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
It follows that our constructed dense, unbounded, linear order hQ, <i constructed in Section 2.6 can be extended to a framehR, <i ∼= hR, <i. Using our intuitions from first order logic as a guide, after extendinghQ, <itohR, <iwe should associate every new point with aQFP maximally consistent set. Then we should extend our Henkin valuation to include these new points, and check that our new modelhR, <i still satisfies ∆∪ {¬χ}. However, in this case our first order intuitions fail us. This is one of the rare practical occurrences where the basic modal language is, in some respects, more expressive than its first order counterpart.
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} In this case the obvious candidate turns out to be the correct one.
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 Since ∆∪{¬χ} ⊆Γq#, it readily follows thatRFPis strongly complete with respect tohR, <i.