[FOM] Correction to "A question about modal models"

[Dana Scott]

I forgot for a moment what "finite meets" has to
entail: the empty meet, which equals 1.

Hence, if L is a complete Boolean algebra, then
the two degenerate sublattices closed under all
finite meets and arbitrary joins are thus:

	H = {0, 1} and H = L.

The modal operators defined by either of these H
are trivial.

Of course -- just as with ordinary topological
spaces -- we can generate an H by closing a given
set, first, under finite meets, then, under arbitrary
joins.  This gives many non-degenerate examples.
Finding "interesting" examples is another question.

	-- Dana Scott