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Continuous First-Order Logic Continuous First-Order Logic Wesley Calvert Calcutta Logic Circle 4 September 2011 Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 1 / 45 Definition A first-order theory T is said to be stable iff there are less than the maximum possible number of types over T , up to equivalence. Theorem (Shelah's Main Gap Theorem) If T is a first-order theory and is stable and . , then the class of models looks like . Otherwise, there's no hope. Model-Theoretic Beginnings Problem Given a theory T , describe the structure of models of T . Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 2 / 45 Theorem (Shelah's Main Gap Theorem) If T is a first-order theory and is stable and . , then the class of models looks like . Otherwise, there's no hope. Model-Theoretic Beginnings Problem Given a theory T , describe the structure of models of T . Definition A first-order theory T is said to be stable iff there are less than the maximum possible number of types over T , up to equivalence. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 2 / 45 Model-Theoretic Beginnings Problem Given a theory T , describe the structure of models of T . Definition A first-order theory T is said to be stable iff there are less than the maximum possible number of types over T , up to equivalence. Theorem (Shelah's Main Gap Theorem) If T is a first-order theory and is stable and . , then the class of models looks like . Otherwise, there's no hope. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 2 / 45 Model-Theoretic Beginnings Example Let T be the theory of vector spaces over some infinite field. Then for each uncountable cardinal κ, there is exactly one model of T , up to isomorphism, with cardinality κ. Moreover, there are only countably many countable models, and we know what they are. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 3 / 45 1 For all x 2 V , we have jxj ≥ 0. 2 If jxj = 0, then x = 0. 3 For all x 2 V and all λ 2 R we have jλxj = jλj · jxj. 4 For all x; y 2 V , we have jx + yj ≤ jxj + jyj. Model-Theoretic Beginnings Definition Let V be a vector space over R.A norm on V is a function j · j : V ! R satisfying the following: Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 4 / 45 2 If jxj = 0, then x = 0. 3 For all x 2 V and all λ 2 R we have jλxj = jλj · jxj. 4 For all x; y 2 V , we have jx + yj ≤ jxj + jyj. Model-Theoretic Beginnings Definition Let V be a vector space over R.A norm on V is a function j · j : V ! R satisfying the following: 1 For all x 2 V , we have jxj ≥ 0. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 4 / 45 3 For all x 2 V and all λ 2 R we have jλxj = jλj · jxj. 4 For all x; y 2 V , we have jx + yj ≤ jxj + jyj. Model-Theoretic Beginnings Definition Let V be a vector space over R.A norm on V is a function j · j : V ! R satisfying the following: 1 For all x 2 V , we have jxj ≥ 0. 2 If jxj = 0, then x = 0. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 4 / 45 4 For all x; y 2 V , we have jx + yj ≤ jxj + jyj. Model-Theoretic Beginnings Definition Let V be a vector space over R.A norm on V is a function j · j : V ! R satisfying the following: 1 For all x 2 V , we have jxj ≥ 0. 2 If jxj = 0, then x = 0. 3 For all x 2 V and all λ 2 R we have jλxj = jλj · jxj. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 4 / 45 Model-Theoretic Beginnings Definition Let V be a vector space over R.A norm on V is a function j · j : V ! R satisfying the following: 1 For all x 2 V , we have jxj ≥ 0. 2 If jxj = 0, then x = 0. 3 For all x 2 V and all λ 2 R we have jλxj = jλj · jxj. 4 For all x; y 2 V , we have jx + yj ≤ jxj + jyj. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 4 / 45 Example Let (V ; j · j) be a normed vector space over R.Then define an equivalence relation x ∼ y iff jxj = jyj, and order the classes by x∼ < y∼ iff jxj < jyj. Thus, (V ; j · j) is unstable, and the class of models of its first-order theory is completely intractable. Problem Then how can we possibly think about Hilbert Spaces? Model-Theoretic Beginnings Proposition A first-order theory is said to be stable iff there is no definable infinite linear ordering in any model. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 5 / 45 Then define an equivalence relation x ∼ y iff jxj = jyj, and order the classes by x∼ < y∼ iff jxj < jyj. Thus, (V ; j · j) is unstable, and the class of models of its first-order theory is completely intractable. Problem Then how can we possibly think about Hilbert Spaces? Model-Theoretic Beginnings Proposition A first-order theory is said to be stable iff there is no definable infinite linear ordering in any model. Example Let (V ; j · j) be a normed vector space over R. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 5 / 45 Thus, (V ; j · j) is unstable, and the class of models of its first-order theory is completely intractable. Problem Then how can we possibly think about Hilbert Spaces? Model-Theoretic Beginnings Proposition A first-order theory is said to be stable iff there is no definable infinite linear ordering in any model. Example Let (V ; j · j) be a normed vector space over R.Then define an equivalence relation x ∼ y iff jxj = jyj, and order the classes by x∼ < y∼ iff jxj < jyj. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 5 / 45 Problem Then how can we possibly think about Hilbert Spaces? Model-Theoretic Beginnings Proposition A first-order theory is said to be stable iff there is no definable infinite linear ordering in any model. Example Let (V ; j · j) be a normed vector space over R.Then define an equivalence relation x ∼ y iff jxj = jyj, and order the classes by x∼ < y∼ iff jxj < jyj. Thus, (V ; j · j) is unstable, and the class of models of its first-order theory is completely intractable. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 5 / 45 Model-Theoretic Beginnings Proposition A first-order theory is said to be stable iff there is no definable infinite linear ordering in any model. Example Let (V ; j · j) be a normed vector space over R.Then define an equivalence relation x ∼ y iff jxj = jyj, and order the classes by x∼ < y∼ iff jxj < jyj. Thus, (V ; j · j) is unstable, and the class of models of its first-order theory is completely intractable. Problem Then how can we possibly think about Hilbert Spaces? Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 5 / 45 Example Let p > 0, and define 1 Z p p jjf jjp := jf j : p We define L (R) to be the set of all functions f : R ! R such that jjf jjp < 1, with pointwise addition and scaling and the norm jj · jjp. Each p space L (R) is a Banach space. Model-Theoretic Beginnings Definition A Banach Space is a normed vector space with the property that every Cauchy sequence converges to a point of the space (i.e. a complete normed vector space). Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 6 / 45 p We define L (R) to be the set of all functions f : R ! R such that jjf jjp < 1, with pointwise addition and scaling and the norm jj · jjp. Each p space L (R) is a Banach space. Model-Theoretic Beginnings Definition A Banach Space is a normed vector space with the property that every Cauchy sequence converges to a point of the space (i.e. a complete normed vector space). Example Let p > 0, and define 1 Z p p jjf jjp := jf j : Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 6 / 45 Model-Theoretic Beginnings Definition A Banach Space is a normed vector space with the property that every Cauchy sequence converges to a point of the space (i.e. a complete normed vector space). Example Let p > 0, and define 1 Z p p jjf jjp := jf j : p We define L (R) to be the set of all functions f : R ! R such that jjf jjp < 1, with pointwise addition and scaling and the norm jj · jjp. Each p space L (R) is a Banach space. Wesley Calvert (SIU / IMSc) Continuous First-Order Logic 4 September 2011 6 / 45 1 For all x; y; z 2 V and all a; b 2 R, we have hax + by; zi = ahx; zi + bhy; zi 2 For all x; y 2 V , we have hx; yi = hy; xi 3 For all x 2 V , we have hx; xi ≥ 0, with equality holding if and only if x = 0.
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