Many-Sorted First-Order Model Theory Lecture 6 2 nd July, 2020 1 / 37
Easy halves: unions of chains Theorem 1 (Chang-� Lo´ s-Suszko: easy direction) Let ( A i ) i <γ be a family of structures that form a chain under embedding. That is, the index set γ is an ordinal and i ≤ j implies A i ≤ A j . Put A = � i <γ A i . Let ϕ ( x ) be a Π 2 formula, and a 0 a tuple from A 0 . If A i | = ϕ ( a 0 ) for every i < γ , then A | = ϕ ( a 0 ) . Proof. ◮ Write ϕ explicitly, as ∀ y · ∃ z · ψ ( x , y , z ). ◮ Take any tuple b from A . Since b is finite it belongs to some A i . ◮ Since A i | = ϕ ( a 0 ) by assumption, we have A i | = ∃ z · ψ ( a 0 , b , z ). ◮ This is an existential formula with parameters from A i , and A i ≤ A . ◮ By the easy half of the � Lo´ s-Tarski Theorem (on substructures), we have A | = ∃ z · ψ ( a 0 , b , z ). ◮ As b was arbitrary, we have A | = ∀ y · ∃ z · ψ ( a 0 , y , z ), as claimed. 2 / 37
Easy halves: unions of chains Theorem 1 (Chang-� Lo´ s-Suszko: easy direction) Let ( A i ) i <γ be a family of structures that form a chain under embedding. That is, the index set γ is an ordinal and i ≤ j implies A i ≤ A j . Put A = � i <γ A i . Let ϕ ( x ) be a Π 2 formula, and a 0 a tuple from A 0 . If A i | = ϕ ( a 0 ) for every i < γ , then A | = ϕ ( a 0 ) . Proof. ◮ Write ϕ explicitly, as ∀ y · ∃ z · ψ ( x , y , z ). ◮ Take any tuple b from A . Since b is finite it belongs to some A i . ◮ Since A i | = ϕ ( a 0 ) by assumption, we have A i | = ∃ z · ψ ( a 0 , b , z ). ◮ This is an existential formula with parameters from A i , and A i ≤ A . ◮ By the easy half of the � Lo´ s-Tarski Theorem (on substructures), we have A | = ∃ z · ψ ( a 0 , b , z ). ◮ As b was arbitrary, we have A | = ∀ y · ∃ z · ψ ( a 0 , y , z ), as claimed. 3 / 37
Unions of chains: algebraically closed fields Example 2 Fix a prime p , and consider the chain GF ( p ) ≤ GF ( p 2 ) ≤ · · · ≤ GF ( p i ) ≤ · · · and let F be its union. For any n consider the sentence ϕ n = ∀ y · ∃ x · y n x n + . . . y 1 x + y 0 = 0. Note that ϕ n is a Π 2 sentence. ◮ We have GF ( p i ) | = ϕ n for i ≥ n . j ≥ i GF ( p j ) for any i ∈ N . ◮ Moreover, F = � ◮ It follows that F | = ϕ n for every n . ◮ Thus, F is an algebraically closed field of characteristic p . Exercise 1 Let F be as above. Fill the gaps in the proof that F is an algebraically closed field of characteristic p. 4 / 37
Unions of chains: algebraically closed fields Example 2 Fix a prime p , and consider the chain GF ( p ) ≤ GF ( p 2 ) ≤ · · · ≤ GF ( p i ) ≤ · · · and let F be its union. For any n consider the sentence ϕ n = ∀ y · ∃ x · y n x n + . . . y 1 x + y 0 = 0. Note that ϕ n is a Π 2 sentence. ◮ We have GF ( p i ) | = ϕ n for i ≥ n . j ≥ i GF ( p j ) for any i ∈ N . ◮ Moreover, F = � ◮ It follows that F | = ϕ n for every n . ◮ Thus, F is an algebraically closed field of characteristic p . Exercise 1 Let F be as above. Fill the gaps in the proof that F is an algebraically closed field of characteristic p. 5 / 37
Exercises Exercise 2 Let ϕ be the sentence ∃ x , y · ∀ z · ¬ ( x < z ) ∨ ¬ ( z < y ) in the language of a binary relation < . Let ψ be the conjunction of ϕ with universal sentences stating that < is a strict linear order. Construct a chain of models ( C n ) n ∈ N such that C n | = ψ but � n ∈ N C n �| = ψ . Conclude that Σ 2 sentences are not preserved under unions of chains. Exercise 3 Prove that positive formulas are preserved under onto homomorphisms. Exercise 4 (Somewhat hard, but instructive) Let { A i : i < γ } be a chain of similar structures, and let A = � i <γ A i . Prove that A ∈ HSP { A i : i < γ } . Hint. Consider “eventually constant” sequences ( u i : i < γ ) , i.e., such if i > i 0 (for some i 0 ), then u i = a i 0 , where a i 0 ∈ | A i 0 | . Next, consider the relation ∼ on these sequences, defined by ( u i : i < γ ) ∼ ( w i : i < γ ) if for all i > j 0 (for some j 0 ) we have u i = w i . 6 / 37
Exercises Exercise 2 Let ϕ be the sentence ∃ x , y · ∀ z · ¬ ( x < z ) ∨ ¬ ( z < y ) in the language of a binary relation < . Let ψ be the conjunction of ϕ with universal sentences stating that < is a strict linear order. Construct a chain of models ( C n ) n ∈ N such that C n | = ψ but � n ∈ N C n �| = ψ . Conclude that Σ 2 sentences are not preserved under unions of chains. Exercise 3 Prove that positive formulas are preserved under onto homomorphisms. Exercise 4 (Somewhat hard, but instructive) Let { A i : i < γ } be a chain of similar structures, and let A = � i <γ A i . Prove that A ∈ HSP { A i : i < γ } . Hint. Consider “eventually constant” sequences ( u i : i < γ ) , i.e., such if i > i 0 (for some i 0 ), then u i = a i 0 , where a i 0 ∈ | A i 0 | . Next, consider the relation ∼ on these sequences, defined by ( u i : i < γ ) ∼ ( w i : i < γ ) if for all i > j 0 (for some j 0 ) we have u i = w i . 7 / 37
Exercises Exercise 2 Let ϕ be the sentence ∃ x , y · ∀ z · ¬ ( x < z ) ∨ ¬ ( z < y ) in the language of a binary relation < . Let ψ be the conjunction of ϕ with universal sentences stating that < is a strict linear order. Construct a chain of models ( C n ) n ∈ N such that C n | = ψ but � n ∈ N C n �| = ψ . Conclude that Σ 2 sentences are not preserved under unions of chains. Exercise 3 Prove that positive formulas are preserved under onto homomorphisms. Exercise 4 (Somewhat hard, but instructive) Let { A i : i < γ } be a chain of similar structures, and let A = � i <γ A i . Prove that A ∈ HSP { A i : i < γ } . Hint. Consider “eventually constant” sequences ( u i : i < γ ) , i.e., such if i > i 0 (for some i 0 ), then u i = a i 0 , where a i 0 ∈ | A i 0 | . Next, consider the relation ∼ on these sequences, defined by ( u i : i < γ ) ∼ ( w i : i < γ ) if for all i > j 0 (for some j 0 ) we have u i = w i . 8 / 37
Preservation under substructures Theorem 3 (� Lo´ s-Tarski) Let T be a theory. If T is preserved under substructures, then T is equivalent to a set of Π 1 formulas. Proof. ◮ Wlog, T is consistent. Let T ∀ be { ϕ ∈ Π 1 : T | = ϕ } . ◮ Let K = { A : A ≤ B for some B ∈ Mod ( T ) } . So K is the class of submodels of models of T. ◮ As T is preserved by substructures, we have Mod ( T ) = K ⊆ Mod ( T ∀ ). We will show that Mod ( T ∀ ) ⊆ K . ◮ Take A | = T ∀ . Claim A: diag ( A ) ∪ T is consistent. ◮ Take a finite D 0 ⊆ diag ( A ) and a finite T 0 ⊆ T . Put δ ( a ) = � D 0 , where a are all diagram constants occurring in D 0 . ◮ If D 0 ∪ T 0 is inconsistent, then T 0 | = ¬ δ ( a ). Note that T 0 does not mention a at all. So T 0 entails ¬ δ ( a ) for arbitrary a. ◮ So, T 0 | = ∀ x · ¬ δ ( x ), and as δ is quantifier free, we have ∀ x · ¬ δ ( x ) ∈ T ∀ . ◮ Thus, in particular, A | = ¬ δ ( a ). ◮ But δ ( a ) ∈ diag ( A ), so A | = δ ( a ). Contradiction. ◮ This proves Claim A. 9 / 37
Preservation under substructures Theorem 3 (� Lo´ s-Tarski) Let T be a theory. If T is preserved under substructures, then T is equivalent to a set of Π 1 formulas. Proof. ◮ Wlog, T is consistent. Let T ∀ be { ϕ ∈ Π 1 : T | = ϕ } . ◮ Let K = { A : A ≤ B for some B ∈ Mod ( T ) } . So K is the class of submodels of models of T. ◮ As T is preserved by substructures, we have Mod ( T ) = K ⊆ Mod ( T ∀ ). We will show that Mod ( T ∀ ) ⊆ K . ◮ Take A | = T ∀ . Claim A: diag ( A ) ∪ T is consistent. ◮ Take a finite D 0 ⊆ diag ( A ) and a finite T 0 ⊆ T . Put δ ( a ) = � D 0 , where a are all diagram constants occurring in D 0 . ◮ If D 0 ∪ T 0 is inconsistent, then T 0 | = ¬ δ ( a ). Note that T 0 does not mention a at all. So T 0 entails ¬ δ ( a ) for arbitrary a. ◮ So, T 0 | = ∀ x · ¬ δ ( x ), and as δ is quantifier free, we have ∀ x · ¬ δ ( x ) ∈ T ∀ . ◮ Thus, in particular, A | = ¬ δ ( a ). ◮ But δ ( a ) ∈ diag ( A ), so A | = δ ( a ). Contradiction. ◮ This proves Claim A. 10 / 37
Preservation under substructures Proof continued. ◮ By Claim A diag ( A ) ∪ T has a model, say, M . ◮ Since M | = diag ( A ), by the diagram lemma A ≤ M . ◮ So, A ∈ K as claimed, finishing the proof. ◮ The “arbitrary constant” trick is a formal version of a common practice of proving a general statement by picking some arbitrary elements. Exercise 5 (Very easy, but instructive) Formally, the arbitrary constant trick is the following statement. ◮ Let Σ be a signature, and C a set of new constants. Let S be a set of Σ[ C ] -sentences in which no constant from c occurs, or, which amounts to the same thing, a set of Σ -sentences in the signature Σ[ C ] . Let ϕ ( c ) be a Σ[ C ] -sentence, with c a sequence of constants from C. Then, S | = ϕ ( c ) implies S | = ∀ x · ϕ ( x ) . Prove it without recourse to completeness, soundness, or proof rules. 11 / 37
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