Answer:
The set of all possible subsets of A is given by A^c = {1, 2, 3, 4, 5} and the set of all non-empty proper subsets of A is given by A\B = {2, 3, 6}.
b) (S∪T)∪(S∩T) = (S∪T) ∪ (S∩T) = S∪(T∪S)∪S∩(T∪S) = S∪T∪S∩T = S∪T = S.
c) S = S\T = {1, 3, 4} and S = S\T = {1, 3, 4} and S = S\T = {1, 3, 4}. This means that the sets S and T are identical.
(a) AUB = {(2,3), (3,2), (2,6), (6,2), (6,3), (3,6), (6,6)}. A\B = {2, 3, 6}.
The power set of A is the set of all possible subsets of A, which can be denoted as P(A). To find the power set of A, we need to use the principle of inclusion-exclusion. The power set of A can be found by taking the union of the complement of A and the set of all possible subsets of A, which can be written as P(A) = A^c \cup P(A^c). Here, A^c is the complement of A, which is the set of all elements in the set A that are not in A. Therefore, we can write P(A) = x ∈ A^c, x ∈ A \cup P(A^c).
The set of all possible subsets of A is given by A^c = {1, 2, 3, 4, 5} and the set of all non-empty proper subsets of A is given by A\B = {2, 3, 6}.
(b) Statement (a) is true. N and Z are both well-ordered sets, meaning that they have a well-defined order. Statement (b) is also true. The union of two sets is equal to the set of all elements that are in either of the sets. Therefore, (S∪T)∪(S∩T) = (S∪T) ∪ (S∩T) = S∪(T∪S)∪S∩(T∪S) = S∪T∪S∩T = S∪T = S.
(c) If S\T = S, then it means that the intersection of S and T is equal to the entire set S. Therefore, any element in S is also an element of T. For example, if S = {1, 2, 3, 4, 5} and T = {2, 4, 5}, then S\T = {1, 3, 4}, which is equal to S. Therefore, S = S\T = {1, 3, 4} and S = S\T = {1, 3, 4} and S = S\T = {1, 3, 4}. This means that the sets S and T are identical.