PS06-solutions

# PS06-solutions - MATH 681 Problem Set#6 Solutions 1(25...

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Unformatted text preview: MATH 681 Problem Set #6 Solutions 1. (25 points) Given posets ( S, S ) and ( T, T ) , let us consider the set S × T subjected to the relation which is such that ( a,b ) ( c,d ) if and only if a S c and b S d . (a) (5 points) Show that is a partial ordering on ( S × T ) . For any s ∈ S and t ∈ T , reflexivity of ≺ S and ≺ T guarantee that s S s and t T t , so ( s,t ) ( s,t ). Thus is reflexive. If ( a,b ), ( c,d ), and ( e,f ) are elements of S × T such that ( a,b ) ( c,d ) and ( c,d ) ( e,f ), the definition of guarantees that a S c and c S e , so by transitivity of S , a S e ; likewise b T d and d T f , so since T is transitive, b f . Thus, ( a,b ) ( e,f ), exhibiting transitivity of . Finally, let us demonstrate antisymmetry. For ( a,b ) and ( c,d ) in S × T , let us explore the circumstances under which it can be true that both ( a,b ) ( c,d ) and ( c,d ) ( a,b ). The former relation requires that a S c and b T d ; the latter that c S a and d T b . By antisymmetry of S , since a S c and c S a , it must be the case that a = c ; likewise, b T d and d T d requires that b = d . Thus, ( a,b ) ( c,d ) and ( c,d ) ( a,b ) only if ( a,b ) = ( c,d ), so is antisymmetric. (b) (5 points) Show that ( a,b ) is a maximal (or minimal) element of S × T if and only if a and b are maximal (or minimal) elements of S and T respectively. We need only prove this for the maximal case; the argument for minimal ( a,b ) is identical with all orderings reversed. If ( a,b ) is maximal, then for all ( c,d ) 6 = ( a,b ), it follows that ( a,b ) ( c,d ); in particular, for all d 6 = b , we know that ( a,b ) ( a,d ). This will be true only if either a S a or b T d . The first possibility is clearly not true (by reflexivity of S ); thus the latter must be true, and since it is true for all d , b must be maximal in T . A similar inspection of the comparison ( a,b ) ( c,b ) will demonstrate maximality of a in S . Conversely, if a is maximal in S and b is maximal in T , then we may show that ( a,b ) is maximal in S × T by considering the circumstances under which ( a,b ) ( c,d ). This can happen only if a c and b d ; however, by maximality of a and b , this is only the case if a = c and b = d . Thus, the only ( c,d ) for which ( a,b ) ≤ ( c,d ) is one equal to ( a,b ), so ( a,b ) is maximal. (c) (5 points) Show that if C is a chain in S × T , then the set C S of first coordinates of elements of C is a chain in S . (similarly, it is true that the set C T of second coordinates of elements of C is a chain in T ). If x and y are elements of C S , then by the construction of C S there must be elements ( x,a ) and ( y,b ) of C . Since C is a chain in S × T , it must be the case that ( x,a ) ( y,b ) or ( y,b ) ( x,a ). Thus, by the definition of , either x S y or y S x . Thus x and y are comparable; since x and y were arbitrarily chosen, it follows that any two members of C S are comparable, so C S is a chain. The argument for C T proceeds along similar lines....
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## This note was uploaded on 01/12/2012 for the course MATH 681 taught by Professor Wildstrom during the Fall '09 term at University of Louisville.

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PS06-solutions - MATH 681 Problem Set#6 Solutions 1(25...

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