On
HS
=
SH
and Duality Theorems of
Intuitionistic Descriptive Frames
Fan Yang
∗Institute for Logic, Language and Computation
Universiteit van Amsterdam
November 8, 2007
Abstract
The variety of Heyting algebras has a nice property that HS=SH. Heyting algebras are the algebraic dual of intuitionistic descriptive frames. The goal of this paper is to define proper dual notions so as to formulate this algebraic properties in the frame language, and to give a frame-based proof of this property and some other duality theorems.
1
Introduction
We know that for any algebraAin a variety, it holds thatSH(A)⊆HS(A) but in general not the other way around. However, the class of Heyting algebras is a variety for which the other direction does hold. This gives the property thatHS(A) =SH(A) for every Heyting algebraA.
Heyting algebras are the algebraic dual of intuitionistic descriptive frames1. So it is possible to express the above property in the framework of descriptive frames. The goal of this paper is to define proper dual notions so as to formulate the above algebraic properties in the frame language, and to give a frame-based proof of this property and some other duality theorems.
2
Intuitionistic Kripke Frames
In order to study descriptive frames, we start with intuitionistic Kripke frames, which is the simplest case of intuitionistic frames. In this section, we prove a theorem which corresponds to the algebraic propertyHS(A) =
SH(A) for Heyting algebras in the intuitionistic Kripke frame case.
2.1
The Definition of Intuitionistic Kripke Frames
Definition 1. Anintuitionistic Kripke frame is a pairF=hW, Ri con-sisting of a nonempty setW and a partial orderRonW.
∗This paper has been written as a MoL project under the advice of Prof. dr. Dick de
Jongh.
Definition 2. A setV ⊆W is called anupward closed subset, if for every
w∈V andu∈W,wRuimpliesu∈V.
Definition 3. An intuitionistic Kripke frameG=hV, Siis called a sub-frame of an intuitionistic Kripke frameF = hW, Ri if V ⊆W and S is the restriction ofRtoV (S=RV, in symbols), i.e.,S=R∩V2. The subframeGis agenerated subframeofFifV is an upward closed subset ofW.
Definition 4. SupposeF=hW, RiandG=hV, Siare two intuitionistic Kripke frames. A mapf fromW toV is called ap-morphism fromFto
Gif the following conditions hold for everyw, u∈W: (P1)wRuimpliesf(w)Sf(u);
(P2)f(w)Sf(u) implies∃v∈W(wRv∧f(v) =f(u)).
2.2
A Frame Instance of HS
(A) =
SH
(A)
The next theorem is actually the dual theorem ofHS(A) = SH(A) of Heyting algebra in terms of intuitionistic Kripke frames. Here we give a constructive proof of this algebraic result.
We will generalize the result of Theorem 5 to the case of descriptive frames in Proposition 44 in Section 4.
Theorem 5. An intuitionistic Kripke frameF0 is a generated subframe of a p-morphic image of an intuitionistic Kripke frame Giff F0 is a p-morphic image of a generated subframe ofG.
Proof. For “⇒”: SupposeF=hV, Siis a p-morphic image ofG=hW, Ri via f, and F0 = hV0, S0i is a generated subframe of F. Define G0 = hf−1(V0
), Rf−1(V0
)iandg=ff−1(V0
) :f−1(V0
)→V0. We prove the following:
(1)G0 is a generated subframe ofG. (2)gis a p-morphism.
For (1): We only need to show thatf−1(V0) is upward closed inW. Supposew∈f−1(V0
) andwRufor someu∈W. Sincefis a p-morphism, we havef(w)Sf(u). Then, fromf(w)∈V0 andV0 being upward closed, we obtainf(u)∈V0, i.e.,u∈f−1(V0
).
For (2): It suffices to show that the mapgfromG0 toF0 satisfies the two conditions of a p-morphism.
(P1) Supposew1Rw2wherew1, w2∈f−1(V0). Sincefis a p-morphism, we havef(w1)Sf(w2), i.e.,g(w1)Sg(w2).
(P2) Supposeg(w)S0g(u). Thenf(w)Sf(u), for somew, u∈f−1(V0). Sincef is a p-morphism, there existsv∈W s.t. f(v) =f(u) andwRv. Thus v ∈ f−1(V0
) since f−1(V0
) is upward closed. Finally we also get
g(v) =f(v) =f(u) =g(u) as required.
For “⇐”: Supposegis a p-morphism fromG0=hW0, R0itoF0=hV0, S0i, whereG0is a generated subframe ofG=hW, Ri. Without loss of gener-ality, we may assumeW∩V0 =∅. We first define a frame F=hV, Siby puttingV =V0∪(W−W0) andS=S0∪S1∪S2, whereS1=R(W−W0) and
S2={(w1, g(w2)) :w1∈W−W0, w2∈W0andw1Rw2}. Note thatS0,S1 andS2 are pairwise disjoint.
In fact, for anyv∈V, ifv∈V0, thenvS0v, sovSv. Ifv∈W −W0, thenvS1v, i.e.,vSv, henceS is reflexive.
SupposexSyandySx. Then by the construction ofS, it is impossible that one ofxandyis inV0but another inW−W0. If bothxandyare inV0, then it follows that xS0yandyS0x, hencex=y. If bothxandy
are inW−W0, thenxRyandyRx, sox=y.
SupposexSyandySz. IfxS0y, thenx, y∈V0, so it must be the case thatz∈V0. Thus by the transitivity ofS0, we havexS0z, hencexSz;
IfxS1y, thenx, y∈W−W0andxRy. Wheneverz∈W −W0, from the transitivity ofR it followsxS1z. Wheneverz ∈ V0, yRw for some
w∈W0 such thatz =g(w), thusxRw, and soxS2z by the definition of
S2.
If xS2y, then x ∈ W −W0, and there exists w ∈ W0 such that
xRwandy = g(w) ∈ V0. So it must be the case that yS0z. Since F0
is a p-morphic image ofG0 viag, by (P2), there existsu∈W0 such that
wRu, z=f(u), thusxRu, and soxS2zby the definition ofS2. Next, we define a mapf:W →V by taking
f(w) =
(
g(w), ifw∈W0 w, ifw∈W−W0.
It is easy to see thatF0is a generated subframe ofFandfis surjective. It remains to show thatf is a p-morphism fromGontoF, i.e.,fsatisfies (P1) and (P2).
(P1) For eachw1, w2∈W s.t.,w1Rw2, from the construction ofS, it is easy to see thatf(w1)Sf(w2).
(P2) For eachw, u∈W s.t.,f(w)Sf(u), we find av∈W s.t.,f(v) =
f(u) andwRv.
Case 1: w∈W0. Thenf(w) =g(w)∈V0, which impliesf(u)∈V0for
V0 is upward closed. Sinceg is a p-morphism, there existsv∈W0⊆W
s.t.,g(v) =g(u) andwR0v. These are equivalent tof(v) =f(u) andwRv. Case 2: w, u∈W−W0. Then fromf(w)Sf(u), we getwSu. It follows thatwS1u. Thus, by the definition ofS1, we havewRuandv=usatisfies (P2).
Case 3: w∈W−W0andu∈W0. Then fromf(w)Sf(u) andu∈W0, we getwSg(u). It follows thatwS2g(u). So by the definition ofS2, there existsv∈W0s.t.,f(v) =f(u) andwRv.
3
Duality Theorems of Intuitionistic
Gen-eral
Frames
In this section, we prove some duality theorems of intuitionistic general frames. These results are crucial in proving some nice properties and duality theorems of descriptive frames.
3.1
The Definition of Intuitionistic General Frames
and Their Duals
upward closed sets, containing∅and closed under∩,∪and the following operation⊃: for everyX, Y ⊆W,
X ⊃Y ={x∈W :∀y∈W(xRy∧y∈X →y∈Y)}
Note 7. From the definition above, it follows thatW ∈ PsinceW=∅ ⊃ ∅.
The next two theorems show that there is a one-to-one correspondence between intuitionistic general frames and Heyting algebras. For the proof of Theorem 8, see Chapter 8 in [5].
Theorem 8. LetF=hW, R,Pibe an intuitionistic general frame. The algebra hP,∩,∪,⊃,∅i is a Heyting algebra and is called the dual of F, denoted byF+.
Theorem 9. LetAbe a Heyting algebra, define A+ =hWA, RA,PAi as
follows:
(i) WA={∇ ⊆A:∇is a prime filter ofA}, (ii) ∇1RA∇2 iff∇1⊆ ∇2,
(iii) PA={ˆa:a∈A}, whereˆa={∇ ∈WA:a∈ ∇}.
ThenA+ is an intuitionistic general frame called the dualofA.
Further-more,A∼= (A+)+=hPA,∩,∪,⊃,∅i.
Proof. We first prove A+ is an intuitionistic general frame. Obviously,
RA is a partial order, andPA is upward closed. We first show that PA
is closed under∩. Since the elements inWA are filters, we have for any
a, b∈A, ˆ
a∩ˆb={∇ ∈WA:a∈ ∇andb∈ ∇}={∇ ∈WA:a∧b∈ ∇}=a[∧b.
Next we show thatPAis closed under the union∪. Indeed, For
arbi-trary ˆa,ˆb∈ PA, sinceWAconsists of all prime filters∇ofA, we have
ˆ
a∪ˆb={∇ ∈WA:a∈ ∇orb∈ ∇}={∇ ∈WA:a∨b∈ ∇}=a[∨b.
To show that PA is closed under the operation⊃, it suffices to prove
that for anya, b∈A,
ˆ
a⊃ˆb=a\→b.
First, we show a\→b ⊆ ˆa ⊃ ˆb. For any ∇ ∈ a\→b and any ∇0 ∈ ˆ
a
such that∇RA∇0, it suffices to show∇0∈ˆbi.e.,b∈ ∇0, but this follows
immediately froma→b∈ ∇ ⊆ ∇0 anda∈ ∇0.
Second, we show ˆa ⊃ˆb⊆a\→b. Suppose∇ ∈ ˆa⊃ˆb. It suffices to show that∇ ∈a\→b. By the definition of operation⊃, for any ∇0⊇ ∇, it holds that
∇0∈ˆa⇒ ∇0∈ˆb.
Consider the prime filter
∇a={x∈A:∃z∈ ∇(z∧a≤x)}.
Clearly, it holds that∇a∈aˆand∇ ⊆ ∇a. Thus we have∇a∈ˆb, which
means∃z ∈ ∇(z∧a≤b), hencez ≤a→b. Therefore a→b∈ ∇i.e., ∇ ∈a\→b.
Lastly we show thatf is injective. Suppose ˆa= ˆb, then ˆ
>=a\→a= ˆa⊃aˆ= ˆa⊃ˆb=a\→b.
From{>} ∈ >ˆ =a\→b, it follows that a→ b∈ {>} i.e.,>= a→ b. Hencea≤b. By a similar argument, we can proveb≤a, hencea=b.
3.2
Duality Theorems about Operations On
Gen-eral
Frames
In the rest of this section, we discuss the relation between two operations on general frames (i.e., the operations of generating generated subframes and forming p-morphic images) and two algebraic operations (i.e., the operations of forming homomorphic images and generating subalgebras) respectively.
Definition 10. An intuitionistic general frameG=hV, S,Qiis called a
generated subframe of an intuitionistic general frame F=hW, R,Piif it satisfies the following conditions:
(S1)hV, Siis a generated subframe ofhW, Ri, (S2)Q={U∩V :U ∈ P}.
The notion of generated subframe of intuitionistic general frames cor-responds to the notion of homomorphic image of Heyting algebras. To see this clearly, we prove the next two theorems.2
Theorem 11. Ifhis an isomorphism of G=hV, S,Qionto a generated subframe ofF=hW, R,Pi, then the maph+ defined by
h+(X) =h−1(X) ={x∈V :h(x)∈X}, for everyX∈P,
is a homomorphism ofF+ ontoG+.
Proof. W.l.o.g., we may assume Gis a generated subframe ofF. Thenh
is an identity map andh+(X) =X∩V.
Clearly, h+ is a surjection. We show it preserves ∩,∪ and ⊃. Let
X, Y ∈ P. Then we have
h+(X∩Y) = (X∩Y)∩V = (X∩V)∩(Y ∩V) =h+(X)∩h+(Y);
h+(X∪Y) = (X∪Y)∩V = (X∩V)∪(Y ∩V) =h+(X)∪h+(Y);
h+(X⊃Y) ={x∈W :∀y∈W(xRy∧y∈X→y∈Y)} ∩V
={x∈V :∀y∈V(xSy∧y∈X∩V →y∈Y ∩V)} (sinceV is upward closed andS=R∩V2)
= (X∩V)⊃(Y ∩V) =h+(X)⊃h+(Y).
Note that in the proof above, we do not need the property that Qis closed under operations. This gives the following corollary.
Corollary 12. LethW0, R0i be a Kripke generated subframe of an intu-itionistic general frameF =hW, R,Pi. Then by taking P0
={U ∩W0 :
U ∈ P}, we can define an intuitionistic general frame F0 =hW0, R0,P0i
which is a generated subframe ofF.
Theorem 13. Supposehis a homomorphism of a Heyting algebraAonto a Heyting algebraB. Then
(i) For any filter∇0
inB,h−1(∇0
)is a filter inA; in particular,h−1(∇0 )
is prime whenever∇0 is prime;
(ii) For any prime filter∇inAsuch thath−1(>)⊆ ∇,∇=h−1(h(∇)), andh(∇)is a prime filter in B;
(iii) The maph+ defined by
h+(∇0) =h−1(∇0), for every prime filter∇0 inB,
is an isomorphisim of B+ onto a generated subframe ofA+.
Proof. (i) Let∇0
be any filter inB. Obviously> ∈h−1(∇0 ).
b, b→a∈h−1(∇0)⇒h(b), h(b→a)∈ ∇0
⇒h(b), h(b)→h(a)∈ ∇0 (Sincehis a homomorphism) ⇒h(a)∈ ∇0(since∇0is a filter)
⇒a∈h−1(∇0).
Henceh−1(∇0
) is a filter. In particular, if∇0
is prime, then
a∨b∈h−1(∇0)⇒h(a∨b) = (h(a)∨h(b))∈ ∇0 (sincehis a homomorphism)
⇒h(a)∈ ∇0 orh(b)∈ ∇0(since∇0 is a prime filter) ⇒a∈h−1(∇0) orb∈h−1(∇0).
Henceh−1(∇0
) is prime.
(ii) It suffices to showh−1(h(∇))⊆ ∇. Indeed,a∈h−1(h(∇)) implies
h(a) =h(b) for someb∈ ∇. Sincehis a homomorphism fromAtoB, we have
h(b→a) =h(b)→h(a) =>.
From the assumption thath−1(>)⊆ ∇, it followsb→a∈ ∇, thusa∈ ∇. Observe that∇=h−1(h(∇)) is equivalent to the following:
a∈ ∇iffh(a)∈h(∇). (1) Supposeh(a), h(a) →h(b)∈h(∇). Sincehis a homomorphism,h(a →
b) =h(a)→h(b)∈h(∇). From (1) it follows thata, a→b∈ ∇, which impliesb∈ ∇since∇is a filter. Henceh(b)∈h(∇) andh(∇) is a filter.
Next we show thath(∇) is prime. For anya, b∈ ∇, we have (h(a)∨h(b))∈h(∇)⇔h(a∨b) = (h(a)∨h(b))∈h(∇)
(sincehis a homomorphism) ⇔a∨b∈ ∇(by (1))
(iii) By (i), h+ is a well-defined map from WB to WA. (1) indicates
thath+is surjective. Obviously,h+is also injective. Soh+is a bijection. Let
W ={∇ ∈WA:h−1(>)⊆ ∇}.
ClearlyW is upward closed inWA.
LetF=hW, R,Pibe a generated subframe ofA+. Note that
U0∈ P ⇔ ∃U ∈ PAs.t. U0=U∩W
⇔ ∃a∈As.t. U0={∇ ∈W:a∈ ∇}.
Then, for anyX⊆WB,
X∈ PB⇔ ∃b∈B, X={∇
0
∈WB:b∈ ∇
0 }
⇔ ∃a∈A, h(a) =b andh−1(X) ={h−1(∇0)∈WA:a∈h
−1(∇0 )}
⇔ ∃a∈A, h−1(X) ={∇ ∈W :a∈ ∇} ⇔h−1(X)∈ P
⇔h+(X)∈ P
To show that h+ is an isomorphism from B+ onto F, it remains to show that for any∇0
1,∇02∈WB,
∇0
1RB∇02 iffh+(∇01)RAh+(∇02). However, this is trivial since∇01 ⊆ ∇
0 2iffh
−1(∇0 1)⊆h
−1(∇0 2).
Definition 14. Given intuitionistic general frames F = hW, R,Pi and
G= hV, S,Qi, we say a map f from W to V is a p-morphism from F
toG, if the following three conditions are satisfied, for allw, u∈W and
X∈ Q:
(P1)wRuimpliesf(w)Sf(u);
(P2)f(w)Sf(u) implies∃v∈W(wRv∧f(v) =f(u)); (P3)f−1(X)∈ P.
As with generated subframes, the notion of p-morphisms between gen-eral frames has its dual notion, i.e., subalgebras of Heyting algebras. We prove the next two theorems.3
Theorem 15. If G=hV, S,Qiis a p-morphic image of F=hW, R,Pi
viaf, then the mapf+ defined by
f+(X) =f−1(X), for everyX∈ Q,
is an isomorphism ofG+ onto a subalgebra ofF+.
Proof. Since f is a p-morphism, we have for anyX ∈ Q, f−1(X)∈ P, which givesf+[Q]⊆ P. We provef+ :Q →f+[Q] is an isomorphism. Clearly,f+ is a bijection. So it suffices to show thatf+ preserves all the operations inG+. LetX, Y ∈ Q. Then we have
f+(X∩Y) =f−1(X∩Y) =f−1(X)∩f−1(Y) =f+(X)∩f+(Y);
x∈f+(X⊃Y)⇔f(x)∈X⊃Y
⇔ ∀f(y)∈V(f(x)Sf(y)∧f(y)∈X →f(y)∈Y) ⇔ ∀y∈W(xRy∧y∈f−1(X)→y∈f−1(Y)) (sincef is a p-morphism)
⇔x∈f−1(X)⊃f−1(Y) ⇔x∈f+(X)⊃f+(Y)
Theorem 16. If f is an isomorphism of a Heyting algebra B onto a
subalgebra ofA, then the mapf+ defined by
f+(∇) =f−1(∇), for every∇ ∈WA,
is a p-morphism fromA+ onto B+.
Proof. W.l.o.g., we may assume B to be a subalgebra of A and sof is the identity map and
f+(∇) =∇ ∩B, for every∇ ∈WA.
Clearly, if∇is a prime filter inAthenf+(∇) is a prime filter inB. So by Lemma 8.70 in [5],f+is a map fromWAontoWB. Next, we showf+ satisfies the three conditions of a p-morphism.
(P1) Suppose∇1RA∇2, for some∇1,∇2∈WA. This means∇1⊆ ∇2, hence∇1∩B⊆ ∇2∩B, i.e.,f+(∇1)RBf+(∇2).
(P2) Suppose f+(∇1)RBf+(∇2), for some ∇1,∇2 ∈ WA. Consider
the filter∇0 inAgenerated by the set
∇1∪(∇2∩B).
And consider the ideal40 inAgenerated byB− ∇2. By Exercise 7.18 in [5], there is a prime filter∇0 inAsuch that∇0⊆ ∇0and40∩ ∇0=∅. By the definition,∇1 ⊆ ∇0 and∇0∩B=∇2∩B. These mean∇1RA∇0
andf+(∇0) =f+(∇2).
(P3) LetX ∈ PB, i.e., there is b∈B such thatX ={∇ ∈WB :b∈
∇}. Then
f+−1(X) ={∇ 0
∈WA:f+(∇ 0
)∈X} ={∇0∈WA:∇
0
∩B∈X} ={∇0∈WA:b∈ ∇
0 }
and sof+−1(X)∈ PA.
4
Descriptive Frames
4.1
The Definition of Descriptive Frames
Definition 17. An intuitionistic general frameF =hW, R,Pi is called
refinedif for anyx, y∈W,
∀X ∈ P(x∈X →y∈X)⇒xRy,
or equivalently,
¬xRy⇒ ∃X∈ P(x∈X∧y6∈X).
Definition 18. A familyX of sets has thefinite intersection propertyif every finite subfamilyX0⊆ X
has a nonempty intersection, i.e.,TX06=∅.
Definition 19. An intuitionistic general frameF =hW, R,Pi is called
compact, if for any familiesX ⊆ P andY ⊆ P ={W−X :X ∈ P}for whichX ∪ Y has the finite intersection property,
\
(X ∪ Y)6=∅.
Definition 20. An intuitionistic general frameF is calleddescriptive iff it is refined and compact.
4.2
Descriptively generated subframes
When trying to define the two operations on descriptive frames, we came to recognize that to restrict all the frames under consideration to be de-scriptive, we cannot simply use the definitions of the intuitionistic general frame case. To see this clearly, we first give an example concerning gen-erated subframes.
Example 21. There exists a generated subframe of a descriptive frame which is not a descriptive frame.
Proof. Define a descriptive frameF=hW, R,Pi, by taking
W =ω∪ {ω},R=≥,P={R(n) :n∈W} ∪ {∅},
whereR(n) ={m∈W:nRm}={0,1, ..., n}(note: R(ω) =W). Obviously P is closed under the two operations ∩ and ∪, and P is refined.
Next we verify thatPis closed under⊃.
Indeed, for anym, n∈W, wheneverm≥n, we have R(n)⊆R(m), hence
R(m)⊃R(n) =R(n)∈ P; wheneverm < n, we haveR(m)(R(n), hence
R(m)⊃R(n) =W ∈ P.
Now we show thatP is compact.
Suppose familyX ∪ Y has the finite intersection property, where
X ={R(j) :j∈J⊆W}={{0,1, ..., j}:j∈J⊆W} ⊆ P,
Y={W−R(i) :i∈I⊆W}={{i+ 1, i+ 2, ..., ω}:i∈I⊆W} ⊆ P.
By the well-orderedness ofW, there exists a least memberk ∈ J, thus {1, ..., k}is the least member ofX with respect to the relation⊆, andk
{1, ..., k}intersects every member {i+ 1, i+ 2, ..., ω} ofY, hence kis in each member{i+ 1, i+ 2, ..., ω}ofY, thereforek∈T
(X ∪ Y). Now, consider the frame
G=hω,≥,Qi,Q={R(n) : n∈ω} ∪ {∅}.
Obviouslyωis an upward closed subset ofW and Gis a generated sub-frame ofF. However,Gis not compact since the family{ω−R(n) : n∈
ω}={{n+ 1, n+ 2, ...}:n∈ω}has the finite intersection property, but
\
{ω−R(n) :n∈ω}=∅.
However, we do have the following result.
Proposition 22. If an intuitionistic general frame G = hV, S,Qi is a generated subframe of a descriptive frameF=hW, R,Pi, thenGis refined. Proof. Suppose¬uSv for someu, v∈V. SinceP is refined, there exists
U ∈ P such thatu ∈ U and v 6∈ U. Let U0 = U ∩V. Then by (S2),
U0∈ Q. And we haveu∈U0 andv6∈U0.
In view of Example 21 and Proposition 22, we need to redefine (or even rename for distinguishing use) “generated subfames” of descriptive frames to guarantee the generated subframes to be compact, thus descriptive as well.
Definition 23.An intuitionistic general generated subframeG=hV, S,Qi (i.e., Gsatisfies (S1) and (S2)) of a descriptive frame F =hW, R,Pi is called adescriptively generated subframe, if it also satisfies the following condition:
(S3)V =T
{U∈ P:V ⊆U}.
We now show that (S3) is a necessary and sufficient condition for a generated subframe of a descriptive frame to be descriptive as well.
Proposition 24. Suppose G = hV, S,Qi is a generated subframe of a descriptive frame F = hW, R,Pi. If G is a descriptive frame, then G
satisfies (S3) in Definition 23, i.e.,Gis a descriptively generated subframe ofF.
Proof. Assume
V 6=\{U ∈ P:V ⊆U}. (2) (2) means V + T{U ∈ P : V ⊆ U}, i.e., there exists w such that
w6∈V and
w∈\{U ∈ P:V ⊆U}. (3) Define
X ={V −U :U∈ P, w6∈U} ⊆ Q.
First, we showXhas the finite intersection property. For allV−U1, ..., V−
Un∈X, we have
(V −U1)∩...∩(V −Un) =V −(U1∪...∪Un). (4) Note thatPis closed under finite unions, so we have (U1∪...∪Un)∈ P. Thus, sincew6∈U1∪...∪Un, by (3), we getV *U1∪...∪Un. It follows that (4)6=∅, henceT
X 6=∅sinceGis compact. Thus there existsu∈V
such thatu∈T
¬uRw. SincePis refined, there existsU ∈ Psuch thatu∈Uandw6∈U. By the definition ofX, it follows thatV −U∈X. Thus, byu∈T
X, we concludeu∈V −U, which contradicts the factu∈U.
Proposition 25. IfG=hV, R,Qiis a descriptively generated subframe of a descriptive frameF=hW, R,Pi, thenGis also a descriptive frame. Proof. By Proposition 22,Gis refined. It remains to show it is compact.
For any families X ⊆ Q and Y ⊆ Q={V −U0 :U0 ∈ Q}, suppose X ∪ Y has the finite intersection property.
Define
X1∗={U ∈ P:U∩V ∈ X }, X2∗={U ∈ P:V ⊆U},
X∗=X1∗∪ X ∗ 2 ⊆ P, and
Y∗={W−U0:U0∈ P, V −(U0∩V)∈ Y} ⊆ P.
Take any
X1∗, ..., X ∗
n∈ X
∗ 1, X
∗
n+1, ..., X ∗
n+m∈ X
∗ 2, andY
∗ 1, ..., Y
∗
k ∈ Y
∗
.
We know that there existX1, ..., Xn∈ X,andY1, ..., Yk∈ Y such that
Xi=Xi∗∩V(1≤i≤n) andYj=Y
∗
j ∩V(1≤j≤k).
Note that by (S3) we haveV =T{
U ∈ P :V ⊆U}=TX∗
2. Observe that n \ i=1 Xi∩ k \ j=1 Yj= n \ i=1
Xi∗∩V ∩ k
\
j=1
Yj∗∩V
=
n
\
i=1
Xi∗∩ k
\
j=1
Yj∗∩
\
X2∗
⊆
n
\
i=1
Xi∗∩ k
\
j=1
Yj∗∩ n+m
\
l=n+1
Xl∗
So, the fact thatX ∪ Y has the finite intersection property implies that X∗∪ Y∗has the finite intersection property.
Similarly, we also have
\
(X ∪ Y) =\X ∩\Y=\X1∗∩
\
Y∗∩\X2∗=
\
(X∗∪ Y∗).
Thus, by the compactness ofP, it holds thatT
(X∗∪ Y∗)6=∅, from which it follows thatT(X ∪ Y)6=∅.
4.3
Descriptively p-morphic Images
Next, we generalize the definition of p-morphisms between intuitionistic general frames to the descriptive frame case.
To this end, we first prove some propositions to analyze the properties of p-morphisms between intuitionistic general frames. The next proposi-tion shows that p-morphisms preserve compactness.
Proof. For any familiesX ⊆ QandY ⊆ Q={V −Y :Y ∈ Q}such that X ∪ Y has the finite intersection property, define
X∗={f−1(X) :X∈ X } ⊆ P,
Y∗={W−f−1(Y) :V −Y ∈ Y} ⊆ P.
ClearlyT
(X∗∪ Y∗) =f−1(T
(X ∪ Y)). In order to showT
(X ∪ Y)6=∅, it suffices to showT(X∗∪ Y∗
)6=∅. Indeed, for any
f−1(Xi)∈ X∗, i= 1,· · ·, n;W−f−1(Yj)∈ Y∗, j= 1,· · ·, m,
since (T
Xi)∩(T
(V −Yj))6=∅and
(\f−1(Xi))∩(\(W−f−1(Yj))) =f−1((\Xi)∩(\(V −Yj))),
we haveT
f−1(Xi)∩(T
W−f−1(Yj) 6= ∅, which means that X∗∪ Y∗ has the finite intersection property. From the compactness ofF, it follows thatT
(X∗∪ Y∗ )6=∅.
We know that the kernelEf of every surjective p-morphismf, defined by
wEfuifff(w) =f(u),
is an equivalence relation. Moreover, the kernel gives rise to a special kind of equivalence relation defined as follows.
Definition 27. LetF=hW, R,Pibe a descriptive frame. An equivalence relationE onW is called abisimulation equivalence onFif the following two conditions are satisfied:
(B1) For everyw, u, v∈W,wEvandvRuimply that there isz∈W such thatwRzandzEu;
(B2) If there is noxsuch that uRx and xEv, then there exists U ∈ P such thatE(U) =U,u∈U andv6∈U.
(B3) IfwRx,xEu,uRyandyEw, thenwEu.
With the bisimulation equivalence, we can define quotient frames of intuitionistic general frames.
Definition 28. Define the quotient frame F/E = hWE, RE,PEi of an
intuitionistic general frameF=hW, R,Piassociated with a bisimulation equivalence relationE by taking
WE ={E(w) :w∈W}, whereE(w) ={u∈W :wEu},
E(w)REE(u) iffw0Ru0 for somew0∈E(w) andu0∈E(u), and
PE={UE:U∈ P, E(U) =U},
where
UE={E(w) :w∈U} and
E(U) =[{E(w) :w∈U}={x:wEx, for somew∈U}.
Note 29. In the above definition,
(i) RE is really a partial order;
Proof. (i) The reflexivity and the antisymmetry ofREfollow immediately from the reflexivity ofRand (B3) respectively.
SupposeE(w)REE(u) andE(u)REE(v). Then by the definition, there existw0∈E(w),u0, u00∈E(u) andv0∈E(v), such thatw0Ru0andu00Rv0. Fromu0Eu00andu00Rv0, by (B1), it follows that there existsz∈W such thatu0Rz and zEv0. Then sincew0Ru0, by the transitivity of R, w0Rz, which meansE(w)REE(v).
(ii) First we showPE is a family of upward closed sets. For anyE(w)∈ UE ∈ PE andE(w)REE(u), there existw0 ∈E(w),u0∈E(u) such that w0Ru0. Since
E(w)⊆E(U) =U,
we havew0 ∈U, henceu0 ∈U since U is upward closed. It then follows that
E(u) =E(u0)∈UE.
SoPE is upward closed.
Since∅ ∈ P andE(∅) =∅,∅=∅E∈ PE.
In order to show thatPEis closed under operations, we first show that
for anyXE, YE∈ PE, the following hold:
E(X∪Y) =E(X)∪E(Y);
E(X∩Y) =E(X)∩E(Y); (5)
E(X⊃Y) =E(X)⊃E(Y).
Letf:W →WE be the natural map defined by
f(w) =E(w).
Observe that for anyU ⊆W,
E(U) =U ⇔ U=f−1(f(U)). (6) Hence, we have
X∪Y =f−1(f(X))∪f−1(f(Y)) =f−1(f(X)∪f(Y)) (7) and
X∩Y =f−1(f(X))∩f−1(f(Y) =f−1(f(X)∩f(Y)). (8) It follows that
f(X∪Y) =f(X)∪f(Y) andf(X∩Y) =f(X)∩f(Y). (9) From (7)-(9) we obtain
X∪Y =f−1(f(X∪Y)) andX∪Y =f−1(f(X∪Y)).
So, by (6), the above gives
E(X∪Y) =X∪Y =E(X)∪E(Y) and
E(X∩Y) =X∩Y =E(X)∩E(Y).
Next we show
i.e.,
x /∈X⊃Y ⇔ f(x)∈/f(X)⊃f(Y).
For anyUE∈ PE, sinceE(U) =U, by (6), we have
w∈U⇔f(w)∈f(U). (11) Supposex /∈X ⊃Y. Then
∃y∈W(xRy∧y∈X∧y /∈Y).
By the definition ofRE and (11), we obtain
∃f(y)∈f(W)(f(x)REf(y)∧f(y)∈f(X)∧f(y)∈/f(Y)),
which meansf(x)∈/f(X)⊃f(Y).
Conversely, supposef(x)∈/f(X)⊃f(Y). Then,
∃f(y)∈f(W)(f(x)Sf(y)∧f(y)∈f(X)∧f(y)∈/f(Y)).
By (B1) and (11), we obtain
∃y0∈W(xRy0∧y0∈X∧y0∈/Y),
which meansx /∈X⊃Y.
Thus, (10) is obtained. It follows that
f(X⊃Y) =f(X)⊃f(Y). (12) By (10) and (12) we obtain
X⊃Y =f−1(f(X⊃Y)),
which by (6) means
E(X ⊃Y) =X ⊃Y =E(X)⊃E(Y).
Now, supposeUE, UE0 ∈ PE. ThenU, U0∈ P,E(U) =U andE(U0) = U0. SincePis closed under∩,∪and⊃, we haveU∩U0, U∪U0, U⊃U0∈ P. Together with (5), we obtain UE∩UE0, UE∪U
0
E, UE ⊃U
0
E ∈ PE.
Thus,PE is closed under operations. This completes the proof.
Now consider the next example.
Example 30. There exists a p-morphic image of a descripitive frame which is not a descripitive frame.
Proof. LetF=hW, R,Pibe any descriptive frame such thatR6=W×W. DefineG=hW, R,Qiby takingQ={∅, W}. Clearly,Gis an intuitionistic general frame and a p-morphic image ofFvia the identity map. However,
Gis not a descriptive frame since it is not refined.
In view of this example, we then define the descriptively p-morphic image which is essentially different from the p-morphic image induced by Definition 14, in the sense that (P4) and (P5) are necessary and sufficient conditions for a p-morphic image of a descriptive frame to be descriptive.
Definition 31. An intuitionistic general frameG=hV, S,Qiis called a
descriptively p-morphic imageof a descriptive frameF=hW, R,Pivia a mapf iff satisfies (P1)-(P3) and the following conditions:
(P4) the kernel Ef satisfies (B2), i.e., if ¬f(u)Sf(v), then there exists
Remark 32. SinceEf(Y) =Y ⇔ Y =f−1(f(Y)), the above (P4) and (P5) are equivalent to the following (P4’) and (P5’) respectively. (P4’) If¬f(u)Sf(v), then there existsY ∈ P such thatY =f−1(f(Y)),
u∈Y andv /∈Y.
(P5’) IfY ∈ PandY =f−1(f(Y)), thenf(Y)∈ Q
Next, we justify that (P4) and (P5) are sufficient conditions.
Proposition 33. IfG=hV, S,Qiis a descriptively p-morphic image of a descriptive frameF=hW, R,Piviaf, thenGis also a descriptive frame. Proof. By Proposition 26, G is compact. It remains to show that Gis refined. Suppose ¬w0Su0 for some w0, u0 ∈ V. Since f is surjective, there existw, u ∈ W such thatf(w) = w0 andf(u) =u0. So we have ¬f(w)Sf(u). Thus, by (P4), there existsU ∈ P such that Ef(U) =U,
w∈U andu 6∈U. By (P5),f(U)∈ Q, and sow0 =f(w) ∈f(U) and
u0=f(u)6∈f(U).
To show that (P4) and (P5) are necessary conditions, we first prove the following lemma.
Lemma 34. Suppose F = hW, R,Pi and F0 = hW, R,Qi are general frames. IfFis descriptive andQ(P, thenF0is not refined, hence is not
descriptive.
Proof. SupposeV ∈ PandV 6∈ Q. Consider the sets X={U∈ Q:V (U},
Y={U0∈ Q:V
(U0},
where
Q={W−U :U∈ Q}.
SinceV ⊆TY
and X is closed under finite intersection,X ∪ Y has the finite intersection property, henceT
(X ∪ Y)6=∅by the compactness ofF.
Case 1. V ∩(T
(X ∪ Y))6=∅. Letw∈V∩(T
(X ∪ Y)). To showF0is not refined, we show that there exists au∈V, hence¬wRuand
∀U ∈ Q(w∈U ⇒u∈U).
That is to show that
(u∈)\(Z ∪ {V})6=∅, (13) whereZ={U ∈ Q:w∈U}.
Claim 1: eachU ∈ Z intersectsV, andZis closed under finite inter-section.
Suppose there existsU ∈ Z such that U∩V =∅. Then V (U, so
U ∈ Y. From w ∈ T(X ∪ Y) it follows that
w ∈ U, which contradicts
w∈U. SinceQis closed under finite intersection,Zis closed under finite intersection.
From claim 1, it follows immediately thatZ ∪ {V}has the finite inter-section property. SinceZ ∪ {V} ⊆ P ∪ P, by the compactness ofF, (13) is obtained.
Case 2. V ∩(T(X ∪ Y
Letw0∈T
(X ∪ Y). Thenw0∈/V. To show thatF0 is not refined, we show that there exists au0∈V, hence¬u0Rw0, and
∀U0∈ Q(u0∈U0⇒w0∈U0),
which is equivalent to
∀U0∈ Q(w0∈U0⇒u0∈U0). That is to show that
(u0∈)\(Z ∪ {V})6=∅, (14) whereZ={U0∈ Q:w0
∈U0}.
Claim 2: each U0 ∈ Z intersects V, and Z is closed under finite intersection.
Suppose there exists some U0 ∈ Z such that U0∩V = ∅. Then
V (U0, soU0∈ X. From w0∈T(X ∪ Y) it follows thatw0∈U0, which
contradictsw0 ∈ U0. Since Q is closed under finite union, Z is closed under finite intersection.
From claim 2, it follows immediately thatZ ∪ {V}has the finite inter-section property. SinceZ ∪ {V} ⊆ P ∪ P, by the compactness ofF, (14) is obtained.
Intuitively, the above lemma means that there are actually not so many descriptive frames. However, with the same domain and relation, if the set of admissible sets are incomparable, two intuitionistic general frames can still both be descriptive.
Example 35. There exist two descriptive framesF=hW, R,PiandF= hW, R,QiwithP andQincomparable.
Proof. Define
W ={0,1,2, ..., ω}, R=∅,
P={U⊆W :U is finite and does not containω, or cofinite inωand containsω},
Q={U0⊆W:U0is finite and does not containωor 0, or cofinite in{2,4, ...}and contains 0,
or cofinite in{1,3,5, ...}and containsω}.
The next proposition shows that (P4) and (P5) are necessary condi-tions.
Proposition 36. If a descriptive frameG=hV, S,Qiis p-morphic image of a descriptive frameF=hW, R,Piviaf, thenGmust be a descriptively p-morphic image ofF.
Proof. Take
Q0={f(U) :U ∈ P, Ef(U) =U}.
By Note 29,Q0
is a set of admissible sets. Claim: G0=hV, S,Q0i
is descriptive.
Clearly, f is also a p-morphism from Fto G0. Then, by Proposition 26,G0is compact. SinceG=hV, S,Qiis refined, to showG0=hV, S,Q0i is refined, it suffices to showQ ⊆ Q0.
Actually, for anyU ∈ Q, it holds that
U =f(f−1(U)), f−1(U)∈ PandEf(f−1(U)) =f−1(U).
Thus,U ∈ Q0
Then, since bothG0=hV, S,Q0iandG=hV, S,Qiare descriptive, by Lemma 34, we are forced to concludeQ0
=Q. In view of the definition of Q0
, obviously f satisfies (P5). Now we show thatf satisfies (P4) as well.
Suppose ¬f(u)Sf(v). Then by the refinedness of G0, there exists
f(U)∈ Q0, such thatf(u)∈f(U) andf(v)6∈f(U). It follows that
U ∈ P,E(U) =U,u∈U andv6∈U. Hence,G=G0 is a descriptively p-morphic image ofF.
So far, we have proved that (P4) and (P5) are sufficient and necessary conditions for a p-morphic image of a descriptive frame to be descriptive. Then, with these two conditions, we can prove some other properties of p-morphic images. Let us first define the isomorphism between general frames.
Definition 37. General frames F = hW, R,Pi and G = hV, S,Qi are
isomorphic(in symbolsF∼=G), if there is an isomorphismffromhW, R,i ontohV, Sisuch thatX ∈ Pifff(X)∈ Q.
Proposition 38. If a descriptive frame G = hV, S,Qi is a p-morphic image of a descriptive frameF=hW, R,Piviaf, thenF/Ef is isomorphic toG.
Proof. Define a maph:WEf →V by taking
h(Ef(w)) =f(w).
First we show that h is well-defined. For any u such that Ef(u) =
Ef(w), according to the definition ofEf,f(u) =f(w), i.e., h(Ef(u)) =
h(Ef(w)). Next, we show thathis an isomorphism fromF/Ef toG0. Clearly, h is surjective, since f is surjective. Assume h(Ef(w)) =
h(Ef(u)). Thenf(w) =f(u), which means Ef(w) =Ef(u). Thus, his injective.
Suppose Ef(w)REEf(u). Then there exist w0 ∈ Ef(w) and u0 ∈
Ef(u) such thatw0Ru0. Sincef is a p-morphism fromFto G, we have
f(w0)Sf(u0). Then from f(w) = f(w0) and f(u) = f(u0), we obtain
f(w)Sf(u). That ish(Ef(w))Sh(Ef(u)).
Supposeh(Ef(w))Sh(Ef(u)), i.e. f(w)Sf(u). By (P2), there exists
v∈W such thatwRv andf(u) =f(v). Thus, by the definition of RE,
E(w)REE(u).
SupposeUEf ∈ PEf. Then U ∈ P andEf(U) =U. By Proposition 36,f satisfies (P5), hence
h(UEf) ={h(Ef(w)) :w∈U} ={f(w) :w∈U} =f(U)∈ Q.
Supposeh(UEf)∈ Q, i.e. f(U)∈ Q. PutU 0
=f−1(f(U)). Sincef is a p-morphism fromFtoG, by (P3),U0∈ P. Obviously it holds that
U0=f−1(f(U0)), i.e.,Ef(U0) =U0,
henceUE0f ∈ PEf. Sincef(U 0
) =f(U) andhis injective, we haveUE0f =
UEf. Hence,UEf ∈ PEf as required. This completes the proof.
(i) For any Q ⊆ ℘(V), if an intuitionistic Kripke framehV, Si is a p-morphic image ofhW, Riviaf, andG=hV, S,Qiis a descriptively p-morphic image of F, thenGis unique.
(ii) If descriptive frames G1 = hV1, S1,Q1i and G2 = hV2, S2,Q2i are
p-morphic images of F =hW, R,Pi viaf1 and f2 respectively, and
Ef1=Ef2, thenG1∼=G2.
Proof. (i) SupposeG=hV, S,QiandG0=hV, S,Q0i
are both descriptive. Then by Proposition 36,Q=Q0 ={f(U) :U ∈ P, Ef(U) =U}. Thus,
G=G0.
(ii) By Proposition 38, we have F/Ef1 ∼=G1 andF/Ef2 ∼=G2. Since
Ef1 =Ef2,G1∼=G2.
The above corollary actually means that the set of admissible sets of a p-morphic image of a descriptive frame is determined uniquely by the p-morphism.
We now prove that there is a one-to-one correspondence between bisim-ulation equivalences onFand descriptively p-morphic images ofF.
Theorem 40. Let F=hW, R,Pi be a descriptive frame. The following hold:
(i) If a descriptive frame Gis a p-morphic image of F via f, then the kernel Ef off is a bisimulation equivalence relation onF;
(ii) IfEis a bisimulation equivalence relation onF, then the natural map
f:F→F/E defined by
f(w) =E(w)
is a p-morphism, and the quotient frameF/Eis a descriptive frame.
Proof. (i) First, we check (B1). SupposewEfvand vRu. Then f(w) =
f(v) and thereforef(w)Sf(u). Sincefis a p-morphism, there exitsz∈W
such thatwRzandf(z) =f(u), which meanszEfu.
By proposition 36 f satisfies (P4), thus Ef satisfies (B2). For (B3), supposef(x) =f(u), f(y) =f(w),wRxanduRy. By (P1),f(w)Sf(x) andf(u)Sf(y). Then sinceS is antisymmetric,f(w) =f(u). Hence,Ef
is a bisimulation equivalence onF.
(ii)(P1) follows immediately from the definition of RE. To provef
satisfies (P2), supposeE(w)REE(u). Then by the definition, there exists
w0 and u0 such that wEw0, u0Eu and w0Ru0. So by (B1), there exists
v∈W such that wRvand vEu0. Since E is an equivalence relation, it follows thatvEu, i.e.,f(v) =f(u). Finally for (P3), supposeUE ∈ PE.
Note thatPE ={UE :U ∈ P, E(U) =U}andE(w) =f−1({E(w)}) for
anyw∈W. Thus,
f−1(UE) =[{f−1({E(w)}) :E(w)∈UE} =[{E(w) :w∈E(U) =U∈ P} =U∈ P.
This proves thatf is a p-morphism.
Note thatEis the kernel off, and sofsatisfies (P4). By the definition ofPE, (P5) is satisfied. By Note 29,F/Eis an intuitionistic general frame.
4.4
Duality Theorems for Descriptive Frames
The next theorem reveals a nice algebraic property of descriptive frames. It means that we can go back and forth between descriptive frames and Heyting algebras. For the proof, see Theorem 8.51 in [5].
Theorem 41. Fis a descriptive frame iff F∼= (F+)+.
Corollary 42. For any Heyting algebra A, A+ is a descriptive frame,
i.e.,A+∼= ((A+)+)+.
Proof. By Theorem 9, we haveA∼= (A+)+, soA+∼= ((A+)+)+.
Theorem 41 gives the following generalized duality theorems for de-scriptive frames as follows.
Theorem 43. LetAandBbe Heyting algebras, andFandGdescriptive frames. Then
1. (a) Ais a homomorphic image ofBiffA+ is isomorphic to a
gen-erated subframe of B+
(b) A is isomorphic to a subalgebra of B iff A+ is a p-morphic
image ofB+
2. (a) Fis isomorphic to a generated subframe ofGiffF+ is a homo-morphic image of G+
(b) Fis a p-morphic image ofGiffF+is isomorphic to a subalgebra of G+
Proof. It follows from Theorem 11,13,15,16 and 41.
4.5
The Generalized Version of Theorem 5
In the last section, we generalize Theorem 5 to the descriptive frames case. Recognizing the facts in Example 21 and 30, we state and prove the generalized descriptive version of Theorem 5 as follows:
Proposition 44. An intuitionistic general frame G0 is a descriptively generated subframe of a descriptively p-morphic image of a descriptive frameF iffG0 is a descriptively p-morphic image of a descriptively gen-erated subframe ofF.
Proof. For “⇒”: SupposeG=hV, S,Qiis a descriptively p-morphic im-age of a descriptive frameF=hW, R,Piviaf, andG0=hV0, S0,Q0iis a descriptively generated subframe ofG.
DefineF0=hf−1(V0
), Rf−1(V0
),P0i, by taking P0={U∩f−1(V0) :U∈ P}.
And define the mapg = f f−1(V0
) : f−1(V0
) → V0. We prove the following:
(i)F0 is a descriptively generated subframe ofF. (ii)G0 is a descriptively p-morphic image ofF0viag.
For (i): By Theorem 5,hf−1(V0), Rf−1(V0)iis a generated subframe of hW, Ri. Clearly, P0
is a set of admissible sets and F0 is a generated subframe ofF.
Since G0=hV0, S0,Q0i
is a descriptively generated subframe ofG, by (S3) we haveV0=T
{U∈ Q:V0⊆U}. Hence by (P3) we obtain
ThereforeF0is a descriptively generated subframe ofF.
For (ii): From (i), by Proposition 25, F0 is a descriptive frame. By Theorem 5,gis a p-morphism fromhf−1(V0), Rf−1(V0)iontohV0, S0i.
Next, we show thatgsatisfies (P3). For anyX0∈ Q0
, by the definition ofQ0, there existsX∈ Qsuch thatX0=X∩V0. Thus
g−1(X0) =f−1(X)∩f−1(V0).
Sincef is a p-morphism fromFonto G,f−1(X)∈ P. Therefore, by the definition ofP0
,g−1(X0 )∈ P0
i.e.,gsatisfies (P3).
Then, G0 is a p-morphic image ofF0. By Proposition 36, G0 is also a descriptively p-morphic image ofF0.
For “⇐”: AssumeG0=hV0, S0,Q0i
is a descriptively p-morphic image of
F0 =hW0, R0,P0i viag, whereF0 is a descriptive generated subframe of
F=hW, R,Pi.
Without loss of generality, we may assumeW∩V0=∅. First, we define
V =V0∪(W−W0) andS=S0∪S1∪S2, whereS1=R(W−W0) and
S2={(w1, g(w2)) :w1∈W−W0, w2∈W0andw1Rw2}. Next, we define a mapf:W →V by taking
f(w) =
(
g(w), ifw∈W0;
w, ifw∈W−W0.
Finally, we define a frameG=hV, S,Qiby putting Q={f(U) :U ∈ P, Ef(U) =U},
whereEf is the kernel off.
Then by Theorem 5,hV0, R0iis a generated subframe ofhV, Ri, andf
is a p-morphism fromhW, RiontohV, Ri.
Clearly,Qis a set of admissible sets. For anyf(U)∈ Q, sinceEf(U) =
U, we have f−1(f(U)) = U ∈ P, i.e., f satisfies (P3). Hencef is a p-morphism fromFtoG.
Next, we show thatG0 satisfies (S2). Note that, sinceG0is a descrip-tively p-morphic image ofF0,
Q0={g(X) :X ∈ P0, Eg(X) =X}.
For anyg(X) ∈ Q0
, sinceF0 is a descriptively generated subframe ofF, there existsY ∈ Psuch thatX=Y∩W0. Note thatEf(Y−X) =Y−X
forf(W−W0) =id. Then we have
Ef(Y) =Ef(X∪(Y −X)) =Eg(X)∪Ef(Y −X) =X∪(Y −X) =Y.
Thus,f(Y)∈ Q. SinceEf(Y) =Y andEf(W0) =W0, we have
g(X) =f(X) =f(Y ∩W0) =f(Y)∩V.
So,G0 is a generated subframe ofG.
Case 1: w0, u0∈V0. For anyw∈f−1(w0
) andu∈f−1(u0
), by (P5’) ofg=fV0, there existsU0∈ P0
such that
f−1(f(U0)) =U0, w∈U0 andu6∈U0.
Since F0 is a generated subframe of F, there exists U ∈ P such that
U0=U∩W0. Then
U =U0∪(U−W0) =f−1(f(U0))∪(U−W0) =f−1(f(U)).
Hence, by the definition,f(U)∈ Q.
Clearly,w∈U which impliesw0=f(w)∈f(U). Sinceu∈W0,u6∈U, and henceu0=f(u)6∈f(U) sincef−1(f(U)) =U.
Case 2: w0, u0 ∈V −V0. Note thatw0=f−1(w0) andu0 =f−1(u0). By (P4’), there existsU ∈ Psuch that
f−1(f(U)) =U, w0∈U andu06∈U.
Thus, by the definition,f(U)∈ Q. Clearly,w0∈U impliesw0=f(w0)∈
f(U). Sinceu06∈U andf−1(f(U)) =U,u0
=f(u0)6∈f(U).
Case 3: w0 ∈V0 and u0 ∈V −V0. For any w∈ f−1(w0) ⊆W0 and
u0=f−1(u0), sinceF0 is a descriptively generated subframe ofF, by (S3) there existsU ∈ Psuch that
u06∈U andU ⊇W0, which meansw∈U. SinceU⊇W0, we have
f−1(f(U)) =f−1(f(W0∪(U−W0))) =f−1(V0∪(U−W0)) =W0∪(U−W0) =U.
Thus, by the definition,f(U)∈ Q. Clearly,w∈U impliesw0 =f(w)∈
f(U). Sinceu06∈U andf−1(f(U)) =U,u0=f(u0)6∈f(U).
Case 4: w0 ∈V −V0 andu0 ∈V0. First note thatf−1(w0) =w0. We prove this case by proving some claims.
Claim 1: There existsU ∈ P, such thatw0∈U andU∩f−1(u0 ) =∅.
Proof of Claim 1. Suppose otherwise. Then, for anyU ∈ P such that
w0∈U, we haveU∩f−1(u0
)6=∅. Consider the set
X ∪Y={X∩W0:w0∈X, X∈ P}∪{f−1(Y) :Y ∈ Q0, u06∈Y} ⊆ P0∪P0. For any two elementsX1∩W0, X2∩W0∈ X, sincew0∈X1∩X2∈ P, we have
(X1∩W0)∩(X2∩W0)∈ X, which implies thatX is closed under finite intersection.
For any two sets f−1(Y
1), f−1(Y2)∈ Y, fromu0 6∈ Y1 ∈ Q0 andu0 6∈
Y2∈ Q0, it follows thatu06∈Y1∪Y2∈ Q0. Observe that
f−1(Y
1)∩f−1(Y2) = (W0−f−1(Y1))∩(W0−f−1(Y2)) =W0−f−1(Y1∪Y2)
=f−1(Y 1∪Y2). This meansY is closed under finite intersections.
For any X ∩W0 ∈ X and f−1(Y) ∈ Y, sinceX ∩f−1(u0
) 6= ∅and
f−1(u0)⊆f−1(Y)⊆W0
, we have
which implies thatX ∪ Y has the finite intersection property. SinceF0is compact, there existsv∈T
(X ∪ Y). Fromv∈T
X, we get
∀X∈ P(w0∈X →v∈X),
which by the refinedness ofFimpliesw0Rv, and so
f(w0)Sf(v),
sincef is a p-morphism. In the meantime, fromv∈T
Y we get
∀Y ∈ Q0(u06∈Y →v6∈f−1(Y)),
i.e.
∀Y ∈ Q0(f(v)∈Y →u0∈Y),
which by the refinedness ofG0meansf(v)S0u0, and so
f(v)Su0.
Thus,
f(w0)Su0, i.e. w0Su0,
which contradicts the assumption that¬w0Su0.
Claim 2: Let U andY be the two sets in Claim 1, then there exists
Y ∈ Q0such thatU∩W0⊆f−1(Y) andu0 6∈Y.
Proof of Claim 2. Suppose otherwise. Then, for any Y ∈ Q0
such that
u06∈Y, we haveU∩W0*f−1(Y)(W0. Hence, for anyf−1(Y)∈ Y, it
holds that
(U∩W0)∩f−1(Y)6=∅. Consider the set
{U∩W0} ∪ Y ⊆ P0∪P0.
SinceYis closed under finite intersection,Y has the finite intersection property. Thus{U∩W0} ∪ Y has the finite intersection property.
Hence, fromF0being compact, it follows that there existsv∈T
({U∩
W0} ∩ Y). Fromv∈T
Y, by a similar argument to the one in the proof of Claim 1, we can prove thatf(v)Su0. By (P2),
vRufor someu∈f−1(u0).
In the meantime, sincev∈U∩W0,v∈U. FromU being upward closed, we concludeu∈U, which contradicts Claim 1.
LetU andY be the sets in Claim 2. SinceF0 is a generated subframe ofF, there exists X ∈ P such thatX∩W0 =f−1(Y). By Claim 2, we haveU∩W0⊆f−1(Y) =X∩W0. Furthermore
f−1(Y) =X∩W0= (U∪X)∩W0,
which leads to
f(U∪X) =f(((U∪X)∩W0)∪((U∪X)−W0)) =f(f−1(Y))∪((U∪X)−W0))
Hence,
f−1(f(U∪X)) =f−1(Y ∪((U∪X)−W0)) =f−1(Y)∪((U∪X)−W0) =U∪X.
ClearlyU∪X∈ P, hence by the definitionf(U∪X)∈ Q.
Sincew0∈U,w0∈U∪X, which impliesw0=f(w0)∈f(U∪X). By Claim 2,u06∈Y, and clearlyu06∈(U∪X)−W0, thus by (15),
u06∈Y ∪(U∪X−W0) =f(U∪X).
This completes the proof of Case 4.
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PhD Thesis, University of Amsterdam, 2006.
[3] P. Blackburn, M. de Rijke, and Y. Venema.Modal Logic,Cambridge University Press, 2001.
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Springer-Verlag, 1981.
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