ON THE LOGIC OF CLOSURE ALGEBRA

Transcription

1 Bulletin of the Section of Logic Volume 40:3/4 (2011), pp Ahmet Hamal ON THE LOGIC OF CLOSURE ALGEBRA Abstract An open problem in modal logic is to know if the fusion S4 S4 is the complete modal logic for the product of any two metric separable dense-in-themselves spaces. This would be settled, positively, if one could prove the following conjecture: When S4 is the complete logic for two complete atomic closure algebras, B and C, then the fusion S4 S4 is the complete modal logic for their product B C. This essay is an effort to give new garments to old results, in the hope that it would lead to a proof of this conjecture. Keywords: Modal logic, Dissectable closure algebra, Completeness, Fusion of modal logics. 1. Introduction The correspondence between elementary topology and the modal logic S4 was first established by McKinsey. In [6], he introduced the topological interpretation of S4, where the necessitation operator box,, is interpreted as the topological interior, and showed that S4 is complete for the class of all topological spaces. Later more mathematically interesting results such as S4 is complete for any dense-in-itself separable metric space where obtained by McKinsey and Tarski [7]. The original algebraic proof given in [7] was very complex, the later more topologic version presented in [10] is not much more accessible. Recently, several attempts were made to simplify the proof in [7] among which we mention Mint s completeness proof of S4 AMS subject classification : 03B45, 06E25, 54A05

2 148 Ahmet Hamal for the Cantor set [8] and two slightly different proofs of the completeness of S4 for the real segment (0, 1) given by Mints and Zhang [9], and Aiello, van Benthem and Bezhanishvili [1]. In this paper, we also made an attempt, in a modern way, to clarify the most important results in [7] whose proofs, mostly implicit and difficult to grasp, are improved by the use of a simple and precise language. 2. Definitions and notations Given a Boolean algebra, we shall use the following notions: (1) x y for the order relation. (2) x y and x y, for the upper and lower bounds respectively. (3) 0 and 1, for the lowest and highest elements respectively. (4) x c, for the complement of x. A closure operation on a Boolean algebra B is a map x x from B to B with the following properties: (1) 0 = 0, (2) (x y) = x y, (3) x x, (4) x = x Closure algebras A Closure algebra is a Boolean algebra with a closure operation. Associated with the closure operation is the interior operation x x = x c c which, therefore, has the following properties: (1) 1 = 1, (2) (x y) = x y, (3) x x, (4) x = x. An element x is said to be closed whenever x = x, open whenever x = x. Clearly then, x is open if and only if x c is closed. A special property. Given a topological space E, the Boolean algebra P(E) of all the subsets of E with its closure operator is, of course a closure algebra. It has the following known ( and quite useful) property: If A is an

3 On the Logic of Closure Algebra 149 open subset, then A B (A B). This is also true in every closure algebra: Let x be open in a closure algebra. We then have x y (x y) for every y. Proof. For more details, see [5]. From this it follows that, if x is open, then x (x y) = x y for each y Relativization The notion of subspace of a topological space generalizes to closure algebras. Given a closure algebra K and one of its elements a, set K a = {x : a x K}. Then K a has a naturel structure of closure algebra induced by the structure on K. In K a, the operations and are unchanged, while the original operations x x c and x x become x a x c and x a x respectively. Moreover, when a is open in K, the original operation x x stays as x x for any x K a. For more details, see [5] Extending a closure operation Let K be a complete Boolean algebra [ i.e., a Boolean algebra in which every subset has a lowest upper bound (l.u.b.) and, therefore, also a greatest lower bound(g.l.b.)]. Let H K be given with an operation from H into H, x x such that (1) 0 H and 0 = 0, (2) if x, y H, then x y H and (x y) = x y, (3) if x H, then x x = x. This looks very much as if it were a closure operation on H, except that H is not really a Boolean algebra because it misses a complement operation. Set g(x) = {z : z H and x z = z }. Then g is a closure operation on K which extends the operation on H; in fact it is the only such closure operation on K. The proof is quite straightforward.

4 150 Ahmet Hamal 2.4. Representation theorems There are two important representation theorems for Boolean algebras and closure algebras. The first is due to M. H. Stone. The second is proved in the paper by McKinsey and Tarski: It can be viewed as a generalization of Stone s Theorem from the discrete case to the general topological case. We first state both theorems. Theorem 2.1 (M. H. Stone). Every Boolean algebra is isomorphic to a subalgebra of the Boolean algebra P(X) of all the subsets of an adequate set X. Theorem 2.2 (Mckinsey-Tarski). Every closure algebra is isomorphic to a subalgebra of the closure algebra of an adequate topological space. For a better understanding of those results, we present the classical notions of filters and ultrafilters for Boolean algebras ( which are not included in [7]). Filters and ultrafilters. Let B be a Boolean algebra. A Boolean filter on B is a nonempty subset F B with the following properties: (i) 0 F, (ii) if x F and x y, then y F, (iii) if x F and y F, then x y F. This is the exact generalization of the notion of filter on a set: A filter on a set X is, precisely, a filter on the Boolean algebra P(X) of all the subsets of X. A Boolean ultrafilter U on B is a nonempty subset U B such that U B and with the following properties: (i) x U and y U iff x y U (ii) x U or y U iff x y U We give here a sketchy of Stone theorem; see [2],[3], for details. Sketch of the proof of Stone Teorem. Let X = Υ(B) be the set of all Boolean ultrafilters on a Boolean algebra B. For each x B, set f(x) = {U : x U Υ(B)}. Using the definition of Boolean ultrafilters,

5 On the Logic of Closure Algebra 151 it is easily seen that the function f : B P(X) is an isomorphism of the Boolean algebra B onto a subalgebra of P(X). The proof of McKinsey-Tarski theorem. Let K be a closure algebra. Using Stone s theorem, we can suppose that K is a subalgebra of P(X) for an adequate set X. Using the extension of closures discussed above, it is immediately seen that the closure operation on K is extended into a closure operation on P(X) which, of course, turns X into a topological space Three classes of closure algebras We will encounter three classes of closure algebras corresponding to three well-known classes of topological spaces. Here are the definitions. Consider the two properties (C) and (W ) for a closure algebra: (C) if x and y are nonempty closed and x y = 1, then x y 0. (W) if x and y are nonempty closed, then x y 0. Closure algebra K is called connected whenever it has property (C), otherwise it is called disconnected. It is called well-connected whenever it has property (W) and 0 1. Similarly, an element a K is called connected whenever the closure algebra K a is connected, otherwise a is called disconnected. The closure algebra K is called totaly disconnected whenever each nonempty open element in K is disconnected. Some peculiarities. Every well-connected closure algebra is connected. The smallest closure algebra, the null algebra {0}, is connected but not well-connected. Moreover, it is also totally disconnected. The second smallest, {0, 1}, is both connected and well-connected. We can reformulate property (W) as: (W1) The lower bound of any finite family of nonempty closed elements is nonempty. Of course (W1)implies (W) but there is a slight difference between the two: Think of the empty family in a closure algebra K; all its elements are nonempty closed and its lower bound is 1. If K satisfies (W1), then 1 must be 0 1. So, the well-connected closure algebras are those that have property (W1), without any proviso!.

6 152 Ahmet Hamal 2.6. Dissectable closure algebras Let K be a closure algebra. In order to make the definitions and statements as simple as possible, we introduce the following terminology. Given a couple of integers r 0 and s 0, an (r, s)-dissection of an element a K is defined to be a set of mutually exclusive nonempty elements a 1, a 2,..., a r and b 1, b 2,..., b s of K such that: (i) a = a 1 a2... ar b1 b2... bs, (ii) the elements a 1, a 2,..., a r all open, (iii) the closures b 1, b 2,..., b s are all equal to a closed element c, (iv) a a c c a i for each i r. The closure algebra K is said to be dissectable whenever each nonempty open x has an (r, s)-dissection for each couple of integers r 0 and s 0. The important fact about dissectable closure algebras is: The closure algebra of a separable metric space without isolated points is dissectable. For more details, see [10] Universal algebras A closure algebra K is called universal for a given class of algebras A whenever each algebra in the class A is isomorphic to a subalgebra of K. The algebra K is called generalized universal for A whenever each algebra in the class A is isomorphic to a subalgebra of the relativized algebra K a for some open a in K. This is taken from [7]. We further introduce the following definitions(implict in [7]). Given two closure algebras H and K together with an element a K, an immersion of H in K relative to a is an isomorphism h of H onto a subalgebra L of the relativized algebra K a such that, for every nonempty element x L, one has a a c x. Any isomorphism of H onto a subalgebra of K is, of course, an immersion (1 1 c = 0). We call the algebra K special universal for the class A whenever, for each algebra H A and each nonempty open a K, there is an immersion of H in K relative to a. Clearly, if K is special universal for the class A, then it is universal for that class. If K is universal for A, then it is generalized universal for A.

7 On the Logic of Closure Algebra Canonical decomposition and main universality results In this section, we define a composition closure algebra and decomposition of a finite closure algebra in order to prove main universality results. Then, we will give main universality results Companion of a closure algebra Let K be a closure algebra. Set K = K {0, 1}. The elements of K are couples (x, e) with x K and e = 0 or 1. In a quite natural way, K is a Boolean algebra, thus (x, e) (y, f) = (x y, e f), (x, e) (y, f) = (x y, e f), (x, e) c = (x c, e c ). Setting (0, 0) = (0, 0) and, otherwise, (x, e) = (x, 1), we obtain a closure operator on K which turns it into a companion closure algebra to K. The main features of the companion algebra K follow: The closed elements are (0, 0) and those of the form (x, 1) with x closed in K. The open elements are (1, 1) and those of the form (x, 0) with x open in K. So (0, 1) is closed and (1, 0) is open. The closure algebra K is well-connected since the closed (0, 1) is included in each (x, 1). Set H = K(1,0). The map x (x, 0) is easily seen to be an isomorphism of the closure algebra K onto the closure algebra H. Since the nonempty open elements of K are the elements (x, 0) with x nonempty open in K, one can easily see that K is dissectable whenever K is dissectable Canonical decomposition in a finite closure algebra Two cases are presented separately: Well-connected algebras and not wellconnected algebras The well-connected case Given a well-connected finite algebra, there is a canonical decomposition

8 154 Ahmet Hamal 1 = c e 1 e2... er with c nonempty closed, (e i ) 1 i r a family of nonempty open elements in H, disjoint from c, and r 0. The construction of this decomposition is by steps. (1) Let A be the set of atoms in H and C = {a : a A}. Since H is well-connected, there is a least element c in C, and we have x c for each nonempty closed x in H. (2) For a A, we have a c iff a = c dir. Indeed, if a c, then a c = c, and a c (because a is nonempty closed) so a = c. The converse is clear. (3) Let B = {a : a A, a c} = {a : a A, a = c}. Then, of course, c = B. (4) Let D = C \ {c}. If D =, let M =, if not, let M be the set of minimal elements in D. In both cases, set M = r so that r 0. In case r > 0, enumerate the elements in M as m 1, m 2,..., m r. (5) For 1 i r, set A i = {a : a A, a m i } and e i = A i. (6) Notice that A is the set-union of B and all the A i s. So c e 1 e2... er = 1. (7) The e i s are open in H, but not necessarily mutually exclusive [see the counter example]. Indeed, Let f i = (A\A i ), then e i = fi c and f i = {a : a A\A i }. We only have to show that f i ei = 0. Suppose, otherwise, that there exists a A\A i such that a e i 0. This means there is an atom d a such that d A i and, therefore m i d a and a A i, a contradiction. Set A = p, A i = p i, M = r, B = s and notice that p i < p, r 0 and s > 0. For simplicity, we shall set e 0 = c and write H i for the relativized subalgebras H ei, 1 i r. Notice that each H i is again a well-connected finite closure algebra: H 0, c is the unique nonempty closed element, and m i ei is a least nonempty closed element in H i for 1 i r. For each x H, define the constituents x i = x e i H i, 1 i r. Then clearly x = x 0 x1 x2... xr.

9 On the Logic of Closure Algebra 155 The counterexample. Let E = {0, 1, 2, 3} be the topological space in which the closure operation is defined by 0 = {0}, 1 = {0, 1}, 2 = {0, 2}, 3 = E. In the closure algebra K of the topologic space E, let the atoms (corresponding to the points) be simply denoted 0, 1, 2, 3. Clearly, K is well-connected and 0 is its least closed nonempty element. We have A 1 = {1, 3} and A 2 = {2, 3}. So 3 belongs to both A 1 and A 2. The not well-connected case Given a not well-connected closure algebra, there is a canonical decomposition 1 = e 1 e2... er with r > 1 and (e i ) 1 i r a family of nonempty open elements in H. [Notice that r > 1 because, otherwise, 1 = e 1 and H would be well-connected.] For simplicity, we shall write H i for the relativized subalgebras H ei, 1 i r. Notice again that each H i is a well-connected finite closure algebra. The construction of this decomposition is a quite similar to that in well-connected case. It can be carried out, mutatis mutandis, with the sole difference that B = Preparation Lemma The well-connected case Let H be well-connected finite closure algebra with its canonical decomposition 1 = c e 1 e2... er We use the notations in 3.2. Let A = p, B = s. Let K be a closure algebra together with a nonempty element a K and an (r, s)- dissection of a in r + s mutually exclusive elements a 1, a 2,..., a r, b 1, b 2,..., b s. We set a 0 = b 1 b2... bs and write, to simple again, K i for the relativized subalgebra K ai, 1 i r. Let c 1, c 2,..., c s be the s atoms in B and set h 0 (c j ) = b j. There is a unique extension of h 0 to an isomorphism h 0 from H 0 onto a subalgebra L 0 of K 0. Notice that we have h 0 (e 0 ) = a 0 and, for each nonempty element

10 156 Ahmet Hamal x H 0, h 0 (x) = k = a 0. Thus, h 0 is an immersion of H 0 into K relative to a 0. Now, for 1 i r, let h i : H i K i be an immersion of H i in K relative to a i. We sew the h i s into a patchwork: For each x H set h(x) = h 0 (x 0 ) h 1 (x 1 )... h r (x r ) to define h : H K. Then h is an immersion of H in K relative to a. Proof. We have to prove that h is an injective Boolean homomorphism such that h(x ) = a h(x) and a a c h(x) for each nonempty x H. (1) If h(x) = h(y) then x = y because the a i s are mutually exclusive, and all the h i s are injective. (2) That h(1) = a and h is an Boolean isomorphism onto L = h(h) is clear enough. (3) For x H, we have have to show that h(x ) = a h(x), the closure h(x) being taken in the algebra K. That is h 0 (x e 0 ) h 1 (x e 1 )... h r (x e r ) = a (h 0 (x 0 ) h 1 (x 1 )... h r (x r ) ), the closures h i (x i ) being taken in the algebra K. We also have to prove, when x is nonempty, that a a c h(x). Of course, all we have to do is to prove the equality and the inclusion for atoms! Here are some details for the proof. If x B, then x = c = e 0, x 0 = x, x e 0 = e 0, and, since h 0 is an isomorphism, a h 0 (x 0 ) = h 0 (x e 0 ). Moreover, for i 0, we have x i = 0 and x e i = 0. If x A i for some i 0, we have x m i, x 0 = 0, x e 0 = e 0, x i = x, x e i = m i ei, and, since h i is an isomorphism, a h i (x i ) = h i (x e i ). While, if x A j for some j 0, then x j = 0 and x e j = 0.

11 On the Logic of Closure Algebra 157 The equality easily follows, just by looking at the formula for h(x ) and h(x). As to the inclusion a a c h(x), it is simply a consequence of the definition of immersions. Indeed, h i is supposed to be an immersion of of H i in K relative to a i. The not well-connected case Let H be a not well-connected finite closure algebra with its canonical decomposition 1 = e 1 e2... er We use the notations in 3.2. Let A = p, B = 0. Let K closure algebra together with a nonempty element a K and r mutually exclusive elements a 1, a 2,..., a r, such that a = a 1 a2... ar. For 1 i r, let h i : H i K i be an isomorphism of H i with a subalgebra of K i. We sew again the h i s into a patchwork: For each x H set h(x) = h 1 (x 1 )... h r (x r ) to define h : H K. Then h is an immersion of H on a subalgebra of K a. The proof is quite similar to the previous one Main universality results We are ready for the main universality results. Theorem 3.1. Every (non null) dissectable closure algebra is special universal (therefore also universal) for the class of well-connected finite closure algebras. Proof. Given a dissectable closure algebra K {0} together with a nonempty open element a K and a well-connected finite closure algebra H, we prove, by induction on the number of atoms in H, the existence of an immersion of H in K relative to a.

12 158 Ahmet Hamal If H has only one atom, the result is clear enough. Given an integer p > 1, suppose the result holds whenever the number of atoms is < p. Let H with its canonical decomposition 1 = c e 1 e2... er have p atoms. All we have to do is to use the Preparation Lemma with an (r, s)- dissection of a where B = s. Indeed, the number of atoms in each H i, for i 0, is < p! so the immersion h i of H i in K relative to a i, and the patchwork produces an immersion h of H in K relative to a. Theorem 3.2. Every (non null) dissectable closure algebra is generalized universal for the class of finite closure algebras. Proof. Given a dissectable closure algebra K {0} and a finite closure algebra H, we have to show that there is an open element a K such that H is isomorphic to subalgebra of the relativized algebra K a. This is easily done using the companion algebra H which is well-connected. By Theorem 3.1, there exist a subalgebra L of K itself together with an isomorphism h : H L. Now, H is isomorphic to the subalgebra H(1,0) of H. Since (1, 0) open in H, the element a = h(1, 0) is open in L and also in K. So that all is said. Theorem 3.3. Every (non null) totaly disconnected dissectable closure algebra is universal for the class of finite closure algebras. Proof. Given a totally disconnected dissectable closure algebra K and a finite closure algebra H, we show that H is isomorphic to subalgebra of K. If H is well-connected, the result follows from Theorem 3.1. If not, consider the canonical decomposition of H and the corresponding relativized algebras H i, each of which is well-connected, for 1 i r. Since K is totally disconnected, there is a set of r mutually exclusive nonempty open elements a 1, a 2,..., a r such that 1 = a 1 a2... ar. The corresponding relativized subalgebras K i are dissectable because each a i is open. Therefore, for each i, there exists a subalgebra L i of K i together with an isomorphism h i : K i L i. We only have to sew those isomorphisms to get an isomorphism h of H onto a subalgebra of K.

13 On the Logic of Closure Algebra 159 This proof is not exactly the one given in [7]. It is an improvement. Indeed, this proof shows a little bit more than is stated : We do not need the algebra K to be totally disconnected. All that is needed is that: (*) For every integer r > 1, a decomposition 1 = a 1 a2... ar exists with r mutually exclusive nonempty open elements. So we have the following stronger result. Theorem 3.4. Every (non null) dissectable closure algebra is universal for the class of finite closure algebras if it has property (*). Remarks. There exist closure algebras that satisfy property (*) without being totally disconnected: The closure algebra K of the topological space E = R\ Z is dissectable but not totally disconnected. Yet E is the union of the mutually exclusive nonempty subsets (n, n + 1) for n Z so that K satisfies condition (*), therefore K is an universal algebra for finite closure algebras. This also leads to the following observation. Can Theorem 3.1 be strengthened to assert that every dissactable closure algebra is universal for all connected finite algebras. The answer is no. Both the question and the answer are given in [7]. Concluding this section, we give two important theorems: Theorem 3.5. S4 is the logic of finite well-connected topological spaces. Proof. See [1]. Theorem 3.6. S4 is the logic of any metric separable dense-in- itself space. Proof. Since the closure algebra of a separable metric space without isolated points is dissectable, proof follows from Theorem 3.1 and Theorem 3.5.

14 160 Ahmet Hamal 4. Products For Closure Algebras 4.1. The Boolean Algebra P(X Y ) Clearly enough, given two sets X and Y, the Boolean algebra of the subsets of their Cartesian product, P(X Y ), is not the product of the Boolean algebras P(X) and P(Y ). Still less : If X and Y are topological spaces, the corresponding closure algebra of the product space X Y cannot be identified with the product of their closure algebras. So what is to be done? Here is tentative answer A special product Let B and C be two complete atomic Boolean algebras whose sets of atoms are H and V respectively. Take B C to be the product B C = v V B v, where B v is a copy of the Boolean algebra B. This, of course, is again a complete Boolean algebra. It is the set of all indexed families x = (x v ) v V with x v B for each v V, partially ordered according to the product relation = v. v V The definition is clearly not symmetric in B and C. Nonetheless, there is a natural isomorphism f between the two Boolean algebras B C and C B. Indeed, given any element x = (x v ) v V in B C, simply define f(x) = (y h ) h H C B by y h = {v V : h x v }. Then, f : B C C B is an isomorphism. Proof. First, show that f is a bijection. Let x x where x = (x v ) v V B C and x = (x v) v V B C. Then there is V V such that x vi x v i for every v i V. Now from B is atomic Boolean algebra we see that x vi = {h H vi : H vi H} and x v i = {h H v i : H v i H} imply H vi H v i. Hence there is an h 0 H such that h 0 H vi and h 0 H v i or h 0 H vi and h 0 H v i. By definition, we have y h0 y h 0, and so (y h ) h H (y h ) h H.

15 On the Logic of Closure Algebra 161 Let y C B with y = (y h ) h H. Then by C is atomic Boolean algebra, we write for all h H. In this case, we set y h = {v V h : V h V }. x v := {h H : v y h } and x := (x v ) v V. Then by definition we have that, f(x) = y and x B C. It is easy to verify that the function f preserves Boolean operations. As an example, Let s show it for the operation. Consider x, x B C. since B is atomic Boolean algebra gives us x v = {h H v : H v H} and x v = {h H v : H v H} for all v V, then we get x v x v = {h : h H v H v}. Thus by definition, we will obtain f(x v x v) = {v V : h x v x v}. Therefore, we deduce easily the required result from {v V : h xv x v} = ( {v V : h x v }) ( {v V : x v}). One way to put it, also, is as follows: h x v v y h ( ) So, B C and C B can be identified, elementwise, thus obtaining one and the same object, a complete atomic Boolean algebra, to be denoted B C, whose set of atoms can be identified with H V. Clearly, if B = P(X) and C = P(Y ), then B C is identified with P(X Y ). Any given subset of the product X Y is an element of the Boolean algebra B C and has two presentations: A vertical presentation as x = (x v ) v V, and a horizontal presentation as y = (y h ) h H, linked together by the relation ( ) above Biclosure product algebra Moreover, if X and Y are topological space, so that B and C are closure algebras, then B C is endowed with a natural structure of biclosure algebra: The two closure operations correspond, respectively, to the horizontal and vertical topologies on X Y. This is quite easily seen, using the two presentations mentioned above. More precisely, let x = (x v ) v V and y = (y h ) h H be the two presentations of the same element a B C. Then, the two closures of a are simply defined as follows: a V = (x v ) v V, a H = (y h ) h H.

16 162 Ahmet Hamal Notice that this algebra is dissectable with respect to the vertical closure operation a V whenever the topological space Y is dissectable. Similarly, this algebra is dissectable with respect to the horizontal closure operation a H whenever the topological space X is dissectable. Let L 1 2 be a bimodal language with modal operators 1 and 2. We recall that the fusion of S4 with itself, denoted by S4 S4, is the least set of formulas containing S4- axioms both 1 and 2, and closed under modus ponens, substitution, ϕ ψ 1 ϕ 1 ψ, and ϕ ψ 2 ϕ 2 ψ. For more details, see [4] Conjecture If S4 is the complete logic for B and for C, two complete atomic closure algebras, then the fusion S4 S4 is the complete modal logic for their product B C. From this would immediately follow that the fusion S4 S4 is the complete modal logic for the product of any two metric separable densein-themselves spaces. Acknowledgement: I m grateful to Professor Labib HADDAD and Professor Mehmet TERZILER for their invaluable contribution and guiding ideas for the manuscript to be complete. References [1] M. Aiello, J. van Benthem, and G. Bezhanishvili, Reasoning about space: the modal way, J. Logic Comput. 13(2003), no. 6, pp [2] P. Blackburn, M. de Rijke and Y. Venema, Modal Logic, Cambridge tracts in theoretical computer science, Vol. 53. CUP, Cambridge, (2001). [3] A. Chagrov and M. Zakharyaschev, Modal Logic, Oxford University Press, volume 35 of Oxford Logic Guides, (1997). [4] D. M. Gabbay, A. Kurucz, F. Wolter and M. Zakharyaschev, Many Dimensional Modal Logics: Theory and Applications. Studies in Logic and the Foundations on Mathematics, Elsevier, Volume 148, (2003). [5] A. Hamal, Spatial Modal Logics, Ph.D. Thesis, Ege University, (2007).

17 On the Logic of Closure Algebra 163 [6] J. C. C. McKinsey, A solution of the decision problem for the Lewis systems S2 and S4, with an application to topology, The Journal of Symbolic Logic, 6 (4)(1941), pp [7] J. C. C. McKinsey and A. Tarski, The algebra of topology, Annals of Mathematics, 45 (1945), pp [8] G. Mints, A completeness proof for propositional S4 in Cantor space, in: E. Orlowska (Ed.), Logic at Work: Essays Dedicated to the Memory of Helena Rasiowa,1999, Physica-Verlag, Heidelberg, pp [9] G. Mints and T. Zhang A proof of topological completeness for S4 in (0, 1), Ann. Pure Appl. Logic, 133(2005), no. 1-3, pp [10] H. Rasiowa and R. Sikorski, The Mathematics of Metamathematics, Panstwowe Wydawnictwo Naukowe, (1963). Department of Mathematics, Ege University, Bornova, Izmir, Turkey ahmet.hamal@ege.edu.tr