A Construction Method for Modal Logics of Space
Acknowledgements
I consider myself very fortunate for having the opportunity to work on this thesis under the supervision of Johan van Benthem and Dick de Jongh. Besides shedding a great deal (of very different!) light upon the problems contained in this thesis, they provided the encour-agement and support needed to see this volume through to its completion. Thank you. Thanks to Nick Bezhanishvili and Yde Venema, for serving as members on my defense com-mittee and for useful suggestions along the way. To Benedikt L¨owe for the same, and for providing inspiring lecture courses, as well as leaving his door open for conversation. To Darko Sarenac for his helpful skepticism and generous hospitality in Palo Alto. And to Niels Molenaar for handling organizational matters right before I was to defend my thesis.
In addition, I would like to thank Thomas, Be, Jill, Charles, Alexandra, Chunlai and Caf´e Reibach for making my time in Amsterdam so enjoyable.
Contents
1 Introduction 4
2 Modal Logics of Space 5
2.1 Syntax and Semantics . . . 5
2.2 Practical Workings . . . 7
2.3 The Historical Results . . . 8
2.4 Modern Extensions . . . 9
2.5 Modal Techniques . . . 11
2.6 The Construction Method . . . 13
2.7 QFP on hQ, <i . . . 13
3 Topological Completeness Proofs 17 3.1 Henkin Models . . . 17
3.2 S4on Q . . . 19
3.3 Remarks on the Construction Method . . . 22
3.4 Q2FP on Q . . . 23
3.5 Extensions ToR: The Temporal Case . . . 30
3.6 Further Extensions toR . . . 33
3.7 QSU on Q . . . 35
3.8 S4⊕S4on Q×Q . . . 41
4 Method Comparison 45 4.1 S4on 2ω, the Model Theoretic Approach . . . 45
4.2 S4on 2ω, the Construction Method . . . 48
1
Introduction
Given the important role of spatial intuitions in cognition, and the apparent unreliability of these intuitions, there is something natural about looking at spatial structures from an axiomatic standpoint. Indeed, it is not surprising that the first and best known application of the axiomatic method was to provide a development of geometry.
In the past century, with the development of formal logic and subsequent discovery that elementary number theory is not axiomatizable, it became both possible and independently interesting to examine spatial structures within given logical formalizations. While the cen-tral framework for examining spatial structures axiomatically has been first order logic, from time to time other logics with spatial interpretations have been considered as well.
Recently, one popular area of investigation has been looking at modal logics with spatial interpretations. This subject can be traced back to McKinsey and Tarksi’s [14] work on boolean algebras with closure operators in the 1940s. However, within the past ten years a more general program of providing a modal analysis of space has emerged.
By and large, the techniques used in investigating modal logics of space have been model theoretic in nature, involving the transfer of geometric or topological structure from the de-sired mathematical object to some Kripke frame. While this works well in cases where the relevant modal logic has nice Kripke frame characterizations, in other cases this way of pro-ceeding can become quite difficult.
In this thesis we will examine a more syntactic approach to establishing completeness re-sults in modal logics of space. The technique we will use has the virtue of constructing the desired mathematical structures directly, rather than working indirectly through Kripke frames. This allows for a good deal of control over what models of the relevant logic look like. In the first chapter, we will use our construction method to give new proofs of the com-pleteness of S4with respect to hQ, τi and S4⊕S4 with respect to hQ×Q, τ1, τ2i. We will
2
Modal Logics of Space
2.1 Syntax and Semantics
Although they would not have been thought of as such at the time, some of the first seman-tic completeness proofs for modal logic date back to McKinsey and Tarski’s [14] work the on foundations of topology in the 1940s . These proofs pre-date the now standard Kripke semantics for modal logic, and in fact are spatial completeness proofs for the modal logicS4 with a topological semantics.
Let L be the basic modal language, consisting of propositional variables p, q, r.., boolean connectives∧,∨,¬ →, and unary operators 2 and 3. The formulas ofL define the least set Γ containing all propositional variables which is closed under boolean connectives and the operators2 and3.
Definition 2.1 A modal logicLis the least set of formulas of Lcontaining all propositional tautologies and the axioms ofL which is closed under substitution and proof rules :
ϕ, ϕ→ψ ψ
ϕ 2ϕ
Definition 2.2 A formula ϕ is said to be provable in L (written `L ϕ) if ϕ ∈L and a set Γ of formulas is said to beL-consistent iff there is no finite set {ϕ1, ϕ2...ϕn} ⊆Γ such that
L` ¬(ϕ1∧...∧ϕn).
The central modal logic for our purposes isS4, which contains the axioms: 2ϕ↔ ¬3¬ϕ
2(ψ→ϕ)→(2ψ→2ϕ)
2ϕ→ϕ 2ϕ→22ϕ
Recall that atopological spaceis a pairhW, τi, whereW is a nonempty set andτ is a collection of subsets ofW satisfying the following properties:
• ∅,W ∈τ;
• ifU, V ∈τ, thenU ∩V ∈τ;
• if{Ui}i∈I ∈τ, then
S
i∈IUi∈τ;
• U isopen ifU ∈τ
• U isclosed ifU /∈τ
• The interior ofU (Int(U)) is the union of all open sets contained inU
• The closure of U (Clo(U)) is the intersection of all closed sets containing U
• Int(U) =W −Clo(W −U)
In the original topological semantics for L proposed by McKinsey and Tarski, the truth of formulas was evaluated at the level of topological models. Thus given a valuationν assigning propositional letters to subsets of W,ν was extended to all formulas of L as follows:
• ν(p)⊆W
• ν(¬ϕ) =W −ν(ϕ)
• ν(ϕ∨ψ) =ν(ϕ)∪ν(ψ)
• ν(ϕ∧ψ) =ν(ϕ)∩ν(ψ)
• ν(ϕ→ψ) = (W −ν(ϕ))∪ν(ψ)
• ν(2ϕ)) =Int(ν(ϕ))
• ν(3ϕ) =Clo(ν(ϕ))
A formula ϕwas said to be true on a topological model hW, τ, νi ifν(ϕ) =W.
Besides working with the notion of truth on a topological model, in this paper we will also make use of the notion of a formulaϕbeing true at points in a topological model. This fine grained topological semantics mirrors the now prevalent Kripke semantics for modal logic, and will facilitate in transferring techniques and results later on.
Given a topological model M = hW, τ, νi, we define a formula ϕ to be true at a point w by induction on the length of ϕ. From now on we will economize on boolean connectives, considering only¬and ∨, the others being definable from these two alone.
• w|=p iffw∈ν(p)
• w|=ψ∨χiff w|=ψorw|=χ
• w|=¬ψ iff notw|=ψ
• w|=2ψ iff ∃U ∈τ such that w∈U and ∀w0 ∈U, w0 |=ψ
• w|=3ψ iff∀U ∈τ, w ∈U implies that∃w0∈U such that w0|=ψ
Definition 2.4 We sayϕis true on a topological modelhW, τ, νi(written hW, τ, νi |=ϕ), if
2.2 Practical Workings
From the perspective of modal logic, the topological semantics proposed above looks quite different from the standard Kripke-style labeled transitions systems. And from the perspec-tive of topology, it is not clear what can be expressed in the modal languageL. To sharpen intuitions on both ends, let us briefly pause to get a sense of the practical workings of the language .
At first glance, it may appear that little of mathematical interest can be expressed in L. For example, there is no obvious way to express familiar topological properties such as con-nectedness, beingT0, etc.. And indeed the results that follow lend support to this belief. They
show, for example, that no formula of Lcan distinguish topological structures as different as
hQ, τi and hR, τi. (So it is true that connectedness can not be expressed in L).
On the other hand, even though many topological properties are undefinable, what formulas can be true at a point in a topological model quickly becomes quite complicated. Just how complicated becomes clear below when we try to construct a modelhR, τ, νiwhich has a real
number modeling an arbitrary consistentS4 formula. But for the moment let us consider a few examples of simple formulas which can be made true at points on the real line.
Definition 2.5 A logicLis said to be complete with respect to a topological spacehW, τiif
`Lϕ⇔ hW, τi |=ϕ.
Much of what follows will be dedicated toproving topological completeness results forS4and its extensions, but for now let us make use of a classical result.
Theorem 2.6 S4is complete with respect to hR, τi.
Viewed contrapositively, the right to left direction of Definition 2.5 says that if ψ is an S4-consistent formula, then it can be made true at a point on the real line.
First consider an easy example, theS4-consistent formula3p∧ ¬p. By completeness, there exists a valuationνsuch thathR, τ, νi, x|=3p∧ ¬pfor somex∈R. And indeed it is not hard to think of such aν andx. For example, ifν(p) ={1/x|x >0}, thenhR, τ, νi,0|=3p∧ ¬p.
A slightly more difficult example is the formula 33p ∧3¬p. Intuitively, this says that there is a sequence of points approachingx and at every point y in this sequence there is a sequence approachingy which models p. Actually, it is easier to give a valuation to describe the situation. If
ν(p) ={ 1
(x+1y) |x∈N/{0}, y ∈N}
A more interesting example is the S4-consistent formula 2(3p∧3¬p). This formula says that every point in a given open neighborhood has both p and ¬p sequences approaching it. Although at first glance it might seem difficult to describe a valuation which makes this formula true at a point inhR, τi, there is actually a quite natural one. The valuationν(p) =Q
makes2(3p∧3¬p) true at every point inhR, τ,i.
All of the examples considered above involve only singleS4-consistent formulas containing one propositional letter. However, by completeness it is guaranteed that every finite consistent set of S4 formulas containing finitely many propositional letters can be made true together at a single point onhR, τi. Now that the inner workings of the language have been explored a bit, it is time to see what properties can be established about the language.
2.3 The Historical Results
In ‘The Algebra of Topology’ McKinsey and Tarski [14] examine the relations between closure algebras and topological spaces. A closure algebra is a five-tuplehK,∪,∩,−, Ci where
• K is a boolean algebra with respect to ∪,∩,− • If x∈K, thenCx is in K
• If x∈K, thenx⊆Cx
• If x∈K, thenCCx=Cx
• If x andy are in K, thenC(x∪y) =Cx∪Cy
• C∅=∅
Clearly every topological spacehX, τican be viewed as a closure algebra C(X) =hP(X),∪,∩,−, Ci, where C is the closure operator on P(X). Conversely, every closure algebra of the form
C(X) =hP(X),∪,∩,−, Cican be viewed as a topological space hX, τisuch that hX, τi |= (i) p→3p
(ii) 33p↔3p
(iii) 3(p∨q)↔3p∨3q (iv) 3⊥↔⊥
Theorem 2.7 (McKinsey and Tarski) For every finite topological space hX, τi and every normal, dense-in-itself topological spacehY, τiwith a countable base, there exists an embedding
C(X)C(Y).
From a logical perspective, the above theorem states that if there is a finite topological space
hX, τisuch that hX, τi 6|=ϕ, then for any normal, dense-in-itself topological spacehY, τiwith a countable base, hY, τi 6|=ϕ. It is known (and proven below in Corollary 3.6) that if 0S4 ϕ, then there is some finite topological space hX, τi such that hX, τi 6|= ϕ. So from a logical perspective, Theorem 2.7 states thatS4 is complete with respect to every normal, dense-in-itself topological space with a countable base. This is a very general result. It shows that, for example, S4 is the complete with respect to the rationals, the reals, the Cantor Space, and the Baire Space.
2.4 Modern Extensions
From the perspective of modal logics of space today, McKinsey and Tarski’s original results are somewhat daunting in means and scope. In means, their arguments are by and large mathematical rather than metamathematical, and make use of facts not necessarily familiar to the modal logician. In scope, their results are sufficiently general to settle almost all ques-tions about the basic modal languageL with a topological semantics.
Thus recent investigations into modal logics of space have taken two directions. One is to make the original results of McKinsey and Tarski more accessible by using the tools built up in modal logic over the past 70 years. This project seems to have started with Mints [15], where a new completeness proof of S4 with respect to the Cantor Space is given. Mints’ paper has in turn inspired several new proofs of completeness ofS4with respect to the real line, including Mints, Zhang [17], Aiello, van Benthem, Bezhanishvili [3] and Bezhanishvili, Gehrke [6].
The other direction is to move to richer modal languages. In the past ten years, many spatial extensions of the basic modal languageLhave been put forward. The first and perhaps most natural extension ofLto be considered was the languageL2F P, originally found in Shehtman
[20]. L2F P contains the 2operator, together with two additional unary operators F and P.
In the intended semantics forL2F P,2is interpreted as interior andF and P are given their
standard temporal interpretation as ’future’ and ’past’.
More formally, given a partial orderhW, <i, the standard order topology τ on hW, <i and a valuationν, the truth of aL2F P formula at a point w∈W is defined as follows:
• w|=F ψ iff∃w0 > w such thatw0|=ψ
• w|=Gψ iff∀w0> w,w0 |=ψ
• w|=Hψ iff∀w0< w,w0 |=ψ
• w|=2ψ iff ∃U ∈τ such that w∈U and ∀w0 ∈U, w0 |=ψ
• w|=3ψ iff∀U ∈τ, w ∈U implies that∃w0∈U such that w0|=ψ
L2F P has the appealing feature of combining two basic properties of mathematical objects:
topological and metric structure. Plus it is topologically more expressive thanL. SinceL2F P
will occur throughout this paper, let us pause for a moment to consider an example.
As mentioned before, it follows from the completeness of S4 with respect to both hR, τi
andhQ, τithat connectedness is not expressible in L. For assume it is, and thatϕexpresses it. ThenhR, τi |=ϕ andhQ, τi 6|=ϕ. But
hR, τi |=ϕ⇔ `S4 ϕ⇔ hQ, τi |=ϕ.
However, connectedness is expressible inL2F P .
Lemma 2.8 hW, τi |=2ϕ∧F¬ϕ→F(3ϕ∧3¬ϕ)⇔ hW, τi is connected.
[⇒] Assume hW, τi is not connected. ThenW can be written as the union of two non empty disjoint open setsU andV. Without loss of generality, assume that there is somex∈U and y∈V such thatx < y. Now consider the valuationν(p) =U.
It is clear that hW, τ, νi, x |= 2p∧F¬p. We want to show hW, τ, νi, x 6|= F(3p∧3¬p), so assume z is an arbitrary point such that x < z. Either z ∈ U or z ∈ V. If z ∈ U, then hW, τ, νi, z |= 2p and if z ∈V, then hW, τ, νi, z |=2¬p. It follows that hW, τ, νi, x|= G(2p∨2¬p).
[⇐] Assume hW, τi 6|= 2ϕ∧F¬ϕ → F(3ϕ∧3¬ϕ). Then there exists a valuation ν and pointx∈W such thathW, τ, νi, x|=2ϕ∧F¬ϕand hW, τ, νi, x|=G(2ϕ∨2¬ϕ). Let
X ={y|y < x}
Y ={z|z < x}
Z =ν(2ϕ) V =ν(2¬ϕ)
Then X and Y are open sets and Z and V are the unions of open sets. It follows that U = X∪Z and T =V ∩Y are two disjoint open sets whose union is W. So hW, τi is not
connected. qed
While L2F P increases topological expressive power by adding metric expressivity, other
additional operator ◦ for one continuous transformation on a topological space. This lan-guage is interpreted into a dynamic topological systemhW, τ, TiwherehW, τiis a topological space and T is a continuous mapping on hW, τi. Given a valuation ν on hW, τ, Ti, a L2◦
formula ϕis defined to be true at a pointw∈W as follows:
• w|=2ψ iff ∃U ∈τ such that w∈U and ∀w0 ∈U, w0 |=ψ
• w|=3ψ iff∀U ∈τ, w ∈U implies that∃w0∈U such that w0|=ψ
• w|=◦ψ iff T(w)|=ψ
InL2◦ it is possible to express not only topological properties, but also properties of functions
on topological spaces. For example, ◦2ϕ→2◦ϕexpresses the continuity of T.
L2◦ was originally proposed by Artemov, Davoren, and Nerode [4] and examined by
Da-voren in her P.h.D thesis [9], where she showed that the logic S4C =
S4+◦2ϕ→2◦ϕ+ the interaction principles◦(ϕ∧ψ)↔ ◦ϕ∧ ◦ψ and◦¬ϕ↔ ¬ ◦ϕ is complete with respect to the class of all topological spaces. Of course given the explicitly mathematical interpretation of the language, this is not a very satisfying result. Recently, Mints, Zhang [18] established that S4C is also complete with respect to the Cantor Space. Otherwise, almost all questions about this language remain open.
There has also been some investigation into the combined language L2◦F P Mints, Kremer
[16], Kremer, Mints [13] . Given the appropriate interpretation ofF andP (namely that they apply only to N), this language is strong enough to express fundamental facts of analysis,
such as properties crucial to proving the Mean Value Theorem. These are only a few of a large number of possible topological extensions of L. For a discussion of some of the other possibilities, see Aiello [1] or Aiello, van Benthem [2].
As a note before moving on, one surprising feature of moving to richer modal languages is how difficult the problems become, even when seemingly increasing the expressive power only slightly. For example,L2F P does not seem like a particularly strong extension ofL. And
it follows from Rabin’s Theorem that hR, τi is axiomatizable inL2F P. Yet no one has been
able to find a complete axiomatization. Even theL2F P axiomatization ofh
Q, τiin Shehtman
[20] is quite difficult, and something of an encyclopedia of modal logic. As mentioned above, matters are worse still in L2◦, where except for the completeness of S4C with respect to the Cantor Space almost all questions remain open. And not even that much is known about
L2◦F P. It seems that by adding only a little bit more expressive power, the proofs swiftly
move beyond strictly logical insight.
2.5 Modal Techniques
hX, τi, the most common strategy has been to find a convenient class of Kripke frames the logic is complete with respect to and establish a modal equivalence between an arbitrary member of this class and the desired topological space.
For example, Aiello, van Benthem, Bezhanishvili [3] uses the fact thatS4is complete with re-spect to the class of all finite transitive and reflexive trees. In outline, the proof then proceeds as follows:
• If 0S4 ϕ, by completeness ϕ can be refuted on some finite transitive and reflexive tree
hW, Ri.
• A labeling is devised between branches of the Cantor space and points in hW, Ri such that every node of hW, Ri is modally equivalent to some branch of the Cantor Space.
• So the point on hW, Ri which refutes ϕ is modally equivalent to some branch of the Cantor Space.
• So ϕin turn can be refuted on the Cantor Space.
In the last chapter we will discuss this example in greater depth.
The line of reasoning sketched above has been used in Shehtman [20] , Mints [15] Aiello, van Benthem, Bezhanishvili [3] and van Benthem, Bezhanishvili, ten Cate, Sarenac [5], among other places, to give streamlined topological completeness proofs. However, the approach is not without difficulty. For one thing, when working in richer modal languages, where more topological properties can be expressed, the Kripke frame characterizations of the logic can become difficult to establish.
2.6 The Construction Method
Curiously, there is a modal technique which works directly on mathematical structures rather than indirectly through Kripke frames, but it has never been applied to topological com-pleteness proofs. This technique was developed by de Jongh and Veltman [8] and Burgess [7], among others, in the 1980s to overcome precisely the difficulties described above, but in this case for axiomatizing structures such ashN, <i,hQ, <i and hR, <i in the basic temporal
languageLF P.
As in the topological case, the original completeness proofs for hN, <i, hQ, <i and hR, <i
in the temporal language (Segerberg [19]) provided a characterization of the relevant logic in terms of Kripke frames, then established a modal equivalence between the desired mathemat-ical structure and its class of Kripke frames. And again as above, the expressive power of the temporal language resulted in Kripke frame characterizations which were not very nice and, as a result, somewhat technical proofs.
Faced with these difficulties, de Jongh and Veltman (among others) developed a more di-rect syntactic approach towards establishing completeness. The basic idea is to construct the desired mathematical structure in stages, at each stage associating every point added with a set of formulas which will be true at that point when the construction is finished. Then if the construction procedure is carried out correctly, there will be a harmony between the set of formulas associated with a point and the formulas which are actually true at that point. Provided that the sets of formulas were selected judiciously, the structure will then serve as a model of the desired state of affairs. Let us illustrate this procedure with our first completeness proof.
2.7 QFP on hQ, <i
Consider the logicQFP containing the axioms: G(ψ→ϕ)→(Gψ→Gϕ)
H(ψ→ϕ)→(Hψ→Hϕ) ϕ→GP ϕ
ϕ→HF ϕ GGϕ→Gϕ Gϕ→GGϕ
F ϕ→G(ϕ∨P ϕ∨F ϕ) P ϕ→H(ϕ∨P ϕ∨F ϕ) F>
P>
and proof rules:
ϕ, ϕ→ψ ψ
ϕ Gϕ
Theorem 2.9 QFP is complete with respect tohQ, <i.
Proof. To show`QFP ϕ⇒ hQ, <i |=ϕit is necessary to show that all of the axioms ofQFP are valid onhQ, <i, and that the proof rules preserve validity on hQ, <i. This involves some
routine fact checking. From now on this direction of completeness will be omitted. To show the converse direction, assume 0QFP ϕ. We will construct hQ, <i (modulo isomorphism) in such a way thathQ, <i 6|=ϕ.
Definition 2.10 A consistent set of formulas Γ is called maximally consistent if there is no
consistent set of formulas properly containing Γ.
Fact 2.11 Every consistent set of formulasΓ can be extended to a maximally consistent set of formulasΓ0,Γ⊆Γ0. Furthermore,Γ0 is maximally consistent if and only if for any formula
ϕ, either ϕ∈Γ0 or ¬ϕ∈Γ0 but not both.
Definition 2.12 Let Γq and Γq0 be QFP maximally consistent sets. We say Γq ≺Γq0 if
Gϕ∈Γq ⇒ϕ∈Γq0
Fact 2.13 Γq≺Γq0 ≺Γq00 ⇒ Γq≺Γq00.
Fact 2.14 Γq≺Γq0 and Γq ≺Γq00 ⇒ Γq0 ≺Γq00 or Γq0 = Γq00 or Γq00 = Γq0. Fact 2.15 The following are equivalent:
1)Γq≺Γq0
2)ϕ∈Γq0 ⇒F ϕ∈Γq
3)Hϕ∈Γq0 ⇒ϕ∈Γq
4)ϕ∈Γq0 ⇒P ϕ∈Γq
The construction ofhQ, <iwill proceed in stages. Each stage n will consist of: 1. a finite set Qn={q0, q1, . . . , qk}
2. an assignment of an QFP maximally consistent set Γqi to each qi ∈Qn
3. a linear ordering <on Qn such thatqj < qh ⇒Γqj ≺Γqh
Letϕ0ϕ1ϕ2... be an enumeration of allF and P formulas in which each formula is repeated
infinitely many times. The construction will proceed as follows:
- Stage 0: LetQ0 ={q#}and associateq#with Γq#, aQFP maximally consistent extension
of{¬ϕ}.
- Stage 3n+1: For all q, q00∈Q3n such thatq00 is the immediate successor of q in Q3n, add
- Stage 3n+2: LetF ϕ be the nextF formula in the enumeration and letq be the greatest element in theQ3n+1 ordering such that F ϕ ∈ Γq and for all q∗ ∈ Q3n+1 such that q < q∗, ¬ϕ ∈ Γq∗. If such a q exists, add a new point q0 immediately after q and associate q0 with Γq0, a QFP maximally consistent extension of Γ0 ={ϕ} ∪ {ψ|Gψ∈Γq}. Let Q3n+1 beQ3n plus any points added by this procedure and let<3n+1 be<3n extended in the obvious way.
- Stage 3n+3. Symmetric to 3n+2 taking the firstP formula in the enumeration that hasn’t been used.
In order to ensure that our eventual model hQ, <, νi comes out right, it must be checked that Conditions 1)−3) above are met at every stage in the construction. As an illuminative example, we will check that Condition 3) is met at Stage 3n+2. Fact 2.13 states that ≺ is transitive, so we only need to check that Γq ≺ Γq0 ≺ Γq00, where q00 is the immediate suc-cessor ofq in the<3n+1 ordering andq0is a new point added betweenqandq00at Stage 3n+2.
It follows from the definition of Γq0 that Γq ≺ Γq0. However, if q has an immediate suc-cessorq00 in Q3n+1, the definition of Γq0 does not immediately guarantee that Γq0 ≺Γq00. On the other hand, it does follow from Fact 2.14 that Γq0 ≺Γq00 or Γq00 = Γq0 or Γq00 ≺ Γq0. If Γq00 = Γq0 holds, then ϕ ∈ Γq00 and this contradicts the assumption that ¬ϕ ∈ Γq∗ for all q <3n+1 q∗. Furthermore, if Γq00 ≺Γq0, then F ϕ ∈ Γq00 and this contradicts the assumption that q is the greatest element in the <3n+1-ordering such that F ϕ ∈ Γq. So it must be the
case that Γq0 ≺Γq00, as required.
To finish off the construction, let hQ, <i = S
{hQn, <ni | n ∈ ω}. We first want to show
our constructed framehQ, <i can be used to refuteϕ. As suggested before, the fundamental idea behind the construction method is that it is possible to build mathematical structures such that the formulas associated with a point in the construction are the set of formulas that will be true at that point when the construction is finished. However, up to now no connection has been made between the syntactic objects associated with each point and the semantic notion of truth at a point. It is the valuation ν which enables us to do this. To get the ball rolling, we will chooseν such that the truth of a propositional letter at a point q corresponds to its membership in Γq. In other words, letν(p) = {q |p ∈Γq}. Now let us
check that we have performed the construction in a such a way that this harmony extends to all formulasϕ.
Lemma 2.16 hQ, <, νi, q|=ϕ⇔ϕ∈Γq.
Proof. By induction on the complexity ofϕ.
• ϕ=p.
By the definition of ν,q|=p⇔p∈Γq.
q |=ψ∧χ⇔q|=ψand q |=χ
q |=ψ andq |=χ⇔ (by induction hypothesis)ψ∈Γq and χ∈Γq
ψ∈Γq and χ∈Γq ⇔ (by maximal consistency)ψ∧χ∈Γq.
• ϕ=¬ψ
q |=¬ψ⇔q6|=ψ
q 6|=ψ⇔ (by induction hypothesis)ψ6∈Γq
ψ6∈Γq⇔ (by maximal consistency)¬ψ∈Γq.
• ϕ=F ψ
[⇐]. Assume F ψ ∈Γq. Then by construction there is a stage j such that q ∈Qj and
F ψ is being treated. At stage j+ 1, there is a q0 ∈Qj+1 such thatq < q0 and ψ∈Γq0. It then follows from the inductive hypothesis that q0 |=ψ, sohQ, <, νi, q|=F ψ.
[⇒]. Assume hQ, <, νi, q|=F ψ. Then there is some q < q0 such that hQ, <, νi, q0 |=ψ. So it follows from the inductive hypothesis that ψ ∈ Γq. Since q < q0, we know by
condition 3) on stages that Γq≺Γq0. SoF ψ ∈Γq.
• ϕ=Gψ.
[⇐] In this case it is easier to argue contrapostively. So assume Gψ /∈ Γq. Then by
maximal consistency F¬ψ∈Γq and it follows from ”[⇐]” above that
hQ, <, νi, q|=F¬ψ. So hQ, <, νi, q6|=Gψ.
[⇒] Once again it is easier to argue contrapositively. So assume hQ, <, νi, q 6|= Gψ. Then hQ, <, νi, q|=F¬ψ and it follows from ”[⇒]” above thatF¬ψ∈Γq. So by
maxi-mal consistency,Gψ6∈Γq.
We will leave the H and P cases to the reader, as they are quite similar to G and
F. qed
Now with the appropriate hindsight, it is easy to see that hQ, <, νi refutesϕ. At Stage 0 we made sure to put ¬ϕin Γq#. So by the above Lemma it follows that hQ, <, νi, q#|=¬ϕ.
To finish the proof, it only remains to be shown that our constructed model is isomorphic to
hQ, <i. To do this we will make use of Cantor’s Theorem.
Theorem 2.17 (Cantor) Every countable dense linear ordering without end points is iso-morphic to Q.
Lemma 2.18 hQ, <i is a countable, dense, linear order without end points.
stage Qk+3 a new pointq0 is added such that q < q0 < q00. For right unboundedness, assume
q is the greatest element in the<j ordering for somej. Then there is a stage` > k such that
the formulaF>is being treated. F>is an axiom ofQFP and so is in everyQFP maximally consistent set. If q does not have a successor at stage `there is noq∗ ∈Q`,q < q∗ such that
> ∈Γq∗. A new point will then be added afterq at this stage. Left unboundedness is similar. qed
It follows from Cantor’s Theorem and the above Lemma that there is an isomorphismf from
hQ, <i tohQ, <i. AssignhQ, <i the valuation ν0(p) =f(ν(p)). It is easy to check that
hQ, <, νi, q|=ψ⇔ hQ, <, ν0i, f(q)|=ψ
SohQ, <, ν0i, f(q#)|=¬ϕand QFP is complete with respect to hQ, <i. qed In de Jongh, Veltman [8], the construction method presented above is used to give simplified completeness proofs for the structures hN, <i, hQ, <i and hR, <i in LF P. The simplicity of
hR, <i is particularly notable (and tempting) for as in the case of hR, τi, the completeness
proofs ofhR, <iworking indirectly through Kripke frames are quite technical.
In the next chapter we will examine how the construction method can be applied to topolog-ical completeness proofs. We will first use the method to axiomatize hQ, τi in the the basic modal language L and the more difficult combined language L2F P. Then we will stop to
consider possible extensions tohR, τi. Finally, we will examine the slightly more expressive
temporal Since/ Until language onhQ, <iand return to the basic modal languageLwith two modal operators21 and 22 on the topological product spacehQ×Q, τ1, τ2i.
3
Topological Completeness Proofs
3.1 Henkin Models
Before moving on to the topological completeness proofs, let us pause for a moment to cor-rectly frame the construction method described above and prove a theorem made use of in Section 2.3.
The construction method is best understood as a means of adding structure to Henkin models. Definition 3.1 Consider the basic modal language L. A Henkin model is a triple M =
hWL, RL, νLi where
• Lis a logic
• WL is the set of all L maximally consistent sets
• RL⊂WL×WL is a binary relation such that RLΓxΓy iff2ϕ∈Γx ⇒ϕ∈Γy
Both Henkin models and the construction method contain the same ingredients: maximally consistent sets, a relation determined by the syntactic structure these sets, and a valuation which connects syntax to semantics. However, while Henkin models are rather unwieldy en-tities, constructed models are as well behaved as we can imagine them to be.
Yet despite their coarseness, Henkin models can be used to establish logically interesting topological completeness proofs.
Theorem 3.2 S4is complete with respect to the class of all topological spaces.
Proof. The proof will make use of the Henkin model for S4. Let us first define a topology onhWS4, RS4, νS4i as follows: let ∆w ={Γz |RS4ΓwΓz}and B={∆w |w∈WS4} serve as
a basis forτS4.
Lemma 3.3 hWS4, τS4i is a topological space.
Proof. It suffices to check the following two properties:
• For any ∆u,∆v ∈ B and any Γx ∈∆u∩∆v, there is some ∆y ∈ B such that
Γx∈∆y ⊆∆u∩∆v
• For any Γx∈WS4, there is some ∆y ∈ B such that Γx∈∆y
Fact 3.4 hWS4, RS4i is transitive and reflexive.
Since hWS4, RS4i reflexive the second property is immediate; for any Γx ∈ WS4, Γx ∈ ∆x.
The first property is also not hard to establish. If Γx∈∆u∩∆v then ∆x ={Γz |RS4ΓxΓz} ⊆
∆u∩∆v. qed
Lemma 3.5 hWS4, τS4, νS4i,Γx|=ϕ⇔ϕ∈Γx.
Proof. By induction on the complexity of ϕ. The propositional and boolean cases are the same as before, and3 is dual to2, so we will only treat the caseϕ=2ψ.
[⇐] Assume 2ψ ∈Γx. We need to show that there is some U ∈τS4 such that Γx ∈U, and
for every Γy ∈ U, Γy |= ψ. Let U = ∆x ={Γz |RS4ΓxΓz}. Then by the definition of RS4,
2ψ ∈Γx ⇒ ψ ∈ Γy for all Γy ∈ U. Furthermore, since hWS4, RS4i is reflexive, Γx ∈ U. It
then follows from the inductive hypothesis thaty |=ψfor all Γy ∈U. So Γx |=2ψ.
[⇒]. We will argue contrapositively. Assume 2ψ 6∈ Γx. Then by maximal consistency
3¬ψ ∈Γx. We want to show that for everyU ∈τS4, if Γx ∈ U then there is some Γy ∈U
such that Γy |= ¬ψ. Since every open set is closed under RS4 successors, it suffices to find
a point Γz such that RS4ΓxΓz and Γz |= ¬ψ. It is clear what any such Γz must look like:
moment we will refrain from checking that Γ is consistent). Then it follows that for allU such that Γx∈U, Γz ∈U and by the inductive hypothesis Γz |=¬ψ. SohWS4, τS4, νS4i,Γx 6|=2ψ.
qed
It is now easy to establish the desired completeness result. Assume 0S4 ϕ. We need to show there exists a topological model hW, τ, νi such that hW, τ, νi 6|=ϕ. Since0S4 ϕ, it follows by definition that ¬ϕis consistent and so is a member of someS4maximally consistent set Γy.
Then by the above Lemma hWS4, τS4, νS4i,Γy |=¬ϕ. qed Corollary 3.6 S4 is complete with respect to the class of all finite topological spaces.
Proof. Every step in the above proof goes through if we restrict the languageLto subformulas of ¬ϕ. Then there are only finitely many maximally consistent sets, each of which is finite.
So the topological model refuting ϕwill be finite. qed
While the above theorem and corollary are logically quite useful, it is not clear how they can be used to establish completeness results for mathematically interesting structures. Corollary 3.6 does show that Henkin models can be modified in certain ways to give more structured completeness results. And indeed there exist far more subtle means for massaging Henkin models into a desired shape than just restricting the cardinality of the language. However, for our purposes such subtleties are not an issue. For one of the fundamental properties of the all the mathematical structures we will be considering, irreflexivity, is not modally definable. This means drastic measures must be taken to guarantee that a Henkin-like model is, say, linearly ordered by a relation <. The construction method can be seen as one such measure. It is the most fine grained means of throwing out points and removing unwanted relations from a Henkin model. Now let us see how the construction method can be used to establish mathematically interesting topological completeness results.
3.2 S4 on Q
The simplest topological space of mathematical interest is hQ, τi. That S4is complete with respect to hQ, τi follows from ’The Algebra of Topology’. Keeping with the program of simplifying McKinsey and Tarski’s results, a new modal proof has been given in van Benthem, Bezhanishvili, ten Cate, Sarenac [5]. Our contribution to this project will be to give another proof using the construction method. However, using the construction method it is possible to strengthen McKinsey and Tarski’s original results.
Definition 3.7 A logic L is said to be strongly complete with respect to a model M =
hW, R, νi if for any set of formulas ∆ and formulaϕ:
M |= ∆⇒M |=ϕ( ∆|=M ϕ) iffϕis provable inL from ∆ ( ∆`Lϕ).
Strong completeness implies completeness (let ∆ =∅), but the converse is not true. Theorem 3.8 S4 is strongly complete with respect to hQ, τi.
Proof. Assume ∆0S4 χ. As in temporal case, we will give a construction (modulo isomor-phism) of the structurehQ, <i. We will then view hQ, <i as the topological space hQ, τi and show that the construction has been carried out such that ∆6|=hQ,τiϕ.
In this case, the construction will be divided into stages and steps. A stage n consists of:
1. a finite set Qn={q0, q1, . . . , qk}
2. a linear ordering<n on Qn
3. an assignment of anS4maximally consistent set Γqi to eachqi∈Qn
Each stagen >0 is divided intok+ 2 steps. A step j, 0≤j≤k+ 1, consists of: 1. a finite set Qjn={q0, q1, . . . , ql}
2. a linear ordering<jn on Qjn
3. an assignment of anS4maximally consistent set Γqi to eachqi∈Q
j n
Note that unlike the temporal construction ofhQ, <i, the relations <n and <jn are
indepen-dent of the syntactic structure of the maximally consistent sets. For someone accustomed to Henkin-style completeness proofs, this probably seems a bit off. But really it is no mystery. When the truth ofL formulas is evaluated in hQ, τi, the ordering of hQ, <iwill play no role.
During the construction we need to keep in mind the eventual topological structure and not the ordering.
With this word of advice in mind, let us begin the construction. Letϕ0ϕ1ϕ2...be an
enumer-ation of all2and 3 formulas in which each formula is repeated infinitely many times. - Stage 0: LetQ0 ={q#} and associateq# with Γq#, an S4 maximally consistent extension
of ∆∪ {¬χ}.
- Stage 2n+1: Let 2ϕ be the next 2 formula in the enumeration and {q1....qk} be an
enumeration of the elements inQ2n ordered by <2n.
- Step0: Q2n=Q02nand <2n=<02n.
- Step j+ 1: If 2ϕ ∈ Γqj let Q
j+1 2n = Q
j
2n∪ {q
∗, q0} and let <j+1 2n be <
j
2n extended to
include new pointsq∗ and q0 immediately before and afterqj. Associate q∗ and q0 with
Γqj. If2ϕ /∈Γqj, letQ
j+1 2n =Q
j
2n and < j+1 2n =<
j
2n.
LetQ2n+1=Qk2+1n and <2n+1=<k2n+1.
- Stage 2n+2: Let3ϕ be the next3 formula in the enumeration and{q1...qk} be an
- Step0 :Q2n+1 =Q02n+1 and<2n+1=<02n+1.
- Step j+1 : If3ϕ∈Γqj, letQ
j+1 2n+1 =Q
j
2n+1∪ {q
0}and<j+1 2n+1 be<
j
2nextended to include
q0 immediately afterqj. Associateq0 with Γq0, an S4maximally consistent extension of Γ ={ϕ} ∪ {2ψ|2ψ∈Γqj}. If3ϕ /∈Γqj, add no points at stepQ
j+1 2n+1.
LetQ2n+2=Qk2+1n+1 and <2n+2=<k2n+1+1.
To finish off the construction, lethQ, <i=S
{hQn, <ni |n∈ω}. Clearly each stage and each
step meets the conditions imposed above. The only thing to check is that the Γq0’s which are claimed to exist in Stage 2n+2 actually do. By Fact 2.11, it suffices to check that Γ is consistent.
Lemma 3.9 Γ is consistent.
Proof. Assume not. Then there are2ψ1, ...,2ψn∈Γqj such that
`S42ψ1∧...∧2ψn∧ϕ→⊥
`S42ψ→ ¬ϕwhere 2ψ=2(ψ1∧...∧ψn)
`S42(2ψ→ ¬ϕ)
`S422ψ→2¬ϕ
`S42ψ→22ψ
`S42ψ→2¬ϕ
Then since2ψ∈Γqj, it follows that2¬ϕ∈Γqj. This contradicts the fact that3ϕ∈Γqj and
Γqj is consistent. qed
Before switching perspectives and viewing our constructed framehQ, <ias a topological space, let us first check it has the desired structure.
Lemma 3.10 hQ, <i is a countable dense unbounded linear order.
Proof. It follows from condition 2) on stages that hQ, <i is linear. And again, since only finitely many points are added at every stage, hQ, <i is countable. To see hQ, <i is dense, consider two arbitrary pointsq, q00 ∈ Q such that q < q00. Then there is some stage k such thatq, q00 ∈Qkand the formula 2>is being treated. 2>is a member of everyS4maximally
consistent set, so in particular2> ∈Γq. Thus by construction, at Stage k+ 1 there will be
a new pointq0 such thatq < q0 < q00. The same argument showshQ, <iis unbounded. qed
Now let us switch perspectives and view our ordered set hQ, <i as a topological space with the standard topology. AssignhQ, τi the expected Henkin valuationν(p) ={q |p∈Γq}.
Proof. By induction on the complexity ofϕ. We will only treat the caseϕ=2ψ.
[⇒] Assume 2ψ /∈ Γq. Then since Γq is an S4 maximally consistent set, 3¬ψ ∈ Γq. We
want to show that for every U ∈τ such that q ∈ U there is somey ∈U such that y|=¬ψ. LetU0 be an arbitrary open containing q and letq00 be a point in U0 such thatq < q00. That such a q00 exists is guaranteed by the density of hQ, <i. Then there is some stage j in the construction such thatq, q00 ∈Qj−1 and 3¬ψ is being treated. When stage j is completed,
there is a pointq0 ∈Qj such thatq < q0 < q00and¬ψ∈Γq0. It then follows from the inductive hypothesis thatq0 |=¬ψ. So hQ, τ, νi, q6|=2ψ.
[⇐] Assume 2ψ ∈ Γq. We want to show that there is some U ∈ τ such that q ∈ U and
for ally∈U,y|=ψ. To find such aU, consider the first stagej such that q∈Qj−1 and 2ψ
is being treated. At stagej, there is some stepiwhere pointsq∗∗andq00are added immediately before and afterqand Γq∗∗ = Γq = Γq00. We claim that for allQpm,m > i,p≥j−1, ifq0 ∈Qmp andq∗∗< q0 < q00, then2ψ∈Γq0. We will argue by induction on the steps of the construction. Assume at step Qrs+1, r+ 1 > i, s ≥ j −1, a point q0 is added between q∗∗ and q00. By inspecting the construction, it is clear that q0 is added immediately before or after the point qr ∈ Qs. So q∗∗ ≤ qr ≤ q00 and it follows from the inductive hypothesis and the fact that
2ψ is in Γq∗∗ and Γq00 that 2ψ ∈ Γq
r. If s is odd then Γq0 = Γqr and if s is even then
{2ψ|2ψ∈Γqr} ⊆Γq0. So 2ψ∈Γq0.
Finally, 2ψ→ψ is an axiom of S4, so for all q0 ∈Q such thatq∗∗< q0 < q00,ψ∈Γq0. By the inductive hypothesis on the complexity formulas, for allq0 ∈Qsuch thatq∗∗< q0 < q00,q0 |=ψ. So if we let U be the open{y |q∗∗< y < q00}we immediately get thathQ, τ, νi, q|=2ψ. qed
To finish the proof, let f be the isomorphism between hQ, <i and hQ, <i. Assign hQ, τi the valuation ν0(p) = f(ν(p)). It is easy to check that hQ, τ, νi, q |= ϕ ⇔ hQ, τ, ν0i, f(q) |= ϕ.
Since we made sure to add ∆∪ {¬χ} to Γq# at the first stage of the construction, it follows
from the above Lemma that hQ, τ, νi, q# |= ∆∪ {¬χ}. So hQ, τ, ν0i, f(q#)|= ∆∪ {¬χ} and
S4is strongly complete with respect tohQ, τi. qed
3.3 Remarks on the Construction Method
The essential feature of the above construction is that once a2ϕformula is treated at a step, a 2ϕ stretch is created which is preserved for the rest of the construction. Since this is the most stringent requirement on the construction, as might be guessed, there is considerable freedom in how formulas are satisfied at a point inhQ, τi.
For example, in the construction given above if 3ϕ ∈ Γq, then maximally consistent sets
both the right and the left. Actually, this last alternative has an interesting property: for all formulas ϕ, ν0(2ϕ) = X has no least upper bound. For assume X does have a least upper bound q∗. Then 2ϕ /∈ Γq∗, so 3¬ϕ∈Γq∗. However, by construction there is a sequence of points modeling ¬ϕ approaching q∗ from left, so q∗ is not the least upper bound of ν0(2ϕ) after all.
This flexibility in how formulas are to be satisfied on the desired topological space is a special feature of the construction method. When proving topological completeness results indirectly through Kripke models, the starting point is an arbitrary S4 modelM refuting a formulaϕ and the end result is that every point in the desired topological space is modally equivalent to some point in M. Thus we know that the desired topological space refutes ϕ, but we don’t have any ideahow it does this.
This flexibility might lead one to be believe that the construction method has an advan-tage over the model theoretic approaches in some cases. For example, by the above result
hQ, τi is a countermodel to an arbitrary non-theorem ϕ of S4. Without knowing anything about how ϕ is refuted on hQ, τ, νi, it seems rather hopeless to extend hQ, τ, νi to a coun-termodel hR, τ, ν0i refuting ϕ. But perhaps by constructing hQ, τi so that ϕ is refuted in a
particular way, it would be possible to extendhQ, τ, νi to a hR, τ, ν0i countermodel . Indeed, extending a QFP countermodel hQ, <, νi to an RFP countermodel hR, < ν0i in the basic
temporal language works quite naturally. We will consider such extensions shortly.
3.4 Q2FP on Q
However, before looking at possible temporal and topological extensions of Q, let us first
examine this structure in the combined temporal and topological language L2F P. As
men-tioned in the previous chapter,L2F P was one of the first spatial extensions of the basic modal
language to be considered. This language can be traced back to Shehtman [20], where an axiomatization ofhQ, τi was given. Since that time, little progress has been made. It turns
out even the next question, axiomatizinghR, τi inL2F P, is quite difficult.
The completeness proof for hQ, τi given in Shehtman [20] works indirectly through Kripke
frames, as sketched in Section 2.5. In the last chapter, we will examine this approach in some detail , but for now it suffices to say that in the case ofL2F P this strategy is quite demanding.
This is one instance where using the construction method improves matters significantly. LetQ2FP be the logic S4+QFP + :
F ϕ→2F ϕ P ϕ→2P ϕ 2ϕ→F ϕ 2ϕ→P ϕ
2ϕ∧Gϕ→2Gϕ 2ϕ∧Hϕ→2Hϕ
Theorem 3.12 Q2FP is strongly complete with respect to hQ, τi.
Proof. Assume ∆0Q2FP χ. As before, we will construct a frame hQ, <i ∼=hQ, <i in steps and stages such thathQ, τi refutes ∆∪ {ϕ}. However, since the ordering of hQ, <i will play a role in evaluating the truth ofL2F P formulas in h
Q, τi, this construction will require some
relation between the maximally consistent sets and <. We will build this interaction into the definitions of a stage and a step.
A stage nconsists of:
1. a finite set Qn={q0, q1, ..., qk}
2. a linear ordering <n on Qn such that qj <nql⇒Γqj ≺Γql
3. an assignment of a Q2FP maximally consistent set Γqi to each qi ∈Qn
Each stagenis divided into ≤k+ 2 steps. A stepj, 0≤j≤k+ 1, consists of: 1. a finite set Qjn={q0, q1, ..., ql}
2. a linear ordering <jn on Qjn such that qh <jnqi⇒Γqh ≺Γqi
3. an assignment of a Q2FP maximally consistent set Γqi to each qi ∈Q
j n
Once again, the salient feature of the construction will be that once a2ψ formula is treated at a step that a2ψ stretch is created which can be preserved for the rest of the construction. However, in the present case preserving 2ψ stretches requires some attention. For example, assume the pointqis in what we intend to be a2ψstretch and thatHϕis in Γq. Furthermore,
assume at some stage we add a pointq00immediately after q andP(¬ϕ∧3¬ψ) is added into Γq00. Nothing so far prevents this situation from occurring. We know by the ordering relation
≺thatHϕmust be in all Γq∗ such thatq∗ < q. So in order to ensure the truth ofP(¬ϕ∧3¬ψ) atq00 when the construction is finished, a point q0 must be added in between q and q00 such that 3¬ψ ∈ Γq0. This ruins our prospective 2ψ stretch. Fortunately, guarding against this sort of occurrence alone is sufficient to preserve2ψ stretches.
Definition 3.13 We say Γq≺2ψ Γq0 if 1) Γq≺Γq0
2)2ψ∈Γq,Γq0
Proposition 3.14 Γq∗ ≺2ψ Γq≺2ψ Γq0 ⇒Γq∗ ≺2ψ Γq0.
Proof. Assume Γq∗≺2ψ Γq≺2ψ Γq0. It is easy to see conditions 1) and 2) of Γq∗≺2ψ Γq0 are met. So assume Hϕ∈Γq∗. By assumption Γq∗ ≺2ψ Γq, so H(ϕ∨2ψ) ∈Γq. Furthermore, Γq ≺2ψ Γq0, so H((ϕ∨2ψ)∨2ψ) ∈ Γq0. It then follows from basic propositional logic and maximal consistency thatH(ϕ∨2ψ)∈Γq0. Condition 4) is similar. qed Fact 3.15 Γq≺2ψ Γq0 and Γq≺2ϕ Γq0 ⇒ Γq ≺2(ψ∧ϕ)Γq0.
Armed with our new condition ≺2ψ to preserve 2ψ stretches, we are ready to begin the construction. Let ϕ0ϕ1ϕ2... be an enumeration of all F, P,2 and 3 formulas in which each
formula is repeated infinitely many times.
- Stage 0: LetQ0 ={q#}with which we associate Γq#, aQ2FP maximally consistent
exten-sion of ∆∪ {¬χ}.
- Stage 5n+1: Let 2ϕ be the next2 formula in the enumeration and let {q1, ..., qk} be an
enumeration of the elements inQ5n ordered by <5n.
- Step 0 :Q5n=Q50n and <5n=<05n
- Step j+ 1 : If2ϕ∈Γqj, letq
∗∗be the immediate predecessor of q
j in the <j5nordering,
provided one exists. Add a new point q∗ immediately before qj and associate q∗ with
Γq∗, a Q2FP maximally consistent extension of: Γ∗={2ϕ} ∪ {2ψ0|Γq∗∗ ≺2ψ0 Γq
j} ∪ {α
0|Hα0 ∈Γ
qj} ∪ {P β
0 |β0 ∈Γ
q∗∗} ∪
{P γ0 |P(γ0∧ ¬2ϕ)∈Γqj} ∪ {Gζ
0|G(ζ0∨2ϕ)∈Γ
qj}
Similarly, if 2ϕ ∈ Γqj let q
00 be the immediate successor of q
j in the <j5n ordering,
provided one exists. Add a new pointq0 immediately afterqj and associateq0 with Γq0, a Q2FP maximally consistent extension of:
Γ0 ={2ϕ} ∪ {2ψ|Γqj ≺2ψ Γq00} ∪ {α|Gα∈Γqj} ∪ {F β|β ∈Γq00} ∪
{H(γ∨2ϕ)|Hγ ∈Γqj} ∪ {F ζ|F(ζ∧ ¬2ϕ)∈Γqj}
Let Qj5+1n be Qj5n plus any points added by the above procedure, and <j5+1n be <j5n extended in the obvious way to include these points.
LetQ5n+1=Qk5+1n and <5n+1=<k5n+1.
The definitions of Γ∗ and Γ0 ensure that for allq∈Q5n, if2ϕ∈Γq then at Stage 5n+ 1q will
have an immediate predecessorq∗ and immediate successorq0 such that Γq∗ ≺2ϕΓq≺2ϕΓq0. It only remains to be checked that Γ0 is consistent.
Proof. Assume not. Then by Proposition 3.15 there is a formula2ψand formulasGα, Hγ1, .., Hγm,
F(ζ1∧ ¬2ϕ), .., F(ζp∧ ¬2ϕ)∈Γqj and β1, .., βk∈Γq00 such that:
`Q2FP 2ϕ∧2ψ∧α∧F β1∧...∧F βk∧H(γ1∨2ϕ)∧...∧H(γm∨2ϕ)∧F ζ1∧...∧F ζp →⊥
`Q2FP α→ ¬(2ϕ∧2ψ∧F β1∧...∧F βk∧H(γ1∨2ϕ)∧...∧H(γm∨2ϕ)∧F ζ1∧...∧F ζp)
(∗)`Q2FP Gα→G¬(2ϕ∧2ψ∧F β1∧...∧F βk∧H(γ1∨2ϕ)∧...∧H(γm∨2ϕ)∧F ζ1∧...∧F ζp) We know that:
- 2ϕ→22ϕis an axiom and 2ϕ,2ψ∈Γqj, so22ϕ∈Γqj and 22ψ∈Γqj
- β1, ..., βk∈Γq00 and Γ≺Γq00, soF β1, ..., F βk ∈Γq00 - F β→2F β is an axiom, so2F β1, ...,2F βk∈Γqj
- 2ϕ∈Γqj, so2(γ1∨2ϕ), ...,2(γm∨2ϕ)∈Γqj
- Hγ1...Hγm ∈Γqj, soH(γ1∨2ϕ), ..., H(γm∨2ϕ)∈Γqj
- 2ψ∧Hψ→2Hψ is an axiom, so 2H(γ1∨2ϕ), ...,2H(γm∨2ϕ)∈Γqj
- F ζ1, ..., F ζp∈Γqj and F ζ→2F ζ is an axiom, so2F ζ1, ...,2F ζp ∈Γqj
Putting all this together it follows that:
2(2ϕ∧2ψ∧F β1∧...∧F βk∧H(γ1∨2ϕ)∧...∧H(γm∨2ϕ)∧F ζ1∧...∧F ζp)∈Γqj
and since2ϕ→F ϕ is an axiom:
F(2ϕ∧2ψ∧F β1∧...∧F βk∧H(γ1∨2ϕ)∧....∧H(γm∨2ϕ)∧F ζ1∧...F ζp)∈Γqj
However,Gα∈Γqj, so by (∗):
G¬(2ϕ∧2ψ∧F β1∧...∧F βk∧H(γ1∨2ϕ)∧...∧H(γm∨2ϕ)∧F ζ1∧...∧F ζp)∈Γqj
This contradicts the consistency of Γqj. qed
Showing Γ∗ is consistent is similar.
- Stage 5n+2: LetF ϕ be the nextF formula in the enumeration.
- Step0 : IfF ϕ∈Γqj and ¬ϕ∧G¬ϕ∈Γqj+1 (or ifF ϕ∈Γqj andj=k) add a new point
q0 immediately afterqj and associate Γq0 with aQ2FP maximally consistent extension of:
Let Q05n+1 be Q5n+1 plus any points added by this procedure and <05n+1 be <5n+1
extended in the obvious way to include any new points. Let Q5n+2=Q05n+1 and <5n+2=<05n+1.
Lemma 3.17 Λ is consistent.
Proof. Assume not. Then there are formulas2ψ and α in Γqj such that:
`Q2FP ϕ∧2ψ∧α→⊥ `Q2FP α→ ¬(2ψ∧ϕ)
`Q2FP Gα→G¬(2ψ∧ϕ)
Since Gα ∈ Γqj, it follows that G(¬2ψ∨ ¬ϕ) ∈ Γqj. Furthermore, since G¬ϕ ∈ Γqj+1
and Γqj ≺2ψ Γqj+1 , it follows that G(2ψ∨ ¬ϕ)∈Γqj. So:
G(2ψ→ ¬ϕ)∧G(¬2ψ→ ¬ϕ)∈Γqj
However, by basic propositional logic it then follows that G¬ϕ∈ Γqj. This contradicts the
fact thatF ϕ∈Γqj. qed
The same argument given in the temporal construction ofhQ, <isuffices to show that Γqj ≺Γq0 ≺Γqj+1.
- Stage 5n+3: Symmetric to 5n+ 2 for the nextP formula in the enumeration.
- Stage 5n+4: Let 3ϕbe the next 3 formula in the enumeration and let{q0, ..., qk} be an
enumeration of the elements inQ5n+3 ordered by <5n+3.
- Step 0 : LetQ5n+3=Q05n+3 and <5n+3=<05n+3.
- Step j+ 1 : If 3ϕ∈ Γqj let q
00 be the immediate successor and q∗∗ be the immediate
predecessor of qj in the<j5n+3 ordering, provided these exist. Let
Λ0 ={ϕ} ∪ {2ψ|Γqj ≺2ψ Γq00} ∪ {α|Gα∈Γqj} ∪ {F β |β∈Γq00}
and Λ∗ ={ϕ} ∪ {2ψ0 |Γqj ≺2ψ0 Γq∗∗} ∪ {α
0 |Hα0∈Γ
qj} ∪ {P β
0|β0∈Γ
q∗∗}
If ϕ ∈ Γqj add no points. If ϕ /∈ Γqj and Λ
0 is consistent, then add a new point
q0 immediately after qj and associate q0 with a maximally consistent extension of Λ0.
Otherwise, add a new pointq∗ immediately beforeqj and associateq∗with a maximally
consistent extension of Λ∗.
Proposition 3.18 Either ϕ∈Γqj, or Λ
0 is consistent, orΛ∗ is consistent.
Proof. Assume not. Then there are formulas 2ψ,2ψ0, Gα, Hα0 ∈ Γqj, β1, ...βn ∈ Γq00 and
β10, ..., βk0 ∈Γq∗∗ such that
- `Q2FP ϕ∧2ψ∧α∧ F β1∧...∧F βn→⊥
- `Q2FP ϕ∧2ψ
0∧α0∧P β0
1∧...∧P βk0 →⊥
As in the previous stages, it follows that:
- `Q2FP Gα→G¬(ϕ∧2ψ∧ F β1∧...∧F βn)
- `Q2FP Hα
0→H¬(ϕ∧2ψ0∧P β0
1∧...∧P βk0)
and thus (**): - ϕ /∈Γqj and
- G¬(ϕ∧2ψ∧ F β1∧...∧F βn)∈Γqj and
- H¬(ϕ∧2ψ0∧P β10 ∧...∧P β0k)∈Γqj
Since Γq∗∗ ≺Γq
j ≺ Γq00, it follows that F β1, ..., F βn, P β
0
1, ..., P βk0 ∈ Γqj. Furthermore, since
2ψ,2ψ0∈Γqj and P ϕ→2P ϕ,F ϕ→2F ϕ,2ϕ→22ϕare axioms, it follows that:
2(2ψ∧2ψ0∧F β1∧...∧F βn∧P β10 ∧...∧P β0k)∈Γqj
It is easy to show that`S42γ∧3ϕ→3(γ∧ϕ), and by substituting 2ψ∧2ψ0∧ F β1∧...∧F βn∧P β10 ∧...∧P βk0
forγ we get:
`Q2FP 2(2ψ∧2ψ
0∧ F β
1∧...∧F βn∧P β10 ∧...∧P βk0)∧3ϕ→
3(ϕ∧2ψ∧2ψ0∧ F β1∧...∧F βn∧P β10 ∧...∧P βk0).
So
3(ϕ∧2ψ∧2ψ0∧ F β1∧...∧F βn∧P β10 ∧...∧P βk0)∈Γqj
3γ →γ∨F γ∨P γ is an axiom, so:
- (ϕ∧2ψ∧2ψ0∧F β1∧...∧F βn∧ P β10 ∧...∧P β0k)∈Γqj or
- P(ϕ∧2ψ∧2ψ0∧F β1∧...∧F βn∧ P β10 ∧...∧P βk0)∈Γqj
This contradicts (∗∗) above. qed
To finish off the construction, let hQ, <i = S
{hQn, <ni | n ∈ ω}. It is not difficult to
check thathQ, <i is a countable dense unbounded linear order (the formula2>ensures both unboundedness and density), so we will move directly on to showing that hQ, τi satisfies ∆∪ {¬ϕ}. Again lethQ, τi have the Henkin valuation ν(p) ={q|p∈Γq}.
Proposition 3.19 hQ, τ, νi, q|=ϕ⇔ϕ∈Γq.
Proof. By induction on the complexity ofϕ. We will treat the casesϕ=2ψand ϕ=F ψ. ϕ=2ψ
[⇒] Assume 3¬ψ ∈ Γq. Let U0 be an arbitrary open containing q and let q∗ and q0 be
points in U0 such that q∗ < q < q0. Then there is a stage j in the construction such that q, q∗, q0 ∈Qj−1and 3¬ψis being treated. When stagej is completed, there is a pointx∈Qj
such that q∗ < x < q0 and ¬ψ ∈ Γx. It then follows from the inductive hypothesis that
x|=¬ψ. SohQ, τ, νi, q|=3¬ψ.
[⇐] Assume 2ψ∈Γq. We know at some stage j and stepi such there are points q∗∗ and q00
immediately before and afterq in the<ij ordering such that Γq∗∗ ≺2ψ Γq≺2ψ Γq00. We claim that for all r, s, t∈Qmp , m > i,p≥j, ifq∗∗ ≤r < s < t≤q00, then Γr ≺2ψ Γs ≺2ψ Γt. We
will argue by induction on the steps of the construction.
Assume at step Qmp +1,m ≥i, p ≥j a point q0 is added between q∗∗ and q00. By inspecting the construction, it is clear that q0 is added immediately before or after the pointqm ∈Qp.
Without loss of generality, assume it is placed immediately after qm. By Proposition 3.14
≺2ψ is transitive, so it suffices to show that Γqm ≺2ψ Γq0 ≺2ψ Γqv, where qv is the
imme-diate successor of qm in the <mp ordering. We will only check that Γqm ≺2ψ Γq0, showing
Γq0 ≺2ψ Γq
v is similar.
By condition 2) on stages, we know that Γqm ≺ Γq0 ≺ Γqv. Furthermore, by the
induc-tive hypothesis we know that Γqm ≺2ψ Γqv. By inspecting each stage of the construction, we
see that {2ϕ|Γqm ≺2ϕ Γqv} ⊆Γq0, so 2ψ∈Γq0. Now assume Hϕ∈Γqm. As Γqm ≺2ψ Γqv,
it follows that H(ϕ∨2ψ)∈Γqv. It is easy to check that this entails HH(ϕ∨2ψ) is also in
Γqv. So by the fact Γq0 ≺Γqv, H(ϕ∨2ψ) ∈Γq0. Showing Gϕ∈Γq0 ⇒ G(ϕ∨2ψ) ∈Γqm is
similar.
So for all r, s, t such that q∗∗ ≤ r < s < t ≤ q00, Γr ≺2ψ Γs ≺2ψ Γt. It follows
we get thathQ, τ, νi, q|=2ψ. ϕ=F ψ
[⇒] Assume F ψ ∈ Γq. Then there is some stage j such that q ∈ Qj−1 and F ψ is being
treated at j . After this stage, there is a point q0 > q such that ψ ∈ Γq0. Then by the inductive hypothesisq0 |=ψ, sohQ, τ, νi, q|=F ψ.
[⇐] Assume hQ, τ, νi, q |= F ψ. Then there is some q0 > q such that q0 |= ψ and by the inductive hypothesis ψ∈Γq0. Since q and q0 are related by≺it follows that F ψ ∈Γq. qed
It follows as before thatQ2FP is strongly complete with respect tohQ, τi. qed
3.5 Extensions To R: The Temporal Case
So far we have seen thatQis complete with respect to the logicsQFP,S4andQ2FP. While this is interesting, Q is not a particularly rich topological structure. From a mathematical
perspective,Ris more interesting. And indeed extensions to Rhave been considered for the
logics QFP, S4, and Q2FP. Working indirectly through Kripke frames, the corresponding cases are found to be technical, difficult, and open, respectively.
As mentioned before, one impressive feature of the temporal construction method is the ease in which completeness onhQ, <i is extended to completeness on hR, <i. Thus in considering topological extensions of hQ, τi tohR, τi forL and L2F P using the construction method, it
is useful to have the original LF P proof in mind. The proof makes use of the following well
known facts.
Fact 3.20 hR, <i is the unique complete linear ordering that has a countable dense subset
isomorphic to hQ, <i.
Fact 3.21 IfhW, <i is a dense unbounded linearly ordered set, then there exists a continuous unbounded linearly ordered set hW0 <0i such that:
- W ⊂W0 and <and <0 agree onW
- W is dense in W0
Theorem 3.22 Every dense unbounded linear order is elementary equivalent.
LetDULO be the first order theory containing axioms for density, unboundedness and lin-earity. Then by the above classical theorem:
hR, <i |=ϕ⇔ `DULO ϕ⇔ hQ, <i |=ϕ
It follows immediately that continuity is not first order definable. However, continuity is definable inLF P. LetAψ denote ψ∧Gψ∧Hψ.
Proposition 3.23 hW, <i |=A(Gϕ→P Gϕ)→(Gϕ→Hϕ)⇔ hW, <i is continuous.
Proof. [⇒] AssumehW, <i is not continuous. Then there is some bounded set X⊆W such that sup X /∈W. Let hW, <i have the valuation ν(p) =W/X. Then for all y such that y < supX,y|=F¬pand for ally such thaty >supX,y|=Gp∧P Gp. Letz be a point greater than supX. Thenz|=A(Gp→P Gp) and z6|=Gp→Hp.
[⇐] Assume hW, <i is continuous and hW, < νi, w |= A(Gϕ → P Gϕ). Furthermore, as-sume towards a contradiction that w |= Gϕ∧P¬ϕ. Then the set X = {x | x < w and x|=¬ϕ} is nonempty. SinceW is continuous, supX=w0 ∈W. So w0 |=Gϕ∧ ¬P Gϕ. But sincew0 ≤w, this means w|=P¬(Gϕ→P Gϕ)∨ ¬(Gϕ→ P Gϕ). So w6|=A(Gϕ→ P Gϕ),
contrary to assumption. qed
It should be clear that with continuity we are already getting more than expected out ofLF P
and that no additional properties ofhR, <ican be expressed. However, we still want to prove this fact. LetRFP be the logicQFP +A(Gϕ→P Gϕ)→(Gϕ→Hϕ).
Theorem 3.24 RFP is strongly complete with respect to hR, <i.
Proof. Assume ∆ 0RFP χ. First carry out the temporal construction procedure forhQ, <i as in Section 2.6, except this time replacing QFP maximally consistent sets with RFP ones. The end result of the construction will be a countable, dense, unbounded linear orderhQ, <i
which satisfies ∆∪ {¬χ}. Now extendhQ, <i to a structurehR, <i ∼=hR, <i. We know many
new points will be added by this procedure, and we want to associate each of them with a RFP maximally consistent set such thathR, <i still satisfies ∆∪ {¬χ}.
An obvious candidate for the irrational newcomerx∈R\Qis Γx, an RFP maximally consis-tent extension of:
Γ ={α| ∃q∈Qsuch that q < xand Gα∈Γq} ∪
{β | ∃q0 ∈Qsuch that x < q0 and Hβ ∈Γq0}
Lemma 3.25 Γ is consistent.
Proof. Assume not. Then there are q1, ...qm < x with Gαi ∈ Γqi for 1 ≤ i ≤ m and
q01, ..., qn0 > xwith Hβj ∈Γq0
j for 1≤j≤nsuch that
`RFP ¬(α1∧...αm∧β1∧...∧βn)
However, sincehQ, <iis dense, there must be someq∗ such thatq1, ...qm< q∗< q01, ..., qn0 and
since the<ordering onhQ, <irespects ≺,α1∧...αm∧β1∧...∧βn∈Γq∗. qed
So we only need to check that extending the Henkin valuation to include the new points preserves the desired harmony between the syntactic and semantic.
Lemma 3.26 For all r ∈ R\Q, if F ψ ∈ Γr then there is some q ∈ Q, r < q, such that
ψ∈Γq.
Proof. Assume F ψ ∈ Γr,r ∈ R\Q, and for all q00 ∈ Q such that r < q00, ¬ψ ∈Γq00. Then we know from the properties of our constructed model hQ, <, νi that for all q00 ∈ Q such thatr < q00, G¬ψ∧P G¬ψ∈ Γq00. Furthermore, we also know that for allq∗ ∈Q such that q∗ < r,F ψ ∈Γq∗, for otherwise G¬ψ∈Γr by the definition of Γr. So for every q ∈Q, either P G¬ψ∈Γq or F ψ ∈Γq. It follows that for allq ∈Q, A(G¬ψ →P G¬ψ)∈ Γq . Now let q0
be a point inQ such that q < q0. From A(G¬ψ→ P G¬ψ)∧G¬ψ ∈Γq0 and the continuity axiom it follows thatH¬ψ∈Γq0. But this contradicts the fact that F ψ ∈Γq∗ for allq∗ < r andhQ, <, νi obeys the ≺ordering. qed Lemma 3.27 hR, <, νi |=ϕ⇔ϕ∈Γr.
Proof. By induction on the complexity ofϕ. We will only treat the caseϕ=F ψ.
[⇐] Assume F ψ ∈ Γr. Then by the above Lemma and the ≺ ordering on hQ, <i, there is
someq∈Qsuch thatr < qandψ∈Γq. By induction hypothesis,q |=ψ, sohR, <, νi, r|=F ψ.
[⇒] Assume F ψ 6∈ Γr. It is not hard to check that ¬ψ ∈ Γq for all q ∈ Q such that
r < q. So assume towards a contradiction that there is some r0 ∈ R\Q such that r < r0 and ψ∈ Γr0. SincehQ, <i is a dense subset of hR, <i, there exists a point q0 ∈ Q such that r < q0 < r0. However, we know G¬ψ ∈Γq0 and thus by the definition of Γr0 that ¬ψ ∈Γr0. This contradicts our assumption. So for all r00 ∈ R such that r < r00, ¬ψ ∈ Γr00. By the inductive hypothesis allr00|=¬ψand hR, < νi, r6|=F ψ. qed
3.6 Further Extensions to R
So we have seen that our constructedLF P model hQ, <, νi can be extended to a new model
hR, <, νi while preserving the truth ofLF P formulas on hQ, <, νi. Now let us consider this
procedure forhQ, τ, νiand hR, τ, νi in the languages L andL2F P.
Let us first consider extendinghQ, τ, νi tohR, τ, νi inL. Unlike the temporal case, we know thatS4is the logic of bothhQ, τiand hR, τi, so we only need to fill in hQ, τi withS4
maxi-mally sets. Otherwise, let us proceed as above.
Assume ∆ 0S4 χ and let hQ, τ, νi be our constructed topological model which satisfies ∆∪ {¬χ}. Extend hQ, <i to a structure hR, <i ∼= hR, <i. Now associate each x ∈ R\Q with Γx, an S4maximally consistent extension of:
Γ ={2ψ| ∃q1, q2 ∈Q such thatq1 < x < q2 and ∀q ∈Qsuch that q1 < q < q2,ψ∈Γq} ∪
{3ϕ| ∀q3, q4 ∈Q such thatq3 < x < q4,∃q0 ∈Q such thatq3 < q0 < q4 and ϕ∈Γq0} Lemma 3.28 hR, τ, νi, r|=ϕ⇔ϕ∈Γr.
Proof. By induction on the complexity ofϕ. We will only treat the caseϕ=2ψ.
[⇐] Assume 2ψ ∈ Γr. We claim that ∃q1, q2 ∈ Q such that q1 < r < q2 and ∀q ∈ Q,
q1 < q < q2,ψ∈Γq. Ifr ∈Q, this holds by the construction of hQ, τi . And ifr ∈R\Q this
follows from the definition of Γr (since3¬ψ /∈Γr).
Now assume towards a contradiction that there is some r0 ∈ R\Q such that q1 < r0 < q2
and¬ψ∈Γr0. Then by maximal consistency 2ψ6∈Γr0. It then follows from the definition of Γr0 that for all q3, q4 ∈Q such thatq3< r0 < q4, there exists some q0 ∈Q , q3< q0 < q4, and
¬ψ∈Γq0. But this means there is aq0∈Qsuch thatq1 < q0< q2 and¬ψ∈Γq0, contradicting the previous paragraph.
So for all x ∈ R such that q1 < x < q2, ψ ∈ Γx. By letting U = {y | q1 < x < q2}
and the inductive hypothesis we get thathR, τ, νi, r|=2ψ.
[⇒] Assume2ψ6∈Γr. We claim that∀q3, q4 ∈Qsuch thatq3 < r < q4,∃q0∈Q,q3 < q0 < q4,
and ¬ψ ∈ Γq0. If r ∈ Q this holds by the construction of hQ, <i and if r ∈ R\Q then this holds by definition of Γr (since 2ψ 6∈Γr). So it follows from the inductive hypothesis that
hR, τ, νi, r6|=2ψ. qed
