Lecture Slides for MAT-60556 Part VIII: Model theory Henri Hansen - PowerPoint PPT Presentation

Lecture Slides for MAT-60556 Part VIII: Model theory Henri Hansen November 5, 2013 1

Preliminaries • We denote by L a first-order logic with strict equal- ity (=) and no functions (but with constants) • We assume a countable set of variables x 0 , x 1 , . . . , predicate symbols { P i | i ∈ I } and constants { c j | j ∈ J } . I and J are usually finite, or at least countable. • If needed, λ ( i ) gives the arity of predicate R i • An L -structure is a triple M = ( A, R , c ) where R is a mapping from I to the set of relations on A , and c is a mapping J �→ A . 2

• Given an L -structure M , we say that L is the lan- guage for M • A sequence σ = a 0 , a 1 , . . . of elements of A is called an assignment; a i is the value given to variable v i . • we write σ ( n | b ) for a 0 , a 1 , . . . a n − 1 , b, a n +1 , . . . , i.e., b is put into the n:th place of σ • We define the satisfying relation, M | = σ φ for for- mulas of L inductively as follows: 1. M | = σ v i = v j iff a i = a j

2. M | = σ c = v j iff a j = c for a constant c 3. M | = σ R i ( t 1 , . . . , t λ ( i ) ) iff ( b 1 , . . . , b n ) ∈ R ( i ) so that for each i M | = σ b i = t i 4. M | = σ ∃ v i φ iff M | = σ ( i | b φ for some b ∈ A • If φ has no free variables (is a closed formula), we write simply M | = φ if M | = σ φ for some σ • M ′ = ( A ′ , R ′ , c ′ ) is a substructure of M iff A ′ ⊆ A and R ′ ( i ) = R ( i ) ∩ A ′ λ ( i ) for each i • A restriction of M into a set B ⊆ A is denoted M | B , it is defined in the natural way; we require that the image of c is contained in B

• An embedding of M into M ′ is a one-to-one map- ping f : A �→ A ′ such that 1. f ( c j ) = c ′ j for all j ∈ J 2. ( a 1 , . . . , a n ) ∈ R i implies ( f ( a 1 ) , . . . , f ( a n )) ∈ R ′ i • If an embedding is a bijection, then it is an isomor- phism . If an isomorphism exists we write M ∼ = M ′ • For an arbitrary embedding f , f [ M ] ∼ = M • M and M ′ are elementarily equivalent iff for every = φ iff M ′ | closed formula φ , M | = φ

More preliminaries • An elementary substructure of M is the structure M ′ such that M ′ ⊆ M and M ′ | = φ ( a 1 , . . . , a n ) if and = φ ( a 1 , . . . , a n ), for every a 1 , . . . , a n ∈ A ′ only if M | • If M ′ is an elementary substructure of M , we write M ′ ≺ M • M ∼ = M ′ does not imply M ≺ M ′ !! • An embedding of M into M ′ is an elementary em- bedding iff for any formula φ we have: M | = φ ( a 1 , . . . , a n ) iff M ′ | = φ ( f ( a 1 ) , . . . , f ( a n )) for all a 1 , . . . , a n . 3

• M is elementarily embeddable to M ′ if there is an elementary embedding from M to M ′ • M is elementary embeddable to M ′ if M is isomor- phic to some elementary substructure of M ′ M ≺ M ′ is equivalent to • Embedding Theorem: saying that for any n and L -formula φ with n + 1 free variables, and any n constants, if M ′ | = φ ( a 1 , . . . , a n , a ) for some a ∈ D ′ then there is some a ′ ∈ A such that M | = φ ( a 1 , . . . , a n , a ′ ) M ′ | – to prove implication: = φ ( a 1 , . . . , a n , a ) im- plies M ′ | = ∃ x ( φ ( a 1 , . . . , a n , x )), and thus M | = ∃ x ( φ ( a 1 , . . . , a n , x )).

– The other direction is by induction on the length of φ . Induction steps for ∧ and ¬ are trivial, and the only trouble remains with ∃ xψ (left as an exercise)

Indexing of sets • Let L K be the language obtained from L by adding constansts c ′ = { c k | k ∈ K } • Then the L K -structure M c ′ = ( A, R , c ∪ c ′ ) is an ex- pansion of ( A, R , c ). If c ′ maps K surjectively to A , then we say that c ′ is an indexing of A by K • Lemma: Let M and M ′ be L -structures, and let c ′ be an indexing of A by K . Then M is elementarily embeddable to M ′ iff there is a mapping c ∗ : K �→ A ′ such that M c ′ is elementarily equivalent to M c ∗ . 4

– Given an elementary embedding f , c ∗ ( k ) = f ( c ′ ( k )) results in M c ∗ ≡ M c ′ – Given c ∗ : K �→ A , define f ( c ′ ( k )) = c ∗ ( k ) results in f being an elementary embedding

L¨ owenheim-Skolem Theorems • We denote by | M | the cardinality of the model (its domain). • General Theorem: Let M be an infinite L -structure, and X ⊆ A Then for any cardinal α ≤ | M | , α ≥ | X | , and α ≥ |L| there is an elementary substructure M ′ of M such that | M ′ | = α and X ⊆ A ′ . – Let h be a choice function for the non-empty subsets of A (use axiom of choice), i.e., h ( Y ) ∈ Y for every Y ⊆ A that is not empty. 5

– Define B 0 , B 1 , . . . as follows: B 0 is any set for which X ⊆ B 0 and | B 0 | = α . – Given a formula φ with m free variables, Y n ( φ ) = { x ∈ A | ∃ a 1 , . . . a m − 1 ∈ B n ( M | = φ ( a 1 , . . . , a m − 1 , x )) } – B n +1 = { h ( Y n ( φ )) | φ ∈ L} – This construction adds to B n +1 elements of A that are needed to make formula in B n with free variables true; and one for each such formula. – B n = α for every n : | B 0 | = α and the there is at most α formulas in L , so B n +1 ≤ α · α = α for all infinite cardinalities

– We define A ′ = B 0 ∪ B 1 ∪· · · , and the Embedding theorem gives us that M ′ ≺ M • The Downward L¨ owenheim Skolem theorem: Let U be a set of closed formulas of L and let U have a model of cardinality α . Then U has a model for all the cardinalities β s.t. max( | U | , | N | ) ≤ β ≤ α • Let M be the model with cardinality α . Let γ be the cardinality of | U | (or | N | which ever is larger). Then L U contains at most γ symbols, i.e., L U is the smallest language that can express U . • Then, by the general theorem, U has a model with cardinality |L U | = | N |

• Corollary: Any countable set of closed formulas has a countable model. • The Upward L¨ owenheim-Skolem theorem: A set of closed formulas U with a model | M | ≥ | N | has a model for every cardinality ≥ | M |

From Ultraproducts to Compactness • For proof-technical reasons, we restrict, for this part, our attention to models that only have the domain and a single binary predicate, i.e. the model is simply of the form( A, R ) • The language L here is a first-order language that uses this predicate and no constants. The exten- sion of this proof to full first-order logic is straight- forward, but the technicalities quickly become ex- tremely tedious. Please try to think at each point, how the concepts would translate to arbitrary pred- icates with constants and functions 6

• We assume we have a family of models, M i = ( A i , R i ) for i ∈ I and ( A, S ) = Π i ∈ I M i is the di- rect product of the whole family. In this case A is the cartesian product of all A i and S = { ( f, g ) | ( f ( i ) , g ( i )) ∈ R i } • The direct product is not very convenient on its own, because the direct product does not share first-order properties of its components. For in- stance, ∀ x ∀ y ( R ( x, y ) ∨ R ( y, x )) may hold for each member, but not for the prouct (show the exam- ple!)

Modified product • We define two mappings, E and R for the product as follows – E ( f, g ) = { i ∈ I | f ( i ) = g ( i ) } – R ( f, g ) = { i ∈ I | ( f ( i ) , g ( i )) ∈ R i } • Let L ( A ) be the language obtained from L by adding a constant symbol for each f ∈ A • We define a mapping [] for closed formulas of L ( A ) into sets of indices inductively as follows 7

– [ f = g ] = E ( f, g ) – [ R ( f, g )] = R ( f, g ) – [ σ 1 ∧ σ 2 ] = [ σ 1 ] ∩ [ σ 2 ] – [ ¬ σ ] = I \ [ σ ] – When x is a free variable in φ , we denote [ ∃ xφ ( x )] = � [ σφ ( f )] f ∈ A • [ · ] assigns a “Boolean truth values” in the boolean algebra of 2 I to sentences if L ( A ), so that the “truth value” of a formula is now a set of indices instead of true or false. The indeces denote which

of the models in the product actually satisfy the formula • Theorem: For a formula φ ( x 1 , . . . , x n ) we have [ φ ( f 1 , . . . , f n )] = { i ∈ I | M i | = φ ( f 1 ( i ) , . . . , f n ( i )) } The proof is by induction, left as an exercise • Lemma: Let φ ( x ) be a formula of L ( A ). Then there exists and f ∈ A such that [ ∃ xφ ( x )] = [ φ ( f )] – Let < be a well-order for A and let X ξ = [ φ ( f ξ )] \ η<ξ [ φ ( f η )], i.e., for the well-ordered set of con- � stants, we take the indices for which those “smaller than” a given f ξ make the formula true.

– If α is an ordinal “big enough” then [ ∃ xφ ( x )] = η<α [ φ ( f η )] � – Because the product’s domains are disjoint, X ξ ∩ X η = ∅ for ξ � = η . We choose f ∈ A so that f | X ξ = f ξ | X ξ and we can deduce that for this f [ ∃ xφ ( x )] = [ φ ( f )] • Let F be some collection of subsets of I . We define – f ∼ F g iff [ f = g ] ∈ F – ( f, g ) ∈ R F iff [ R ( f, g )] ∈ F • Lemma: Let F be a filter over I . Then ∼ F is an

equivalence relation on A , and: f ∼ F f ′ and g ∼ F g ′ , and ( f, g ) ∈ R F imply that ( f ′ , g ′ ) ∈ R F – Recall that [ f = g ] denotes the set { i ∈ I | f ( i ) = g ( i ) } – We need to show that ∼ F is transitive, i.e., that [ f = g ] ∈ F and [ g = h ] ∈ F imply that [ f = h ] ∈ F . – The meet in I is intersection so [ f = g ] ∩ [ g = h ] ∈ F , and it is clear that [ f = g ] ∩ [ g = h ] ⊆ [ f = h ] and because ⊆ is the partial order of F the result follows.