Page 51 - Textos de Matemática Vol. 41

6.7. Complemented sublocales 37
6.7. Complemented sublocales. All (classical) subspaces of a (classical) space constitute, for trivial reasons, a Boolean algebra: they are determined by subsets of the space in question.
The reader hardly expects that the same should hold for the lattice Sl(L) of all sublocales of a locale L. Rather, the fact that the open and closed sublocales complement each other (which, of course, leads to complementarity of many other ones) comes as a pleasant special fact.
6.7.1. A sublocale S ⊆ L is complemented if it has a complement T, that is, there is a sublocale T such that
S∩T =O and S∨T =L.
Thus for instance, all finite combinations (in ∩ and ∨) of open and closed sublo-
cales are complemented. We have the obvious
Observation. The complemented sublocales of a locale L constitute a Boolean
sublattice Slcomp(L) of Sl(L).
Note that by 6.5 each sublocale is a meet of complemented ones. Thus, it is seldom the case that Slcomp(L) is a complete sublattice: indeed it happens only if Slcomp(L) = Sl(L).
This fact of 6.5 has an important interpretation in C(L) (dual to Sl(L), see 5.6): this locale is zero-dimensional (using the obvious extension of the term from classical topology) in the sense that each element is a join of complemented ones.
6.7.2. There are many complemented locales that are not finite combinations of open and closed ones. A very involved descriptive characterization has been presented by Isbell in [33]. This is beyond our means here. We will present instead a simple external (algebraic) characterization (also following from much more general Isbell’s results [30]).
Proposition. A sublocale S is complemented in Sl(L) iff it satisfies, for any system (Ti)i∈J, the distributive property
( Ti) ∩ S = (Ti ∩ S). (∗) i∈J i∈J
Proof. ⇒: Let T be the complement of S. Then we have
(Ti)∩S∨S=S=(Ti∩S)∨S and i∈J i∈J
(Ti)∩S ∨T=(Ti)∨T= (Ti∩S) ∨T. i∈J i∈J i∈J