7. Let A = {0, 1, {0,1}}. Give an example of a partition S of the power set P(A) such that SI = 3.

Answer:[tex] S=\Bigg\{~~\bigg\{\emptyset\bigg\},~~\bigg\{ \{0\},\{1\},\{\{0,1\}\} \bigg\},~~\bigg\{ \{0,1\},\{0,\{0,1\}\},\{1,\{0,1\}\},\{0,1,\{0,1\}\} \bigg\}~~ \Bigg\}[/tex]Step-by-step explanation:The power set P(A) is the set made of ALL subsets of A. In this case it will be a set of 8 elements:[tex] \mathcal{P}(A)=\Bigg\{~ \emptyset,~ \bigg\{0\bigg\},~ \bigg\{1\bigg\},~ \bigg\{\{0,1\}\bigg\},~ \bigg\{0,1\bigg\}, ~\bigg\{0,\{0,1\}\bigg\},~ \bigg\{1,\{0,1\}\bigg\}, ~\bigg\{0,1,\{0,1\}\bigg\}~ \Bigg\}[/tex](Notice A has 3 elements on it, and P(A) is made of all subsets of it, we first listed those subsets of A with 0 elements (the empty set only), then we listed those with 1 element, then those with 2 elements, and finally those with 3 elements (which is the set A itself) )A partition of P(A) is distributing all the elements of P(A) into nonempty disjoint sets. Think of it as simply distributing the elements of P(A) into different groups. In our case we want the partition to have 3 elements, so we want to distribute the 8 elements of P(A) into 3 groups. On the first group we just put the empty set, on the second group we put the elements of P(A) made of 1 single element, and on the third group we put the remaining 4 elements of P(A).