ABSTRACT
INVESTIGATION OF THE TOPOLOGICAL INTERPRETATION OF MODAL LOGICS
This thesis studies interpretations of modal logics in topological spaces.
We prove that adding a new axiom (P → P ) to S4 gives us a sound and complete axiom system over all discrete topological spaces. We show non-completeness of S4 over three families of topological spaces: particular point, excluded point, and another natural generalization of the Sierpi´nski spaces. Namely, we find extensions of S4 that are sound over these families, although their completeness remains an open question. We show that given any set X and any interpretation of in X that satisfies S4, the image of this interpretation is a topology on X. We also study the influence of the modal axioms of S4 on topological properties of the image.
Bing Xu May 2016
INVESTIGATION OF THE TOPOLOGICAL INTERPRETATION OF MODAL LOGICS
by
Bing Xu
A thesis submitted in partial fulfillment of the requirements for the degree of Master of Arts in Mathematics in the College of Science and Mathematics California State University, Fresno
May 2016 APPROVED
For the Department of Mathematics:
We, the undersigned, certify that the thesis of the following student meets the required standards of scholarship, format, and style of the university and the student’s graduate degree program for the awarding of the master’s degree.
Bing Xu Thesis Author
Maria Nogin (Chair) Mathematics
Katherine Kelm Mathematics
Marat Markin Mathematics
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Signature of thesis author: ACKNOWLEDGMENTS
First, I would like to thank my advisor, Dr. Maria Nogin, for all the meetings we had throughout the past two semesters and all the hours you spent on reading and commenting on my drafts. I am very grateful for all the valuable advice and timely feedback. Thank you for being so patient listening to all my thoughts and answering all the questions I asked. Also, I really appreciated all the encouragement you gave me to pursue the thesis option every time I thought about writing a project. Thank you for pushing me through this great accomplishment. I could never have finished my thesis without your guidance and support. It was a great honor and pleasure for me to work with you. I would also like to thank Drs. Marat Markin and Katherine Kelm for being on my thesis committee. Thank you for reading all the drafts of my work and for your invaluable mathematics expertise. I also want to thank my mother and my husband for being so supportive to my education. It was my mother who watcedh my baby all the hours in which I was writing the thesis, and my husband who believes there is nothing greater to advance the family that to get advanced degrees. Last but not least, I would like to apologize to my two-year-old daughter Medleyana for not spending much time with you lately. Thank you for being so understanding and not being mad at me at all. You are the reason that everything I did was meaningful! TABLE OF CONTENTS
Page LIST OF TABLES ...... vi LIST OF FIGURES ...... vii INTRODUCTION ...... 1 The Classical Propositional Logic ...... 1 Interpretations of Formulas in Subsets ...... 5 Modal Logics ...... 6 Interpretations of Modal Logics in Topological Spaces ...... 9 THE LOGIC OF DISCRETE TOPOLOGICAL SPACES ...... 12 NON-COMPLETENESS OF S4 OVER SOME FAMILIES OF TOPO- LOGICAL SPACES ...... 18 THE RELATIONSHIP BETWEEN THE TOPOLOGICAL PROPER- TIES AND COMMON MODAL LOGICS ...... 26 CONCLUSIONS ...... 39 REFERENCES ...... 40 LIST OF TABLES
Page Table 1. The Truth Table of ¬F,F ∧ G, F ∨ G, F → G, and F ↔ G 1 Table 2. The Truth Table of the Formulas F → G and ¬F ∨ G . . . 2 Table 3. Some Axioms in Modal Logics ...... 7 Table 4. Common Modal Logics and the Corresponding Added Axioms 8 Table 5. The Effect of Modal Axioms on Topological Properties . . . 27 LIST OF FIGURES
Page
Figure 1. The Topological Spaces Xi for i = 2, 3, and n ...... 20
Figure 2. The Topological Space X∞ ...... 21 Figure 3. The Sublogic Relationship among Some Modal Logics . . . 28 INTRODUCTION
The Classical Propositional Logic
Classical propositional logic is based on propositional variables and logical connectives. A propositional variable is a variable that can be assigned a truth value: true or false. Symbols ¬, ∧, ∨, → and ↔ are called logical connectives. The logical connectives enable compound formulas to be built from simpler formulas. A formula is defined as follows: (i) Every propositional variable is a formula. (ii) Given a formula F , the negation ¬F (not F ) is a formula. (iii) Given two formulas F and G, the conjunction F ∧ G (F and G), the disjunction F ∨ G (F or G), the implication F → G (F implies G), and the biconditional F ↔ G (F if and only if G) are formulas. The truth value of the compound formula is defined as a certain function of the truth values of the simpler formulas. Namely, the truth values of ¬F,F ∧ G, F ∨ G, F → G, and F ↔ G are defined as shown in Table 1.
Table 1. The Truth Table of ¬F,F ∧ G, F ∨ G, F → G, and F ↔ G
F G ¬F F ∧ G F ∨ G F → G F ↔ G T T F T T T T T F F F T F F F T T F T T F F F T F F T T
In classical propositional logic, an interpretation of a formula is an assignment of truth values to its component variables. It is possible that a formula has different truth values under different interpretations. A formula F 2 is valid (also called a tautology) if F is true under every interpretation. If G is false under every interpretation, then G is called a contradiction. For example, the formula S ∨ ¬S is a tautology and the formula S ∧ ¬S is a contradiction. A truth table of a formula is a table that lists all its interpretations. For example, Table 2 is the truth table of the formulas F → G and ¬F ∨ G.
Table 2. The Truth Table of the Formulas F → G and ¬F ∨ G
F G ¬F ¬F ∨ G F → G T T F T T T F F F F F T T T T F F T T T
Notice from the truth table above that (F → G) and (¬F ∨ G) have the same truth values for any combination of truth values of formulas F and G. We say that F → G and ¬F ∨ G are logically equivalent, and denote this by (F → G) ≡ (¬F ∨ G). Notice also that for formulas A and B, A ≡ B if and only if A ↔ B is a tautology. An axiom system consists of a set of formulas (called axioms) and some rules (called rules of inference). We say that an axiom system is sound if every formula that is derivable from this axiom system is valid. An axiom system is complete if every valid formula can be derived from this axiom system. The following is one of the sound and complete axiom systems for classical propositional logic: Axioms (1) (X ∧ Y ) → X 3
(2) (Y ∧ X) → X (3) X → (X ∨ Y ) (4) X → (Y ∨ X) (5) ¬¬X → X (6) X → (Y → X) (7) X → (Y → (X ∧ Y )) (8) ((X → Y ) ∧ (X → ¬Y )) → ¬X (9) ((X → Z) ∧ (Y → Z)) → ((X ∨ Y ) → Z) (10) ((X → Y ) ∧ (X → (Y → Z))) → (X → Z) Rule of inference X,X → Y Modus Ponens Y Remark 1. The rule of Modus Ponens means that if X and X → Y are derivable from the axiom system, then so is Y .
Example 2. It is easy to show using a truth table that (A ∧ B) → (B ∧ A) is a tautology. Let us show how it can be derived from the above the axiom system.
1. B → (A → (B ∧ A)) axiom (7) 2. (A ∧ B) → B axiom (2) 3. (((A ∧ B) → A) ∧ ((A ∧ B) → (A → (B ∧ A))) → ((A ∧ B) → (B ∧ A)) axiom (10) by substituting X with (A ∧ B), Y with A, and Z with (B ∧ A) 4. (((A ∧ B) → B) ∧ ((A ∧ B) → (B → (A → (B ∧ A))))) → ((A ∧ B) → (A → (B ∧ A))) axiom (10) by substituting X with (A ∧ B), Y with B, and Z with [A → (B ∧ A)] 5. (B → (A → (B ∧ A))) → ((A ∧ B) → (B → (A → (B ∧ A)))) 4
axiom (6) by substituting X with (B → (A → (B ∧ A))) and Y with (A ∧ B) 6. (A ∧ B) → (B → (A → (B ∧ A))) Modus Ponens, 1, 5 7. ((A ∧ B) → B) → (((A ∧ B) → (B → (A → (B ∧ A)))) → (((A ∧ B) → B) ∧ ((A ∧ B) → (B → (A → (B ∧ A))))) axiom (7) by substituting X with ((A ∧ B) → B) and Y with ((A ∧ B) → (B → (A → (B ∧ A)))) 8. ((A ∧ B) → (B → (A → (B ∧ A)))) → (((A ∧ B) → B) ∧ ((A ∧ B) → (B → (A → (B ∧ A)))) Modus Ponens, 2, 7 9. ((A ∧ B) → B) ∧ ((A ∧ B) → (B → (A → (B ∧ A)))) Modus Ponens, 6, 8 10. (A ∧ B) → (A → (B ∧ A)) Modus Ponens, 4, 9 11. (A ∧ B) → A axiom (1) 12. ((A ∧ B) → A) → (((A ∧ B) → (A → (B ∧ A))) → (((A ∧ B) → A) ∧ ((A ∧ B) → (A → (B ∧ A))))) axiom (7) by substituting X with ((A ∧ B) → A) and Y with ((A ∧ B) → (A → (B ∧ A))) 13. ((A ∧ B) → (A → (B ∧ A))) → (((A ∧ B) → A) ∧ ((A ∧ B) → (A → (B ∧ A)))) Modus Ponens, 11, 12 14. ((A ∧ B) → A) ∧ ((A ∧ B) → (A → (B ∧ A))) Modus Ponens, 10, 13 15. (A ∧ B) → (B ∧ A) Modus Ponens, 3, 14 Also, the following two topologies will be very useful last in this text.
Lemma 3 (contrapositive). (A → B) → (¬B → ¬A).
Lemma 4 (transitivity). ((A → B) ∧ (B → C)) → (A → C). 5
Both of the above tautologies can be easily verified using truth tables, or derived from axioms. Note that together with the Modus Ponens, these imply that if A → B and B → C can be derived, then ¬B → ¬A and A → C can be derived.
Interpretations of Formulas in Subsets
In the last section, we defined the interpretation of a formula in propositional logic. In a general sense, an interpretation of a formula is an assignment of meaning to the symbols that comprise this formula. We divide symbols in a formula into two kinds: the logical symbols (also called logical constants) and the non-logical symbols. The logical constants (like logical connectives) are always given the same meaning in every interpretation, and the non-logical symbols (like propositional variables) are assigned with different values for different interpretations. Let X be a set. We will interpret logical connectives as set operations: negation (¬) as complement (C ), conjunction (∧) as intersection (∩), and disjunction (∨) as union (∪). Note that the implication and the biconditional can be expressed in terms of the other operations:
F → G ≡ ¬F ∨ G
F ↔ G ≡ (F → G) ∧ (G → F ), so they are interpreted as the corresponding set operations. 6
Let f be any mapping (interpretation) that sends propositional variables to subsets of X. Let P and Q be propositional variables and f(P ) = A, f(Q) = B, where A, B ⊆ X. Then f(¬P ) = (f(P ))C = AC , f(P ∧ Q) = f(P ) ∩ f(Q) = A ∩ B, and f(P ∨ Q) = f(P ) ∪ f(Q) = A ∪ B. Note that every formula is mapped to a subset of X. If a formula F is mapped to the whole set X for any mapping f, i.e., ∀f f(F ) = X, then we say that F is valid with respect to any interpretation in X, or, shorter, F is valid in X.
Lemma 5. Let F and G be formulas and X be a set. Then F → G is valid with respect to any interpretation in X if and only if f(F ) ⊆ f(G) for any mapping f that sends propositional variables to subsets of X.
Proof. Observe that F → G is valid with respect to any interpretation in X if and only if X = f(F → G) = f(¬F ∨ G) = f(¬F ) ∪ f(G) = (f(F ))C ∪ f(G) for any f, which is true if and only if f(F ) ⊆ f(G) for any f.
Modal Logics
A modal logic is an extention of the classical propositional logic with one or more logical connectives. In this thesis we consider the following two additional logical connectives (modal operators): necessity and possibility, which are denoted by and ♦, respectively. Let F be a formula. Then F (necessarily F , or square F , or box F ) is a formula. Similarly, ♦F (possibly F , or diamond F ) is a formula defined as ¬¬F (it is not necessary that not 7
F ). Observe that F is then equivalent to ¬♦¬F (it is not possible that not F ). Modal logics are used in many different contexts, and there is a variety of interpretations of and ♦ in different fields and areas. For example, in philosophy, can be interpreted as “it is true in every accessible possible world” and ♦ as “it is true in some accessible possible world”. In computer science, can be interpreted as “has to hold on the entire subsequent path” and ♦ as “has to hold somewhere on the subsequent path”. Of course, when we introduce additional operations (modal connectives), we may want them to have certain properties, so we may want to introduce additional axiom(s) and rule(s) of inference. The axioms and rules of inference depend on the context. Some of the most meaningful and useful axioms are listed in Table 3.
Table 3. Some Axioms in Modal Logics
Name Axiom K (P → Q) → (P → Q) D P → ♦P T P → P 4 P → P B P → ♦P 5 ♦P → ♦P
The following rule of inference is commonly associated with the necessity connective: P Necessitation Rule . P Adding the Necessitation Rule and some of the above axioms to any axiom system of the classical propotional logic, we get the axiom systems for some common modal logics (see Table 4). 8
Table 4. Common Modal Logics and the Corresponding Added Axioms
Logic Added Axioms K K D K,D T K,T K4 K, 4 B K,T,B D4 K,D, 4 S4 K,T, 4 S5 K,T, 5 or K,T, 4,B
The logic K is the weakest modal logic of those that proved to be the most useful. The axiom systems of most other modal logics that are studied in the literature are extensions of that of the logic K. However, we will consider a couple of other logics in section “The Relationship between the Topology Properties and Common Modal Logics”.
Theorem 6. (X ∧ Y ) ↔ (X ∧ Y ) is derivable in logic K.
A proof is given on page 69 in [1]. Notice that all the formulas that are valid in K4 are also valid in S4. Thus, K4 is called a sublogic of S4. In a similar way, B is a sublogic of S5 and K is a sublogic of all the others. Finally, D is a sublogic of T:
Theorem 7. Axiom D can be derived in logic T.
Proof. 1. P → P axiom T 2. ¬P → ¬P axiom T 3. P → ¬¬P contrapositive of 2 4. P → ¬¬P 1, 3, transitivity 9
Interpretations of Modal Logics in Topological Spaces
In this thesis we study interpretations of and ♦ in topological spaces. Let us recall some concepts in point set topology.
Definition 8. Let X be a set. A topology T on X is a collection of subsets of X with the following properties: (i) The empty set and X are in T ; (ii) The union of the elements of any subcollection of T is in T ; (iii) The intersection of the elements of any finite subcollection of T is in T . A set X for which a topology T has been specified is called a topological space. The elements in T are called open sets. A subset of X is called closed if its complement is open. If every subset of X is open, then we say that X is a discrete topological space. Given a topological space X and a subset A of X, the interior of A is the largest open set that is contained in A, denoted by int(A). The closure of A is the smallest closed set that contains A, denoted by A. A topological space X is called connected if ∅ and X are the only simultaneously closed and open subsets of X; it is called well-connected if there exists a least nonempty closed subset of X (see [3]).
For any interpretation f that maps propositional variables to subsets of X and for any propositional variable P , we interpret the logical connective
as follows: f(P ) = int(f(P )). 10
Then
C C C C f(♦P ) = f(¬¬P ) = (f(¬P )) = (int(f(¬P ))) = (int((f(P )) )) , which is the closure of the set f(P ). It was shown by McKinsey and Tarski in [2] that every finite well-connected topological space is an open image of a metric separable dense-in-itself space. This implies that S4 is complete with respect to any metric separable dense-in-itself space such as Rn. The original proof of McKinsey and Tarski was quite tedious and technical. Simpler, shorter, more geometric proofs were given in [4], [5], and [6]. It follows that S4 is sound and complete over all interpretations in all topological spaces. This means that a formula is derivable from the axiom system of S4 if and only if it is valid in all topological spaces. Note that if a formula is valid in some topological space, this does not mean that this formula is derivable from S4. For example, the formula (P → P ) is valid in any discrete topological space X because f(P ) = int(f(P )) = f(P ) for any interpretation f that sends propositional variables to subsets of X. But it is not derivable from S4 because it is not valid in all topological spaces. For example, consider the standard topology of
R with the interpretation f such that f(P ) = [0, 1]. Then
f(P ) = int[f(P )] = int([0, 1]) = (0, 1) + f(P ).
The rest of this thesis is organized as follows. In the second section, we show that adding a new axiom (P → P ) (which we name as Axiom X) to 11
S4 gives us a sound and complete axiom system over all discrete topological spaces (we name this new modal logic S5X). In the third section, for some families of finite topological spaces, we find an extension of S4 that is sound over them, although its completeness is an open question. In the fourth section, we will study this topic from a different direction: given any set X and any interpretation of in X that satisfies S4, we show that the image of this interpretation is a topology on X. We also study the influence of the modal axioms of S4 on topological properties of the image. THE LOGIC OF DISCRETE TOPOLOGICAL SPACES
Let S5X be S4 plus P → P . Notice that Axiom 5 (♦P → ♦P ) is derivable from Axiom X (P → P ) , and thus the axiom system of S5 is derivable from that of S5X. In other words, S5 is a sublogic of S5X. Further, from the first section we know the S4 axiom system consists of all the axioms and rules of the classical propositional logic with an addition of Axiom
K ((P → Q) → (P → Q)), Axiom T (P → P ), Axiom P 4 (P → P ), and the Necessitation Rule . Notice that adding Axiom P X to S4 makes (P ↔ P ) derivable in this new modal logic S5X. From the classical propositional logic and (P ↔ P ) we can derive Axioms K,T, 4,X and the Necessitation Rule. So the logic S5X is the same as classical propositional logic plus the axiom (P ↔ P ). In this section we will show that S5X is sound and complete over discrete topological spaces. Theorem 9 (the Soundness of S5X for All Discrete Topological Spaces). If a formula F is derivable from the axiom system S5X, then F is valid in every interpretation in every discrete topological space.
Proof. We know that S4 is sound for all topological spaces and thus is sound for all discrete topological spaces. So we only need to show that (P → P ) is valid in every discrete topological space. Let f be any interpretation in X, 13 where X is a discrete topological space. Observe that
f(P → P ) = f(¬P ∨ P )
= f(¬P ) ∪ f(P ) = (f(P ))C ∪ int(f(P ))
= (f(P ))C ∪ f(P )
= X.
Therefore, (P → P ) is valid in X.
Theorem 10 (the Completeness of S5X for All Discrete Topological Spaces). If F is valid in every discrete topological space, then F is deriable from the axiom system S5X.
Proof. Let Σ be the set of all axioms of S5X and let F be a formula. Assume that F is not derivable from S5X. We will show that F is not valid in some discrete topological space, i.e., there exists an interpretation f in a discrete topolgical space X such that f(F ) 6= X. Namely, consider (1) the set
X = {x| x is a maximal set of non-contradicting formulas containing Σ} with discrete topology; (2) f is an interpretation (mapping) such that any propositional variable A is mapped to the subset of X consisting of all points that include A. 14
Lemma 11. Let x ∈ X and F be any formula. Then exacly one of formulas F , ¬F is in x.
Proof. First, F and ¬F can not both be in x because F and ¬F are contradicting. Now we will show that it is not possible that F/∈ x and ¬F/∈ x. Assume that neither F or ¬F is in x. Then there exist two subsets of x, {A1,A2, ..., An} and {B1,B2, ..., Bm} such that
(A1 ∧ A2 ∧ ... ∧ An ∧ F ) → C and
(B1 ∧ B2 ∧ ... ∧ Bm ∧ ¬F ) → C for some contradiction C. Observe that
(A1 ∧ A2 ∧ ... ∧ An) ∧ (B1 ∧ B2 ∧ ... ∧ Bm)
→ (A1 ∧ A2 ∧ ... ∧ An ∧ B1 ∧ B2 ∧ ... ∧ Bm) ∧ (F ∨ ¬F )
→ (A1 ∧ A2 ∧ ... ∧ An ∧ F ) ∨ (B1 ∧ B2 ∧ ... ∧ Bm ∧ ¬F ) → C ∨ C → C.
This is impossible because A1,A2, ..., An,B1,B2, ..., Bm are in x and thus are non-contradicting.
Lemma 12. For each formula F , f(F ) = {x ∈ X|F ∈ x}.
Proof. We will prove this by induction on the complexity of F . Here we define the complexity degree of a formula to be the least number of the logical connectives ¬, ∧, ∨, this formula contains. For example, all propositional variables have complexity degree 0 and ¬(P ∧ Q) has complexity degree 2. 15
Initial Step: As defined, f(P ) = {x ∈ X|P ∈ x} for any propositional variable P .
Inductive Step: Assume f(F1) = {x ∈ X|F1 ∈ x} for any F1 that has a smaller complexity degree than F .
Case 1: F = F1 ∧ F2,
Since F1 and F2 have smaller complexity degrees than F , we have f(F1) = {x ∈ X|F1 ∈ x} and F2 = {x ∈ X|F2 ∈ x}. So
f(F ) = f(F1 ∧ F2)
= f(F1) ∩ f(F2)
= {x ∈ X|F1 ∈ x} ∩ {x ∈ X|F2 ∈ x}
= {x ∈ X|F1 ∈ x and F2 ∈ x}
We will show that F1,F2 ∈ x if and only if F1 ∧ F2 ∈ x. If F1,F2 ∈ x,
F1 and F2 are non-contradicting with the other formulas in x. Then so is
(F1 ∧ F2). Since x is a maximal set of noncontradicting formulas, F1 ∧ F2 ∈ x.
For the other direction, if (F1 ∧ F2) ∈ x, then F1 ∧ F2 does not contradict with the other formulas in x. Hence both F1 and F2 are non-contradicting with the other formulas in x. So F1,F2 ∈ x. Therefore
f(F ) = f(F1 ∧ F2) = {x ∈ X|(F1 ∧ F2) ∈ x} = {x ∈ X|F ∈ x}.
Case 2: F = F1 ∨ F2.
Notice that F1 ∈ x or F2 ∈ x if and only if F1 ∨ F2 ∈ x. Then 16
f(F ) = f(F1 ∨ F2)
= f(F1) ∪ f(F2)
= {x ∈ X|F1 ∈ x} ∪ {x ∈ X|F2 ∈ x}
= {x ∈ X|F1 ∈ x or F2 ∈ x}
= {x ∈ X|(F1 ∨ F2) ∈ x}
= {x ∈ X|F ∈ x}.
Case 3: F = ¬F1. We have
f(F ) = f(¬F1)
= X \ (f(F1))
= X \{x ∈ X|F1 ∈ x}
= {x ∈ X|F1 ∈/ x}
= {x ∈ X|¬F1 ∈ x} (by Lemma 11).
Case 4: F = F1. 17
Since (P ↔ P ) ∈ x for any x ∈ X, it follows that F1 ∈ X if and only if F1 ∈ x. Then we have
f(F ) = f(F1)
= int(f(F1))
= f(F1) (since X is discrete)
= {x ∈ X|F1 ∈ x}
= {x ∈ X|F1 ∈ x} = {x ∈ X|F ∈ x}.
Therefore, f(F ) = {x ∈ X|F ∈ x} for any formula F .
Proof of Theorem 10 continued. Since F is not derivable from S5X, it follows that the axioms in Σ and ¬F are non-contradicting. Hence, there exists y ∈ X such that ¬F ∈ y. By Lemma 11, F/∈ y. Then y∈ / f(F ). So f(F ) 6= X and thus F is not valid in X. NON-COMPLETENESS OF S4 OVER SOME FAMILIES OF TOPOLOGICAL SPACES
Recall that S4 is complete over all topological spaces. Namely, if a formula is valid in every topological space, then it is derivable from S4. In this section, we show that the formulas (♦P ) ∨ (♦¬P ) (we name it Axiom Y ) and (♦P ) ∨ (♦¬P ) (we name it Axiom Z) are not derivable from S4 but is valid in some families of topological spaces, i.e. S4 is not complete over these families of topological spaces.
Theorem 13. Axiom Y ((♦P ) ∨ (♦¬P )) is not derivable from S4.
Proof. Since S4 is sound in R with the standard topology, it follows that if a formula is derivable from S4, then it is valid in R. We will show that Axiom Y is not valid in standard R, i.e., there exists an interpretation f such that f((♦P ) ∨ (♦¬P )) 6= R. Consider f(P ) = Q ⊆ R. Then f((♦P ) ∨ (♦¬P )) = int(Q) ∪ int(QC ) = ∅ ∪ int(R \ Q) = ∅ ∪ ∅ = ∅.
Theorem 14. Axiom Z ((♦P ) ∨ (♦¬P )) is not derivable from S4.
Proof. We will also show that Axiom Z is not valid in standard R. Let a, b ∈ R and a < b. Consider f(P ) = [a, b]. Then
f((♦P ) ∨ (♦¬P )) = int(f(P )) ∪ int(f(P ))C ) = int([a, b]) ∪ int((−∞, a) ∪ (b, ∞))
= (a, b) ∪ int((−∞, a] ∪ [b, ∞))
= (a, b) ∪ ((−∞, a) ∪ (b, ∞))
= R \{a, b}= 6 R. 19
So Axiom Z is not valid in standard R, and thus is not derivable from S4.
Definition 15. Let X be any nonempty set and p ∈ X. The collection
Tp = {S ⊆ X| p ∈ S or S = ∅} of subsets of X is called a particular point topology on X (see page 44 in [7]).
Remark 16. Since for any paricular point topological space X with the particular point p ∈ X, any nonempty open subset includes p, it follows that every proper closed set of X does not include p.
Definition 17. Let X be any nonempty set and e ∈ X. The collection
Te = {S ⊆ X| e∈ / S or S = X} of subsets of X is called an excluded point topology on X (see page 47 in [7]).
Remark 18. Since for any excluded point topological space X with the excluded point e ∈ X, every proper open subset of X that excludes e, it follows that every nonempty closed set includes e.
Let {Xn}n∈N be a family of finite topological spaces where
Xn = {x1, x2, ..., xn} with the topology
Tb = {S ⊆ X|S = ∅ or S = {x1, x2, ..., xk} for 1 ≤ k ≤ n} (see Figure 1). 20
. . .
Figure 1. The Topological Spaces Xi for i = 2, 3, and n
Remark 19. For each Xn, Tb is a subtopology of both a particular point topology (with the particular point being x1) and an excluded point topology
(with the excluded point being xn).
Remark 20. The family of topological spaces {Xn} arises in a natural way in algebraic topology, namely, these are quotient spaces of the standard
CW-complexes for RPn (real projective planes).
Remark 21. Any set Xof cardinality larger than 1 with a particular point topology, an excluded point topology, or a topology Tb is T0, i.e., for any two distinct points x, y ∈ X, there is an open set U such that x ∈ U and y∈ / U, or y ∈ U and x∈ / U, but is not T1, i.e., it is not true that for any two distinct 21 points x, y ∈ X there exist two open sets U and V such that x ∈ U and y∈ / U, and y ∈ V and x∈ / V (see page 67 in [8]).
Definition 22. Sierpi´nskispace is a topological space with two points, only one of which is closed (see page 44 in [7]).
Remark 23. Notice that the Sierpi´nskispace (see the first graph in Figure 1) is a topological space with paricular point topology, excluded point topology, and topology Tb.
Theorem 24. Axiom Y is valid in (1) any particular point topological space; (2) any excluded point topological space;
(3) any finite topological space with topology Tb;
∞ (3 ) the countably infinite topological space X∞ = {x1, x2, ...} with the topology Tb∞ = {S ⊆ X|S = X or S = ∅ or S = {x1, x2, ..., xk} for k ∈ N} (see Figure 2).
. . .
Figure 2. The Topological Space X∞
Proof. Let f be any interpretation and P be any formula. Let Y = f(P ). We will show that
f((♦P ) ∨ (♦¬P )) = int(Y ) ∪ int(Y C ) = X. 22
(1) Let X be a particular point topological space with particular point p. Case 1.1: p ∈ Y . Then Y is open. So int(Y ) = Y . Since the only closed set that includes p is X, int(Y ) = Y = X. Therefore int(Y ) ∪ int(Y C ) = X. Case 1.2: p∈ / Y . Then p ∈ Y C . Similarly to Case 1.1, we get int(Y ) ∪ int(Y C ) = X. (2) Let X be a nonempty set with the excluded point topology. Let p ∈ X be the excluded point. Case 2.1: Y = X. Then int(Y ) = X = X and thus int(Y ) ∪ int(Y C ) = X. Case 2.2: Y = ∅. Then Y C = X. Similarly to Case 2.1, int(Y ) ∪ int(Y C ) = X. Case 2.3: Y = X \{e}. Then int(Y ) = Y and thus int(Y ) = Y = X. So int(Y ) ∪ int(Y C ) = X. Case 2.4: Y = {e}. Then int(Y C ) = X \{e}. Similar to Case 2.3, int(Y ) ∪ int(Y C ) = X. Case 2.5: e ∈ Y ⊂ X and Y 6= {e}. Then int(Y ) = Y \{e} = Y and int(Y C ) = Y C = Y C ∪ {e}. So int(Y ) ∪ int(Y C ) = Y ∪ Y C ∪ {e} = X. Case 2.6: e∈ / Y , Y 6= ∅, and Y 6= (X \{e}). Then e ∈ Y C , Y C ⊂ X, and Y C 6= {e}. Similarly to Case 2.5, int(Y ) ∪ int(Y C ) = X.
(3) Let X = {x1, x2, ..., xn} and
Tb = {S ⊆ X|S = ∅ or S = {x1, x2, ..., xk} for 1 ≤ k ≤ n}. Then the collection 23 of all closed sets in X is
{C ⊆ X|C = X or C = X \{x1, x2, ..., xk} for 1 ≤ k ≤ n}
={C ⊆ X|C = X or C = {xk+1, xk+2, ..., xn} for 1 ≤ k ≤ n}.
Notice that the only closed set that includes x1 is X. So if x1 ∈ Y , then
C x1 ∈ int(Y ), and thus int(Y ) = X. So int(Y ) ∪ int(Y ) = X. If x1 ∈/ Y , then
C C C x1 ∈ Y . Similarly, we get int(Y ) = X. Then int(Y ) ∪ int(Y ) = X.
∞ (3 ) Since collection of closed sets in X∞ is
{C ⊆ X∞|C = ∅ or C = X∞ or C = {xk+1, xk+2, ...} for k ∈ N},
the only closed set in X∞ that includes x1 is X∞. Similarly to Case 3, for
C both cases x1 ∈ X∞ and x1 ∈/ X∞, we get int(Y ) ∪ int(Y ) = X.
Theorem 25. Axiom Z is valid in (1) any particular point topological space;
(2) any finite topological space with topology Tb;
∞ (2 ) the countably infinite topological space X∞ = {x1, x2, ...} with the topology Tb∞ = {S ⊆ X|S = X or S = ∅ or S = {x1, x2, ..., xk} for k ∈ N} (see Figure 2).
Proof. Let f be any interpretation and P be any formula. Let Y = f(P ). We will show that
f((♦P ) ∨ (♦¬P )) = int(Y ) ∪ int(Y C ) = X. 24
(1) Let X be a particular point topological space with particular point p. Case 1.1: p ∈ Y . Since the only closed set that includes p is X, Y = X. Then int(Y ) = int(X) = X. So int(Y ) ∪ int(Y C ) = X. Case 1.2: p∈ / Y . Then p ∈ Y C . Similarly to Case 1.1, int(Y ) ∪ int(Y C ) = X.
(2) Let X = {x1, x2, ..., xn} and
Tb = {S ⊆ X|S = ∅ or S = {x1, x2, ..., xk} for 1 ≤ k ≤ n}. Let Y = f(P ).
Notice that the only closed set that includes x1 is X. So if x1 ∈ Y , then
C int(Y ) = int(X) = X, and thus int(Y ) ∪ int(Y ) = X. If x1 ∈/ Y , then
C C C x1 ∈ Y . Similarly, we get int(Y ) = X. Then int(Y ) ∪ int(Y ) = X.
∞ (2 ) We know that the only closed set in X∞ that includes x1 is X∞.
Similarly to Case 2, for both cases x1 ∈ X∞ and x1 ∈/ X∞, we get int(Y ) ∪ int(Y C ) = X.
Theorem 26. There exists an excluded point topological space where Axiom Z is not valid.
Proof. Consider X = {a, b, c} with an excluded point topology where the point c is the excluded point. Let f(P ) = {b, c} for some interpretation f and formula P . Then f((♦P ) ∨ (♦¬P )) = int(f(P )) ∪ int((f(P ))C ) = int({b, c}) ∪ int({a}) = int({b, c}) ∪ int{a, c} = {b} ∪ {a} = {a, b}= 6 X.
Remark 27. Axiom Z distinguishes the particular point topology and the excluded point topology.
Remark 28. Axiom Z is not derivable from S4+Axiom Y . 25
Theorem 29. Axiom Y is not derivable in S4+Axiom Z.
Proof. Consider X = {a, b} with indiscrete topology, i.e., only X and ∅ are open in X. We will show that Axiom Z is valid in this topological space but Axiom Y is not. First we will show Axiom Z is valid in X. Let Y = f(P ). We will show that
f((♦P ) ∨ (♦¬P )) = int(Y ) ∪ int(Y C ) = X.
Case 1: Y = X. Then int(Y ) ∪ int(Y C ) = X ∪ ∅ = X ∪ ∅ = X. Case 2: Y = ∅. Similarly to Case 1, int(Y ) ∪ int(Y C ) = X. Case 3: Y = {a}. Then int(Y ) ∪ int(Y C ) = int({a}) ∪ int({b}) = int(X) ∪ int(X) = X ∪ X = X. Case 4: Y = {b}. Similarly to Case 3, int(Y ) ∪ int(Y C ) = X. Now we will show that Axiom Y is not valid in X. Consider f(P ) = {a}. Then f((♦P ) ∪ (♦¬P )) = int(f(P )) ∪ int((f(P ))C ) = int({a}) ∪ int({b}) = ∅ ∪ ∅ = ∅= 6 X.
Remark 30. Axiom Y and Axiom Z are independent, i.e., Axiom Y does not imply Axiom Z, and Axiom Z does not imply Axiom Y .
As mentioned in Introduction, McKinsey and Tarski’s result in [2] that every finite well-connected topological space is an open image of a metric separable dense-in-itself space implies that S4 is complete with respect to any metric separable dense-in-itself space. However, this does not imply that S4 is complete with respect to any finite well-connected space itself because each
finite member in {Xn} is well-connected but S4 is not complete in it. THE RELATIONSHIP BETWEEN THE TOPOLOGICAL PROPERTIES AND COMMON MODAL LOGICS
As we saw in the first section, S4 is sound with respect to any interpretation in any topological space. In this section, we reverse the question. Suppose we are given a set X and a mapping i : P(X) → C, where P(X) is the power set of X and C = i(P(X)) ⊆ P(X). For any mapping f : {P |P is a propositional variable} → P(X) and any formulas F and G, define f(F ∧ G) = f(F ) ∩ f(G), f(F ∨ G) = f(F ) ∪ f(G), f(¬F ) = (f(F ))C ,
f(F ) = i(f(F )).
We say that a formula F is valid in X if for every such mapping f, f(F ) = X. Suppose that such mapping i makes S4 sound, i.e., all formulas derivable in S4 are valid in X. We ask the following questions. Is the set C = i(P(X)) a topology of X? If so, does i have to satisfy all axioms and rules of S4 to guarantee that C = i(P(X)) is a topology of X? We show that the answers to both questions above are “yes”. If any axiom of S4 is dropped, then C is not necessarily a topology. Moreover, we determine which axioms of modal logic are responsible for which axioms of the topological spaces. The following table (Table 5) shows us which axioms of topology ((A) ∅ ∈ C; (B) X ∈ C; (C) X1,X2 ∈ C ⇒ (X1 ∩ X2) ∈ C; (D) S {Xj}j∈J ⊆ C ⇒ j∈J Xj ∈ C) the set C = i(P(X)) has when i satisfies the axiom system of logic L, where L= (1) the classical logic equipped with a modal operator and the Necessitation Rule (denoted CL below), (2) CL 27 plus Axioms T, 4, (3) K, (4) K4, (5) D, (6) D4 (7) T, (8)S4, (9)S5. The symbol ”×” means that C does not have the property with the assigned logic, and ”X” means it does. Of course, unless we put some interpretation on , i can be any mapping, so we can not expect it to have any topological properties. However, as we will see in Proposition 33, requiring as little as the Necessitation Rule already guarantees one of the properties.
Table 5. The Effect of Modal Axioms on Topological Properties
A. B. C. D. X ∈ C ∅ ∈ C X1,X2 ∈ C ⇒ {Xj}j∈J ⊆ C ⇒ S (X1 ∩ X2) ∈ C j∈J Xj ∈ C 1. CL X × × × 2. CL+ Axioms T, 4 X X × × 3. Logic K (CL+ X × X × Axiom K) 4. Logic K4 (CL+ X × X × Axioms K, 4) 5. Logic D (CL + X X X × Axioms K,D 6. Logic D4 (CL+ X X X × Axioms K, 4,D ) 7. Logic T (CL + X X X × Axioms K,T ) 8. Logic S4 (CL + X X X X Axioms K, 4,T ) 9. Logic S5 (CL + X X X X Axioms K,T, 5)
Figure 3 is a map of the sublogic relationship among all the modal logics we discuss in Table 5. 28
Figure 3. The Sublogic Relationship among Some Modal Logics
Observe in Table 5 that no proper sublogic of S4 guarantees that C is a topology. Also observe that Axiom D is weaker than Axiom T , but logic D guarantees the same properties as logic T. It is also interesting that logic D plus Axiom 4 (i.e., D4) guarantees the same properties as logic D, but the logic T plus Axiom 4 (i.e., S4) guarantees more properties than logic T.
Lemma 31. If Axiom K is satisfied, then i is an increasing mapping. 29
Proof. Let A ⊆ B ⊆ X. We will show that i(A) ⊆ i(B). Let f(F ) = A and f(G) = B for some interpretation f and formulas F and G. Then
i(A) = i(A ∩ B) = i(f(F ) ∩ f(G)) = i(f(F ∧ G)) = f((F ∧ G))
= f(F ∧ G) = f(F ) ∩ f(G) = i(f(F )) ∩ i(f(G)) = i(A) ∩ i(B) ⊆ i(B).
Lemma 32. Let i satisfy logic S4. If A ∈ C, then i(A) = A.
Proof. First we will show that i(A) ⊆ A. Let A = f(F ) for some interpretation f and formula F . Since F → F , it follows that i(A) = i(f(F )) = f(F ) ⊆ f(F ) = A. For the other direction, since A ∈ C = i[P(X)], there exists B ∈ P(X) such that i(B) = A. Now let B = f(G) for some interpretation f and formula G. Then
i(A) = i(i(B)) = f(F ) ⊆ f(F ) = i(B) = A
. Therefore, i(A) = A.
First we show that all logics listed in Table 5 guarantee the property X ∈ C (see Column A of Table 5).
Proposition 33. The property X ∈ C holds for any set X and any mapping i : P(X) → C that satisfies the axiom system of the logic L, where L= CL, CL plus Axioms T, 4, K, K4, D, D4, T, S4, or S5. 30
Proof. It is sufficient to prove this is true for CL because it is a sublogic of all the other logics. We will show that i(X) = X ∈ C. Let A be any formula derivable in this logic. Then f(A) = X for any interpretation f. By the
Necessitation Rule, A is valid. Therefore, i(X) = i(f(A)) = f(A) = X.
Now we show that CL, logic K, or logic K4 does not guarantee the property ∅ ∈ C (see boxes B1, B3, B4 of Table 5).
Proposition 34. Let X be a nonempty set. Then there exists a mapping i : P(X) → C = i(P(X)) such that i satisfies logic L, where L= CL, K, or K4, however, ∅ ∈/ C.
Proof. Define i(Y ) = X for any Y ⊆ X. In this case, C = {X} and thus the property ∅ ∈ C does not hold. However, this mapping i satisfies the Necessitation Rule, Axiom K, and Axiom 4. It satisfies the Necessitation
Rule (if F 7→ X, then F 7→ X) because for any formula F , f(F ) = i(f(F )) = X. The Axiom K ((P → Q) → (P → Q)) is valid in X because for any formulas P , Q and for any interpretation f, 31
f((P → Q) → (P → Q)) = f(¬(P → Q) ∨ (P → Q))
= f(¬(¬P ∨ Q) ∨ (¬P ∨ Q))
= f(¬(¬P ∨ Q)) ∪ f(¬P ∨ Q)
C = (f((¬P ∨ Q))) ∪ f(¬P ) ∪ f(Q)
C C = (i(f(¬P ∨ Q))) ∪ (f(P )) ∪ i(f(Q)) = (i((f(P ))C ∪ f(Q)]}C ∪ (i(f(P )))C ∪ i(f(Q))
= XC ∪ XC ∪ X
= X.
The Axiom 4 (P → P ) is valid in X because for any formula P and any interpretation f,
C C f(P → P ) = f(¬P ∨ P ) = (i(f(P ))) ∪ i(i(f(P ))) = X ∪ X = X.
The following proposition shows that the logics 2, 5, 6, 7, 8, and 9 listed in Table 5 guarantee the property ∅ ∈ C (see boxes B2, B5, B6, B7, B8, and B9 of Table 5).
Proposition 35. If i satisfies the axiom system of logic L, where L= CL plus Axioms 4, T , D, D4, T, S4, or S5, then for every set X and each mapping i : P(X) → C, ∅ ∈ C. 32
Proof. Since each logic in the proposition includes CL and Axiom D (P → ♦P ), it is sufficient to show the case that i satisfies CL plus Axiom D. Suppose that i satisfies CL plus Axiom D. Let F be a contradiction in the classical logic, e.g. F = S ∧ ¬S. Then f(F ) = ∅ for any interpretation f. So
C C C C i(f(F )) = f(F ) ⊆ f(♦F ) = f(¬¬F ) = (i((f(F )) )) = (i(∅ )) = (i(X))C = XC (As proved in the proof of Proposition 33, i(X) = X.)
= ∅.
Therefore, ∅ ∈ C.
Now we discuss the results of C1 and C2 in Table 5.
Proposition 36. There exists a set X and a mapping i : P(X) → C that satisfies CL or CL plus Axioms T and 4, but the property
X1,X2 ∈ C ⇒ (X1 ∩ X2) ∈ C does not hold.
Proof. Consider the set X = {a, b, c} and the mapping i such that
∅ if Y = {b} i(Y ) = Y if Y 6= {b}
The mapping i satisfies the necessitation rule because if f(F ) = X, then f(F ) = i(f(F )) = i(X) = X. It satisfies Axiom 4 because for any subset Y of X such that Y 6= {b}, i(i(Y )) = i(Y ) and for Y = {b}, i(i({b})) = i(∅) = ∅ = i({b}). It satisfies Axiom T because i(Y ) ⊆ Y for any 33
Y ⊆ X. However, the property X1,X2 ∈ C ⇒ (X1 ∩ X2) ∈ C does not hold because i({a, b}) ∩ i({b, c}) = {a, b} ∩ {b, c} = {b} ∈/ C.
The following proposition concerns the results C3, C4, C5, C6, C7, C8, and C9 in Table 5.
Proposition 37. The property X1,X2 ∈ C ⇒ (X1 ∩ X2) ∈ C holds for any mapping i : P(X) → C that satisfies logic K, K4, D, T, ,D4, S4, or S5.
Proof. It is sufficient to prove this is true for logic K because it is a sublogic of all the other listed logics. Let X1,X2 ∈ C. Then X1 = i(A),X2 = i(B) for some A, B ⊆ X. Let A = f(F ) and B = f(G) for some interpretation f and formulas F and G. Then
X1 ∩ X2 = i(A) ∩ i(B) = i(f(F )) ∩ i(f(G)) = f(F ) ∩ f(G)
= f(F ∧ G) = f((F ∧ G)) (by Theorem 6) = i(f(F ∧ G)) = i(f(F ) ∩ f(G)) = i(A ∩ B) ∈ C.
Now let us look at the results D1, D2, D3, D4, D5, D6, and D7.
Proposition 38. There exists a set X and a mapping i : P(X) → C that satisfies logic L (where L =CL, CL plus Axioms 4,T , K, K4, D, T, or S D4), but the property {Xj}j∈J ⊆ C ⇒ j∈J Xj ∈ C does not hold for the set C.
Proof. Results D1 and D2: L =CL, or CL plus Axioms 4,T . 34
Consider X = {a, b, c} and the mapping i such that
{a} if Y = {a, b}, {a, c}, i(Y ) = Y if Y 6= {a, b}, {a, c}.
The mapping i satisfies the necessitation rule because if f(F ) = X for some formula F , then
f(F ) = i(f(F )) = i(X) = X.
It satisfies Axiom T because i(Y ) ⊆ Y for any Y ⊆ X. Finally, it satisfies Axiom 4 because i(i(Y )) = i(Y ) for any Y ⊆ X. The union property does not hold because i({a}) ∪ i({b}) = {a, b} ∈/ C
Results D3, D4, D5, and D6: L= K, K4, D, or D4. Consider the set X = {a, b, c} and the mapping i such that
i(∅) = i({c}) = ∅, i({a}) = i({a, c}) = {a}, i({a, b}) = i(X) = X, i({b}) = i({b, c}) = {b}.
Then
(i) Axiom K ((P → Q) → (P → Q) is valid in X since
i(Y C ∪ Z) ⊆ (i(Y ))C ∪ i(Z) for any Y,Z ⊆ X. 35
Case 1: Y = X. Then i(Y C ∪ Z) = i(Z) ⊆ (i(Y ))C ∪ i(Z) for any Z ⊆ X. Case 2: Y = ∅. Then i(Y C ∪ Z) = i(X) = (i(Y ))C ∪ i(Z) for any Z ⊆ X. Case 3: Y = {a}. We need to show that i({b, c} ∪ Z) ⊆ {b, c} ∪ i(Z) for any Z ⊆ X. If Z = ∅, {b}, {c}, or {b, c}, then
i({b, c} ∪ Z) = i({b, c}) = {b} ⊆ {b, c} ∪ i(Z).
If Z = {a}, {a, b}, {a, c}, or X, then
i({b, c} ∪ Z) = i(X) = X = {b, c} ∪ i(Z).
Case 4: Y = {b}. The proof in this case can be obtained from that in Case 3 by switching a and b. Case 5: Y = {c}. Then
i(Y C ∪ Z) = i({a, b} ∪ Z) ⊆ X = (i(Y ))C ∪ i(Z).
Case 6: Y = {a, b}. Then
i(Y C ∪ Z) = i({c} ∪ Z) = i(Z) ⊆ (i(Y ))C ∪ i(Z).
Case 7: P = {b, c}. We will show that i({a} ∪ Z) ⊆ {a, c} ∪ i(Z) for any Z ⊆ X. 36
If Z = ∅, {a}, {c}, or {a, c}, then
i({a} ∪ Z) = {a} ⊆ {a, c} = {a, c} ∪ i(Z).
If Z = {b}, {a, b}, {b, c}, or X, then
i({a} ∪ Z) = X = {a, c} ∪ i(Z).
Case 8: P = {a, c}. The proof in this case can be obtained from that in Case 7 by switching a and b. (ii) Axiom 4 is valid in X because i(Y ) = i(i(Y )) for any Y ⊆ X.
C C (iii)Axiom D(P → ♦P ) is satisfied, namely, i(Y ) ⊆ (i(Y )) for all Y ⊆ X. This is equivalent to i(Y C ) ⊆ (i(Y ))C for all Y ⊆ X. We know that Axiom K is satisfied. So we have i(Y C ∪ ∅) ⊆ (i(Y ))C ∪ i(∅) for all Y ⊆ X. Therefore, i(Y C ) ⊆ (i(Y ))C for all Y ⊆ X. (iv) The union property does not hold because
i({a}) ∪ i({b}) = {a} ∪ {b} = {a, b} ∈/ C.
Result D7: L= T. Consider the set X = {a, b, c} and the mapping i such that 37
{a} if Y = {a, b} {b} if Y = {b, c} i(Y ) = {c} if Y = {a, c} X if Y = X ∅ otherwise
(i) Axiom K is valid in X, namely, i(Y C ∪ Z) ⊆ ((i(Y ))C ∪ i(Z)) for any Y,Z ⊆ X. Case 1: Y = X. Then i(Y C ∪ Z) = i(∅ ∪ Z) = i(Z) = ∅ ∪ i(Z) = (i(Y ))C ∪ i(Z) for any Z ⊆ X. Case 2: Y is a subset of X such that i(Y ) = ∅. Then i(Y C ∪ Z) ⊆ X = (i(Y ))C ∪ i(Z) for any Z ⊆ X. Case 3: Y = {a, b}, or {b, c}, or {a, c}. Without loss of generality, let Y = {a, b}. We need to show that i({c} ∪ Z) ⊆ {b, c} ∪ i(Z). If Z = X, then i({c} ∪ Z) = i(X) = X = {b, c} ∪ i(Z). If Z = ∅, then i({c} ∪ Z) = i({c}) = ∅ ⊆ {b, c} ∪ i(Z). If Z = {a} or {a, c}, then i({c} ∪ Z) = i({a, c}) = {c} ⊆ {b, c} = {b, c} ∪ i(Z). If Z = {b} or {b, c}, then i({c} ∪ Z) = i({b, c}) = {b} ⊆ {b, c} = {b, c} ∪ i(Z). If Z = {c}, then i({c} ∪ Z) = i({c}) = ∅ ⊆ {b, c} ∪ i(Z). If Z = {a, b}, then i({c} ∪ Z) = i(X) = X = {b, c} ∪ i(Z). (ii) Axiom T is valid in X since for all Y ⊆ X, i(Y ) ⊆ Y. 38
(iii) The union property does not hold because
i({a, b}) ∪ i({b, c}) = {a} ∪ {b} ∈/ C.
Proposition 39. If i : P(X) → C satisfies logic S4 or S5, then the property S {Xj}j∈j ⊆ C ⇒ j∈J Xj ∈ C holds for the set C.
Proof. It is sufficient to prove this is true for logic S4 because it is a sublogic of logic S5. Let {Xj}j∈J ⊆ C. By Lemma 32, we have
[ Xj = i(Xj) ⊆ i( Xj) j∈J
for each Xj. Then [ [ Xj ⊆ i( Xj). j∈J j∈J
By Axiom T (P → P ), we have
[ [ i( Xj) ⊆ Xj. j∈J j∈J
Therefore, [ [ Xj = i( Xj) ∈ C. j∈J j∈J CONCLUSIONS
We began with studying the logic of discrete topological spaces. Knowing that S4 is sound and complete over all topological spaces, we showed that S4 plus P → P (we called it S5X) is a sound and complete axiom system over all discrete topological spaces. In order to prove the completeness, we constructed a discrete topological space X by mimicking the classical canonical model. Then we showed that any formula that is not derivable from S5X is not valid in X. Next, we extended S4 with independent formulas Axiom Y
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