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# handout012 - Math 5020 Handout 1.2 Proofs of Theorems in...

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Math 5020 Handout 1.2 Proofs of Theorems in Handout 1 Proposition A B iff | A | ⊆ | B | and for every constant symbol c L , n -ary relation symbol R L and function symbol f L , c B = c A , R B ∩ | A | n = R A , f B | A | n = f A . Proof ( ) Suppose A B . Then by definition | A | ⊆ | B | and the identity map e : | A | → | B | is an embedding. By the definition of embedding, we have that if c is a constant symbol, then e ( c A ) = c B ; but e ( c A ) = c A , and so c A = c B . if f L is an n -ary function symbol, then f B ( e ( a 1 ) , . . . , e ( a n )) = e ( f A ( a 1 , . . . , a n ) if a 1 , . . . , a n ∈ | A | ; since e is the identity map, we have f B ( ⃗a ) = f A ( ⃗a ), and therefore f B | A | n = f A . if R L is an n -ary relation symbol, then R B ( ⃗a ) ⇐⇒ R A ( ⃗a ) for ⃗a ∈ | A | n ; thus R B ∩ | A | n = R A . ( ) All the steps in the above proof can be reversed. Lemma Suppose A 0 A 1 ⊆ · · · ⊆ A n . . . . . . Then there is a unique L -structure A such that | A | = n | A n | and A n A for all n N . Proof We tacitly assume that the substructure relation is transitive (see Homework Problem 1.2). To define the structure A we only need to interpret the symbols in L . Let c L be a constant symbol. Note that c A 0 = c A 1 = . . . and we must define c A = c A 1 in order for A 1 A . Since they all agree we do have c A n = c A for all n N . Next let f L be a k -ary function symbol. To define f A we consider an arbitrary k -tuple a 1 , . . . , a k | A | = n | A n | . Let n 0 be the smallest natural number such that a 1 , . . . , a k ∈ | A n 0 | . Then for all n n 0 , a 1 , . . . , a k ∈ | A n | , and f A n ( a 1 , . . . , a k ) = f A n 0 ( a 1 , . . . , a k ) since A n 0 A n . In order for A n 0 A we must define f A ( a 1 , . . . , a k ) = f A n 0 ( a 1 , . . . , a k ) . We do have that for all n N and k -tuples a 1 , . . . , a k ∈ | A n | , f A ( a 1 , . . . , a k ) = f A n ( a 1 , . . . , a k ) . Finally let R L be a k -ary relation symbol. To define R A we consider an arbitrary k -tuple a 1 , . . . , a k | A | = n | A n | . Let n 0 be the smallest natural number such that a 1 , . . . , a k ∈ | A n 0 | . Then for all n n 0 , a 1 , . . . , a k ∈ | A n | , and R A n ( a 1 , . . . , a k ) ⇐⇒ R A n 0 ( a 1 , . . . , a k ) since A n 0 A n . In order for A n 0 A we must define R A ( a 1 , . . . , a k ) ⇐⇒ R A n 0 ( a 1 , . . . , a k ) . We do have that for all n N and k -tuples a 1 , . . . , a k ∈ | A n | , R A ( a 1 , . . . , a k ) ⇐⇒ R A n ( a 1 , . . . , a k ) .

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We have by now defined the structure A so that | A | = n | A n | and checked along the way that A n A for all n N . From our definitions and remarks it is clear that the definition is unique. Tarski’s Elementary Chain Theorem Suppose A 0 A 1 ≺ · · · ≺ A n ≺≺ · · · · · · Let A = n A n . Then A n A for all n N . Proof By the preceding lemma the structure A is well defined. We show the following statement by induction on the L -formula φ : For any L -formula φ ( x 1 , . . . , x k ), any n N and k -tuple a n 1 , . . . , a n k ∈ | A n | , A n | = φ [ a n 1 , . . . , a n k ] ⇐⇒ A | = φ [ a n 1 , . . . , a n k ] . In the following we omit the superscripts for simplicity, and in fact write ⃗a whenever the arity is not important.
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handout012 - Math 5020 Handout 1.2 Proofs of Theorems in...

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