Let's go through the solution step-by-step:

Step 1: Finding the Power Set $P(A)$

Given the set $A = \{a, b, c\}$, the power set $P(A)$ consists of all possible subsets of $A$. The subsets of $A$ are:

• The empty set: $\emptyset$
• Single-element subsets: $\{a\}, \{b\}, \{c\}$
• Two-element subsets: $\{a, b\}, \{a, c\}, \{b, c\}$
• The full set: $\{a, b, c\}$

Therefore, the power set $P(A)$ is: $P(A) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\}$

Step 2: Finding the Cartesian Product $A \times P(A)$

The Cartesian product $A \times P(A)$ consists of all ordered pairs where the first element is from the set $A$ and the second element is from the set $P(A)$. We list each element in $A$ paired with each subset from $P(A)$.

(a, \emptyset), (a, \{a\}), (a, \{b\}), (a, \{c\}), (a, \{a, b\}), (a, \{a, c\}), (a, \{b, c\}), (a, \{a, b, c\}),$$ $$(b, \emptyset), (b, \{a\}), (b, \{b\}), (b, \{c\}), (b, \{a, b\}), (b, \{a, c\}), (b, \{b, c\}), (b, \{a, b, c\}),$$ $$(c, \emptyset), (c, \{a\}), (c, \{b\}), (c, \{c\}), (c, \{a, b\}), (c, \{a, c\}), (c, \{b, c\}), (c, \{a, b, c\}) \}$$ This set contains all ordered pairs formed by combining each element of $$A$$ with each subset in $$P(A)$$. ### Final Answer - The power set $$P(A) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\}$$. - The Cartesian product $$A \times P(A)$$ is listed as all the ordered pairs above. Would you like more details or have any questions? Here are some related questions for further exploration: 1. What is the size of the power set of a set with $$n$$ elements? 2. How is the Cartesian product of two sets defined when they have different sizes? 3. Can the Cartesian product of more than two sets be computed, and if so, how? 4. How do power sets relate to binary representation of numbers? 5. What is the significance of the Cartesian product in set theory? **Tip:** The number of subsets in the power set of a set with $$n$$ elements is always $$2^n$$.