Comparing the profinite completion and the MacNeille completion of a modal algebra
by
Jacob Vosmaer
Institute for Logic, Language and Computation (University of Amsterdam)

Consider a modal algebra A (a Boolean algebra with an additional additive unary operator \Diamond). In general its profinite completion [^A] (a subalgebra of the product of all finite quotients of A) and MacNeille completion [`A] (an expansion of a completion of the underlying partial order of A) are not isomorphic. One simple reason is that the profinite completion [^A] is always atomic while the MacNeille completion [`A] on the other hand shares its atoms with A. Thus for any non-atomic A, it will be the case that [^A] \not ≅ [`A]. Moreover, A cannot always be embedded in its profinite completion [^A]: in fact, A can be embedded into [^A] iff A is residually finite. Since the map from A to its MacNeille completion [`A] is always injective, this is another reason while in general [^A] and [`A] are not isomorphic.

This sets the stage for our search for conditions under which the profinite completion and the MacNeille completion are isomorphic: we should at least assume that our modal algebra A is residually finite and atomic. It turns out that the atomicity is implicit if we assume that A is residually finite, finitely generated and transitive (satisfies \Diamond\Diamond a ≤ \Diamonda). Using duality with Kripke frames and a result of Chagrov and Zakharyaschev we can prove the following:
Theorem If A is a residually finite, finitely generated transitive modal algebra then the profinite completion [^A] is isomorphic to the lower MacNeille completion [`A].

References
Alexander Chagrov and Michael Zakharyaschev: Modal Logic, Clarendon Press, Oxford, 1997.

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Date received: May 14, 2007