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

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Unformatted text preview: 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 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 = 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 be the smallest natural number such that a 1 , . . . , a k | A n | . Then for all n n , a 1 , . . . , a k | A n | , and f A n ( a 1 , . . . , a k ) = f A n ( a 1 , . . . , a k ) since A n A n . In order for A n A we must define f A ( a 1 , . . . , a k ) = f A n ( 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 be the smallest natural number such that a 1 , . . . , a k | A n | . Then for all n n , a 1 , . . . , a k | A n | , and R A n ( a 1 , . . . , a k ) R A n ( a 1 , . . . , a k ) since A n A n . In order for A n A we must define R A ( a 1 , . . . , a k ) R A n ( 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 ) . 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....
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This note was uploaded on 04/11/2011 for the course MATH 5020 taught by Professor Staff during the Spring '11 term at North Texas.

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

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