JR: Ella mando a mi hijo “mamá vamos” y ¿mi hermana? Dijo no, dijo no que “la de la

Before getting to work, a brief remark. There is another way of thinking about axiomatic exten- sions of K. Instead of viewing them as giving rise to brand new modal logics, we can simply view them as theories constructed over the minimal logic K in much the same way as a first-order the- ory (of say, linear orders) is constructed over the set of first-order validities. Nothing of substance hangs on this shift of perspective, but it fits more naturally with our focus on expressivity.

So, bearing this in mind, let’s pose the first question: what can modal formulas say about frames? A natural way to approach this is to introduce the concept of frame definability. We shall say that a modal formulaϕdefines a class of framesFiff it is valid on precisely the frames inF. That is, not only mustϕbe valid on every frame inF, it must also be possible to falsifyϕ

on any frame that is not inF. So, what classes of frames can modal languages define? Here are some simple examples:

PROPOSITION 27.

1. 2p→22pdefines the class of transitive frames; that is, frames such that∀xyz(Rxy∧

Ryz→Rxz).

2. 3p→2pdefines the class of frames where the frame relationRis a partial function; that is, frames such that∀xyz(Rxy∧Rxz→y=z).

3. p ↔ 2pdefines the class of frames which consist of isolated reflexive points; that is, frames such that∀xy(Rxx∧(Rxy∧Ryx→x=y)).

4. 2⊥defines the class of frames which consist of isolated irreflexive points; that is, frames such that∀xy¬Rxy.

Proof. We have already asked the reader to check that these formulas are valid on the class of

frames in question. So to complete the proofs of these definability claims we need merely check that each formula can be falsified on any frame that does not belong to the relevant class.

Let’s deal with the second example. Suppose (W, R) is a frame whereR is not a partial function. This means that there is a pointw∈W that has two distinctR-successors, sayuand

v. It follows that we can falsify3p→2pon(W, R)atw. For letV be the valuation that makes

ptrue atuand nowhere else. Then(W, R, V), w |=3pbut(W, R, V), w|=2p, sincepis not true atv. So we have falsified3p→2pon(W, R)as required. a

A remark on terminology. Instead of saying, for example, that2p→22pdefines the class of transitive frames, we often simply say that2p→22pdefines transitivity. It is also usual to say that2p→22pcorresponds (at the level of frames) to∀xyz(Rxy∧Ryz →Rxz), or that ∀xyz(Rxy∧Ryz→Rxz)is a frame correspondent for2p→22p.

Now for an important question: how do we go about showing that a class of frames cannot be modally defined? Answering such questions is typically more demanding than proving the type of result noted in Proposition 27, for instead of checking that a given formula defines a given frame class, we now have to prove that no modal formula is capable of this. How can we prove such general results? By finding ways of transforming frames that preserve frame validity. For if we can show that a class of framesFis not closed under such a transformation, it follows thatF is not modally definable. Let’s take a closer look.

The first step is to find transformations that preserve frame validity. Three lie close to hand: the formation of disjoint unions, generated submodels, and bounded morphic images. In Section 3.2 we defined these constructions at the level of models, and they can be lifted to the level of frames simply by ignoring the requirements imposed on the valuations. For example, a bounded morphism between frames(W, R)and(W0, R0)is a functionf fromwtoW0 that satisfies the morphism condition (if Rwv, then R0f(u)f(v)) and the zag condition (if R0w0v0, then there exists avsuch thatf(v) =v0andRwv), and we say that frame(W0, R0)is a bounded morphic image of frame (W, R)if there is a surjective bounded morphism from (W, R) to(W0, R0). Lifting these constructions to the level of frames immediately gives us three validity preservation results:

THEOREM 28.

For all basic modal formulasϕwe have that:

1. Let{Fi | i ∈ I}be a family of frames. Then ifFi |= ϕfor everyiinI, we have that U_F

i|=ϕtoo. That is, frame validity is preserved under the formation of disjoint unions. 2. LetF0 _{be a generated subframe ofF. Then ifF |=ϕ, we have thatF}0 _{|=ϕtoo. That is,}

frame validity is preserved under the formation of generated subframes.

3. LetFandF0be frames andf a surjective bounded morphism fromFtoF0. Then ifF|=ϕ, we have thatF0 |= ϕtoo. That is, frame validity is preserved under the formation of bounded morphic images.

Proof. We prove the result for bounded morphisms. Let F = (W, R)andF0 = (W0, R0)be frames, and suppose for the sake of a contradiction thatF |=ϕbutF0 6|=ϕ. This means that for some valuationV0onF0and some pointw0 ∈W0 we have that(F0, V0), w0 6|=ϕ. LetV be the valuation onFdefined byV(p) = {u∈ W | f(u)∈ V0(p)}, for all proposition lettersp. Furthermore, letwbe any point such thatf(w) =w0; there must be at least one such point as

f is surjective. Then the model(F0, V0)is a bounded morphic image of the model(F, V), and hence(F, V), w6|=ϕ. But this contradicts our assumption thatF|=ϕ, hence we conclude that

F0 |=ϕafter all. a

Applying this theorem immediately gives rise to a crop of non-definability results. Here are some simple ones. Basic modal languages cannot define the class of simply connected frames, that is, the class of frames such that ∀xy(Rxy ∨Ryx). Why not? Because this class is not closed under the formation of disjoint unions: taking the disjoint union of two frames with this property clearly results in a frame without it. It also follows that basic modal languages cannot

define the class of frames containing an isolated reflexive point. Why not? Because this class is not closed under the formation of generated subframes. For consider a frame consisting of two isolated points, one reflexive, the other irreflexive. This frame belongs to the required class, however the subframe generated by the irreflexive point does not. Nor is the class of irreflexive frames modally definable. Why not? Because it is not closed under the formation of bounded morphic images (recall the bounded morphism of Figure 13 which collapses the natural numbers to a single reflexive point). But frame validity is preserved under this transformation, hence no modal formula can define irreflexivity. For more sophisticated applications of these validity preservation results, see van Benthem [125].

As we shall soon see, the three frame transformations just introduced all play a role in the Goldblatt-Thomason Theorem, a characterisation of modally definable classes of elementary frames. But a fourth transformation, namely the formation of ultrafilter extensions, is also needed to complete the statement of this celebrated result, so let’s take this opportunity to define this (somewhat more complex) frame construction. First we recall a standard mathematical concept. Given a non-empty setW, a filterF over W is any subset of2W (the power set ofW) that containsW and is closed under finite intersection (that is, ifX, Y ∈F thenX ∩Y ∈F) and set-theoretic inclusion (that is, ifX ∈FandX ⊆Y ⊆W thenY ∈F). A filter is called proper if it is distinct from2W_{. An ultrafilter is a proper filterU such that for allX ∈2}W_{,X ∈U iff} (W\X)6∈U. A standard result assures us that any proper filter can be extended to an ultrafilter. Bearing this in mind, we make the following definition:

DEFINITION 29 (Ultrafilter Extensions of Frames). Let F = (W, R)be a frame. For any

X ⊆W we definel(X)to be{w∈W |for allv∈W, ifRwvthenv∈X}.Then the ultrafilter extension ue(F)of F is defined to be the frame (uf(W), Rue_{), where uf(W)is the set of all} ultrafilters onW andRue_{is the relation consisting of all pairs of ultrafiltersU, U}0_{such that for}

allX⊆W, ifl(X)∈U, thenX∈U0.

We can now state the required theorem. Note that the direction of validity preservation is the reverse of that found in Theorem 28. That is, here frame validity is preserved from the transformed frame (here the ultrafilter extension) back to the original one:

THEOREM 30. For any basic modal formulaϕ, if ue(F)|=ϕthenF|=ϕdoes too. That is, frame validity reflects ultrafilter extensions.

Proof. The use of ultrafilter extensions in modal logic traces back to Goldblatt [52, 53], van

Benthem [118], and Fine [38]. For a detailed proof of this theorem, see Proposition 2.59 and Corollary 3.16 of Blackburn, de Rijke and Venema [13]. a Although this transformation is harder to visualise than the previous three, it too gives rise to some simple non-definability results. Here’s a nice example, taken from Goldblatt and Thoma- son [54], showing that the class of frames satisfying∀x∃y(Rxy∧Ryy)is not modally definable. We can see this as follows. The ultrafilter extension of (_N, <), the natural numbers in their usual order, looks a bit like a gigantic lolly-pop. It has an infinite handle, an isomorphic copy of (_N, <), consisting of all the principal ultrafilters (that is, those ultrafilters which contain a singleton set{n}, wherenis a natural number). This is followed by the lolly: an uncountable collection of non-principal ultrafilters which are all related to one another and reflexively related to themselves. Henceue(N, <)has the property∀x∃y(Rxy∧Ryy). Why? Because every point

in the frame is related to the reflexive points in the lolly. However this formula is clearly not valid on the original frame(N, <). As frame validity reflects ultrafilter extensions, it follows that

the class of frames satisfying∀x∃y(Rxy∧Ryy)is not modally definable. For further discus- sion of ultrafilter extensions from a model-theoretic perspective, see Chapter 5 of this handbook. There is also an important algebraic perspective on ultrafilter extensions, which is discussed in Chapter 6.