoked to assure existence of every increasing orthocomplement free set is contained in a maximal increasing orthocomplement free set. In each maximal orthocomplement-free increasing set there is precisely one of each pair of orthocomplements. These are called M-bases in Katrnoska and Marlow and orthosections finale in Mayet. $\endgroup$ – Walter Bruce Sinclair Oct 10 '19 at 2:03
  • $\begingroup$ You know that every regular open set is clopen because the interior of every closed set is clopen, so any set that is regular open, the closure of its interior, will be clopen. $\endgroup$ – Walter Bruce Sinclair Oct 10 '19 at 2:14
  • $\begingroup$ Okay so it's analogous to how the Stone space of a complete Boolean algebra is extremally disconnected (every regular open is clopen). I forgot that my original question was about complete orthocomplemented lattices. $\endgroup$ – James Hanson Oct 10 '19 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.