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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 , nary 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 nary 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 nary 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 Lstructure 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 kary function symbol. To define f A we consider an arbitrary ktuple 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 ktuples 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 kary relation symbol. To define R A we consider an arbitrary ktuple 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 ktuples 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.
 Spring '11
 staff
 Math, Set Theory

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