Equivalence relations A sample theory The models of Eq are equivalence relations (over some X). Consider a language with the binary predicate symbol ‘≡’ Formally: Structures (X, ∼), where for all a,b,c ∈ X: (written infix) as its only non-logical symbol. (E1) Reflexivity: a ∼ a, We want to study (and count) models of Eq = Re ∧ Sy ∧ Tr: (E2) Symmetry: If a ∼ b then also b ∼ a, ⋄ Re : ∀x x ≡ x (E3) Transitivity: If a ∼ b as well as b ∼ c, then a ∼ c. ⋄ Sy : ∀x∀y x ≡ y → y ≡ x

⋄ Tr : ∀x∀y∀z (x ≡ y ∧ y ≡ z) → x ≡ z Equivalence relations can be obtained from partitions. DEF: A partition Π of X is a set of nonempty subsets of X, s.t. [As a ‘warm up’: (P1) Disjointness: A, B ∈ Π implies either A = B or A ∩ B = ∅, describe two non-isomorphic 2-element models of Eq] (P2) Exhaustiveness: Every a ∈ X belongs to some A ∈ Π.

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IMPORTANT: We consider only denumerable models M of Eq. DEF: The ∼Π induced by the partition Π The signature of M is a mapping σM : ω 7→ ω ∪ {∞}, where is defined by σM(n) is the number of equivalence classes of size n, n ≥ 1, and σM(0) is the number of equivalence classes of infinite size. a ∼Π b iff for some A ∈ Π: a ∈ A and b ∈ A We denote signatures as infinite vectors [Check that ∼Π fulfills (E1), (E2), and (E3).] (σM(0), σM(1), σM(2), σM(3),...) FACT: Example: (A promiscuous model) Every equivalence relation is induced by a partition. Let M be a model of Eq and also of To see this let [a] denote the equivalence of a w.r.t ∼, Eprom : ∀x∀y x ≡ y. i.e., [a]= {x ∈ X : a ∼ x}. Then a ≡M b holds for all a, b ∈ |M|. It suffices to check that {[a]: a ∈ X} is a partition of X, Therefore there is only one eqivalence class (which, by i.e., fulfills (P1) and (P2). definition, is of infinite size); i.e., σM = (1, 0, 0, 0,...). All models of Eq ∧∀x∀y x ≡ y are isomorphic to M; i.e., there is only one type.

35 36 Example: (An eremitic model) Example: (n isomorphism types)

Add the following sentence to Eq: Analogously to Eux we can enforce n-element equivalence classes for every n ≥ 1. E : ∀x∀y x ≡ y ↔ x = y. erem For every k > 0, exactly k isomorphism types can be enforced by

All models M have signature σM = (0, ∞, 0, 0,...). disjunction over corresponding sentences. Again, there is only one isomorphism type. Example: (denumerably many isomorphism types) Example: (Two isomorphism types) Eden : ∀x∀y(∃u(u 6= x ∧ u ≡ x ∧∃v(v 6= y ∧ v ≡ y)) → x ≡ y) Eq ∧ (Eprom ∨ Eerem) is made true only by the eremitic and the promiscuous model. has models, where all elements that are not ‘isolated’ are Hence there are exactly two isomorphism types. in a single equivalence class (which can be of any size). The following signatures arise: Example: (An uxorious model) (0, ∞, 0, 0,...) (1, 0, 0, 0,...) , , , ,... , , , ,... Eux : ∀x∃y(x 6= y ∧ x ≡ y ∧∀z(z ≡ x → (z = x ∨ z = y))) (1 ∞ 0 0 ) (1 1 0 0 ) (0, ∞, 1, 0,...) (1, 2, 0, 0,...) is made true only by models M with only 2-element equivalence (0, ∞, 0, 1,...) (1, 3, 0, 0,...) classes; i.e. σM = (0, 0, ∞, 0,...). ··· ···

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The L¨owenheim-Skolem theorem Remember: We can enforce models of all finite sizes, Example: (Non-enumerably many isomorphism types) but also models of infinite size. We count isomorphism types of denumerable models of Eq itself. Can we enforce non-enumerable models? (Think, e.g., of R.) For every infinite set S ⊆ ω of we can define a model Obviously yes, if we allow non-enumerably many constants M where S and non-enumerably many sentences. σMS (i) = 1 if i ∈ S and However: We forbid non-enumerable languages σMS (i) = 0 if i 6∈ S. (but don’t restrict sets of sentences otherwise). The models MS are non-isomorphic for different S. Since there are non-enumerably many infinite subsets of ω it The answer then is no : follows that there are non-enumerably many isomorphism types. Theorem (downward L¨owenheim-Skolem theorem) Every satisfiable set of sentences has an enumerable model. [Before proving this we explore its consequences]

39 40 Consequences of the L¨owenheim-Skolem theorem 1 Consequences of the L¨owenheim-Skolem theorem 2 An immediate application of the L¨owenheim-Skolem theorem is Corollary: (Canonical-domains lemma 1) the existence of certain non-standard models . Let Γ be a sentence or a set of sentences. Corollary: (Skolem’s paradox) If Γ is satisfiable, then it has a model with domain either The first-order theory of R, i.e., the set all of sentences {0, 1,...,n − 1} or ω. R (over some enumerable language) that are made true by , has Corollary: (Canonical-domains lemma 2) enumerable models. Let ∆ be a sentence or a set of sentences without identity and Consequently: without symbols. The reals cannot be characterized canonically by any set of If ∆ is satisfiable, then it has a model with domain ω. R sentences: There are always models R, non-isomorphic to . Proof: Similar remarks hold for first-order . In to L¨owenheim-Skolem one needs the following For cognoscenti: observation: ‘Powers sets’ need not be interpreted to ‘contain really’ all Without identity and function symbols we can map all ‘too large’ subsets, but only those that can be ‘described’ within the theory. integers into a particular element of a finite model. (See [BBJ])

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Consequences of the L¨owenheim-Skolem theorem 3 Γ . . . set of sentences DEF: Γ is (implicationally) complete if for every sentence A of its language either Γ |= A or Γ |= ¬A. DEF: Γ is denumerably categorical if all denumerable models of Γ are isomorphic. Corollary: (Vaught’s test). If Γ is denumerably categorical but not finitely satisfiable, then Γ is complete. [[ Proof on blackboard ]]

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