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Revista de la Unión Matemática Argentina
versión impresa ISSN 0041-6932versión On-line ISSN 1669-9637
Rev. Unión Mat. Argent. v.49 n.1 Bahía Blanca ene./jun. 2008
Saturated neighbourhood models of Monotonic Modal Logics
Sergio Arturo Celani
Abstract. In this paper we shall introduce the notions of point-closed, point-compact, and m-saturated monotonic neighbourhood models. We will give some characterizations, and we will prove that the ultrafilter extension and the valuation extension of a model are m-saturated.
Key words and phrases. monotonic modal logic, neighbourhood frames and models, m-saturated models, ultrafilter extension, valuation extension.
2000 Mathematics Subject Classification. 06D05, 06D16, 03G10.
Monotonic neighbourhood semantics (cf. [1] and [6]) is a generalization of Kripke semantics. It is also the standard tool for reasoning about monotonic modal logics in which some (Kripke valid) principles such as , do not hold. A monotonic neighbourhood model, or monotonic model, is a structure
where
is closed under supersets for each
, and
is a valuation defined on
.
The main objective of this paper will be the identification and study of some properties of saturation of monotonic models. It is also intented to prove that the ultrafilter and valuation extension of a monotonic model is -saturated. We will define the image-compact, point-compact, point-closed, and modally saturated (or
-saturated) models. These notions are defined in topological terms. For each monotonic model
we will define a topology
in the set
![KR = {Y ⊆ X : ∃x ∈ X (Y ∈ R (x))},](/img/revistas/ruma/v49n1/1a1014x.png)
using the Boolean algebra , and taking as sub-basis the collection of all sets of the form
![L = {Y ∈ K : Y ∩ V (φ) ⁄= ∅} . V (φ) R](/img/revistas/ruma/v49n1/1a1016x.png)
The topological space is called the hyperspace of
relative to
. The notions of point-compact, point-closed, and
-saturated monotonic models are defined relative to this space. For instance, a model
is
-saturated if
is a compact subset of
for each
, and for each
, there exists a compact set of
of
such that
and
. With this notion of
-saturation we will be able to prove that the ultrafilter extension of a monotonic model is
-saturated. This question has already been addressed in [6], but with a different notion of
-saturation.
In Section 2 we will recall the principal results on the relational and algebraic semantics for monotonic modal logic. In Section 3 we will introduce the notions of compact, image-compact, point-compact, point-closed, and -saturated monotonic models. We will give a characterizations of point-compact models. We will show that some of the notions introduced are invariant by surjective bounded morphisms. Section 4 is the core of this paper. We shall prove that for a monotonic model
there exists at least two saturated extensions: the valuation extension and the ultrafilter extension. The valuation extension of a model is a bounded image of the ultrafilter extension, and as the property of
-saturation is invariant under surjective bounded morphism, we will deduce that the valuation extension is also
-saturated. Saturated extensions of monotonic models may be seen as a completation of the underlying frame structure. For Kripke models, the saturated extensions are modally saturated structures, which implies that modally states are bisimilar (see [7] for Kripke models and [6] for monotonic models).
A monotonic algebra is a pair where
is a Boolean algebra, and
is a monotonic function, i.e. if
then
, for all
The dual operator
is defined as
. The filter (ideal) generated by a set
will be denoted by
(
. The set of all prime filters or ultrafilters of
is denoted by
.
Given a set , we denote by
the powerset of
, and for a subset
of
, we write
for the complement
of
in
. We will call a space a pair
where
is a set, and
a subalgebra of the Boolean algebra of
. We note that
is a basis of a topology
on
whose open sets are the unions of subsets of
. All the elements of
are clopen (closed and open) subsets of
, because
is a Boolean algebra, but an arbitrary clopen set does not to be need an element of
. Given a space
and
we will use the notation
to express the closure of
The set of all closed subsets (compact subsets) of
will be denoted by
(
). We note that
and
are posets under the inclusion relation. Some topological properties of
can be characterized in terms of the map
given by
. The map
is called the insertion map in [2]. A space
is called a Boolean space if it is compact and totally disconnected. If
is a Boolean space, then the family
of clopen subsets is a basis for
.
To each Boolean algebra we can associate a Boolean space whose points are the elements of
with the topology determined by the basis
, where
. It is known that
is a Boolean subalgebra of
. By the explanation above we have that, if
is a Boolean space, then
, by means of the map
, and if
is a Boolean algebra, then
, by means of the map
. Moreover, it is known that the map
establishes a bijective correspondence between the lattice of all filters of
and the lattice
of all closed subsets of
.
Definition 1. Let be a space. The lower topology
on a subset
of
has as sub-base the collection of all sets of the form
for
. The pair
is called the hyperspace of
relative to
.
Let be a set. A neighbourhood relation, or multirelation, defined on
is a relation
.
Definition 2. A monotonic neighbourhood frame, or monotonic frame, is a structure where
is a multirelation on
such that
is an increasing subset of
for all
.
Every monotonic frame gives rise to a monotonic algebra of sets. Consider the monotonic map
defined by:
![♢R(U ) = {x ∈ X : ∃Y ⊆ X (Y ⊆ U and (x, Y) ∈ R )},](/img/revistas/ruma/v49n1/1a10126x.png)
for each . It is clear that the pair
is a monotonic algebra. Using the notation introduced in Definition 1 the map
can be defined also as
for each
. The dual map
is defined by
![□R (U ) = {x ∈ X : R (x) ⊆ LU } ,](/img/revistas/ruma/v49n1/1a10133x.png)
for each .
Next we show that any monotonic algebra with monotone gives rise to a monotonic frame by invoking the basic Stone representation. In other words, we represent the elements of a monotonic algebra as subsets of some universal set, namely the set of all ultrafilters, and then define a multirelation over this universe.
Let be a monotonic algebra. Let us define a multirelation
by:
![]() | (2.1) |
where . We note that for any filter
and for all
,
![(x, ˆF ) ∈ R ♢ iff F ⊆ ♢ -1(x).](/img/revistas/ruma/v49n1/1a10141x.png)
Theorem 3. Let be a monotonic algebra. Then
is a monotonic frame.
for all
Proof. (1) Clearly is an increasing subset of
, for each
.
(2) We prove that for all
, and for all
![♢ (a ) ∈ x iff ∃F ∈ Fi(A ) : (x, ˆF ) ∈ R ♢ and a ∈ F.](/img/revistas/ruma/v49n1/1a10152x.png)
If , then
. So,
. If there exists
such that
and
then
Thus,
. As consequence of this we have that
for all
. Now it is easy to prove
. □
Let us consider a propositional language defined by using a denumerable set of propositional variables
, the connectives
and
the negation
, the modal connective
, and the propositional constant
We shall denote by
the operator defined as
, for
. The set of all well formed formulas will be denoted by
.
Definition 4. A monotonic model in the language is a structure
where
is a monotonic frame, and
is a valuation.
Every valuation can be extended to by means of the following clauses:
The notions of truth at a point, validity in a model and validity in a frame for formulas are defined as is usual. A formula is valid at point
in a model
, in symbols
if
The formula
is valid in a model
, in symbols
, if
Let be a monotonic model. As
is a Boolean algebra of sets, and
, for every
,
is a monotonic subalgebra of the algebra
. We shall denote by
the space generated taking
as a basis for a topology defined in
Let
![KR = {Y ⊆ X : ∃x ∈ X ((x, Y) ∈ R )}](/img/revistas/ruma/v49n1/1a10203x.png)
the range of the multirelation . Let
the hyperspace of
relative to
. Recall that
denotes the set of all compact subsets of
.
Definition 5. Let be a model. We shall say that:
is compact if the space
is compact,
is image-compact if for all
and for all
, there exists
such that
and
.
is point-compact if
is compact subset in the topological space
, for each
,
is point-closed, if
is a closed subset of the space
for each
.
is modally saturated (or m-saturated) if it is image-compact and point-compact.
Remark 6. In [6] Hansen defines the notion of -saturated model as follows:
A model is
-saturated if the following conditions hold:
- (m1) For any
,
and
such that
, if
, for all finite subset
of
, then
.
- (m2) For any
, and
if for every finite subset
of
there exists an
such that
, then there exists an
such that
.
The condition (m2) is equivalent to say that the model is point-compact, as we will see. But the condition (1) is not equivalent to the notion of image-compact. Hansen's definition has the disadvantage that the ultrafilter extension of a model (see definition 15) fails to meet the condition (m1). Now we will see that the condition (m2) is equivalent to require that
should be point-compact.
Proposition 7. Let be a model. Then
is point-compact iff it satisfies the condition (m2).
Proof. Let
and
such that for every finite subset
of
there exists an
such that
. Suppose that
, for any
. Then for each
there exists
such that
. So,
![⋃ { } R (x) ⊆ LV (¬φy) : φy ∈ Γ .](/img/revistas/ruma/v49n1/1a10264x.png)
As is a compact subset of
there exists a finite subset
such that
![]() | (3.1) |
By hypothesis, there exists such that
which is a contradiction to (3.1). Thus, there exists an
such that
.
Let
. We prove that
is compact subset in the topological space
. Let
such that
. Suppose that
![⋃ { } R (x) ⊈ LV (φ) : φ ∈ Γ i ,](/img/revistas/ruma/v49n1/1a10279x.png)
for any finite subset of
. So for each finite subset
of
there exists
such that
. From condition (m1), there exists
such that
. Thus,
, which is a contradiction. □
Let be a model. Recall that the insertion map
of
is defined by
, for each
. We write
by
.
Lemma 8. Let be a model. If
is point-closed, then
![⋂ FɛV(Y) = {ɛV(y) : y ∈ Y } ⊆ ♢-R 1(ɛV (x)) implies that (x,Y ) ∈ R,](/img/revistas/ruma/v49n1/1a10298x.png)
for all and for all
.
Proof. Let and
such that
. Suppose that
. Since
is point-closed, there exists
such that
and
Since
, there exists
such that
, which is a contradiction, because
. Therefore
□
Let be a model. Let us consider the following property:
(P): For all and for all
, if
then there exists
such that
and
,
where is the topological closure in the space
.
Proposition 9. Let be a model. If
is point-compact, then
satisfies the property (P).
Proof. Let and let
. Assume that
![]() | (3.2) |
Suppose that for all
. So for each
there exists
such that
and
, i.e.
Thus,
![⋃ { } R (x) ⊆ LV(¬φi) : Y ⊆ V (φi) .](/img/revistas/ruma/v49n1/1a10336x.png)
Since is a compact subset of
, there exists a some finite set of formulas
such that
![R (x) ⊆ LV (¬φ1) ∪ ⋅⋅⋅ ∪ LV (¬φn).](/img/revistas/ruma/v49n1/1a10340x.png)
Then,
![x ∕∈ ♢R (V (φ1) ∩ ...∩ V (φn)) = V (♢(φ1 ∧ ...∧ φn )),](/img/revistas/ruma/v49n1/1a10341x.png)
and By (3.2),
i.e.,
which is a contradiction. Therefore, there exists
such that
□
Proposition 10. Let be a compact model. Then
satisfies the property (P) iff
is point-compact.
Proof. Suppose that satisfies the property (P). Let
. Suppose that for every finite subset
of
![]() | (3.3) |
We prove that . Suppose the contrary. Then
. As
is compact,
, for some
. From (3.3), there exists
such that
, i.e.,
, which is impossible. Thus,
Let
![⋂ Z = {V (φ )c : V (φ) ∈ W } .](/img/revistas/ruma/v49n1/1a10364x.png)
It is clear that is a closed subset of
, and by compacity,
is compact. We prove that
![⋂ -1 {ɛV (z) : z ∈ Z} ⊆ ♢R (ɛV (x)) .](/img/revistas/ruma/v49n1/1a10368x.png)
If , then
. By compacity, there exists a finite set
such that
![c c V (φ1) ∩ ... ∩ V(φn ) ⊆ V (ψ ).](/img/revistas/ruma/v49n1/1a10372x.png)
Then . From (3.3), there exists
such that
i.e.,
Thus,
. So,
![⋂ -1 {ɛV (z) : z ∈ Z} ⊆ ♢R (ɛV (x)) ,](/img/revistas/ruma/v49n1/1a10378x.png)
and by hypothesis there exists such that
Then
, for every
Thus,
![⋃ R (x) ⊈ {L : V(φ ) ∈ W } . V(φ)](/img/revistas/ruma/v49n1/1a10383x.png)
The other direction is followed by Proposition 9. □
The maps between monotonic frames and monotonic models which preserve the modal structure will be referred to as bounded morphisms. These have previously been studied in [6] (see also [5]).
Definition 11. ([6]) A bounded morphism between two monotonic models and
is a function
such that
, for each propositional variable
,
- If
, then
, and
- If
then there exists
such that
and
.
It follows that truth of modal formulas is invariant under bounded morphisms ([6]).
Proposition 12. Let and
be monotonic models. If
is a bounded morphism from
to
then for each formula
,
.
The following technical lemma is needed in the next results.
Lemma 13. Let be a model. Then
is compact iff
is surjective.
Proof. Let
. We prove that
Let us suppose the opposite. Then
Since
is compact,
for some formulas
. So,
, which is impossible. It follows that there exists
. Now, it is easy to see that
.
Let
Suppose that for any finite subset
of
,
Let us consider the filter
of
generated by the set
It is not difficult to prove that
is proper. It follows that there exists an ultrafilter
of
such that
Since the map
is onto, there exists
such that
Then,
for all
So,
which is a contradiction. Thus, there is a finite subset
of
, such that
. □
The concepts of compact and point-compact models are preserved by surjective bounded morphisms.
Proposition 14. Let be a bounded morphism between the monotonic models
and
.
- If
is compact, then
is compact.
- If
is surjective and
is compact, then
is compact.
- If
is point-compact, then
is point compact.
- If
is surjective and
is point-compact
then
is point compact.
Proof. (1) By Lemma 13 we need to prove that the map is surjective. Let
. Consider the set
. It is easy to prove that
. Since
is surjective, there exists
such that
. Let
. Then, it is easy to see that
.
(2) By Lemma 13 we need to prove that the map is surjective. Let
. Consider
. It is easy to see that
. As
is surjective, there exists
such that
. Since
is surjective, there exists
such that
. We prove that
. For all
,
iff
iff
iff
. Thus,
.
(3) Let and let
Suppose that
![⋃ { } R1 (x ) ⊆ LV1(φ) : V1 (φ) ∈ Γ .](/img/revistas/ruma/v49n1/1a10478x.png)
Taking into account that , for all
, it is easy to see that
![⋃ { } R2 (f(x)) = R2(y ) ⊆ LV2(φ) : V1(φ ) ∈ Γ .](/img/revistas/ruma/v49n1/1a10481x.png)
As is point-compact, there exists a finite set
such that
Let
. Then
. So there exists
such that
. Thus
.
(4) Let and let
Suppose that
![⋃ { } R2 (y ) ⊆ LV2(φ) : V2 (φ ) ∈ Γ .](/img/revistas/ruma/v49n1/1a10492x.png)
As is surjective there exists
such that
. We prove that
![]() | (3.4) |
Let . Since
is a bounded morphism,
. So, there exists
such that
, i.e.
. Thus (3.4) is valid. As
is point-compact, there exists
such that
. We prove that
![R2 (y) ⊆ LV2(φ1) ∪ ...∪ LV2(φn).](/img/revistas/ruma/v49n1/1a10506x.png)
Let . Since
is a bounded morphism, there exists
such that
and
. As
, there exists
such that
.i.e.
. It follows
. Thus,
. So,
is point-compact. □
4. Ultrafilter and model extension
Let be a model. Let
. Let
the Boolean space of the algebra
. Let us consider the ultrafilter frame
of the monotonic algebra
. Recall that
is defined as:
![(P, Y ) ∈ R iff ∃C ∈ C (Ul(X )) (C ⊆ Y and F ⊆ ♢ -1(P )), P(X) C R](/img/revistas/ruma/v49n1/1a10526x.png)
where and
. We write
. We note that if
is a closed subset of
, then
![⋂ (P, Y ) ∈ RU iff FY = {Q : Q ∈ Y } ⊆ ♢ -1(P). R](/img/revistas/ruma/v49n1/1a10532x.png)
Definition 15. The ultrafilter extension of a monotonic model is the structure
![U e (M ) = ⟨U l(X ),RU ,VU ⟩,](/img/revistas/ruma/v49n1/1a10534x.png)
where is a map defined by:
![VU (p) = {P ∈ U l(X ) : V (p) ∈ P} ,](/img/revistas/ruma/v49n1/1a10536x.png)
for every
It is easy to see that , for each
(see [6]).
Given a model we can define another extension taking the set
as the base set of a model. Let
be the Boolean space of
. Let us consider the ultrafilter frame
of the monotonic algebra
, where
is defined by
![-1 (P, Y ) ∈ RDV iff ∃C ∈ C(U l(DV )) (C ⊆ Y and FC ⊆ ♢ R (P)),](/img/revistas/ruma/v49n1/1a10547x.png)
where and
.
Definition 16. The valuation extension, or valuation model, of a model is the structure
![V e(M ) = ⟨U l(DV ),RDV ,VDV ⟩ ,](/img/revistas/ruma/v49n1/1a10551x.png)
where the function is defined by
![V (p) = {P ∈ U l(D ) : V (p) ∈ P }, DV V](/img/revistas/ruma/v49n1/1a10553x.png)
for every
We note that for each
.
Theorem 17. Let be a monotonic model. Then for any
and
,
iff
and
iff
Proof. We prove , for any
. The other proof is very similar. The proof is by induction on the complexity of
We consider the case
Let
. Let
Let us consider the filter
in the Boolean algebra
Then it is easy to see that
For the other direction, suppose that
Then there exists a filter
of
such that
![ˆ ˆ (P, F) ∈ RDV and F ⊆ VDV (φ).](/img/revistas/ruma/v49n1/1a10577x.png)
By inductive hypothesis we have . Thus,
, i.e.,
.
(2). We prove iff
The other proof is similar. Assume that
. Then,
. But
belong to every ultrafilter of
. Then,
, i.e.,
. Now, if
, then there exists
such that
Then,
. Thus,
□
Theorem 18. Let be a model. Then the model
is
-saturated.
Proof. Let be a monotonic model. We prove that
is compact i.e.,
is a compact space. From Lemma 13 it is enough to prove that the map
is surjective. Let
. Consider the filter
in
generated by
. It is clear that
is proper. So there exists
such that
. By construction
. So,
, and consequently
is surjective.
We prove that is point-compact in the hyperspace
, where
. As
is a compact space we will apply Proposition 10. Let
and let
be a subset of
Suppose that
![⋂ -1 {ɛU (Q ) : Q ∈ Y } ⊆ ♢ RU(ɛU (P )).](/img/revistas/ruma/v49n1/1a10618x.png)
We need to prove that there exists a subset of
such that
and
, where
is the topological closure of
in the topological space
.
Consider the filter generated by the set
. We prove that
![(P,Fˆ) ∈ RU .](/img/revistas/ruma/v49n1/1a10628x.png)
Let . Then there exists
such that
![Y ⊆ VU (φ ) ∩ ...∩ VU (φn) and V (φ ) ∩ ...∩ V (φn) ⊆ W.](/img/revistas/ruma/v49n1/1a10631x.png)
So, . Then
![♢ (V (φ ∧ ...φ ) = V (♢(φ ∧ ...∧ φ )) ∈ ɛ (P ), RU U n U 1 n U](/img/revistas/ruma/v49n1/1a10633x.png)
i.e. . Then
. Thus
. Taking into account that the topological closure of
in the space
is
, it is easy to prove that
. Therefore
is
-saturated. □
Theorem 19. Let be a model. Then the model
is
-saturated.
Proof. Since is a subalgebra of the algebra
, we can define a map
by
. By well-established results in the duality theory of Boolean algebras the map
is the dual map of the inclusion homomorphism between
and
. From the results given by H. Hansen [6] we get that
is a surjective bounded morphism between the ultrafilter extension
and the valuation extension
. Since
is
-saturated
by Proposition 14 we get that
is also
-saturated. □
[1] B. F. Chellas, Modal Logic: an introduction, Cambridge Univ. Press,1980. [ Links ]
[2] R. Goldblatt, Maps and Monads for Modal Frames, Studia Logica Volume 83, Numbers 1-3 (2006), pp.309-331. [ Links ]
[3] R. Goldblatt, Mathematics of Modality, CSLI Lectures Notes 43, 1993. [ Links ]
[4] R. Goldblatt, Logics of Time and Computation, CSLI Lectures Notes 7, Second Edition, 1992. [ Links ]
[5] H. H. Hansen and C. Kupke. A Coalgebraic Perspective on Monotone Modal Logic, Electronic Notes in Theoretical Computer Science, 106, pages 121-143, Elsevier, 2004. [ Links ]
[6] H. H. Hansen. Monotonic modal logic (Master's thesis). Preprint 2003-24, ILLC, University of Amsterdam, 2003. [ Links ]
[7] M. J. Hollenberg, Hennessey-Milner Classes and Process Algebra. In M. de Rijke A. Ponse and Y. Venema, editors, Modal Logic and Process Algebra: a Bisimulation Perspective, volume 53 of CSLI Lecture Notes, pages 187-216. CSLI Publications, 1995. [ Links ]
Sergio Arturo Celani
CONICET and Departamento de Matemáticas,
Facultad de Ciencias Exactas,
Universidad Nacional del Centro, Pinto 399,
7000 Tandil, Argentina
scelani@exa.unicen.edu.ar
Recibido: 10 de abril de 2008
Aceptado: 28 de abril de 2008