Topologies on pseudoinnite paths

Topologies on pseudoinnite paths Andrey Kudinov Institute for Information Transmission Problems, Moscow National Research University Higher School of Economics, Moscow Moscow Institute of Physics and Technology June 14, 216

PLAN 1. Language 2. Topological semantics 3. Products of logics 4. Neighborhood frames 5. Without seriality 6. Dense topological spaces 7. Logic S5 8. Ideas of the proof

Language and logics φ ::= p φ φ φ iφ, i = 1, 2., and i are expressible in the usual way. Normal modal logic. K n denotes the minimal normal modal logic with n modalities and K = K 1. L 1 and L 2 two modal logics with one modality then the fusion of these logics is dened as L 1 L 2 = K 2 + L 1 + L 2 ; where L i is the set of all formulas from L i where in all formulas is replaced by i.

Topological semantics We can dene topology on set X by specifying a closure operator C : 2 X 2 X, satisfying the Kuratowski axioms: 1. C( ) =, 2. A C(A), for A X, p p 3. C(A B) = C(A) C(B), for A, B X, (p q) p q 4. C ( C(A) ) C(A). A X, p p Logic S4 In topological semantics for modal logic closure operator correspond to. Topological model (X, θ), where X = (X, C) topological space: p θ(p) X θ(φ ψ) = θ(φ) θ(ψ) θ( φ) = X \ θ(φ) θ( φ) = C(θ(φ)). X = φ θ(θ(φ) = X), Log(X) = {φ X = φ}.

Alexandro topology On a transitive reexive Kripke frame F = (W, R) we can dene topology, i.e. closure operator: C F (A) = R 1 (A). It will be an Alexandro topology (any intersection of open sets is open, all points have minimal neighborhood). Lemma F = φ (W, C F ) = φ. We dene T op(f ) = (W, C F ) Completeness of S4 w.r.t. all Alexandro spaces. Many topologies are not Alexandro: R n, Cantor space, Q or any metric space.

Derivational semantics We can dene topological space using derivative operator d : 2 X 2 X, where d(a) is the set of all limit points of A. In a similar way we dene derivational semantics: θ( φ) = d(θ(φ)) The logic of all topological space is wk4 = K + p p p (Esakia'1981). The logic of Q, Cantor space (or any dense-in-itself zero-dimensional metric space) is D4 = K + p p + (Shehtman'199).

The product of Kripke frames For two frames F 1 = (W 1, R 1) and F 2 = (W 2, R 2) F 1 F 2 = (W 1 W 2, R 1, R 2), where (a 1, a 2)R 1(b 1, b 2) a 1R 1b 1 & a 2 = b 2 (a 1, a 2)R 2(b 1, b 2) a 1 = b 1 & a 2R 2b 2 For two logics L 1 and L 2 L 1 L 2 = Log({F 1 F 2 F 1 = L 1 & F 2 = L 2}) (Shehtman, 1978) For two classes of frames F 1 and F 2 Log({F 1 F 2 F 1 F 1 & F 2 F 2}) Log(F 1) Log(F 2)+ + 1 2p 1 2p + 1 2p 2 1p. K K = K K + 1 2p 1 2p + 1 2p 2 1p S4 S4 = S4 S4 + 1 2p 1 2p + 1 2p 2 1p.

The product of topological spaces (van Benthem et al, 25) For two topological space X 1 = (X 1, τ 1) and X 2 = (X 2, τ 2) X 1 X 2 = (X 1 X 2, τ 1, τ 2 ), where τ 1 has base {U 1 x 2 U 1 τ 1 & x 2 X 2} τ 2 has base {x 1 U 2 x 1 X 1 & U 2 τ 2}

The product of topological spaces (van Benthem et al, 25) For two topological space X 1 = (X 1, τ 1) and X 2 = (X 2, τ 2) X 1 X 2 = (X 1 X 2, τ 1, τ 2 ), where τ 1 has base {U 1 x 2 U 1 τ 1 & x 2 X 2} y τ 2 has base {x 1 U 2 x 1 X 1 & U 2 τ 2} x

The product of topological spaces (van Benthem et al, 25) For two topological space X 1 = (X 1, τ 1) and X 2 = (X 2, τ 2) X 1 X 2 = (X 1 X 2, τ 1, τ 2 ), where τ 1 has base {U 1 x 2 U 1 τ 1 & x 2 X 2} For two logics L 1 and L 2 τ 2 has base {x 1 U 2 x 1 X 1 & U 2 τ 2} L 1 t L 2 = Log({X 1 X 2 X 1 = L 1 & X 2 = L 2} S4 t S4 = Log(Q Q) = S4 S4 (van Benthem et al, 25) Log(R R) S4 S4 (Kremer, 21?) Log(Cantor Cantor) S4 S4 d-logic of product of topological spaces was considered by L. Uridia (211). Log d (Q Q) = D4 D4 Generalization to neighborhood frames was done by K. Sano (211).

Known results Theorem (212) Let L 1 and L 2 be from the set {D, T, D4, S4} then L 1 n L 2 = L 1 L 2. Not straightforward but still a Corollary In derivational semantics 1. D4 d D4 = D4 D4. 2. [Uridia'211] Log d (Q Q) = D4 D4

Topological semantics based on closure operator or derivative operator can be generalized in the neighborhood semantics. We can consider neighborhood function τ : X 2 2X. For x X τ(x) is a set of neighborhoods of x. It connected with C is the following way: A τ(x) x I(A), where I(A) = X \ C(X \ A). And for derivational semantics A τ(x) x d(a), where d(a) = X \ d(x \ A).

Neighborhood frames A (normal) neighborhood frame (or an n-frame) is a pair X = (X, τ), where X ; τ : X 2 2X, such that τ(x) is a lter on X; τ neighborhood function of X, τ(x) neighborhoods of x. Filter on X: nonempty F 2 X such that 1) U F & U V V F 2) U, V F U V F (lter base) The neighborhood model (n-model) is a pair (X, V ), where X = (X, τ) is a n-frame and V : P V 2 X is a valuation. Similar: neighborhood 2-frame (n-2-frame) is (X, τ 1, τ 2) such that τ i is a neighborhood function on X for each i. Validity in model: M, x = iψ V τ i(x) y V (M, y = ψ). M = ϕ X = ϕ X = L Log(C) = {ϕ X = ϕ for some X C} nv (L) = {X X is an n-frame and X = L}

Connection with Kripke frames Denition Let F = (W, R) be a Kripke frame. We dene neighborhood frame N (F ) = (W, τ) as follows. For any w W τ(w) = {U R(w) U W }. Lemma Let F = (W, R) be a Kripke frame. Then Log(N (F )) = Log(F ).

Bounded morphism for n-frames Denition Let X = (X, τ 1,...) and Y = (Y, σ 1,...) be n-frames. Then function f : X Y is a bounded morphism if 1. f is surjective; 2. for any x X and U τ i(x) f(u) σ i(f(x)); 3. for any x X and V σ i(f(x)) there exists U τ i(x), such that f(u) V. In notation f : X Y. Lemma If f : X Y then Log(Y) Log(X).

Not always fusion It is not the case for logic K! Lemma For any two n-frames X 1 and X 2 X 1 X 2 = 1 2 1. And even more, for any closed 1-free formula φ and any closed 2-free formula ψ X 1 X 2 = φ 1φ, X 1 X 2 = ψ 2ψ. Proof. X 1 X 2, (x, y) = 1 τ 1(x, y) τ 1(x) y X 2 ( τ 1(x, y )) y X 2 (X 1 X 2, (x, y ) = 1 ) = X 1 X 2, (x, y) = 2 1. Hence, X 1 X 2 = 1 2 1.

Not always fusion It is not the case for logic K! Lemma For any two n-frames X 1 and X 2 X 1 X 2 = 1 2 1. And even more, for any closed 1-free formula φ and any closed 2-free formula ψ X 1 X 2 = φ 1φ, X 1 X 2 = ψ 2ψ. Proof. Since ψ does not contain neither 2, nor variables, its value does not depend on the second coordinate. Let F = X 1 X 2. So F, (x, y) = ψ, then y (F, (x, y ) = ψ), hence, F, (x, y) = 2ψ.

Not always fusion Lemma For any two n-frames X 1 and X 2 X 1 X 2 = 1 2 1. And even more, for any closed 1-free formula φ and any closed 2-free formula ψ X 1 X 2 = φ 1φ, X 1 X 2 = ψ 2ψ. Denition For two unimodal logics L 1 and L 2, we dene L 1, L 2 = L 1 L 2 +, where = {φ 2φ φ is closed and 2-free} {ψ 1ψ ψ is closed and 1-free}. Lemma For any two normal modal logics L 1 and L 2 L 1, L 2 L 1 n L 2. Note that if L 1 L 2 then L 1 L 2 =.

Cantor space and innite paths Standart construction: Cantor space as the set on innite paths on innite binary tree T 2. The base of topology is the sets of the following type: where α and β are two innite paths: U m(α) = {β a 1 = b 1,... a m = b m}. α = a 1a 2a 3..., β = b 1b 2b 3.... In order to proof completeness of S4 w.r.t. Cantor space we need to construct p-morphism from Cantor space to arbitrary nite S4-frame. [Mints, 1998]

Constructing dense topologies from Kripke frames We need to construct a dense topological space based on a Kripke frame. This becomes important in studying of products of topological spaces.

Example r q 111 222 h i j k l m n p d e f g b c a

Example r q 111 222 h i j k l m n p d e f g b c a a bej ω

Example r q 111 222 h i j k l m n p d e f g b c b a a bej ω

Example r q 111 222 h i j k l m n p d e f g e b c b a a bej ω

Example r q 111 222 h i j k l m n p j d e f g e b c b a a bej ω

Example r q 111 222 h i j k l m n p j d e f g e b c b a a bej ω a bej ω

Example r q 111 222 h i j k l m n p j d e f g e b c b a a bej ω a bej

Example r q 111 222 h i j k l m n p j d e f g e b c b a α = a bej ω f F (α) = j

N ω (F ) Denition Sets U n(α) form a lter base. So we can dene τ(α) the lter with base {U n(α) n N} ; N ω(f ) = (W ω, τ) is a dense n-frame based on F. Frame N ω(f ) is dense in a sense that the intersection of all neighborhoods of a point is empty. So, there are no minimal neighborhoods unlike in T op(f ). Lemma Let F = (W, R) be a Kripke frame with root a, then f F : N ω(f ) N (F ). Corollary For any frame F Log(N ω(f )) Log(N (F )) = Log(F ).

Counterexample It possible that Log(N ω(f )) Log(F ). Consider: Obviously G = p p. Lemma N ω(g) p p G = ({1}, S), 1 n S1 m m = n + 1. Proof. Consider valuation θ(p) = { 2n 1 ω n N }. Then in any neighbourhood of point ω there are points where p is true and there are points where p is false. Hence, N ω(g) = p p.

Completeness results Theorem (214) K n K = K, K. Theorem If logics L 1 and L 2 are axiomatizable by closed formulas and by axioms like k p p then L 1 n L 2 = L 1, L 2. Corollary K4 d K4 = K4, K4.

Logic S5 We put 1 = {φ 2φ φ is closed and 2-free}, com 12 = 1 2p 2 1p, com 21 = 2 1p 1 2p, chr = 1 2p 2 1p. Theorem If logic L is axiomatizable by closed formulas and by axioms like k p p then L n S5 = L S5 + 1 + com 12 + chr.

How to prove PLAN We have two loogic L 1 and L 2 Canonicity of the logic L 1, L 2. Construct F 1 = L 1 and F 2 = L 2 and F 1, F 2 F L1,L 2. Construct N ω Γ 1 (F 1) N ω Γ 2 (F 2) N ( F 1, F 2 Γ 1 Γ 2 ). Check that N ω Γ 1 (F 1) = L 1 and N ω Γ 2 (F 2) = L 2

11 12 1a 1b 21 22 2a 2b a1 a2 aa ab b1 b2 ba bb 1 2 a b

1b 11 12 1a 1b 21 22 2a 2b a1 a2 aa ab b1 b2 ba bb 1 1 2 a b Λ (α, β) α = 1... β = b...

11 12 1a 1b 21 22 2a 2b a1 a2 aa ab b1 b2 ba bb 1 2 a b (α, β) α = 1... β = b...

11 12 1a 1b 21 22 2a 2b a1 a2 aa ab b1 b2 ba bb 1 2 a b 1 (α, β) α = 1... β = b...

11 12 1a 1b 21 22 2a 2b a1 a2 aa ab b1 b2 ba bb b 1 2 a b 1 (α, β) α = 1... β = b...

11 12 1a 1b 21 1b 22 2a 2b a1 a2 aa ab b1 b2 ba bb b 1 2 a b 1 (α, β) α = 1... β = b...

How to prove for S5 PLAN We have two loogic L and S5 Canonicity of the logic L, S5]. Construct F 1 = L and F 2 = (R, R 2 ) and F 1, F 2 F L,S5]. ( Construct N Γ ω (F 1) N ω(f 2) N F 1, F 2] Γ). Check that N Γ ω (F 1) = L

We dene C 12 = {ab ba a W 1, b W 2} We also dene three Kripke frames: F 1, F 2 = (F 1 F 2, R < 1, R < 2 ) F 1, F 2] = (F 1 F 2, R < 1, R 2) ar < 1 b u W 1( b = au) ar < 2 b v W 2( b = av) ar 2 b b ( ar < 2 b & b == C12 b) Lemma For F 1 and F 2 dened above F 1, F 2] = com 12, chr 1.

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