Download Duality between modal algebras and neighbourhood frames

KOs
A Duality Between Modal Algebras
Oo
EN and Neighbourhood Frames
Abstract. This paper presents duality results between categories of neighbourhood frames for
modal logic and categories of modal algebras (i.e. Boolean algebras with an additional unary
operation). These results extend results of Goldblatt and Thomason about categories of relational
frames for modal logic.
The representation theory for Boolean algebras with additional unary
operations started with the work of J6nsson and Tarski [6]. In the seventies,
modal logicians became aware of this algebraic theory, and they incorporated
it in the theory of so-called Kripke relational frames for modal logic. Two
standard results from this period are to be found in [4] and I1 i], where it is
shown that certain categories of relational frames are dual to categories of
Boolean algebras with operations called normal modal algebras (see Sections
8-10,16 below); when we say that two categories are dual, we mean that they are
equivalent by two contravariant functors. Some notions which appear in these
duality results seem to have been formulated independently by modal logicians
and algebraists, as witness the notion of frame homomorphism from [4] (p. 53;
see Section l0 below), whose analogues we can find in [8] (p. 37) and [12] (p.
2). In general, one might say that on the logical side a greater stress is made on
the set-theoretical aspects of the underlying Stone duality, whereas on the
algebraic side topological aspects are more prominent (see, for example, [5];
however, the related duality theory for Heyting algebras, worked out by the
logician Esakia in [2] and [3], stresses also topological aspects).
In this paper we shall present in a rather self-contained manner what,
starting from a logical background, can be said concerning duality for Boolean
algebras with an arbitrary additional unary operation. These algebras, which
we shall call modal algebras, form a wider class than normal modal algebras.
For this duality we shall consider a more general class of frames which occur in
modal logic, viz. neighbourhood frames. The underlying Stone duality will be
manifested by its set-theoretical aspects.
In the first part of the paper we show what category of neighbourhood
frames is dual to the category defined by modal algebras and their homomorphisms. These neighbourhood frames, and the required frame morphisms, are
a generalization of the descriptive relational frames and frame homomorphisms
of Goldblatt [4]. In general, the duality results we present are an extension of
ideas in [4]. We shall compare our results with Goldblatt's, and show how they
are related. In the second, shorter part of the paper, we consider what category
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K. Dogen
of modal algebras is dual to the category of full neighbourhood frames and
appropriate frame morphisms. The results of this part build on corresponding
results of Thomason [11] about categories of full relational frames. At the end,
we shall compare our results with Thomason's.
The general background of this paper is provided by [4]. Neighbourhood
frames for modal logic (sometimes called Scott-Montagueframes) are treated in
[10] and [1] (Part IIl; there these frames are called minimal). Basic notions
which we presuppose about Boolean algebras and categories may be found in
[9] and [7].
A number of proofs, which we consider straightforward, will be omitted.
The remaining proofs are mostly sketches, but the reader should have no
difficulty in reconstructing complete proofs.
General neighbourhood frames
1. The category M A . A modal algebra is an algebra A = (D, ^ , v ,
~ , 1,0, l--I),where (D, A, V, ~ , 1, 0) is a non-degenerate Boolean algebra,
and the domain D is closed under the unary operation []. For elements of D we
shall use b, b I , b 2 . . . . . and for elements of the power set .~"D of D we shall use
X, X l , X 2. . . .
The category M A of modal algebras is defined by the following:
objects: modal algebras,
morphisms: homomorphisms.
We shall use h, h 1, h2, ... for homomorphisms between modal algebras, and
Id a will be the identity homomorphism of A.
2. Descriptive neighbourhood flames. A general neighbourhood frame, or
general frame for short, is F = (C, N, Dr), where:
(l)
(2)
(3)
(4)
C # O (C is the carrier of F),
D F _ ~ C is closed under ~ (finite set intersection), - (set complementation with respect to C), and a unary operation L, and D r # 0,
N : C - - , ~ D r (if x e C , then N(x) is the set of neighbourhoods of x),
(V B e D v) (x e LB r B e S (x)).
For elements of C we shall use x, x 1, x2, ..., and for elements of ~ C we
shall use B, B 1, B 2. . . .
For every general frame F let J F = (D r, n , w, - , C, 0 , L). It is clear
that the following holds:
THEOREM 1A.
~F
is a modal algebra.
Intuitively, a general frame is a set modal algebra spread over a carrier C,
and by (4) we can take that either L is defined in terms of N, or N in terms of L.
(To define a valuation from a modal propositional language with a single
modal operator into a modal algebra, we proceed in the usual obvious way. In
Duality between modal algebras ...
221
virtue of the connexion between L and N, a valuation on .~r serves as
a valuation on F.)
Had we allowed modal algebras to be based also on degenerate Boolean
algebras (with a single element), then in general frames we should take C to be
any set, including O, and nothing would change essentially in the results which
follow. (However, we do not allow degenerate algebras and frames with an
empty carrier, because in them we could validate the inconsistent logic made of
all formulae.)
As over a general frame F, i.e. over its carrier, we have spread a modal
algebra d F , so over a modal algebra A we can spread a general frame
,~A = (C A, N A, D~A>, where:
(1)
(2)
C A = {X _= D: X is an ultrafilter of A},
q:D~(~D)
is defined by q ( b ) = { X e C A : beX}, and D~A={q(b):
beD},
(3) N A : C ' I ~ D '~A is defined by NA(X)= {q(b): D b e X } ,
(4) (VbED) LA(q(b)) = {XeC'~: q(b)~NA(X)}.
(By an ultrafilter we understand as usual a maximal proper filter.)
To confirm that .~-A is indeed a general frame we must verify that D*A is
closed under ~, - and LA.
The mapping q:D--*D ~A is a mapping from A to ~r
It is not difficult
to prove the following for q:
THEOREM 2A.
The mapping q is an isomorphism from A to zC(,~A).
Analogously to the mapping q we define a mapping p from F to J~(.zCF), i.e.
from the carrier of F to the carrier of ff(~CF). Let p:C--,g2(.~
be defined by
p(x) = {B e Dr: x e B}. It is easy to check that p(x) is an ultrafilter of ~r hence,
p : C ~ C ~r.
N o w we can define descriptive neighbourhood frames, or descriptive frames
for short, as general frames F which satisfy:
(I)
(II)
p(x~) = p(x2)='xl = x2,
(VXeC~F)(qxeC) X = p(x).
In other words, (I) p is one-one, and (II) p is onto.
3. Frame morphisms. The objects of the category DF of descriptive frames
will be descriptive frames. N o w we shall define the morphisms of this category.
Let F t = (C1, N t, DF~> and F 2 = (C2,N2,D F2> be general frames. Let
f:Cl ~C2, and for every BzeD r2, let (.~r
{xl: f(xl)cB2}. Then f is
a frame morphism from F 1 to F 2 iff for every B2eD ~'2 and every x l e C l :
(i) (~r
(ii)
D F',
(s~cf)(B2)e Nl(Xt)c:,B2~ N2(f (xl)).
We shall use f, ]'1, f2,-.- for frame morphisms.
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Logica
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K. Dogen
It is a straightforward matter to prove the following theorem, where d . f is
as above:
THEOREM 3A. The mapping fi C 1 --. C z is a frame morphism from Ft to F a
iff s l f is a homomorphism from d F 2 to d F r
So, if in the definition of general frames we were to take L as primitive, and
N as defined, it would be natural to define a frame morphism from F 1 to F z as
an f : C 1-*C 2 such that sCf is a homomorphism from ~r 2 to ~r r
A frame morphism f from F t to F 2 is a frame isomorphism i f f f is one-one
and onto, a n d f -1 is a frame morphism from F 2 to F 1. (In general, there is no
guarantee that if f is one-one and onto, then f - J is a flame morphism: for
example, if D F # ~C, the identity map f from (C, N, .~C> to (C, N, Dr> is
a frame morphism, but f-1 does not satisfy (i); this example is from [4], p. 53.)
An alternative definition of frame isomorphisms is provided by the following.
Let f be a frame morphism from F1 to F 2 and for every B l e D r' let
f i B 1 ] = {f(xt): x t e B t } . Then f is an embedding i f f f is one-one and:
(iii)
(VB t e D r l ) ( 3 B 2 e D r2) f i B 1 ] = f [ C l ] n B 2.
We have the following lemma:
LEMMA 1. Suppose the frame morphism f from F t to F z is one-one and onto.
Then f -t is a frame morphism from F 2 to F t iff f is an embedding.
PROOF. (=:-) Let f - ~ be a frame morphism from F 2 to F t. Then, by
Theorem 3A, ~r
t): D r, __,Dr2 is a homomorphism. It is not difficult to check
that f [ B 1 ] = f [ C l ] ~ ( ~ r
(r
Let (iii). hold for f To check (i) for f - t
we show that
(zg(f-a))(Bt) = B z, It is not difficult to check (ii) for f - i
q.e.d.
4. The category DF. It is easy to verify that the identity mapping on C is
a frame morphism from F to F; we call this mapping Id r. To verify that the
composition of frame morphisms is a frame morphism we just check that
,~r
= (zaCf0o(zaCf2),and use Theorem 3A. So, we may define the category
D F of descriptive frames by the following:
objects: descriptive neighbourhood frames,
morphisms: frame morphisms.
5. The funetor ,d. In Sections 2 and 3 we have shown how to associate
with each general frame F a modal algebra ~r
and with each frame
m o r p h i s m f f r o m F t to F z a homomorphism ~r from ~r 2 to sCF r It is easy
to check the following theorem:
THEOREM 4A.
(a) ~ l d F = Ida, F;
(b) ..~r
(~f,)~
So, .~g defines a contravariant functor from DF to MA.
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Duality between modal algebras ...
We can also easily check the following theorem for frame morphisms f:
(a)
(b)
THEOREM 5A.
l f f is an embedding, then ~ f is onto.
l f f is onto, then ~ f is one-one.
6. The functor ~-J. In Section 2 we have shown how to associate with each
modal algebra A a general frame ~-~A. For this general frame we can prove the
following:
THEOREM I F.
~ A is a descriptive frame.
PROOF. (I) Suppose p ( X 1 ) = p(X2),
b e X , ~ q ( b ) e p ( X ) , it follows that X t =
(II) I f X ' is an ultrafilter of ~r
to check that X is an ultrafilter of A,
i.e. q(b)ep(Xt)c:,q(b)ep(X2). Since
X 2.
then let X = {b: q(b)eX'}. It is easy
and that X ' = p(X). q.e.d.
If h is a homomorphism from A t to A2, and X 2 is an ultrafilter of A 2, let
( ~ h ) ( X 2 ) = {bl: h(bl)~X2}. For ~-h we can prove the following theorem:
THEOREM 3F. The mapping h: D 1 ~ D 2 is a homomorphism from A t to A 2
iff ~ h is a frame morphism from ~ A 2 to ~ A r
PROOF. (=>) Suppose h: D t --*D2 is a homomorphism from A t to A 2, and
X 2 is an ultrafilter of A 2. It is easy to check that (~-h)(X2) is an ultrafilter of At;
so, ~:h:Ca2--*C al. It is also easy to check that ~r
s~al ~ D ~ a 2 . Let ql be
the isomorphism from A m to d ( ~ A l ) and q2 the isomorphism from A z to
sC(~-A2). Then we can prove that ~r
q2ohoq{ t, and so, ~r
is
a homomorphism. It follows, by Theorem 3A, that ,~-h is a frame morphism
from ~-A 2 to ~-A 1.
(~=) I f ~ - h is a frame morphism from ~ A z to ~ A 1, it follows, by Theorem
3A, that ~ ( ~ h ) is a homomorphism from ~r
to ,~r
It remains to
show that h = q ~ l o d ( ~ h ) o q l . So, h is a homomorphism from A t to
A 2. q.e.d.
So, with each modal algebra A we can associate a descriptive frame ~-A,
and with each homomorphism h from A1 to A 2 we can associate a frame
morphism ~-h from ~-A 2 to ~ A r It remains to check the following theorem:
THEOREM 4F.
(a) ~ I d A = Ida:A;
(b) ~-(h2ohl) = (~ht)o(~-h2).
So, .~- defines a contravariant functor from M A to DF.
We can prove the following analogue of Theorem 5A:
(a)
(b)
THEOREM 5F.
I f h is onto, then ~ h is an embedding.
I f h is one-one, then ~ h is onto.
PROOF. (a) Suppose h:DI--,D 2 is onto. It is easy to check that ~-h is
one-one. To check (iii) for ~ h we proceed as follows. If B 2 e D ~a~, then for some
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K. Do.qet~
b 2 E D 2 w e have B 2 = q ( b 2 ) , and since h. is onto, for some b ~ D 1 we have
bE = h(bl). It is not difficult to prove that (,~h)[B2] =(,~h)[CA:]c~q(bt).
(b) Suppose h:DI-~D 2 is one-one, and X I e C A'. Let X 2 ={b2~D2:
(:lb~Xl) h(b) <~ b2}. We can check that X z is a proper filter of A 2. Hence, X 2
can be extended to an ultrafilter X~ of A 2. Using the fact that h is one-one
and that ( V b I ~ X I ) ( h ( b i ) ~ X 2 or - h ( b i ) ~ X 2 ) , we can show that
Xt--h-~[XE]=(~h)(X~),
where h - t [ X 2 ] is {b~eDt:h(b~)eX2}. q.e.d.
(The proof of 5F (b) is not exactly parallel to the analogous proof of
Theorem 10.9 (2) in [4], p. 71. l was unable to prove 10.9 (2) the way Goldblatt
did. Instead, one can take y' = {a: (3ai ~x)d/(al) <<,a}, prove that y' is a proper
filter, and extend it to an ultrafilter y, for which we have ~h~(y)= x.)
7. Duality between M A and DF.
Theorem 2A:
THEOREM 2F.
:~ (~/ F).
First we prove the following analogue of
I f F is descriptive, then p is a frame isomorphism from F to
PROOF. It is trivial that p:C--.C ~F is one-one and onto. I~ remains to
prove that p is a frame morphism and that p is an embedding.
To prove that p is a frame morphism, we prove first that if q is the
isomorphism from . ~ F to m/(~(<utF)), then .~p = q- 1. And to prove that, it is
enough to prove (.~p)(q(B)) = B and q((dp)(B)) = B. For the first identity we
use p(x)~ q(B)c~x ~ B, whereas for the second we use the fact that p is one-one
and onto. Since .~p = q - l , it follows, by Theorem 3A, that p is a frame
morphism.
To show that p is an embedding, we must prove:
(iii)
(V B 1~ DF)(3B2 ~ D ~ r ) ) p[BI] = p[C] c~ B 2.
This holds because for B2 we can take q(B1), where q is the isomorphism from
,c~F to ,~(<~(~4F)), and p[Bl] = q(B~), q.e.d.
We are now ready to prove the duality between M A and DF:
THEOREM 6.
The categories M A and DF are dual by the functors d and ~ .
PROOF. To prove this theorem, we show that the following diagrams
commute:
h
A1
f
)
A2
F1
ql
By Theorem 2A we have that ql and q2 are isomorphisms, and by Theorem 2F
we have that p~ and Pz are frame isomorphisms, q.e.d.
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Duality between modal algebras ...
8. Normal modal algebras and filter frames. A modal algebra A is normal
iff for every bl, b 2 e D we have R ( b 1 ^ b2)= [-]b t ^ []b 2 and []1 = 1. The
category N M A of normal modal algebras is defined by the following:
objects: normal modal algebras,
morphisms: homomorphisms.
A general frame F is a filter frame iff for every x e C the set N(x) is a filter
(not necessarily proper) of ~/F. The category D F F of descriptive filter frames is
defined by the following:
objects: descriptive filter frames,
morphisms: frame morphisms.
It is not difficult to prove the following theorems:
THEOREM 7A.
F is a filter frame iff ~ F
THEOREM 7F.
~ A is a f l t e r frame iff A is normal.
is normal.
These two theorems, together with the results of the preceding sections, enable
us to prove that the categories N M A and D F F are dual by the functors ~t/and
9. Reducible frames and hyperfiiter frames. A general relational frame is
Fr = (C, R, Dr), where:
(1) c ~ o ,
(2) D r ~_ / ~ C is closed under r - and a unary operation L, and D r # f~,
(3) R ~_ C 2, and S ( x ) = df{Xz: x R x l } ,
(4) ( V B ~ D r ) ( x ~ L B o S ( x ) ~ B).
(General relational frames are sometimes called first-order frames.)
In the definition above S (the set S(x) is the set of successors of x) is defined
in terms of R; but, alternatively, we could take S as primitive and define R in
terms of S by x I R x 2 o x 2 E S(xl). Clause (4) enables us to take L as defined in
terms of R (or S), but, in contradistinction to what we had with general
neighbourhood frames, it does not give a definition of R in terms of L.
We shall say that a general relational frame Fr is a reducible frame iff for
every x 1, x 2 ~ C we have:
(5)
(V B e D r) (x i e LB =~ x 2 e B) =>x I R x 2.
This clause (which is called Axiom II in [4], p. 64) enables us to take R as
defined in terms of L in reducible frames, since the converse of (5) holds for all
general relational frames. In terms of S, the clause corresponding to (5) would
read:
(5') N { B ~ O r : x ~ L B } ~_S(x).
The converse inclusion of (5') holds for all general relational frames.
Reducible frames are "intertranslatable" with a special kind of general
neighbourhood frames which we proceed to define. Let us call a general
neighbourhood frame F = (C, N, DF~ a hyperfilter frame iff for every x~ C:
(6) ( V B r
~- B=~Br
K. DoWn
226
The converse of (6) holds trivially. (Note that a hyperfilter frame is not an exact
analogue of what is called an augmented frame in [1], p. 220, because 0 N(x)
need not belong to D r, and hence it need not belong to N(x); in other words,
N(x) is not necessarily a complete filter.) It is easy to check the following
lemma:
LEMMA 2.
Every hyperfilter frame is a filter frame.
The converse of this lemma does not hold necessarily if N(x) is infinite (a
counterexample from [1], pp. 220-221, takes F with C the set of real numbers,
D r = t~C, and N(x) = (B ~_ C: (3y > x) the open interval (x, y) ___B}; this F is
a filter frame which is not a hyperfilter frame).
The following lemmata show that reducible frames are intertranslatable
with hyperfilter frames:
LEMMA 3.1.
If in a hyperfilter frame F = (C, N, DF) we define R by:
(7) x1Rx2c:,x2E('~N(Xl),
then Fr = ( C, R, Dr) is a reducible frame in which we have:
(8) N(x) = {BeDF: S(x) =__B).
If in a reducible frame Fr = ( C, R, DF) we define N by (8),
then F = (C, N, Dr) is a hyperfilter frame in which we have (7).
LEMMA 3.2.
It is clear that with the interdefinability of R and S in relational frames, clause
(7) amounts to:
(7')
S(x) = ~ N(x).
It is also clear that descriptive hyperfilter frames are intertranslatable with
reducible frames where p is one-one and onto, i.e. with descriptive reducible
frames. These last frames are the frames which Goldblatt calls descriptive frames
(see [4], pp. 63ff; Axioms I and III of [4] state that p is one-one and onto,
whereas Axiom II is built into the definition of reducible frames).
Although not every filter frame is a hyperfilter frame, for descriptive frames
we have the following lemma:
LEMMA 4.
Every descriptive filter frame is a hyperfilter frame.
PROOF. Suppose F = (C, N, D F) is a descriptive filter frame, and for some
x e C and some B e D r we have Br
The set N(x)w ( - B } has the finite
intersection property, i.e. every finite subset of this set has a nonempty
intersection (otherwise, since N(x) is a filter, for some B~ e N(x) we would have
B 1 c~ - B = 0, and this yields BeN(x), a contradiction). So, N(x)w { - B } can
be extended to an ultrafilter X 1 of ~r Since F is descriptive, for some x t e C
we have X 1 = p ( x l ) . It remains to check that x l e ~ N ( x ) and xl~B. So,
A N ( x ) ~=B. q.e.d.
(Note that this proof shows that every filter frame where p is onto is
a hyperfilter frame.)
Duality between modal algebras ...
227
Hence, descriptive filter frames are intertranslatable with descriptive
reducible frames.
I0. Frame homomorphisms and frame morphisms. Next we show that
Goldblatt's frame homomorphisms between descriptive reducible frames
amount to our frame morphisms between descriptive filter frames. In [4] (p. 53)
a f l a m e homomorphism between Fr 1 = (C1, R1, D F'> and Fr 2 = (C2, RE, D F2>
is a mapping f : C l - , C 2 such that for every B z e D F', every x l , x 2 ~ C 1, and
every x e C 2 :
(i)
(~ff) (B2) e D F',
(ii.i) x I R 1 x 2 =~f (xl) R 2 f (x2) ,
(ii.ii) f ( x l ) R 2 x =*.3x2(x -- f ( x 2 ) & x 1 R 1 x2);
where z l f and z~tFr are defined analogously to what we had in Sections 2
and 3.
Clause (i) above is exactly as in our definition of frame morphisms in
Section 3, whereas clauses (ii.i) and (ii.ii) can be replaced by:
(ii) f ( x l ) g 2 x c : , 3 x 2 ( x = f ( x 2 ) & x 1 R 1 x2).
Given the interdefinability of R and S, we can replace (ii) by:
(ii')
S2(f(xl) ) = { f ( x 2 ) : x 2 ~ S l ( x l ) } ,
which can be compared with the following reformulation of clause (ii) in the
definition of frame morphisms:
N2(_f(x,) ) = {B2ED/~2: ( ~ ] f ) ( B 2 ) e N , ( x , ) }.
For frame homomorphisms we can prove the following analogue of
Theorem 3A:
LEMMA 5. I f F r 1 = ( C 1, R 1, Dr'> and Fr 2 = <C 2, R 2, DF2> are descriptive
reducible frames, then the mapping f'. C 1 --. C 2 is a f l a m e homomorphism from Fr 1
to Fr 2 iff s g f is a homomorphism from ~iCFr2 to MFrs.
PROOF. If f is a frame homomorphism, then d f is a homomorphism (see
[4], pp. 53-54). For the converse, we sketch only the proof of (ii.ii) (analogous
to the proof of 10.9 (1) in [4], p. 71).
S u p p o s e f ( x l ) R 2 x . The sets ]11 = {BI eDF':'xl eLB1} and Y2 = {('~r
x e B~} are closed under finite ~(Y~ because zCFr~ is a normal modal algebra,
and Y2 because ~ f is a homomorphism). The set Y1 u Y2 has the finite
intersection property, i.e. every finite subset of ]'1 w Y2 has a nonempty
intersection (otherwise, for some B 1 e Y1 and (._~r
Y2, we could prove in
a few steps L B 1 c_ ( ~ f ) ( L ( - B 2 ) ) ; from that we obtain f ( x ~ ) e L ( - B 2 ) , and
since f ( x l ) R 2 x , we have xe-B2, i.e. x r
but (,.qff)(B2)e Y2, and hence,
XE B2, a contradiction). So, Y1 w Y2 can be extended to an ultrafilter of zgFr r
Since p in onto, this ultrafilter is of the form P(X2) for some x 2 e C x. It remains
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K. Dogen
to check that Y2 ~ p(x2) implies x = f ( x 2 ) (using the fact that p is one-one), and
that YI ~- P(X2) implies x 1 R t x 2. q.e.d.
With the help of this lemma and Theorem 3A we can easily establish the
following lemmata:
LEMMA 6.1. Let F 1 = ( C 1, N 1, D r ' ) and F 2 = ( C 2, N 2, D e2) be descriptire filter frames, let R 1 in F 1 and R E in F 2 be defined by (7) of Lemma 3.1, and let
f: C 1 --, C 2 be a frame morphism from F 2 to F 2. Then f is a frame homomorphism
from the descriptive reducible frame Fr 1 = ( C 1, R l, D r ' ) to the descriptive
reducible frame Fr 2 = (C2, R 2, DP2).
LEMMA 6.2. Let Fr I = (C t, R1, D r ' ) and F r 2 = ( C 2 , R E, D r2) be descriptive reducible farmes, let N 1 in Fr t and N 2 in Fr 2 be defined by (8) of Lemma 3.1,
and let f: C l ~ C 2 be a frame homomorphism from Fr I to Fr 2. Then f is a .frame
morphism from the descriptive filterframe F t = (Cl, N 1, D ~'~) to the descriptive
filter frame F 2 = (C 2, S 2, Dr2).
(As a matter of fact, Lemma 6.2 holds if we replace "descriptive reducible" by
"reducible", and "descriptive filter" by "hyperfilter', as can easily be checked by
a direct proof of (,z/f)(B2) r N l ( x l ) ~ B 2 r N2(f(xl)). )
So, the category D F F of descriptive filter frames with frame morphisms is
isomorphic with the category D R F of descriptive reducible frames with frame
homomorphisms. The functors by which this categorial isomorphism is
established are the functors whose definition we can infer from Lemmata 3.1
and 3.2; according to Lemmata 6,1 and 6.2, these functors assign to frame
morphisms and frame homomorphisms the very same mapping f : C 1 --,C 2.
Now, for the category DRF Goldblatt has shown in [4] (pp. 69-72) that it is
dual to the category N M A of normal modal algebras of Section 8. The functors
by which this duality is established are the functor d and a functor ~- which
is like ~-, save that we have:
(3,)
(4,)
X I R a , X 2 c ~ { b I : D b l e X I } ~_X2,
(VbED)LA,(q(b)) = { X ~ C ~ : SA,(X) ~_ q(b)}.
The functor ~ . conceived as a functor which assigns to a normal modal
algebra A a descriptive filter frame, rather than a descriptive reducible frame,
has as the corresponding clauses:
(3.)
(4,)
NA.(X) = {q(b):S~(X) c_ q(b)}
= {q(b):(VX~ eC'4)('Cbt(Rb, e X ~ b ~ ~ X , ) ~ b e X ~ ) } ,
( V b e D ) L ~ ( q ( b ) ) = {X eCa: O NA,(X) ~ q(b)}"
Let N'* and LA be defined as for o~(see Section 2) by:
(3)
(4)
n a ( x ) = {q(b): l N b e X } ,
(VbeD)LA(q(b)) = { X e C a : q ( b ) e N a ( X ) } .
The following lemma shows that N a ( X ) = N~ (X):
Duality between modal algebras ...
LEMMA 7.
X e CA:
229
In every normal modal algebra A, for every b e D , and every
D b ~ X c~(V X 1 e CA)(Vbl ([]bl e X =:'b I ~ X l)=~b ~ X l).
PROOF. From left to right the proof is obvious. For the other direction the
proof is analogous to the proof of Lemma 4. Suppose [] b r X. Then we show
that {b~: l-lbl e X} u {,-~b} has the finite intersection property. The ultrafilter
X t to which this set can be extended is such that Vb~([]b~ e X ~ b ~ eXx) and
b r X 1. q.e.d.
Since NA(X) = Na,(X), and since in .~"~,A, which
have ( ] N a ( X ) ~ q(b)~',,q(b)eSa(X), it follows that
the functors ~ and .~-, do not differ, and our duality
in Section 8 is essentially the same as Goldblatt's
and DRF.
is a hyperfilter frame, we
La(q(b))= L~(q(b)). So,
result for N M A and DFF
duality result for N M A
Full neighbourhood frames
11. The category CAA. A modal algebra A = (D, ^ , v , ---, 1, 0, F-q) is
complete iff for every X _ D the infimum of X, denoted by A X, and the
supremum of X, denoted by v X, belong to D. An element b e D is an atom iff
b -r 0 and (Vb I <~ b)(b I = b or b 1 = 0). For the atoms of a modal algebra we
shall use a, al, a2, ..., and for sets of atoms we shall use Z, Z 1, Z 2 . . . . . A modal
algebra is atomic iff (Vb # O)(3a)a <<.b.
A complete homomorphism from A 1 to A 2 is a homomorphism h:D~--.D 2
which satisfies:
h ( ^ { b F i ~ l } ) = ^ {h(b,):iEl},
h ( v {b,:i~I})= v {h(bi):i~l},
The category CA A of complete atomic modal algebras is defined by the
following:
objects: complete atomic modal algebras,
morphisms: complete homomorphisms.
To abbreviate, we shall call the objects of CAA full algebras.
12. Full neighbourhood flames. A general neighbourhood frame F = (C,
N, D r) is full iff D r = ~C. To abbreviate, we shall call such frames full frames.
Full frames coincide with the usual notion of neighbourhood frames.
If d F is defined as in Section 2, it is clear that the following holds:
I f F is a full frame, ~r is a full algebra.
Over a full algebra A we define a frame ffA = ( C A, N A, D~A), where:
THEOREM 7A.
(1)
(2)
(3)
(4)
CA= {a:aED and a is an atom},
O~A = ~ C A,
N A : C A - , ~ ( ~ C A) is defined by NA(a) = {Z ~_ CA:a ~< [] v Z},
( V Z ~_ CA)LAZ = {a~CA:Zr
230
K. DoWn
We can immediately infer the following:
THEOREM 7F.
I f A is a full algebra, ~ A is a full frame.
The mapping s:D-~D ~A from A to M(ffA), analogous to q, is defined by
s(b) = {a: a <<.b}. We easily infer the following theorem:
THEOREM 8A.
I f A is full, s is a complete isomorphism from A to d ( ~ A ) .
The mapping r : C ~ C ~'F from a full F to ff(,~'F), analogous to p, is defined
by r(x) = {x}. We immediately have that r is one-one and onto. (So, every full
frame is "descriptive" with respect to r.)
For a frame m o r p h i s m f : C ~ - , C 2 from a full frame F 1 to a full frame F2,
clause (i) of Section 3 is trivially satisfied, and we easily obtain the following
theorem:
THEOREM 9A. The mappingf:C~ ~ C 2 is a frame morphismfrom a full F 1 to
a full F 2 iff d f is a complete homomorphism from s~F 2 to d F 1.
Every one-one frame m o r p h i s m f f r o m a full F~ to a full F 2 is an embedding,
sinceJ~Bl] = f [ C 1 ] c~ f [ B l ] . So, by Lemma l, a frame m o r p h i s m f f r o m a full
F 1 to a full F 2 is a frame isomorphism ifffis one-one and onto. Then we prove
the following theorem:
THEOREM 8F.
I f F is full, r is a frame isomorphism from F to (#(~F).
PROOF. It is clear that r : C ~ C ~ r is one-one and onto. It remains only to
show that for every W_~ C ~ r we have ( d r ) ( W ) ~ S(x),:~ W e N~e(r(x)). Since
r(x) = {x} and
( . ~ r ) ( W ) = { x : { x } e W } = U W, we
need to show
U W e N ( x ) c * , W e N d r ( { X } ) , and for that we have:
L(U W)
L(U W)
~
WeN(x).
q.e.d.
13. The category F N F . The category F N F of full neighbourhood frames
is defined by the following:
objects: full neighbourhood frames,
morphisms: frame morphisms.
It is clear that ~1, defined as in Sections 2 and 3, determines a contravariant
functor from F N F to C A A .
14. The functor f~. In Section 12 we have shown how to associate with
each full algebra A a full frame (~A. It remains to show how we associate with
each complete homomorphism h from a full A 1 to a full A 2 a frame morphism
@h. For a 1 ~ C A' and a 2 e C A" let (C~h)(a2) = a I r 2 ~ h(al). First we show that
this equivalence defines a genuine function (~h from C A2 to CA':
LEMMA 8. l f h is a complete homomorphism from a full A I to a full A 2, then
for every atom a2~D 2 there is a unique atom a l e D 1 such that a 2 <<.h(al).
Duality between modal aloebras ...
231
PROOF. Suppose h is as in the lemma, and there is no a I e D x such that
a 2 <~ h(al). We infer that (Va I eD1)a 2 <~ ~ h ( a t ) , since for every a t o m a 2 either
a 2 ~< b or a 2 ~< -,-b. So, a 2 ~< A { ~ h ( a t ) : a l e D t } , and then by using the fact
that h is a complete h o m o m o r p h i s m and that A 2 is full, we obtain a 2 = 0,
which is a contradiction. Suppose now for some a'~,a'~eD 1 such that a'~ # a'~
we have a 2 ~< h(a'O and a 2 ~< h(a~). We again easily infer the contradiction
a 2 = 0. q.e.d.
Then we have the following theorem, for which, using T h e o r e m 9A, we can
give a proof analogous to the proof of Theorem 3F:
THEOREM 9F. The mappin9 h:D x -*D 2 is a complete homomorphism from
a full A x to a full A 2 iff ~ h is a frame morphism from f f A 2 to ffA 1.
We can also easily check the following theorem:
THEOREM 10F.
~ I d a = Idea;
(b) (~(h2ohl) = (~hl)~
(a)
So, ~ defines a contravariant functor from C A A to F N F . Next we prove the
following analogue of Theorem 5F for complete h o m o m o r p h i s m s h from a full
A I to a full A2:
(a)
(b)
THEOREM l l F .
I f h is onto, then @h is one-one.
I f h is one-one, then ~h is onto.
In order to prove (a) of this theorem, we first establish the following lemma:
LEMMA 9.
I f h is onto, then (Va2~D2)(]a I E O l ) a 2 = h(at).
PROOF. Since h is onto, ( V a z 6 D 2 ) ( 3 b t ~ D 1 ) a 2 = h(bl). It easily follows
that a 2 = v {h(aa): a t ~< b~}. Let X 2 = {h(at): a t <<.bt&h(at) # 0}. It is easy to
see that X 2 # O and that a z = .V X 2. So, for an h ( a t ) E X 2 we have h(at) ~< a 2.
Since h ( a l ) # 0, and a 2 is an atom, h(ax)= 02. q.e.d.
PROOF OF THEOREM 11F. (a) Suppose h is onto, and suppose (ffh)(a~) =
= (~h)(a'~). It follows that (Vat~D1)(a'2 ~ h(at)~,a~ <<.h(at)). By L e m m a 8,
this means that (Va26D2)(a'2 ~ a2*x,a~ ~ a2), and this implies a~ = a'~.
(b) Suppose h is one-one. We want to show that ( V a t , D r ) (3a2~D2)
a z ~ h(a O. First we show that (Va t ~ D t ) h ( a l ) ~ 0 (otherwise, h(at) = h(0), and
since h is one-one, a t = 0 , a contradiction). Since A 2 is atomic, ( ' ] a 2 ~ D 2 )
a z <<.h(al), q.e.d.
The analogue for M of Theorem 10F, and the analogue for f of Theorem
l IF, hold in virtue of Theorems 4A and 5A.
15. Duality between C A A and F N F . N o w we can prove the duality
between C A A and FNF:
232
K. Do~en
THEOREM 12.
and f#.
PROOF.
The categories C A A and F N F are dual by the functors
We have to show that the following diagrams commute:
h
f
A1
9
A2
F1
"
F2
/
s1
s 2r
)
)
rl
9 ,.~,(~A2,)
~(4F1, }
r2
~(4f]
By Theorem 8A we have that s I and s z are complete isomorphisms, and by
Theorem 8F we have that r~ and r 2 are frame isomorphisms, q.e.d.
Analogously to what we had in Section 8, we can establish duality between
the following two categories. On the algebraic side we have the category
defined by the following:
objects: complete atomic normal modal algebras,
morphisms: complete homomorphisms,
and on the frame side we have the category defined by the following:
objects: full filter neighbourhood frames,
morphisms: frame morphisms.
To prove this duality, which holds by the functors M and ~, we use the
following analogue of Theorem 7F:
THEOREM 13F.
For a full A we have that (~A is a filter frame iff A is normal.
16. Full relational flames and full hyperfilter frames. A general relational
frame (see Section 9)-Fr = (C, R, Dr> is full iff D r = .~C. These frames are the
usual Kripke frames for modal logic. Every full relational frame is reducible,
since we have:
(V B E ~ C ) ( x I ~ LB=~ x 2 e B) =~(x I ~ L(S(Xl))=~ x 2 e S(xt) )
='(S(xl) =- S(xl)=
x2 e S(xl))
~ x 1 R x 2.
In a full hyperfilter frame the clause:
(6) ( V B e ; ~ c ) ( O N ( x )
c_ B=~ B e N ( x ) )
is replaceable by the requirement that our frame be a filter frame in which for
every x we have ~ N ( x ) e N ( x ) (so, full hyperfilter frames are augmented in the
sense of [1], p. 220; i.e. for every x we have that N(x) is a complete filter).
However, a full filter frame need not be a hyperfilter frame (the counterexample
mentioned after Lemma 2 in Section 9 is based on a full filter frame).
According to Lemmata 3.1 and 3.2, full relational frames are intertranslatable with full hyperfilter frames. It is now easy to prove directly Lemmata 6.1
233
Duality between modal algebras ...
and
and
full
full
6.2 where "descriptive filter frame" is replaced by "full hyperfilter frame",
,,descriptive reducible frame" by "full relational frame". So, the category of
hyperfilter frames with frame morphisms is isomorphic to the category of
relational frames with frame homomorphisms.
F o r this last category T h o m a s o n asserts in [!1] that it is dual to the
category whose objects are completely normal modal algebras, i.e. complete
atomic normal modal algebras which satisfy:
[]^
{b,:iel}=
^ {9
and whose morphisms are complete homomorphisms. The functors by which
this duality holds are ar and a functor fq,, which is like f#, save that we have:
(3,)
a , R ' ~ a 2 c ~ a , ^ - . . [ ] . , . a2 :/:O
r
<~ ~ [] ... az,
(4,)
LA,Z = { a : S a , ( a ) c Z } .
The functor N , conceived as a functor which assigns to a completely normal
modal algebra A a full hyperfilter frame, rather than a full relational frame, has
as the corresponding clauses:
(3,)
(4,)
N'~(a) = { Z : V a l ( a <~ .,~ [] ~ a I =*.a I eZ)},
L ,A Z = { a : ~ N a , ( a ) c Z } .
N o w , N a and LA for ~ were defined in Section 12 by:
(3) N*(a)= {z: a.< [] vZ},
(4)
LAZ = { a : Z e N A ( a ) } .
The following l e m m a t a show that N A ( a ) = N,a(a):
LEMMA 10.1. In every complete atomic normal modal algebra A, for every
a e D , and every Z c D:
a <~ [] v Z = > g a l ( a <.N ~ [] " ~ a l = ~ a l ~ Z ) .
PROOF. Suppose a ~ < [ ] v Z ,
a ~ < , - ~ [ ] . - - a 1 and alq~Z. Since a I is an
atom, from a I ~ Z
it follows that a I ~< ~ v Z, and this entails
[] v Z ~< [ ] ~ al.
Hence,
a ~< []--- a~,
which
is
impossible
since
a~< ~ [ ] - - . a I. q.e.d.
LEMMA 10.2.
In every completely normal modal algebra A, f o r every a ~ D,
and every Z c D:
V a l ( a <~ .,. [] .,~ at=a.al ~Z)=a,a <~ [] v Z.
PROOF. Suppose not a<<. [] v Z. Next suppose not (3al <~ ~ v Z)
a ~< .-- [] ~ a t.
Then
(Va I ~< --- v Z ) a <~ [] .,. al,
and
hence,
a <<. ^ {[] ",- a~:a~ ~< --. v Z}. Since our algebra is completely normal, we have
a ~< [] ^ { ~ a , : a~ ~< --. v Z}, and this easily yields a ~< [] v Z, a contradiction. So, for some al ~< ,-- v Z we have a ~< -,- [] -,- a I . This al does not belong
to Z (otherwise, a 1 ~< v Z). q.e.d.
234
K. Do~en
(The proof of these lemmata is essentially the proof we must go through in
order to show that s is a homomorphism from a completely normal A to
M(~,A).)
Since NA(a)= N2(a), and since in ~ , A , which is a hyperfilter frame, we
have NNA(a)~_ Z<::.Z~NA(a), it follows that LAZ= L,Az. So, the functors
fq and @, are essentially the same functor, and Thomason's duality result
mentioned above amounts to the assertion that the category of completely
normal modal algebras with complete homomorphisms is dual by the functors
and ~ to the category of full hyperfilter neighbourhood frames with frame
morphisms.
References
[1] B. F. CHELLAS,Modal Logic:. An Introduction, Cambridge University Press, Cambridge,
1980.
[2] L. L. ESAKIA,On topological Kripke models (in Russian), Doklady Akademii Nauk SSSR 214
(1974), pp. 298301.
[3] L. L. ESAKtA,Heyting Algebras !: Duality Theory (in Russian), Metsniereba, Tbilisi, 1985.
[4] R. I. GOLDaI.ATT,Metamathematics of modal logic, Reports on Mathematical Logic 6 (1976),
pp. 41-77; 7 (1976), pp. 21-52 (all page references in the text are to the first part, in 6 (1976)).
[5] G. HANSOUL,A duality for Boolean algebras with operators, Algebra Universalis 17 (1983),
pp. 34 49.
[6] B. J6NSSON and A. TARSKI,Boolean algebras with operators: Part I, American Journal of
Mathematics 73 (1951), pp. 891 939.
[7] B. PAREIGIS,Categories and Functors, Academic Press, New York, 1970.
[8] R. S. PIEgCE, Compact Zero-Dimensional Metric Spaces of Finite Type, Memoirs of the
American Mathematical Society 130 (1972).
[9] H. RASIOWA and R. SIKORSKI, The Mathematics of Metamathematics, Pafistwowe
Wydawnictwo Naukowe, Warsaw, 1963.
FI0] K. SEGERBERG,An Essay in Classical Modal Logic, University of Uppsala, Uppsala, 1971.
El 1] S. K. TSOMASON,Categories of frames for modal logic, The Journal of Symbolic Logic 40
(1975), pp. 439-442.
[I 2] J. WILLIAMS,Structure diagrams for primitive Boolean algebras, Proceedings of the American
Mathematical Society 47 (1975), pp. 1-9.
MATEMATI~KI INSTITUT
KNEZ MIHAII.OVA35
11001 BEOGRAO, P.F. 367
YUGOSLAVIA
Received October 7, 1987
Studia Logica XLVIII, 2