1. Introduction: Modal Logic Language and Its Interpretation in Topological Spaces

Definition 1.1. Let be a modal language consisting of propositional variables, conjunction , disjunction negation , and modality Implication and biconditional are defined in the traditional way ( and ). Modality is defined as the dual to , i.e.

Modal logics have a variety of interpretations and applications in difference sciences, from philosophy to computer science. Depending on the context and interpretation of , its different properties may be desired. Some of the most meaningful, useful, and studied axioms are listed below.

Table 1. Some Axioms in Modal Logics

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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:

Table 2. Common Modal Logics

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Definition 1.2. A topological model of is a pair , where

1. is a topological space, and

2. is a function mapping formulas in to subsets of . It assigns each propositional variable an arbitrary subset and satisfies the following conditions for every :

3. where is the topological interior operator.

Remark 1.3. Note that the modality then maps to the topological closure operator, i.e.

Definition 1.4. Let be any formula. We say that is valid in a topological space if for any topological model , we have .

Theorem 1.5. (Topological completeness of S4, ^{[3, 4, 5, 6, 7]}) For any formula , the following two statements are equivalent:

• is derivable in S4;

• is valid in all topological spaces.

In the rest of this paper, we will study this topic from a different direction: given any set and any interpretation of in that satisfies S4 (this concept is defined in the next section), we show that the image of this interpretation is a topology on . Moreover, in section 2 we summarize the influence of the modal axioms of S4 on topological properties of the image. Proofs are given in section 3.

2. The Relationship between the Topological Properties and Common Modal Logics

Suppose we are given a set and a mapping , where is the power set of and . For any mapping is a propositional variable} and any formulas and , define

We say that a formula is valid in if for every such mapping , . We say that a mapping satisfies a modal logic L, or equivalently, makes L sound, if all formulas derivable in L are valid in .

Suppose that such a mapping makes S4 sound. We ask the following questions. Is the set a topology of ? If so, does have to satisfy all axioms of S4 to guarantee that is a topology of ? We show that the answers to both questions above are “yes". If any axiom of S4 is dropped, then is not necessarily a topology. Moreover, we determine which axioms of modal logic are responsible for which axioms of the topological spaces. More precisely, Table 3 below shows which axioms of topology,

•

•

•

•

the set has when i satisfies the axiom system of logic L, where L is one of the following:

(1)

(2) plus Axioms and 4,

(3) K,

(4) K4,

(5) D,

(6) D4,

(7) T,

(8) S4,

(9) S5.

Table 3. The Effect of Modal Axioms on Topological Properties

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In this table, the symbol "×" means that does not have to have the property with the assigned logic, and "√" means it does. Of course, unless we put some interpretation on , can be any mapping, so we can not expect it to have any topological properties. However, as we see (in cell 1A), requiring as little as the Necessitation Rule already guarantees one of the properties.

Observe that no proper sublogic of S4 guarantees that is a topology. Also observe that Axiom D is weaker than Axiom , but logic D guarantees the same properties as logic T. It is also interesting that adding Axiom 4 to K or D (and obtaining K4 or D4, respectively) does not add any topological property, but adding Axiom 4 to T (and obtaining S4) does.

3. Proofs

The following two lemmas are proved e.g. in [^[2], p.69] and [^[1],p. 38].

Lemma 3.1. The formula is derivable in logic K.

Lemma 3.2. Axiom D follows from Axiom T.

Lemma 3.3. If Axiom is satisfied, then i is an increasing mapping.

Proof. Let We will show that Let and for some interpretation and formulas and Then

Lemma 3.4. Let satisfy logic S4. If then

Proof. First we will show that Let for some interpretation and formula Since it follows that For the other direction, since there exists such that Now let for some interpretation and formula Then

Therefore,

Proposition 3.5. Column A in Table 3. The property holds for any set X and any mapping that satisfies the axiom system of the logic L, where plus Axioms T, 4, K, K4, D, D4, T, S4, or S5.

Proof. It is sufficient to prove this is true for because it is a sublogic of all the other logics. We will show that Let be any formula derivable in this logic. Then for any interpretation By the Necessitation Rule, is valid. Therefore,

Proposition 3.6. Results B1, B3, and B4 in Table 3. Let be a nonempty set. Then there exists a mapping such that i satisfies logic L, where K, or K4, however,

Proof. Define for any In this case, and thus the property does not hold. However, as we will show below, this mapping satisfies the Necessitation Rule, Axiom , and Axiom 4.

It satisfies the Necessitation Rule (if then ) because for any formula F such that we have

Axiom is valid in because for any formulas and for any interpretation

Axiom 4 is valid in because for any formula and any interpretation

Proposition 3.7. Results B2, B5, B6, B7, B8, and B9 in Table 3. If satisfies the axiom system of logic L, where plus Axioms 4, D, D4, T, S4, or S5, then

Proof. Since each logic in the proposition includes and Axiom it is sufficient to show the case when i satisfies plus Axiom Suppose that satisfies plus Axiom Let be a contradiction in the classical logic, e.g. Then for any interpretation So

Therefore,

Proposition 3.8. Results C1 and C2 in Table 3. There exists a set and a mapping that satisfies or plus Axioms T and 4, but the property does not hold.

Proof. Consider the set and the mapping i such that

The mapping satisfies the necessitation rule because if then It satisfies Axiom 4 because for any subset of such that we have and for It satisfies Axiom because for any However, the property does not hold because

Proposition 3.9. Results C3, C4, C5, C6, C7, C8, and C9 in Table 3. The property holds for any mapping 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 Then for some Let and for some interpretation and formulas and Then

Proposition 3.10. Results D1, D2, D3, D4, D5, D6, and D7 in Table 3. There exists a set and a mapping that satisfies logic L (where plus Axioms K, K4, D, T, or D4), but the property does not hold for the set

Proof. Results D1 and D2: or plus Axioms

Consider and the mapping such that

The mapping satisfies the necessitation rule because if for some formula F, then It satisfies Axiom because for any Finally, it satisfies Axiom 4 because for any The union property does not hold because

Results D3, D4, D5, and D6: L= K, K4, D, or D4.

Consider the set and the mapping i such that

Then

(i) Axiom is valid in since, as we will see below,

Case 1: Then

for any

Case 2: Then

for any

Case 3: We need to show that for any

If or then

If or then

Case 4: The proof in this case can be obtained from that in Case 3 by switching and

Case 5: . Then

Case 6: Then

Case 7: We will show that for any

If or then

If , or then

Case 8: The proof in this case can be obtained from that in Case 7 by switching and

(ii) Axiom 4 is valid in because for any

(iii) Axiom is valid. Indeed, we know that Axiom is satisfied. So we have for all Therefore, for all we have which is equivalent to

(iv) The union property does not hold because

Result D7: L= T.

Consider the set and the mapping i such that

(i) Axiom is valid in namely, for any

Case 1: Then

for any

Case 2: is a subset of such that Then

for any

Case 3: or or . Without loss of generality, let We need to show that

If then

If then

If or then

If or then

If then

If then

(ii) Axiom is valid in since for all

(iii) The union property does not hold because

Proposition 3.11. Results D8 and D9 in Table 3. If satisfies logic S4 or S5, then the property holds for the set

Proof. It is sufficient to prove this is true for logic S4 because it is a sublogic of logic S5. Let By Lemma 4, we have

for each Then

By Axiom we have

Therefore,

Acknowledgments

Both authors would like to thank the College of Science and Mathematics of the California State University, Fresno for supporting this work.

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