Ultravaluations and their Applications in $\textsf{CPL}$

Abstract

This paper introduces the construct of an ultravaluation inspired by the well-known ultraproduct. Basic properties and exemplary applications of this notion are shown: for compactness and definability theorems. We also use ultravaluations to check failure of compactness and undefinability.

1 Introduction and Overview

The ultraproduct is a semantical construction of an immense importance both in model theory as well as abstract algebra. Besides its wide range of applications in the above mentioned fields, it has not been of any use in Classical Propositional Logic. In the present paper, we wish to keep some of the basic concepts behind the ultraproduct and propose new construction which will enable us to prove theorems for $\textsf{CPL}$ analogous to those which can be found in first order model theory and abstract algebra. This way, we shall extend the universality of some of the fundamental Jerzy Łoś’s ideas.

The article introduces the concept of an ultravaluation, to some extent similar to the Łoś’s ultraproduct. The difference lies in the fact that in our proposal we do not use direct product operation and do not formulate congruence in the domain of a model—we consider valuations in $\textsf{CPL}$, so there is no reason to refer to any domain of a model. Furthermore, for our construction we prove theorem analogous to the Łoś’s theorem [4], which allows us to obtain several results. We start with an alternative proof of compactness for $\textsf{CPL}$. Then we move to the theorem saying that the set of valuations is definable iff it is closed under ultravaluations. We conclude that the set of valuations is definable iff it is compact. We also show a certain property of an ultravaluation which is in a sense analogous to the property of an ultraproduct. We use this property to show that any finite set of valuations is definable. Moreover, we give an example of a set of valuations which is not compact (definable) and use it to identify two types of sets of valuations which are not compact (definable).¹ Finally we pose an open question: to spot a set of valuations which is not of any of the two types and still is not compact (definable).

2 Background

We assume basic connectives to be $\lnot ,\wedge$. $\mathcal {L}$ is set of propositional letters (can be uncountable). Formulas are built in a standard way. We shall use p, q, r, ... as variables for elements from $\mathcal {L}$, $\varphi ,\psi ,\chi ,...$ for arbitrary formulas and $\Gamma ,\Sigma ,...$ for sets of formulas. Valuation for $\mathcal {L}$ is a function $\textsf{v}:\mathcal {L}\longrightarrow \{0,1\}$. The set of all $\mathcal {L}$ valuations will be denoted by $\{0,1\}^\mathcal {L}$. Truth in a valuation is defined in a standard way: $\textsf{v}\vDash p$ iff $\textsf{v}(p)=1$; $\textsf{v}\vDash \lnot \varphi$ iff $\textsf{v}\nvDash \varphi$; $\textsf{v}\vDash \varphi \wedge \psi$ iff $\textsf{v}\vDash \varphi$ and $\textsf{v}\vDash \psi$. If $\Sigma$ is a set of formulas we write $\textsf{v}\vDash \Sigma$ when $\textsf{v}\vDash \sigma$ for all $\sigma \in \Sigma$. Let $\textsf{K}\subseteq \{0,1\}^\mathcal {L}$, we say that a formula (set of formulas) $\varphi$ $(\Gamma )$ is satisfiable in $\textsf{K}$ iff there is $\textsf{v}\in \textsf{K}$ such that $\textsf{v}\vDash \varphi$ $(\textsf{v}\vDash \Gamma )$. If $\varphi$ $(\Gamma )$ is satisfiable in $\{0,1\}^\mathcal {L}$, we simply say that $\varphi$ $(\Gamma )$ is satisfiable. If for any $\textsf{v}\in \{0,1\}^\mathcal {L}$, $\textsf{v}\vDash \Gamma$ implies $\textsf{v}\vDash \varphi$, we write $\Gamma \vDash \varphi$.

Let us state an important fact expressing $\textsf{CPL}$’s perfect expressive power over its models, i.e. elementary equivalence in the set of valuations comes down to simple identity.

Fact 2.1

(Perfect expressive power). Let $Th(\textsf{v})=\{\varphi :\textsf{v}\vDash \varphi \}$. $Th(\textsf{v}_1)=Th(\textsf{v}_2)$ iff $\textsf{v}_1=\textsf{v}_2$.

Proof

Right to left is trivial. For left to right assume $\textsf{v}_1\ne \textsf{v}_2$. Hence $\textsf{v}_1(p)\ne \textsf{v}_2(p)$ for some p which means $Th(\textsf{v}_1)\ne Th(\textsf{v}_2)$. $\square$

This fact will turn out to be useful in later investigations. Also note that if the theory of a given valuation is included in a satisfiable set of formulas, then such set is identical to that theory, i.e.:

Fact 2.2

(Maximality of valuational theories). Let $\Sigma$ be a satisfiable set of formulas and $Th(\textsf{v})\subseteq \Sigma$ for some valuation $\textsf{v}$. Then $Th(\textsf{v})=\Sigma$.

Proof

For any $\textsf{v}$, $Th(\textsf{v})$ is maximally consistent set of formulas. $\square$

For the sake of later investigations, we need to recall the notion of an ultrafilter. For the more exhaustive exposition of filters and boolean algebras in general, an exemplary reference is Burris [1]. Let W be a non-empty set. By $\mathcal {P}(W)$ we will denote the set of all subsets W. We say that $F\subseteq \mathcal {P}(W)$ is a filter over W if the following stipulations are met: i)$W\in F$, ii)$X,Y\in F$ implies $X\cap Y\in F$, iii) if $X\in F$ and $X\subseteq Y\subseteq W$, then $Y\in F$. Filter F over W is said to be proper iff $F\ne \mathcal {P}(W)$. Here we give three equivalent definitions of an ultrafilter:

Definition 2.3

Ultrafilter U over W is a proper filter that meets any of the following equivalent condition:

• (i) for any $X\subseteq W$ we have $X\in U$ iff $W\setminus X\notin U$,

• (ii) for any $X,Y\subseteq W$ $X\cup Y\in U$ iff $X\in U$ or $Y\in U$,

• (iii) is maximal among proper filters.

Let $A\subseteq \mathcal {P}(W)$. We say that A has the finite intersection property iff the intersection of any finite number of elements of A is non-empty: $X_1\cap ...\cap X_n\ne \emptyset$ for any $n\in \mathbb {N}$ such that $X_1,...,X_n\in A$. The filter generated by A is the set $\bigcap \{F\subseteq \mathcal {P}(W): A\subseteq F\,\,\text {and}\,\,F\,\, \text {is a filter over}\,\,W\}$. It can be easily proven that such set is indeed a filter. We shall also recall the well-known fact about ultrafilters often referred to as the ultrafilter theorem. We will use this fact extensively in later investigations.

Fact 2.4

If $A\subseteq \mathcal {P}(W)$ has the finite intersection property, then it can be extended to an ultrafilter.

Proof

To sketch a proof, it is enough to note that $F=\{X\subseteq W:\text { there are }Y_1$ $,...,Y_n\in A \text { such that } Y_1\cap ...\cap Y_n\subseteq X\}$ is a filter generated by A. Also by finite intersection property, $\emptyset \notin F$. Since F is proper, it can be shown using Kuratowski–Zorn Lemma and Definition 2.3(iii) that F is included in an ultrafilter. $\square$

An ultrafilter $U\subseteq \mathcal {P}(W)$ is called principal if it is generated by a singleton set $\{w\}$ for some $w\in W$. By $[\{w\})$ we shall denote the set $\{X\subseteq W: w\in X\}$. It is routine to check that $[\{w\})$ is an ultrafilter generated by $\{w\}$.

3 Ultravaluation

Here, we will define the key concept of the paper: the ultravaluation. This notion will be incorporated in most proofs that will be presented.

Definition 3.1

(Ultravaluation). Let $(\mathsf {v_i})_{i\in I}$ be an indexed family of valuations where I is non-empty. Let $U\subseteq \mathcal {P}(I)$ be an ultrafilter over I. Ultravaluation of $(\mathsf {v_i})_{i\in I}$ under U, shortly $\mu _{i\in I}\mathsf {v_i}/U$ is given by the definition:

$$\begin{aligned} \underset{i\in I}{\mu }\mathsf {v_i}/U(p)= {\left\{ \begin{array}{ll} 1 &{}\text {if }\{i\in I: \mathsf {v_i}(p)=1\}\in U,\\ 0&{}\text {if }\{i\in I: \mathsf {v_i}(p)=1\}\notin U. \end{array}\right. } \end{aligned}$$

Obviously, ultravaluation is well defined, hence $\mu _{i\in I}\mathsf {v_i}/U\in \{0,1\}^\mathcal {L}$. In the face of the above definition, ultravaluation can be seen as a method of transforming a whole class of valuations into a single valuation. We can easily prove the analogon of Łoś’s theorem:

Lemma 3.2

(Ultravaluation lemma). For any $(\mathsf {v_i})_{i\in I}$, any ultrafilter U over I:

$$\begin{aligned} \mu _{i\in I}\mathsf {v_i}/U\vDash \varphi \text { iff } \{i\in I: \mathsf {v_i}\vDash \varphi \}\in U\end{aligned}$$

Proof

We obtain base case by definition. Let $\varphi =\lnot \psi$. $\mu _{i\in I}\mathsf {v_i}/U\vDash \lnot \psi$ iff $\mu _{i\in I}\mathsf {v_i}/U\nvDash \psi$ iff (by inductive hypothesis) $\{i\in I: \mathsf {v_i}\vDash \psi \}\notin U$ iff [by Definition 2.3(i)] $I {\setminus } \{i\in I: \mathsf {v_i}\vDash \psi \} \in U$ iff $\{i\in I: \mathsf {v_i}\nvDash \psi \} \in U$ iff $\{i\in I: \mathsf {v_i}\vDash \lnot \psi \}\in U$. Let $\varphi = \psi \wedge \chi$. $\mu _{i\in I}\mathsf {v_i}/U\vDash \psi \wedge \chi$ iff $\mu _{i\in I}\mathsf {v_i}/U\vDash \psi$ and $\mu _{i\in I}\mathsf {v_i}/U\vDash \chi$ iff (by inductive hypothesis) $\{i\in I: \mathsf {v_i}\vDash \psi \}\in U$ and $\{i\in I: \mathsf {v_i}\vDash \chi \}\in U$ iff (since it is a filter) $\{i\in I: \mathsf {v_i}\vDash \psi \} \cap \{i\in I: \mathsf {v_i}\vDash \psi \}=\{i\in I: \mathsf {v_i}\vDash \psi \ \wedge \chi \} \in U$. $\square$

Let us state an important property of ultravaluations, which says that an ultravaluation of an indexed family of valuations, whose underlying set is finite, is already included in that set.

Theorem 3.3

If $(\mathsf {v_i})_{i\in I}$ is such that for each $i\in I$ $\mathsf {v_i}$ is one of the valuations from the finite set $\{\textsf{v}_1,...,\textsf{v}_m\}$, then $\mu _{i\in I}\mathsf {v_i}/U= \mathsf {v_i}$ for some $i\in I$.

Proof

Let $(\mathsf {v_i})_{i\in I}$ be such that for each $i\in I$ $\mathsf {v_i}\in \{\textsf{v}_1,...,\textsf{v}_m\}$. Let U be an ultrafilter over I. For each $k\le m$, let $S_k=\{i\in I: \mathsf {v_i}=\textsf{v}_k\}$. $S_1\cup ...\cup S_m=I\in U$, hence from Definition 2.3(ii) $S_k\in U$ for some $k\le m$. For any p such that $\textsf{v}_k(p)=1$, we have $S_k\subseteq \{i\in I: \mathsf {v_i}(p)=1\}\in U$, hence $\mu _{i\in I}\mathsf {v_i}/U(p)=1$. For any q such that $\textsf{v}_k(q)=0$ we have $S_k\cap \{i\in I: v_i(q)=1\}=\emptyset$, hence $\{i\in I: v_i(q)=1\}\notin U$ (since otherwise U would be improper filter) which means $\mu _{i\in I}\mathsf {v_i}/U(q)=0$. We conclude that $\mu _{i\in I}\mathsf {v_i}/U=\textsf{v}_k$. $\square$

As a corollary of the previous theorem, we can state that an ultravaluation of the same valuations—the notion analogous to an ultrapower—is always identical to a single component.

Corollary 3.4

Let $(\mathsf {v_i})_{i\in I}$ be such that for each $j,k\in I$, $\mathsf {v_i}=\textsf{v}_j$. Then $\mu _{i\in I}\mathsf {v_i}/U=\mathsf {v_i}$ for any $i\in I$.

Note that Corollary 3.4 can be also proven directly from Fact 2.1 and Lemma 3.2. To see that, let $i\in I$. $\mathsf {v_i}\vDash \varphi$ iff $\{i\in I: \mathsf {v_i}\vDash \varphi \}=I\in U$ iff (by Lemma 3.2) $\mu _{i\in I}\mathsf {v_i}/U\vDash \varphi$. Hence $Th(\mathsf {v_i})=Th(\mu _{i\in I}\mathsf {v_i}/U)$. By Fact 2.1, $\mathsf {v_i}=\mu _{i\in I}\mathsf {v_i}/U$.

Theorem 3.5

Let $j\in I$. Let $U=[\{j\})$ be a principal ultrafilter. Then $\mu _{i\in I}\mathsf {v_i}/U=\textsf{v}_j$.

Proof

Assume $v_j(p)=1$. Then $j\in \{i\in I: \mathsf {v_i}(p)=1\}\in U$. Hence $\mu _{i\in I}\mathsf {v_i}/U(p)=1$ by Lemma 3.2. For the other direction assume $\mu _{i\in I}\mathsf {v_i}/U(p)=1$. This means by Lemma 3.2 that $\{i\in I: \mathsf {v_i}(p)=1\}\in U$. But for each $X\in U$, $j\in X$, so $\textsf{v}_j(p)=1$. $\square$

4 Compactness and Definability

Compactness of $\textsf{CPL}$ is usually proven as a corollary of completeness theorem with respect to some finite deductive aparatus. One of the exceptions is [3, pp. 7–8], where the method is purely model-theoretic. Here we will present an alternative model-theoretic proof, which uses the construct of an ultravaluation. It is important to note that our method covers both countable and uncountable languages.

Theorem 4.1

(Compactness of $\textsf{CPL}$). Let $\Sigma$ be a set of formulas. If all finite subsets of $\Sigma$ are satisfiable, then $\Sigma$ is satisfiable.

Proof

Let I be a set of all finite subsets of $\Sigma$ and $(\mathsf {v_i})_{i\in I}$ be such that $\mathsf {v_i}\vDash i$ for each $i \in I$. Let $\overline{\varphi }= \{i\in I: \mathsf {v_i}\vDash \varphi \}$ for each $\varphi \in \Sigma$. Notice that $\{\overline{\varphi }: \varphi \in \Sigma \}$ has the finite intersection property, since $\{\varphi _1,..., \varphi _n\} \in \overline{\varphi _1}\cap ...\cap \overline{\varphi _n}$. By Fact 2.4 it can be extended to an ultrafilter. Let U be such an ultrafilter. For each $\varphi \in \Sigma$, $\overline{\varphi }= \{i\in I: \mathsf {v_i}\vDash \varphi \} \in U$, hence by Lemma 3.2$\mu _{i\in I}\mathsf {v_i}/U\vDash \varphi$ for each $\varphi \in \Sigma$, which means $\Sigma$ is satisfiable. $\square$

In order to prove further theorems, we need some definitions. Let us start with the closure under ultravaluations—the key concept which will allow us to obtain further results.

Definition 4.2

(Closure under ultravaluations). Let $\textsf{K}$ be a set of valuations. $\textsf{K}$ is closed under ultravaluations iff for any I, any ultrafilter U over I, if for each $i\in I$, $\mathsf {v_i}\in \textsf{K}$, then $\mu _{i\in I}\mathsf {v_i}/U\in \textsf{K}$.

We also need to generalise the notion of compactness for arbitrary sets of valuations.

Definition 4.3

(Relative compactness). Let $\textsf{K}$ be a set of valuations. We say that $\textsf{K}$ is compact iff for any set of formulas $\Sigma$, if each finite subset of $\Sigma$ is satisfiable in $\textsf{K}$, then $\Sigma$ is satisfiable in $\textsf{K}$.

The last definition states what does it mean for a given set of valuations to be definable.

Definition 4.4

(Definability of a set of valuations) Let $\textsf{K}$ be a set of valuations. $\textsf{K}$ is definable iff there is a set of formulas $\Gamma$ such that for any valuation $\textsf{v}$, $\textsf{v}\vDash \Gamma$ iff $\textsf{v}\in \textsf{K}$.

Now we can move on to our key results. Starting with relative compactness:

Theorem 4.5

(Relative compactness). Let $\textsf{K}$ be a set of valuations. $\textsf{K}$ is compact iff $\textsf{K}$ is closed under ultravaluations.

Proof

For right to left we reason as in Theorem 4.1. For the opposite direction, assume $\textsf{K}$ is not closed under ultravaluations. Let $\mathsf {v_i}\in \textsf{K}$ for each $i\in I$ and $\mu _{i\in I}\mathsf {v_i}/U\notin \textsf{K}$ for some ultrafilter U. Let $\Sigma = Th(\mu _{i\in I}\mathsf {v_i}/U)$. From Lemma 3.2 for each $\varphi \in \Sigma$ we have $\{i \in I: \mathsf {v_i}\vDash \varphi \} \in U$. Let $\Gamma = \{\varphi _1,...,\varphi _n\} \subseteq \Sigma$. Observe that $\{i \in I: \mathsf {v_i}\vDash \varphi _1\} \cap ...\cap \{i \in I: \mathsf {v_i}\vDash \varphi _n\} = \{i \in I: \mathsf {v_i}\vDash \Gamma \} \in U$, which means $\{i \in I: \mathsf {v_i}\vDash \Gamma \} \ne \emptyset$. Hence every finite subset of $\Sigma$ is satisfiable in $\textsf{K}$. Assume for reductio that $\Sigma$ is satisfiable in $\textsf{K}$. Let $\textsf{v}\in \textsf{K}$ and $\textsf{v}\vDash \Sigma$. Hence $Th(\mu _{i\in I}\mathsf {v_i}/U)\subseteq Th(\textsf{v})$ which by Fact 2.2 means $Th(\mu _{i\in I}\mathsf {v_i}/U)=Th(\textsf{v})$. By Fact 2.1$\textsf{v}=\mu _{i\in I}\mathsf {v_i}/U\notin \textsf{K}$—contradiction, hence $\Sigma$ is not satisfiable in $\textsf{K}$. $\square$

It turns out that the similar result holds when we investigate definability.

Theorem 4.6

(Definability). $\textsf{K}$ is definable iff $\textsf{K}$ is closed under ultravaluations.

Proof

For left to right assume $\Gamma$ defines $\textsf{K}$. Let $(\mathsf {v_i})_{i \in I}$ be such that, for any $i \in I$, $\mathsf {v_i}\in \textsf{K}$. Let U be an ultrafilter over I. Note that for all $\varphi \in \Gamma$ $\{i \in I: \mathsf {v_i}\vDash \varphi \} = I \in U$, so by Lemma 3.2$\mu _{i\in I}\mathsf {v_i}/U\vDash \varphi$, for any $\varphi \in \Gamma$ which means $\mu _{i\in I}\mathsf {v_i}/U\in \textsf{K}$. For the opposite direction assume $\textsf{K}$ is closed under ultravaluations. Let $\Gamma = \bigcap \{Th(\textsf{v}): \textsf{v}\in \textsf{K}\}$. We will show that $\Gamma$ defines $\textsf{K}$. Obviously if $\textsf{v}\in \textsf{K}$ then $\textsf{v}\vDash \Gamma$ for any $\textsf{v}$. For the opposite direction let $\textsf{v}$ be such that $\textsf{v}\vDash \Gamma$. Observe that each finite subset of $Th(\textsf{v})$ is satisfiable in $\textsf{K}$. To see that assume for reductio that there is finite $\Sigma = \{\varphi _1,...,\varphi _n\} \subseteq Th(\textsf{v})$ which is not satisfiable in $\textsf{K}$. Hence, for each $\textsf{v}' \in \textsf{K}$ $\textsf{v}' \vDash \lnot \varphi _1 \vee ...\vee \lnot \varphi _n$, which means that $\lnot \varphi _1 \vee ...\vee \lnot \varphi _n \in \Gamma$. So $\textsf{v}\nvDash \Sigma$—contradiction. Let I be a set of all finite subsets of $Th(\textsf{v})$ and $(\mathsf {v_i})_{i\in I}$ be such that $\mathsf {v_i}\vDash i$ for each $i \in I$. Let $\overline{\varphi }= \{i\in I: \mathsf {v_i}\vDash \varphi \}$ for each $\varphi \in Th(\textsf{v})$. Notice that $\{\overline{\varphi }: \varphi \in Th(\textsf{v})\}$ has the finite intersection property, since $\{\varphi _1,..., \varphi _n\} \in \overline{\varphi _1}\cap ...\cap \overline{\varphi _n}$. By Fact 2.4 it can be extended to an ultrafilter. Let U be such an ultrafilter. For each $\varphi \in Th(\textsf{v})$, $\overline{\varphi }= \{i\in I: \mathsf {v_i}\vDash \varphi \} \in U$, hence by Lemma 3.2$\mu _{i\in I}\mathsf {v_i}/U\vDash \varphi$ for each $\varphi \in Th(\textsf{v})$. By Facts 2.2 and 2.1$\textsf{v}= \mu _{i\in I}\mathsf {v_i}/U\in \textsf{K}$. This means that $\Gamma$ defines $\textsf{K}$. $\square$

4.1 Corollaries

Now we will state some important corollaries of the above proven results. As a consequence of Theorems 4.1 and 4.6, we obtain the following theorem.²

Corollary 4.7

$\textsf{K}$ is definable iff $\textsf{K}$ is compact.

By Theorems 3.3, 4.1 and Corollary 4.7 we obtain:

Corollary 4.8

If $\textsf{K}$ is finite, then $\textsf{K}$ is definable/compact.

We can also construct the defining set of formulas for finite set of valuations.

Fact 4.9

Let $\textsf{K}=\{\textsf{v}_1,...,\textsf{v}_n\}$ be a finite set of valuations. $\Gamma =\{\varphi _1\vee ...\vee \varphi _n:\textsf{v}_1\vDash \varphi _1,...,\textsf{v}_n\vDash \varphi _n\}$ defines $\textsf{K}$.

Proof

Let $\textsf{K}$ and $\Gamma$ be as above. Obviously if $\textsf{v}\in \textsf{K}$, then $\textsf{v}=\mathsf {v_i}$, $1\le i\le n$, so $\textsf{v}\vDash \varphi _i$ for each $\varphi _1\vee ...\vee \varphi _n\in \Gamma$, which means $\textsf{v}\vDash \Gamma$. For the opposite direction assume $\textsf{v}\notin \textsf{K}$. Hence $\textsf{v}\ne \mathsf {v_i}$ for each $1\le i\le n$. By Fact 2.1, there are $\varphi _1,...,\varphi _n$ such that $\mathsf {v_i}\vDash \varphi _i$ and $\textsf{v}\nvDash \varphi _i$ for each $1\le i\le n$. This means $\textsf{v}\nvDash \varphi _1\vee ...\vee \varphi _n$, so $\textsf{v}\nvDash \Gamma$. $\square$

Note that a set $\Gamma$ defined as above can be in many cases a proper subset of $\bigcap \{Th(\textsf{v}):\textsf{v}\in \textsf{K}\}$. For this reason an additional remark may appear valuable: that for any definable set of valuations $\textsf{K}$, $\bigcap \{Th(\textsf{v}):\textsf{v}\in \textsf{K}\}$ is the biggest (inclusion-wise) set of formulas which define $\textsf{K}$. To see that for any definable set of valuations $\textsf{K}$, $\bigcap \{Th(\textsf{v}):\textsf{v}\in \textsf{K}\}$ defines $\textsf{K}$, note that $\textsf{K}$ is closed under ultravaluation by Theorem 4.6 and then simply repeat the reasoning from the right-to-left part of the proof of Theorem 4.6. To prove that $\Gamma \subseteq \bigcap \{Th(\textsf{v}):\textsf{v}\in \textsf{K}\}$ for any $\Gamma$ which defines $\textsf{K}$, note that if $\varphi \notin \bigcap \{Th(\textsf{v}):\textsf{v}\in \textsf{K}\}$, then $\textsf{v}\nvDash \varphi$ for some $\textsf{v}\in \textsf{K}$, hence $\varphi \notin \Gamma$. Furthermore, we can prove that $Cn(\Gamma ):=\{\varphi :\Gamma \vDash \varphi \}=\bigcap \{Th(\textsf{v}):\textsf{v}\in \textsf{K}\}$ for any $\Gamma$ that defines $\textsf{K}$. Assume $\varphi \in Cn(\Gamma )$. Therefore $\Gamma \vDash \varphi$. Since $\Gamma$ defines $\textsf{K}$, $\textsf{v}\in \textsf{K}$ implies $\textsf{v}\vDash \varphi$. Hence, $\varphi \in \bigcap \{Th(\textsf{v}):\textsf{v}\in \textsf{K}\}$. For the opposite inclusion, let $\varphi \notin Cn(\Gamma )$. In consequence, $\Gamma \nvDash \varphi$. Again using the fact that $\Gamma$ defines $\textsf{K}$, $\textsf{v}\in \textsf{K}$ and $\textsf{v}\nvDash \varphi$, so $\varphi \notin \bigcap \{Th(\textsf{v}):\textsf{v}\in \textsf{K}\}$.

The last corollary that we are going to present formulates necessary and sufficient conditions for a set of valuations to be definable by a single formula. By $\textsf{K}^-$ we shall denote $\{0,1\}^\mathcal {L}\setminus \textsf{K}$.

Theorem 4.10

$\textsf{K}$ is definable by a single formula iff both $\textsf{K}$ and $\textsf{K}^-$ are closed under ultravaluations.

Proof

For left to right assume that $\varphi$ defines $\textsf{K}$. Hence, $\lnot \varphi$ defines $\textsf{K}^-$. By Theorem 4.6 both $\textsf{K}$ and $\textsf{K}^-$ are closed under ultravaluations. For the opposite direction let both $\textsf{K}$ and $\textsf{K}^-$ be closed under ultravaluations. By Theorem 4.6 both are definable. Let $\Gamma$ define $\textsf{K}$ and $\Sigma$ define $\textsf{K}^-$. Note that $\Gamma \cup \Sigma$ is not satisfiable since otherwise there would be $\textsf{v}$ such that $\textsf{v}\in \textsf{K}$ and $\textsf{v}\in \textsf{K}^-$. By compactness Theorem 4.1 there are finite $\Gamma '\subseteq \Gamma$, $\Sigma '\subseteq \Sigma$ such that $\Gamma ' \cup \Sigma '$ is not satisfiable. If $\Gamma '$ is empty, then $\Sigma '$ and also $\Sigma$ is unsatisfiable, which means $\textsf{K}^-$ is empty. Then $\lnot (p \wedge \lnot p)$ defines $\textsf{K}= \{0,1\}^\mathcal {L}$. Similarly, if $\Sigma '$ is empty, then $\lnot p \wedge p$ defines $\textsf{K}= \emptyset$. Assume that $\Gamma ' = \{\gamma _1,...,\gamma _n\} \ne \emptyset$ and $\Sigma '= \{\sigma _1,...,\sigma _m\} \ne \emptyset$ are separately satisfiable. Note that $\gamma _1 \wedge ...\wedge \gamma _n \vDash \lnot \sigma _1 \vee \cdots \vee \lnot \sigma _m$. We claim that $\gamma _1 \wedge \cdots \wedge \gamma _n$ defines $\textsf{K}$. For if $\textsf{v}\vDash \gamma _1 \wedge \cdots \wedge \gamma _n$, then $\textsf{v}\vDash \lnot \sigma _1 \vee \cdots \vee \lnot \sigma _m$ which means $\textsf{v}\notin \textsf{K}^-$, thus $\textsf{v}\in \textsf{K}$. Conversely, if $\textsf{v}\in \textsf{K}$, then $\textsf{v}\vDash \Gamma$, so $\textsf{v}\vDash \gamma _1 \wedge \cdots \wedge \gamma _n$. $\square$

4.2 Counterexamples

Let us provide an example of a set of valuations which is not definable (compact). By $\mathbb {N}$ we shall denote the set of natural numbers.

Example 4.11

Let $\mathcal {L}=\{p_n:n\in \mathbb {N}\}$. Let $\textsf{K}=\{\textsf{v}_n:n\in \mathbb {N}\}\subseteq \{0,1\}^\mathcal {L}$ be such that for each $\textsf{v}_n$, $\textsf{v}_n(p_n)=1$ and $v_n(p_k)=0$ for any $k\ne n$. $\textsf{K}$ is not definable (compact).³

Proof

Let $G=\{\mathbb {N}\setminus \{n\}:n\in \mathbb {N}\}$. $\emptyset \ne \mathbb {N}\setminus \{n_1,...,n_k\}=\mathbb {N}\setminus \{n_1\}\cap ...\cap \mathbb {N}\setminus \{n_k\}$, hence G has finite intersection property. By Fact 2.4, it can be extended to an ultrafilter. Let U be such an ultrafilter. Observe that for any $n\in \mathbb {N}$, $\{n\}\notin U$, since $\mathbb {N}\setminus \{n\}\in U$. For each m, $\{n\in \mathbb {N}:\textsf{v}_n(p_m)=1\}=\{m\}\notin U$, hence $\mu _{n\in \mathbb {N}}\textsf{v}_n/U(p_m)=0$. For any m, $\mu _{n\in \mathbb {N}}\textsf{v}_n/U\ne \textsf{v}_m$, because $\textsf{v}_m(p_m)=1\ne \mu _{n\in \mathbb {N}}\textsf{v}_n/U(p_m)$, so $\mu _{n\in \mathbb {N}}\textsf{v}_n/U\notin \textsf{K}$. By Theorems 4.6 (4.5), we conclude that $\textsf{K}$ is not definable (compact). $\square$

We can generalise our example to some kind of a theorem. To do that, we need to define the notion of the finite satisfaction property.

Definition 4.12

We say that a set of valuations $\textsf{K}$ has the finite satisfaction property iff for each $p\in \mathcal {L}$, a set $\{\textsf{v}\in \textsf{K}:\textsf{v}(p)=1\}$ is finite.

Let the zero valuation $\textsf{v}^\bot$ be such that for each p, $\textsf{v}^\bot (p)=0$. Analogously, we shall refer to $\textsf{v}^\top$: $\textsf{v}^\top (p)=1$ for each p—as the top valuation.

Theorem 4.13

Let $\textsf{K}$ be a set of valuations such that $\textsf{v}^\bot \notin \textsf{K}$. If $\textsf{K}$ has an infinite subset $\textsf{K}'$ which has the finite satisfaction property, then $\textsf{K}$ is not definable (compact).

Proof

We reason in an analogous way. Let $(\textsf{v}_n)_{n\in \mathbb {N}}\subseteq \textsf{K}'$. Let U be an ultrafilter generated from $\{\mathbb {N}\setminus \{n\}:n\in \mathbb {N}\}$. For each p set $\mathbb {P}=\{n\in \mathbb {N}:\textsf{v}_n(p)=1\}$ is finite. So $\mathbb {N}{\setminus } \mathbb {P}=\bigcap _{m\in \mathbb {P}}(\mathbb {N}{\setminus }\{m\})\in U$, so $\mathbb {P}$ is not included in U for each p, hence $\mu _{n\in \mathbb {N}}\textsf{v}_n/U=\textsf{v}^\bot \notin \textsf{K}$. $\square$

Dually we can define valuations $\textsf{K}$ which have the finite falsification property, i.e. each propositional atom is false in a finite set of valuations $\textsf{K}'\subseteq \textsf{K}$.

Theorem 4.14

Let $\textsf{K}$ be a set of valuations without the top valuation. If $\textsf{K}$ has an infinite subset $\textsf{K}'$ with the finite falsification property, then $\textsf{K}$ is not definable (compact).

Proof

Completely analogous as in Theorem 4.13. $\square$

We shall end with an open question: to spot (if there is such) an undefinable set of valuations which is neither top-free with an infinite subset that has finite satisfaction property, nor zero-free with an infinite subset that has finite falsification property.