f. . Prove false by creating a counterexample: 9(AUB) C P(A) U P(B)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Problem Statement:**
*f. Prove false by creating a counterexample:*
\[
\mathcal{P}(A \cup B) \subseteq \mathcal{P}(A) \cup \mathcal{P}(B)
\]
**Explanation:**
This mathematical statement suggests that the power set of the union of sets \(A\) and \(B\) is a subset of the union of their power sets. The task is to demonstrate that this statement is incorrect by providing a counterexample.
**Steps to Prove False:**
1. **Define Sets \(A\) and \(B\):**
- Let \(A = \{1\}\)
- Let \(B = \{2\}\)
2. **Find the Power Sets:**
- \(A \cup B = \{1, 2\}\)
- \(\mathcal{P}(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}\)
- \(\mathcal{P}(A) = \{\emptyset, \{1\}\}\)
- \(\mathcal{P}(B) = \{\emptyset, \{2\}\}\)
3. **Union of Power Sets:**
- \(\mathcal{P}(A) \cup \mathcal{P}(B) = \{\emptyset, \{1\}, \{2\}\}\)
4. **Validate the Subset Relation:**
- Check if \(\mathcal{P}(A \cup B) \subseteq \mathcal{P}(A) \cup \mathcal{P}(B)\)
- The element \(\{1, 2\}\) from \(\mathcal{P}(A \cup B)\) is not present in \(\mathcal{P}(A) \cup \mathcal{P}(B)\).
**Conclusion:**
The counterexample shows that the statement is false because not all elements of \(\mathcal{P}(A \cup B)\) are included in the union of \(\mathcal{P}(A)\) and \(\mathcal{P}(B)\). Thus, \(\mathcal{P}(A \cup B)\) is not a subset of \(\mathcal{P}(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb34753c0-065a-4307-ab8c-7afc3888779e%2F6e974ee9-80bc-4b46-8c49-70e5be03853e%2Fuwhxno9_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
*f. Prove false by creating a counterexample:*
\[
\mathcal{P}(A \cup B) \subseteq \mathcal{P}(A) \cup \mathcal{P}(B)
\]
**Explanation:**
This mathematical statement suggests that the power set of the union of sets \(A\) and \(B\) is a subset of the union of their power sets. The task is to demonstrate that this statement is incorrect by providing a counterexample.
**Steps to Prove False:**
1. **Define Sets \(A\) and \(B\):**
- Let \(A = \{1\}\)
- Let \(B = \{2\}\)
2. **Find the Power Sets:**
- \(A \cup B = \{1, 2\}\)
- \(\mathcal{P}(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}\)
- \(\mathcal{P}(A) = \{\emptyset, \{1\}\}\)
- \(\mathcal{P}(B) = \{\emptyset, \{2\}\}\)
3. **Union of Power Sets:**
- \(\mathcal{P}(A) \cup \mathcal{P}(B) = \{\emptyset, \{1\}, \{2\}\}\)
4. **Validate the Subset Relation:**
- Check if \(\mathcal{P}(A \cup B) \subseteq \mathcal{P}(A) \cup \mathcal{P}(B)\)
- The element \(\{1, 2\}\) from \(\mathcal{P}(A \cup B)\) is not present in \(\mathcal{P}(A) \cup \mathcal{P}(B)\).
**Conclusion:**
The counterexample shows that the statement is false because not all elements of \(\mathcal{P}(A \cup B)\) are included in the union of \(\mathcal{P}(A)\) and \(\mathcal{P}(B)\). Thus, \(\mathcal{P}(A \cup B)\) is not a subset of \(\mathcal{P}(
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