3. Let \( A=\{b, c, d\} \) and \( B=\{a, b\} \). Find the following: (a) \( (A \times B) \cap(B \times B) \) (b) \( (A \times B) \cup(B \times B) \) (c) \( \mathcal{P}(A) \cap \mathcal{P}(B) \) (d) \( \mathcal{P}(A \cap B) \)

Let's solve each part step by step. Given: \( A = \{b, c, d\} \) \( B = \{a, b\} \) **(a)** \( (A \times B) \cap (B \times B) \) First, we find \( A \times B \) and \( B \times B \). \( A \times B = \{(b, a), (b, b), (c, a), (c, b), (d, a), (d, b)\} \) \( B \times B = \{(a, a), (a, b), (b, a), (b, b)\} \) Now, we take the intersection: \( (A \times B) \cap (B \times B) = \{(b, b)\} \) --- **(b)** \( (A \times B) \cup (B \times B) \) We already have \( A \times B \) and \( B \times B \). Now we take the union: \( (A \times B) \cup (B \times B) = \{(b, a), (b, b), (c, a), (c, b), (d, a), (d, b), (a, a), (a, b), (b, a), (b, b)\} \) Removing duplicates gives us: \( \{(b, a), (b, b), (c, a), (c, b), (d, a), (d, b), (a, a), (a, b)\} \) --- **(c)** \( \mathcal{P}(A) \cap \mathcal{P}(B) \) The power set \( \mathcal{P}(A) \) and \( \mathcal{P}(B) \) are as follows: \( \mathcal{P}(A) = \{\emptyset, \{b\}, \{c\}, \{d\}, \{b, c\}, \{b, d\}, \{c, d\}, \{b, c, d\}\} \) \( \mathcal{P}(B) = \{\emptyset, \{a\}, \{b\}, \{a, b\}\} \) The intersection is: \( \mathcal{P}(A) \cap \mathcal{P}(B) = \{\emptyset, \{b\}\} \) --- **(d)** \( \mathcal{P}(A \cap B) \) First, we find \( A \cap B \): \( A \cap B = \{b\} \) Now the power set of \( A \cap B \) is: \( \mathcal{P}(\{b\}) = \{\emptyset, \{b\}\} \) So the final results are: (a) \( \{(b, b)\} \) (b) \( \{(b, a), (b, b), (c, a), (c, b), (d, a), (d, b), (a, a), (a, b)\} \) (c) \( \{\emptyset, \{b\}\} \) (d) \( \{\emptyset, \{b\}\} \)