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Statement-1 P(A)nnP(B)=P(AnnB), where P(...

Statement-1 `P(A)nnP(B)=P(AnnB)`, where P(A) is power set of set A.
Statement-2 `P(A)uuP(B)=P(AuuB)`.

A

Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1

B

Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1

C

Statement-1 is true, Statement-2 is false

D

Statement-1 is false, Statement-2 is true

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze both statements regarding power sets and their operations. ### Step 1: Analyze Statement 1 **Statement 1:** \( P(A) \cap P(B) = P(A \cap B) \) 1. **Understanding the Power Set:** The power set \( P(A) \) is the set of all subsets of \( A \). Similarly, \( P(B) \) is the set of all subsets of \( B \). 2. **Right-Hand Side (RHS):** The right-hand side \( P(A \cap B) \) is the power set of the intersection of sets \( A \) and \( B \). This means it contains all subsets of the intersection of \( A \) and \( B \). 3. **Left-Hand Side (LHS):** The left-hand side \( P(A) \cap P(B) \) consists of all subsets that are in both \( P(A) \) and \( P(B) \). This means these subsets must be subsets of both \( A \) and \( B \), which implies they are subsets of \( A \cap B \). 4. **Conclusion for Statement 1:** Since every subset in \( P(A \cap B) \) is also in \( P(A) \cap P(B) \) and vice versa, we conclude that \( P(A) \cap P(B) = P(A \cap B) \). Thus, Statement 1 is **True**. ### Step 2: Analyze Statement 2 **Statement 2:** \( P(A) \cup P(B) = P(A \cup B) \) 1. **Right-Hand Side (RHS):** The right-hand side \( P(A \cup B) \) is the power set of the union of sets \( A \) and \( B \). This contains all subsets of the union of \( A \) and \( B \). 2. **Left-Hand Side (LHS):** The left-hand side \( P(A) \cup P(B) \) consists of all subsets that are either in \( P(A) \) or in \( P(B) \). This means it includes all subsets of \( A \) and all subsets of \( B \), but not necessarily all subsets of \( A \cup B \). 3. **Counterexample:** Let's take \( A = \{1\} \) and \( B = \{2\} \). - \( P(A) = \{\emptyset, \{1\}\} \) - \( P(B) = \{\emptyset, \{2\}\} \) - \( P(A) \cup P(B) = \{\emptyset, \{1\}, \{2\}\} \) - \( A \cup B = \{1, 2\} \) and \( P(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\} \) - Clearly, \( P(A) \cup P(B) \neq P(A \cup B) \). 4. **Conclusion for Statement 2:** Since we found a counterexample, Statement 2 is **False**. ### Final Conclusion - **Statement 1:** True - **Statement 2:** False
To solve the problem, we need to analyze both statements regarding power sets and their operations. ### Step 1: Analyze Statement 1 **Statement 1:** \( P(A) \cap P(B) = P(A \cap B) \) 1. **Understanding the Power Set:** The power set \( P(A) \) is the set of all subsets of \( A \). Similarly, \( P(B) \) is the set of all subsets of \( B \). 2. **Right-Hand Side (RHS):** The right-hand side \( P(A \cap B) \) is the power set of the intersection of sets \( A \) and \( B \). This means it contains all subsets of the intersection of \( A \) and \( B \). 3. **Left-Hand Side (LHS):** The left-hand side \( P(A) \cap P(B) \) consists of all subsets that are in both \( P(A) \) and \( P(B) \). This means these subsets must be subsets of both \( A \) and \( B \), which implies they are subsets of \( A \cap B \). ...
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