Wednesday, September 10th, 1997 ------------------------------- Answer 1.a. and 1.b. and 1.d. Signal the instructor when your group has finished solving the three problems or if you have any questions or comments. 1. For each collection of sets, find the smallest set A such that the collection is a subset of power(A). a. { {a}, {b,c} } b. { {a}, { emptySet } } c. { {a}, { {a} } } Answer from Instructors manual for 7c. A = { a, {a} } d. { {a}, { {b} }, {a,b} } Answer from Instructors manual for 7d. A = { a, b, {b} } 1.c. explained. If power(A) is just a collection of every possible subset of A, then it has sets as members. { {a}, { {a} } } is a set and it has 1 2 two members, {a} and { {a} } {a} must be one of its members. Thus a is an element of A, in order to make {a} one of the possible subsets of A, i.e. one of the power(A) members is {a}. { {a} } has to be another one of the sets that is in power(A), which means {a} is a member of A. This is enough to do the job. A = a set containing two members, with a as 1st member or element and {a} as the 2nd element, i.e. power( { a, {a} } ) would then be: { emptySet, {a}, {{a}}, { a, {a} } } subset 1st 2nd 3rd 4th You can see here that the 2nd and the 3rd members of power(A) are the elements we were originally given. Since 2nd is element of power(A) and 3rd is element of power(B), we have the answer. You have to have a be in A and {a} also be in A. --------- ---------------- a is distinct from {a} and these two members are DIFFERENT entities. - ---