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4Логические исследования 2023. Т. 29. № 1. С. 147-163 УДК 16
Logical Investigations 2023, Vol. 29, No. 1, pp. 147-163 DOI: 10.21146/2074-1472-2023-29-1-147-163
Weijun Shi
A Topological-algebraic Approach to the Compactness Theorem of Classical Logic
Weijun Shi
Department of Philosophy, School of Humanities and Arts, Xidian University,
266 Xinglong Section of Xifeng Road, 710126, Shaanxi, the People's Republic of China.
E-mail: shiweijun@xidian.edu.cn
Abstract: There are some methods of proof of the compactness theorem for classical logic which bypass the completeness theorem. Among them are the purely topological one, the purely algebraic one, and the hybrid one. These methods make essential use of either Ty-chonoff's Theorem, the concept of ultraproducts or the concept of Cantor sets as topological spaces. Instead of these conceptual tools, the paper provides the theorem with a method of proof that appeals to the concept of Stone spaces of Boolean algebras. In connection with a classical logical system (a propositional calculus or a predicate calculus), the method consists of five components. Firstly, the problem of the compactness of the logical system is reduced to that of the compactness of some topological space. Secondly, what is called the Lindenbaum algebra of the system is set up, which is in fact a Boolean algebra. Thirdly, it has to be shown that the Stone space of the Boolean algebra is compact. Fourthly, the set of sentences whose equivalent classes are members of the Stone space is shown to be satisfiable or simultaneously true. Finally, a homeomorphism is constructed between the topological space and the compact Stone space. Additionally, the method admits of a natural generalisation to the proof of the compactness theorem for modal logic.
Keywords: Topology, compactness, Boolean algebras, Stone spaces, classical logic
For citation: Shi WJ. "A Topological-algebraic Approach to the Compactness Theorem of Classical Logic", Logicheskie Issledovaniya / Logical Investigations, 2023, Vol. 29, No. 1, pp. 147-163. DOI: 10.21146/2074-1472-2023-29-1-147-163
1. Introduction
The compactness theorem is the cornerstone of model theory. In connection with classical logic, this theorem is usually proven with the help of the completeness theorem, but the latter turns out to be unnecessary to the proof of the former. There are several methods of proving the former in the literature which dispense with the latter, for example, the renowned Henkin's method [Marker, 2003, pp. 35-39], and the less well-known method [Cowen, 1970], based on a combinatorial lemma [Rado, 1949]. Besides them, there are a purely algebraic one [Bell et al., 2006, p. 102], [Paseau, 2022, sec. 4c], a purely topological
© Shi WJ., 2023
one [Bell et al., 2006, p. 46], [Paseau, 2022, sec. 5], [Beth, 1951], and a hybrid one [Frayne et al., 1962], which uses the algebraic notion of ultraproducts and the topological notion of Stone spaces.1
The paper shall provide a proof of the compactness theorem for classical logic which falls into the category of the just-mentioned hybrid method. However, our proof is distinguished from the one using this hybrid method in [Frayne et al., 1962] by the fact that it makes no use of the algebraic notion of ul-traproducts. Moreover, it is also distinguished from the purely topological proofs in [Bell et al., 2006, p. 46], [Paseau, 2022, sec. 5], [Beth, 1951] by the fact that it appeals neither to Tychonoff's Theorem (or some principle weaker than the theorem) nor to the notion of compact Cantor set (as used in topology). In connection with classical logic, our proof uses the conceptual tool of Stone spaces of Boolean algebras instead. The proof can be adapted for use as a proof of the compactness theorem for modal logic, too, although I shall not offer a detailed proof of this. What runs through all of these proofs is the conceptual tool and this is what makes them stand out among others. Hence, our proofs are in themselves interesting from a methodological point of view although the results to be proven are altogether not completely original.
The paper is organised as follows. A number of preliminary notions and results concerning deductive systems, Boolean algebras, and topological spaces are stated in section 2 and section 3. I shall provide the compactness theorem of propositional logic with a proof using the above-mentioned conceptual tool in section 4, and that of (first-order) predicate logic in section 5. In section 6, besides some conclusive remarks, I shall give an outline of how to generalise the method used in our proof to modal logic.
2. Algebraic aspects
Suppose (X, is a partially ordered pair, where X is a nonempty set and ^ is a binary partial relation on X, i.e. a reflexive, antisymmetric, and transitive relation. For any subset Y of X, we use sup(Y) to denote an element in X that is the supremum (the least upper bound) of Y if there is any. Similarly, if there is an element in X that is the infimum (the greatest lower bound) of Y, we use inf (Y) to denote it.
Definition 1. (X, is a lattice if and only if for each two element subset Y of X, sup(Y) and inf (Y) exist.
1For the development and history of the compactness theorems, see [Dawson, 1993], and for a detailed investigation of some of these methods and the related philosophical significance, see [Paseau, 2011].
Given a lattice (X, and any Y = {x, y} C X, let x U y, called the meet of x and y, and x n y, called the join of x and y, be the supremum and infimum of Y, respectively. That is to say, x U y = sup(Y) and x n y = inf (Y).
For a partially ordered pair (X, if there is an element x e X such that y ^ x for all y e X, then x is said to be a maximum of X. Similarly, if there is an element x e X such that x ^ y for all y e X, then x is said to be a minimum of X. We use 1 and 0 to denote the maximum and minimum of X, respectively, if they exist. As is easy to show, if a lattice has a maximum, then it is unique; in the same vein, if it has a minimum, it is also unique.
Definition 2. A lattice (X, is complemented if and only if
(1)X has a maximum 1 and a minimum 0;
(2) for any x e X, there is some y e X such that x U y = 1 and x n y = 0.
Such an element y is called a complement of x. A member of a complemented lattice does not necessarily have a unique complement. However, their complements are indeed unique if the lattice is distributive:
Definition 3. A lattice (X, is distributive if and only if for any x, y, z e X,
(1) (x U y) n z = (x n z) U (y n z).
(2) (x n y) U z = (x U z) n (y U z).
In fact, if a lattice satisfies clause (1) of Definition 3, then it satisfies clause (2) thereof and verse versa. We use x* to denote the unique complement of x in a distributive complemented lattice. Now a Boolean algebra is a distributive complemented lattice.
Besides the notion of Boolean algebras, we also need the notions of filters, ultrafilters, and quotient algebras. Suppose we have a lattice (X, and a non-empty subset F of X.
Definition 4. 1. F is a filter in (X, if and only if
(1.1) for any x, y e X, if x, y e F, then x n y e F.
(1.2) for any x e F and any y e X, if x ^ y, then y e F. 2. F is an ultrafilter in (X, if and only if
(2.1) F is a filter in the lattice.
(2.2) for any filter G in the lattice, F is not a proper subset of G.
Any filter F in the lattice (X, is said to be proper if F = X. Similarly, any ultrafilter F in the lattice is said to be proper if F = X. From now on, whenever speaking of filters and ultrafilters, we always mean proper ones. It stands to reason that any ultrafilter in a Boolean algebra does not contain the minimum 0. For otherwise it would not be an (proper) ultrafilter. There is a lemma which tells if a filter is an ultrafilter.
Lemma 1. If F is a filter in a Boolean algebra (X, then F is an ultrafilter iff for each x £ X, either x £ F or x* £ F, but not both.
Let F be a filter in a Boolean algebra A = (X, We define a relation on the algebra A as follows:
For all x, y £ X, x y iff x n f = y n f for some f £ F.
is a congruence relation on A. Now for x £ X, let |x| be the equivalence class to which x belongs under the relation . Moreover, let A/F = {|x| : x £ X}. The set can be given a structure of Boolean algebras as follows:
| x| n | y| = | x n y| , |x| U |y| = |x U y|,
| x| * = | x* | .
This set with the structure is called the the quotient algebra of A modulo F. The function which maps each element x in X to | x| is called the canonical homomorphism of A onto A/F. The following is a lemma concerning quotient algebras to be used later:
Lemma 2. For any filter F in A, |x| = 1 in A/F, i.e.. |x| is the maximum in the quotient algebra, iff x £ F.
Define a relation ^ on the set {0,1} as follows: 0 ^ 0, 0 ^ 1, and 1 ^ 1. 2 = ({0,1}, is a Boolean algebra. Two Boolean algebras (X, and (Y, are called to be isomorphic if there is a bijection f from X to Y such that f (x n y) = f (x) n f (y), f (x U y) = f (x) U f (y), and f (x*) = f (x)*. Now we can formulate a very important theorem to be used in our proof of Theorem 7:
Theorem 1. For any filter F in A, A/F and 2 are isomorphic iff F is an ultrafilter.
For a detailed proof, see [Bell et al., 2006, p. 20].
3. Topological aspects
We have already introduced all required concepts and propositions in Boolean algebras. Now we need some concepts in topology. Let X be a set and T be a family of subsets of X.
Definition 5. T is a topology for X if T has the following three properties:
(1) X £ T and 0 £ T.
(2) Any union of sets in T belongs to T, i.e. for any subset T0 of T, IJ U £ T.
u er0
(3) Any intersection of two sets in T belongs to T, i.e. U1 n U2 £ T for any Ui,U2 £ T.
The pair (X, T) is called a topological space. For convenience, we refer to X as a topological space when its topology is clear from the context. The sets in T are called open sets, and any subset Y of X is called a closed set if the complement of Y in X is in T, i.e. X \ Y e T.
Let (X, T) be a topological space and B be a family of open subsets of X.
Definition 6. B is base for the topology T of X if every open subset of X is a union of sets in B, i.e. for every U e T, there is a subset B0 of B such that
u = u v .
V eB0
An appeal can be made to the following theorem to judge if a given family of subsets of X is a base for some topology of X [Gamelin et al., 1999, p. 70].
Theorem 2. A family X of subsets of the set X is a base for some topology of X if and only if X has the following two properties:
(1) Each member x e X lies in at least one set in X.
(2) For any x e X and any U, V e X, if x e U n V, then there is some W e X such that x e W C U n V.
If X satisfies these two clauses, then it uniquely determines a topology for X, that is, there is some topology for X for which X is a base.
Separation properites are used to measure if there are enough open sets in a topological space. For our purpose, we only need the notion of T2-spaces or Hausdorff spaces.
Definition 7. A topological space (X, T) is a Hausdorff space iff for each pair of distinct points x, y e X, there are disjoint open sets U, V e T (i.e. UnV = 0) such that x e U and y e V.
A subset T0 of the topology T is called an open cover of X if X is the union
of all sets in the subset, i.e. X = IJ V. An open cover T0 of X is a finite
V eT0
open cover if it has only finite members.
Definition 8. A topological space (X, T) is compact iff, if X has an open cover, then X has a finite open cover.
Definition 9. A topological space (X, T) is a Boolean space iff it is a compact Hausdorff space with a base of open and closed sets.
There are two theorems about compact topological spaces to be used later in section 4 and 5. One of them is as follows:
Theorem 3. (X, T) is compact if and only if any collection of its closed sets having the finite intersection property has nonempty intersection, i.e. for any X = {U : X \ U £ T}, if the intersection of every finite subset of X is not empty, then the intersection of X is not empty.
The other is that compactness is a topological property. To articulate it, we need the notion of homeomorphism:
Definition 10. Given two topological spaces (X,T) and (Y,T7), a function f from X to Y is a homeomorphism iff it has the following properties:
(1) f is a bijection.
(2) For any subset U of X, U £ T iff f [U] £ T', where f [U] = (J {f (x)}.
xeu
With this notion, we can put the just-mentioned theorem as follows:
Theorem 4. If there is a homeomorphism from a topological space X to another topological space Y, then X is compact iff Y is compact [Gamelin et al., 1999, p. 79].
The notion of Stone spaces is a combination of topological and algebraic ones. Suppose A = (X, is a Boolean algebra. Let S(A) be the set of all ultrafilters in A. For x £ X, let u(x) = {U £ S(A) : x £ U}, that is, u(x) is the set of all ultrafilters in A containing x. So u[X] is the set of all u(x), that is, u[X] = {u(x) : x £ X}. According to Theorem 2, there is a topology Ts for S(A) such that u[X] is a base for the topology. S(A) is called the Stone space of the Boolean algebra A. Regarding the relation between Boolean algebras and Stone spaces, we have the following theorem:
Theorem 5. Given any Boolean algebra A, its Stone space S(A) is a Boolean space.
The Stone space is a Hausdorff space and is compact. For a proof of this fact, see [Bell et al., 2006, p. 26], [Dunn et al., 2001, p. 435]. Hence, it is, by Definition 9, a Boolean space. This theorem plays a pivotal role in our proofs to come in the following two sections.
4. The compactness of propositional logic
Let L be a language of propositional logic. We assume that it can have arbitrarily many k (any cardinal number larger than 0) propositional variables. It has - and ^ as its only primitive connectives and the other familiar logical connectives are introduced in the canonical manner. Let Sent be the set of all
sentences (well-defined formulas) of L. We set up a propositional calculus SC with the following as its axioms:
0 ^ (^ ^ 0),
(x ^ (0 ^ ^ ((x ^ 0) ^ (x ^ m
(-0 ^ ((-0 ^ ^ ^ 0),
where 0, e Sent. The only rule of reference of SC is Modus Ponens. The notions of proof, satisfiability, and theoremhood are defined in the usual way. The notation 'h 0' is used to mean that 0 is a theorem of SC.
If v is a valuation of the set of propositional variables, i.e. a function from the set to the set {0,1} of truth-values, then it assigns a truth-value to any sentence of L. In other words, it can be extended to a function from Sent to {0,1}. Let Vk = {v : v : Sent ^ {0,1}} be the set of all such functions.2
Now let us equip the set Vk with a topology as follows. For any 0 e Sent, define U< = {v e Vk : v(0) = 1}. So Bk = {U< : 0 e Sent} is the set of all such U<. It can easily be shown that Bk satisfies the conditions listed in Theorem 2. Firstly, for any v e Vk, there is some sentence 0 such that v(0) = 1; hence, v e U<. Secondly, if x e U< n U^ then x e C U< n U^. Hence, there is a topology Tk for Vk with Bk as its base .
The compactness theorem of SC is this: a set £ of sentences in Sent is satisfiable iff each finite subset of £ is satisfiable. The direction from left to right is obvious. Now the other direction is equivalent to the statement that if £ is unsatisfiable, then there is an unsatisfiable finite subset £fm. £ is unsatisfiable iff for every v e Vk there is some 0 e £ such that v(-0) = 1. v(-0) = 1 iff v e U-,<. If follows that £ is unsatisfiable iff for every v e Vk there is some 0 e £ such that v e U-<. Hence, £ is unsatisfiable iff {U-< : 0 e £} is an open cover of Vk. Thus, to show the compactness of SC, it suffices to prove the following statement:
If {U-< : 0 e £} is an open cover of Vk, then there is a finite subset £fm of £ such that {U-< : 0 e £fm} is an open cover of Vk.
Now clearly, if the topological space Vk is compact, then by Definition 8, the antecedent of the statement implies its consequence. So the task of showing the compactness of SC boils down to showing the following:
Theorem 6. The topological space (Vk, Tk) is compact.
One topological way of proving Theorem 6 is as presented in [Bell et al., 2006, p. 46], [Paseau, 2022, sec. 5].3 The gist of such a proof is as follows. Since L has k propositional variables, i.e. the cardinality of the set Var of
2The notation 'v : Sent ^ {0,1}' means that v is a function from Sent to {0,1}.
3The topological proof in [Beth, 1951] uses the notion of compact Cantor sets.
all propositional variables is k, the cardinality of Sent is the same as that of lJ{Var^ : £ < w}, which is k.4 Now we equip the set {0,1} with the discrete topology consisting of all subsets of {0,1}. Trivially, the topological space is compact. Now consider the set {0,1}k with the product topology [Gamelin et al., 1999, pp. 91-92]. This topological space is compact according to Ty-chonoff's Theorem. Moreover, Vk and {0,1}k are homeomorphic and therefore, Vk is compact. So Theorem 6 gets proven. The proof is also viable if Tychon-off's Theorem is replaced by a weaker version of it, namely, that the product of compact Hausdorff spaces is also compact.
Let us provide a proof of Theorem 6 without using Tychonoff's Theorem or the just-mentioned weaker version. To do this, we have to construct a Boolean algebra by imposing a Boolean structure on the set Sent of formulas of L. Define a relation ~ on the set as follows:
Definition 11. For any 0,0 £ Sent, 0 ~ 0 iff h 0 o 0, i.e. 0 o 0 is a theorem of SC.
This relation is an equivalence relation. Let [0] be the equivalence class of 0 under the relation, i.e. [0] = {0 £ Sent : h 0 o 0}. Moreover, let Sent~ = {[0] : 0 £ Sent} be the set of equivalence classes of all sentences of L. We shall partially order the set by defining a partial relation ^ on it.
Definition 12. For all 0,0 £ Sent, [0] ^ [0] iff h 0 ^ 0.
The ordered pair (Sent~, is called the Lindenbaum algebra of SC. According to Definition 1, Definition2, Definition3, and the properties of SC, (Sent~, is a Boolean algebra [Bell et al., 2006, pp. 41-42]. With respect to this algebra, the following lemma holds:
Lemma 3. For any member [0] and [0] in Sent~, the following hold:
n [0] = [0 A 0]. U [0] = [0 V 0].
* = [-0]. = 1 iff h 0. = 0 iff h-0.
Let A be (Sent~, As said before, (S(A), Ts) is then the Stone space of A, whose base is u[Sent~] = {u([0]) : [0] £ Sent~}, where u([0]) = {U £ S(A) : [0] £ U}. The Stone space is, by Theorem 5, a Boolean space, which is, by Definition 9, compact.
4Here u is the first limit ordinal and £ is any ordinal number less than it. Var^ is the set of all sequences whose domain is i.e. the set of all functions from £ to Var.
Now the compactness of SC will be shown if the following theorem is true: Theorem 7. There is a homeomorphism from (Vk, Tk) to (S(A), Ts).
Proof. Let h : Vk ^ S (A) be such that for each v £ Vk, h(v) = {[0] : v(0) = 1}. We claim that h is a bijection.
Firstly, we show that it is a function. For each v, it can be shown that h(v) is an ultrafilter in A. By Lemma 1, it suffices to show that for each member [0] £ Sent~, either [0] £ h(v) or [0]* £ h(v) but not both. By Lemma 3, [0]* = [-0]. Since either v(0) = 1 or v(-0) = 1 but not both, either [0] £ h(v) or [0]* £ h(v) but not both.
Secondly, we show that h is one-one. If vi = v2, then there is some 0 such that v1 (0) = v2 (0). Hence, h(v1) = h(v2).
Finally, we show that h is onto. Each U £ S(A) is an ultrafilter, so by Theorem 1, there is an isomorphism f from the quotient algebra A/U of A to 2; moreover, there is a homomorphism g from A to A/U such that g([0]) = | [0] |, i.e. the equivalence class of [0] under the relation ~u; so there is a homomorphism f o g (the composition function of f and g) from A to {0,1}. We claim that (*) is true:
(*) For each ultrafilter U, there is a valuation vU such that h(vU) = {[0] : vu (0) = 1} = U.
Let vU be the composition f o g of f and g. Then, Let us show that h(vU) = {[0]: vu(0) = 1} = U. [0] £ h(vu) iff [0] £ U. vu(0) = 1 iff [0] £ U. f o g([0]) = 1 iff [0] £ U. This is, by Lemma 2, equivalent to that f(|[0]|) = 1 iff |[0]| = 1.5 This statement is true because f, as an isomorphism, sends the maximum in A/U onto the maximum in 2. Hence, h is onto.
To show that the two spaces are homeomorphic, by Definition 10, it suffices to show that h maps open sets onto open sets. Suppose Y £ Tk is an open set. Then there must be a subset B of the base Bk for the topology Tk such that
Y = U U<. So h[Y] = U h[U<].
u^eB u^eB
We claim that (**) is true: (**) h[U<] = u([0]).
Suppose v £ U< and h(v) £ h[U<]. To show that h[U<] C u([0]), it suffices to show that [0] £ h(v), i.e. v(0) = 1. But this is obviously true because v £ U<.
5Note that the 1 on the right is the maximum in the Boolean algebra, while the 1 on the left is a truth-value.
Suppose U e u([0]). Hence, U is an ultrafilter and [0] e U. By (*), there is a valuation v such that v(0) = 1 iff [0] e U. So v(0) = 1 because [0] e U. Consequently, [0] e h(v) e h[U<] and U e h[U<]. So u([0]) C h[U<].
Since u([0]) is an open set in Ts, by Definition 5, Y e 7S.
Suppose h[Y] e TS. Then there is some subset B of the base for Ts such that h[Y] = U u([0]). Y = h-1 o h[Y] = (J h-1[u([0])]. By (**),
u([<])eB u([<])eB
Y = |J U<. Since U< is an open set in Tk, by Definition 5, it follows that
u([<])eB
Y e Tk. ■
By Definition 9 and Theorem 5, S(A) is compact. Theorem 6 follows from Theorem 4 and the fact that (Vk,Tk) and (S(A),TS) are homeomorphic.
Our proof applies to any propositional system P as long as it is at least as strong as SC, in the sense that both have the same theorems in common. This proof may not work any longer if P is weaker than SC; for the Lindenbaum algebra of P might not be a Boolean algebra. Indeed, as can be seen from the proof that the Lindenbaum algebra (Sent~, of SC is a Boolean algebra [Bell et al., 2006, pp. 41-42], the former would not be a Boolean algebra should some sentences, say, 0 A 0 ^ 0, 0 A 0 ^ 0, 0 V -0, or -(0 A -0), not be theorems of SC.
Unlike our proof, the purely topological proof mentioned above always works no matter what propositional systems are set up in the language L: it is entirely independent of systems. This is its advantage over our proof using the concept of Stone spaces of Boolean algebras.
5. The compactness of first-order logic
The method we used for the compactness of propositional calculus SC can be generalised to first-order predicate logic. Before showing how this generalisation can be carried out, we have to specify a first-order language L and some strong predicate calculus PC in it. There are no function letters in L, but it has a (countable or empty) set of predicate letters Pn, a (countable or empty) set of individual constants an, and a nonempty set of individual variables xn. These are all its extralogical constants and its logical constants are the familiar ones.
As in the case of propositional logic, PC must be strong enough for its Lindenbaum algebra to be a Boolean algebra. This strongness will be guaranteed if the axioms of SC are also axioms of PC: the results of replacing 0, ■0, and x occurring in the axioms of SC by any well-defined formulas of L are axioms of PC. Of course, the rule of inference of PC should contain at least modus ponens.
The notions of proof, satisfiability, and theoremhood in connection with PC are defined in the usual way. The notation 'hPC 0' is used to mean that the (well-defined) formula 0 of L is a theorem of PC.
The compactness theorem of PC says that a set of sentences (well-defined formulas without free variables) of L is satisfiable iff every finite subset of it is so. A formula 0 of L is satisfiable iff there are a L-structure M and a valuation v such that M =v 0, i.e. v satisfies 0 in M.6 If 0 is a sentence, then M =v 0 iff M = 0, i.e. 0 is true in M. Suppose £ is a set of sentences thereof. Hence, the compactness of PC says that £ is satisfiable iff every finite subset £fm of £ is so. The direction from left to right of this biconditional is, by the definition of satisfaction, obvious, so the compactness of PC follows if the following holds:
(Com) For every finite subset £fm of £, if there is a structure M such that M = £fm, then there is also a structure M such that M' = £.
The standard proof of (Com) is by dint of the famous Henkin constructions [Marker, 2003, pp. 35-36], which is neither algebraic nor topological. The purely algebraic proof of (Com) makes use of the Boolean algebraic notion of ultraproducts, as well as Los ultraproduct theorem [Bell et al., 2006, p. 102], [Paseau, 2022, sec. 4c]. The purely topological proof thereof [Beth, 1951] resorts to the compactness of Cantor set to show the compactness of such spaces after introducing the so-called 'reduced logic' (equivalent to propositional logic) into the language L. The hybrid one [Frayne et al., 1962] makes an essential appeal to the Boolean algebraic notions of ultraproducts and ultralimits in its dealing with the compactness of certain related topological spaces. As is to be shown, a proof is available which dispenses with both the just-mentioned algebraic notions and compact Cantor sets.
There are two ways of constructing topological spaces required to prove (Com) [Baldwin et al., 1974]. The first is as follows. Let [M] be the equivalent class of a L-structure M under the relation of isomorphism7. Further, let M be the class of all [M]. For any sentence 0 of the language, let U< be the class {[M] : M = 0}. It is easy to show that the class of all U< is a base
6Instead of structures of languages [Marker, 2003, p. 8], logicians also use the notion of interpretations of languages exactly in the same sense; they are the same. The notion is a commonplace in model theory. A L-structure M consists of a pair (D,I), where D is called the 'domain' of M and I is an assignment function that assigns every constant of L some set-theoretic constructions of appropriate type on D. So, for example, if an is an individual constant, than I assigns it an element of D; if Pn is a n-place predicate letter, I assigns it a n-ary relation on D. Additionally, the notion of satisfaction is a commonplace, too. For its explicit definition, see, say [Ibid., p. 11].
7Here instead of isomorphism, the relation of elementary equivalence is also viable.
for M. Moreover, each U< is a clopen (closed and open) set: U< is a closed set because M \ U< = U-< is an open set. Now suppose M is compact. It goes without saying that {U< : 0 £ Xfin} is a set of closed sets. Assume every ttfin is satisfiable. Then, there is some M such that M N Xfin; consequently, P|{U< : 0 £ Xfin} = 0 because M is a member thereof. Therefore, by Theorem 3, P|{U< : 0 £ X} = 0. It follows that there is some M such that M N X. So (Com) holds if M is compact.
The second way is as follows. A L-theory T, which is just a set of sentences of L, is said to be maximal if for every sentence 0, either 0 £ T or -0 £ T. Let Th be the set of all maximal, satisfiable L-theories. Moreover, for every sentence 0 in L, let U< be the set {T £ Th : 0 £ T}. Then, as is easy to show, the set of all such U< is a base for Th. We use Tth to denote the topology for Th uniquely determined by the base. Moreover, each U< is a clopen set.
Now we show that the following lemma is true:
Lemma 4. If T is a satisfiable L-theory, then there is a maximal, satisfiable L-theory r such that T C r.
Proof. Suppose T is satisfiable. So there is some L-structure M such that M N T. We claim that for each sentence 0 in L, either T U {0} or T U {-0} is satisfiable. If T U {0} is unsatisfiable, then M ¥ 0. So M N -0; consequently, T U {-0} is satisfiable. Similarly, if T U {-0} is unsatisfiable, then T U {0} is satisfiable. Let {0 : i < j} (j ^ k) be the set of all sentences of L (k is some ordinal number). We define recursively as follows: r0 = T.
ri+1 _ U {0i} if r U {0i} is satisfiable;
[P U {-0i} if ri U {-0i} is satisfiable. Then, let r be the union IJ< P. Due to its constructions, r is a maximal, satisfiable theory such that T C r. ■
Assume Th is compact. Suppose each Xfin is satisfiable. By Lemma 4, there is a maximal, satisfiable theory r such that Xfin C r £ Th. Hence, P|{U< : 0 £ Xfin} = 0 because r is a member thereof. By Theorem 3, this implies that P|{U< : 0 £ X} = 0. Therefore, there is some T £ Th such that 0 £ T for every 0 £ X. Since T £ Th is satisfiable, so is X. So (Com) holds.
The two just-mentioned proofs of (Com) are premised on M and Th being compact. Now the problem is how to show their compactness. As promised, our way of showing this uses the notion of Stone spaces of Boolean algebras instead of those in the existing literature mentioned previously.
For that purpose, let us first construct the Lindenbaum algebra of PC in the same way that we used for SC. Let Fml be the set of all formulas (in-
cluding sentences) of L. Define an equivalence relation ^ on Fml in the same way as Definition 11: 0 ^ 0 iff hPC 0 o 0. Similarly, for 0 e Fml, let [0] be {0 : 0 ^ 0}, i.e. its equivalence class under the equivalence relation. Let Fml^ be the class of all [0]. Then, define a relation ^ on Fml^ in the same way as Definition 12: [0] ^ [0] iff hPC 0 ^ 0. As can easily be shown, the Lindenbaum algebra (Fml^, of PC is a Boolean algebra.
Now we construct a subalgebra of (Fml^, Let Sent be the set of all sentences of L. Then, Sent^ is a proper subset of Fml^. It is not hard to see that Sent^ is closed under the operations of join, meet, and complementation.8 Therefore, (Sent^, is a subalgebra of (Fml^, and is a Boolean algebra, too. Let S(Sent^) be the set of all ultrafilters in Sent^. For [0] e Sent^, let u([0]) = {U e S(Senr) : [0] e U}. Then, the set {u([0]) : [0] e Sent~} is a base for S(Sent^) with some topology, say, T (this topology is unique). So we have a Stone space (S(Sent^), TS).
There does not appear to be any one-one correspondences between S(Sent^) and M. For there are at most k sentences in our language L because it contains at most k symbols. Nevertheless, the cardinality of M cannot be pinned down. In contrast, there indeed exists a one-one correspondence between S(Sent^) and Th, as shall be shown below.
In the case of SC, for any ultrafilter U, there is a valuation v such that v(0) = 1 if and only if [0] e U, as is shown in (*). This has a counterpart in PC:
Lemma 5. For any U e S(Sent^) and 0 e Sent, the set V = {0 : [0] e U} is satisfiable.
Proof. We claim that V is consistent. For otherwise there would be a sentence 0 such that 0 and -0 are derivable from V in S. Hence, there would be some finite 01,..., 0n e V such that {01,..., 0n} hPC 0 A -0. This together with the fact that [01 A ... A 0n] e U implies that [0] and [-0] are members of U. This, however, contradicts the fact that U is an ultrafilter.
At this point, we might be tempted to derive the satisfiability of V from its consistency straightforwardly. This is, however, not legitimate because the theorem of completeness is not presupposed here. We have to construct a model M for it instead.
Let D be the countable set of individual constants of L. For any n-place predicate letter P (n is a natural number larger than 0) of the language, we assign it the n-ary relation R which is defined as follows: R holds between a1,..., an, if and only if [P(a1,..., an)] e U. Moreover, for every constant a of L,
8Suppose [4>], [rf] e Sent". Then, [0] n [rf] = [0 A rf] e Sent". Similarly, [0 V rf] e Sent", and [—0] e Sent".
we assign a to itself. The set D and the two assignments constitute a countable
L-structure M. Now we claim that (+) is true:
(+) M N p holds for every 0 £ V i.e. M is a model of V.
This can be shown by induction on num(0)-the number of logical symbols (A, -, and V) occurring in 0 £ V.
1. num(0) = 0.
Since 0 is a sentence, 0 is of the form P(a1,..., an). M N 0 iff (a1,..., an) £ R iff [P(a1,..., an)] £ U. But since 0 £ V, [P(a1,..., an)] £ U.
2. num(0) = 1.
In this case, 0 is of the form (2.1) 0 Ax, or (2.2)-0, or (2.3) VxP(a1,..., an, x), where 0 and x are atomic sentences.
In case 2.1, [0 A x] £ U. Since [0 A x] is less than [0] and [x] and U is a filter, both [0] and [x] are members of U. Hence, M is a model of 0.
In case 2.2, 0 is -0. M is a model of 0 iff [0] £ U. Since U is an ultrafilter,
[0] £ u .
In case 2.3, M is a model of 0 iff for every a £ D, (a1,..., an, a) £ F, iff for every n, [P(a1,..., an, a)] £ U. Since VxP(a1,..., an,x) ^ P(a1,..., an, a) is a theorem of PC and U is a filter, [P(a1,..., an, a)] £ U.
3. Suppose the claim (+) is true if num(0) is less than k. We show that it is true when num(0) is k.
3.1 0 is 0 A x, where the sum of 0 and x is k.
[0] and [x] are members of U. So by the supposition, M is a model of 0 and x; hence, it is a model of 0 A x.
3.2 0 is -0, where num(0) is k — 1.
[0] £ U. So by the supposition, M is not a model of 0; hence, it is a model of -q.
3.3 0 is Vx0(x), where if there is any free variable occurring in the formula 0, it is x, and num(0(x)) = k — 1. Recall that D is the set of all individual constants of L. Thus, M is a model of 0 iff it is a model of every 0(a), where a £ D and 0(a) is the result of substituting a for all occurrences of x. Since [0(a)] £ U holds for every a £ D and num(0(a)) = k — 1, by the supposition, M is model of 0(a). ■
With all the topological and algebraic resources at our disposal, we come to prove the following theorem implying (Com):
Theorem 8. There is a homeomorphism from (Th, 7th) to (S(Sent^), 7S).
Proof, let h be such that for any T £ Th, h(T) = {[0] : 0 £ T}. We claim that h(T) £ S(Sent^), that is, it is an ultrafilter in S(Sent^). For every [0] £ Sent", either [0] £ h(T) or [0]* £ h(T). For if neither [0] £ h(T) nor [0]* £ h(T), then 0 £ T and -0 £ T, which contradicts the fact that T is a maximal theory. Moreover, it cannot be the case that both [0] £ h(T) and [0]* £ h(T). For otherwise, 0 £ T and -0 £ T, which contradicts the fact that T is satisfiable. So by Theorem 1, h(T) is an ultrafilter. Thus, h is a function from Th to S(Sent").
As can easily be shown, h is one-one (injective). We show that it is also onto, i.e. surjective. For any U £ S(Sent"), the set T = {0 : [0] £ U} is mapped by h onto U if it is a member of Th. As is already shown in Lemma 5, T is consistent. Additionally, T is maximal. For if there were some sentence 0 such that neither 0 nor -0 are in it, then neither [0] nor [-0] would be in U. Nevertheless, this would contradict the fact that U is an ultrafilter. Since T is a maximal consistent set, it is, by Lemma 5, satisfiable. Hence, T Th.
Finally, we show that h maps open sets to open sets, that is, for any subset X of Th, X £ 7th iff h[X] £ 7S. We just prove the direction from left to right of the biconditional and the other direction can be shown in a similar way. Suppose X £ 7th. Then, there is subset B of the base of the topology such that X = U U*. h[X] = U h[U*]. We claim that h[U*] = u([0]). For every
T £ U*, h(T) £ u([0]). This is so because h(T) is an ultrafilter and [0] £ h(T). So h[U*] C u([0]). Additionally, for every U £ u([0]), U is an ultrafilter and [0] U. As shown in the last paragraph, {0 : [0] U} Th. We claim that it is a member of U*, namely, 0 £ {0 : 0 £ U}. This is so because [0] £ U. Due to the construction of h, h({0 : [0] £ U}) = U. Therefore, U £ h[U*]. So u([0]) C h[U*]. Thus, h[U*] is an open set in 7S; consequently, by Definition 5, X £ Ts. ■
By Definition 9 and Theorem 5, S(Sent") is compact. Since there is a homeomorphism from Th to S(Sent"), Th is, by Theorem 4, compact. So (Com) holds.
6. Concluding remarks
Our proof of the compactness theorem for classical logic consists of several crucial steps. Firstly, the problem of the compactness of the logical system (SC or PC) is reduced to that of the compactness of some topological space. Secondly, the Lindenbaum algebra of the system is set up and is then shown to be a Boolean algebra. Thirdly, the Stone space of the algebra is shown to be
compact. Fourthly, the set of sentences whose equivalent classes are members of an ultrafilter in the algebra is shown to be satisfiable (or all sentences of the set can be assigned the truth-value 1 simultaneously). Finally, a homeomorph-ism is constructed between the topological space and the compact Stone space.
This approach is easy to be generalized to the proof of the compactness theorem for (both propositional and predicate) modal logic. However, instead of providing a detailed argument in support of the viability of the generalisation, I shall rest content with giving an outline of the proof by taking any strong modal predicate system S as an example.9 Such proof proceeds following the just-mentioned steps. Firstly, we reduce the issue of the compactness of S to that of the compactness of some topological space, which is the set of all maximal, satisfiable theories in the language of S. Secondly, we construct the Lindenbaum algebra of S in the same way that we used for classical logic previously. Thirdly, the Stone space of the algebra (which is a Boolean algebra) can be shown to be compact. Fourthly, we have to prove that the set of sentences whose equivalent classes are ultrafilters in the algebra is satisfiable. This can be done by appealing to the notion of canonical models [Cresswell et al., 1996, p. 261]. Finally, we show that the Stone space and the topological space are homeomorphic.
Compared with the already existing methods of proof of the compactness theorem for classical logic in the literature, which also bypass the completeness theorem, our method of proof has two merits. Firstly, it appeals neither to Tychonoff' Theorem nor to the notion of ultraproducts nor to the compactness of Cantor sets. Instead, it makes uniform use of the concept of Stone spaces of Boolean algebras. Secondly, it admits of a natural generalisation to modal logic. Nevertheless, our method also has a disadvantage: it presupposes strong logical systems.
Acknowledgements. The author disclosed receipt of the following financial support for the research of this article: This work was supported by "the Fundamental Research Funds for the Central Universities", grant No. 20103227626.
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