Many-Sorted First-Order Model Theory Lecture 4 12 th June, 2020 1 / - PowerPoint PPT Presentation

Many-Sorted First-Order Model Theory Lecture 4 12 th June, 2020 1 / 18

Ordinals Ordinals are special sets constructed as follows: ◮ 0 = ∅ ◮ ω + 1 = { 0 , 1 , 2 , . . . , ω } ◮ 1 = 0 ∪ { 0 } = { 0 } ◮ ω + 2 = { 0 , 1 , 2 , . . . , ω + 1 , ω + 2 } ◮ 2 = 1 ∪ { 1 } = { 0 , 1 } . . . . . . ◮ ω + n +1 = { 0 , 1 , 2 , . . . , ω, ω +1 , ω +2 , . . . , ω + n } ◮ n + 1 = n ∪ { n } = { 0 , 1 , . . . , n } . . . . . . ◮ ω ′ = { 0 , 1 , 2 . . . , ω, ω + 1 , ω + 2 . . . } ◮ ω = { 0 , 1 , 2 , . . . } All ordinal numbers greater than 0 are produced in this way, ◮ either by taking the successor of the last produced ordinal, or ◮ if there is no such last ordinal, by taking the set of all the ordinals produced so far (as in the case of ω ) which yields a new limit ordinal. Fact 1 ◮ For all sets A there exists an ordinal i and a bijective mapping f : i → A. ◮ Any subclass of ordinals has a least element. ◮ For any ordinal i, the pair ( i , ∈ ) is a total ordering. ◮ One cannot take the set of all ordinals, since this set would be a new limit ordinal, which is impossible, as we already defined them all (the collection of all ordinals is a class). 2 / 18

Transfinite induction Convention Whenever it is convenient, we denote the membership relation ∈ among ordinals by < . Ordinals support the following principle of Transfinite/Ordinal Induction : ◮ Let P ( i ) be a property defined for all ordinals i . ◮ Suppose that whenever P ( j ) is true for all j < i then P ( i ) is also true. ◮ Then transfinite induction tells us that P is true for all ordinals. Usually the proof is broken down into three cases: 1. Zero case Prove that P (0) is true. 2. Successor case Prove that P ( i ) implies P ( i + 1). 3. Limit case Prove that for any limit ordinal i , P ( i ) follows from P ( j ) for all j < i . 3 / 18

Cardinals ◮ A cardinal number (or simply cardinal) say how many of something there are, for example, one, two, three, four, five, six. ◮ Cardinal numbers are equivalence classes, or representatives of equivalence classes, of sets under the bijection relation. ◮ The representative of each equivalence class is the least ordinal in the equivalence class, which exists since any set of ordinals has a least element. ◮ card ( A ) = card ( B ) iff there exists a bijection f : A → B . ◮ card ( A ) ≤ card ( B ) iff there exists an injection f : A → B . ◮ card ( A ) ≥ card ( B ) iff there exists a surjection f : A → B . Lemma 2 For all sets A there is no bijection f : A → P ( A ) . Proof. Suppose there exists a bijective mapping f : A → P ( A ). Let A 0 := { a ∈ A | a �∈ f ( a ) } ∈ P ( A ). Since f is bijective, there exists a 0 ∈ A such that f ( a 0 ) = A 0 . ◮ If a 0 ∈ f ( a 0 ) then a 0 �∈ A 0 , that is a 0 �∈ f ( a 0 ). ◮ If a 0 �∈ f ( a 0 ) then a 0 ∈ A 0 , that is a 0 ∈ f ( a 0 ). 4 / 18

More on cardinals ◮ If card ( A ) = α then α < card ( P ( A )) = 2 α . ◮ For each cardinal α , the least cardinal greater than α is denoted by α + . ◮ The Generalized Continuum Hypothesis states that 2 α = α + for all infinite cardinals α . In particular ω + = 2 ω = c , the cardinal of real numbers. Exercise 1 ◮ Any language defined over a countable alphabet is countable. Hint: prove that there exists an injection f : ω ∗ → ω . ◮ Prove that ω n is countable for all n ∈ ω . 5 / 18

More on cardinals ◮ Countable union of countable sets is countable too. Let { A n } n ∈ ω be an ω -sorted set such that card ( A n ) = ω for all n ∈ ω . Let f : ω 2 → � n ∈ ω A n defined by f ( n , m ) = a n , m for all n , m ∈ ω , where A n = { a n , 0 , a n , 1 . . . } for all n ∈ ω . One can easily notice that f is surjective. Hence, � n ∈ ω A n is countable. . . . . . . . . . . . . . . . A 2 a 2 , 0 a 2 , 1 a 2 , 2 . . . A 1 a 1 , 0 a 1 , 1 a 1 , 2 . . . A 0 a 0 , 0 a 0 , 1 a 0 , 2 . . . ◮ Since ω ∗ = � n ∈ ω ω n , we have ω n ⊆ ω ∗ for all n ∈ ω ; it follows that card ( ω n ) = ω for all n ∈ ω . ◮ For any set A of infinite cardinality, let’s say α , we have card ( A ∗ ) = α . n ∈ ω card ( A ) n = ω ∗ α n = α . n ∈ ω A n ) = � card ( A ∗ ) = card ( � ◮ Let α and β be two infinite cardinals, and let { A i } i ∈ β be a family of sets such that (a) β < α and (b) card ( A i ) < α for all i ∈ β . Then card ( � i ∈ β A i ) = � i ∈ β card ( A i ) < β ∗ α = α . 6 / 18

� � � �� Signature extensions Lemma 3 Let Σ be a signature of power α , and C an S-sorted set of new constants such that card ( C s ) = α for all sorts s ∈ S. ι C : Σ ֒ → Σ[ C ] For all enumerations { ρ i ∈ Sen (Σ[ C ]) | i < α } there exists a chain of signature morphisms χ 0 , 1 χ 1 , 2 χ i , i +1 Σ[ C 0 ] ֒ → Σ[ C 1 ] ֒ → . . . Σ[ C i ] ֒ → . . . Σ[ C α ] such that ◮ C i ∈ P α ( C ) for all i < α , and C α = C, Σ[ C i , X i ] X i ◮ ρ i = χ i ,α ( γ i ) for some γ i ∈ Sen (Σ[ C i ]) , ι Xi and ϑ i ϑ i ◮ if γ i is of the form ∃ X i · γ ′ i then there Σ[ C i ] � � χ i , i +1 � Σ[ C i +1 ] C i +1 \ C i exists an injective mapping ϑ i : X i → C i +1 \ C i . 7 / 18

Maximally consistent sets Lemma 4 Let Σ be a signature of power α , and C an S-sorted set of new constants such that card ( C s ) = α for all sorts s ∈ S. Any consistent set Γ of Σ -sentences can be ‘extended’ 1 to a maximally consistent set Γ α of Σ[ C ] -sentences such that ◮ if ∨ E ∈ Γ α then e ∈ Γ α for some e ∈ E, and ◮ if ∃ X · ρ ∈ Γ α then ϑ ( ρ ) ∈ Γ α for some injective mapping ϑ : X → C. 1 Here ‘extended’ means that ι C (Γ) ⊆ Γ α , where ι C : Σ ֒ → Σ[ C ]. 8 / 18

Proof of Lemma 5. Given an enumeration { ρ i ∈ Sen (Σ[ C ]) | i < α } of the Σ[ C ]-sentences, consider the chain of χ 0 , 1 χ 1 , 2 χ i , i +1 signature morphisms Σ[ C 0 ] ֒ → Σ[ C 1 ] ֒ → . . . Σ[ C i ] ֒ → . . . Σ[ C α ] from Lemma 4 such that for all i < α , we have ρ i = χ i ,α ( γ i ) for some Σ[ C i ]-sentence γ i . We construct a chain of presentations morphisms γ 0 γ i γ i +1 χ 0 , 1 � . . . χ i , i +1 � (Σ[ C i +1 ] , Γ i +1 ) � � χ i +1 , i +2 � . . . (Σ[ C 0 ] , Γ 0 ) � � (Σ[ C i ] , Γ i ) � � (Σ[ C α ] , Γ α ) with the following properties: 1. (Σ[ C i ] , Γ i ) is consistent for each i ≤ α , 2. χ i , j (Γ i ) ⊆ Γ j for all ordinals i and j such that i < j ≤ α , 3. χ i , i +1 ( γ i ) ∈ Γ i +1 or ¬ χ i , i +1 ( γ i ) ∈ Γ i +1 for all ordinals i < α , and 4. if χ i , i +1 ( γ i ) ∈ Γ i +1 and γ i is of the form ∃ X i · γ ′ i then ϑ i ( γ ′ i ) ∈ Γ i +1 . We proceed by induction on ordinals. 9 / 18

� �� � Proof of Lemma 5. ◮ i = 0 Γ 0 := ι C 0 (Γ), where ι C 0 : Σ ֒ → Σ[ C 0 ] is an inclusion. Since Γ is consistent, Γ 0 is consistent as well. ◮ i ⇒ i + 1 There are two subcases. 1. Γ i ∪ { γ i } �⊢ ⊥ There are two subcases. 1.1 γ i = ∃ X i · γ ′ We have ι X i (Γ i ∪ {∃ X i · γ ′ i } ) ∪ { γ ′ i } �⊢ ⊥ ; i Σ[ C i , X i ] indeed if ι X i (Γ i ∪ {∃ X i · γ ′ i } ) ∪ { γ ′ i } ⊢ ⊥ then by ( Quant I ), Γ i ∪ {∃ X i · γ ′ i } ⊢ ⊥ which is a ι Xi ϑ i contradiction. Since ϑ i is injective, we have ϑ i ( ι X i (Γ i ∪ {∃ X i · γ ′ i } ) ∪ { γ ′ i } ) �⊢ ⊥ . Γ i +1 := Σ[ C i ] � � χ i , i +1 � Σ[ C i +1 ] χ i , i +1 (Γ i ) ∪ { χ i , i +1 ( γ i ) , ϑ i ( γ ′ i ) } is consistent. 1.2 γ i is not existentially quantified Γ i +1 := χ i , i +1 (Γ i ) ∪ { χ i , i +1 ( γ i ) } which is consistent, as Γ i ∪ { γ i } �⊢ ⊥ and χ i , i +1 is injective. 2. Γ i ∪ { γ i } ⊢ ⊥ Γ i +1 := χ i , i +1 (Γ i ) ∪ {¬ χ i , i +1 ( γ i ) } . Since Γ i ⊢ ¬ γ i and Γ i is consistent, Γ i ∪ {¬ γ i } is consistent too. Since χ i , i +1 is injective, Γ i +1 = χ i , i +1 (Γ i ) ∪ {¬ χ i , i +1 ( γ i ) } is consistent. ◮ i ≤ α is a limit ordinal Γ i := � j < i χ j , i (Γ j ). Suppose that Γ i is not consistent. By compactness, χ j , i (Γ j ) is not consistent for some j < i . Since χ i , j is injective, by ( Cons ), Γ j is not consistent, which is a contradiction. Hence, Γ i is consistent. 10 / 18