Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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![### Proposition Analysis
**Problem:**
Examine the provided "proof" of the given "proposition." Is this "proposition" true? If not, identify a counterexample.
#### Proposition
Consider sets \( A \), \( B \), and \( C \) such that \( A \subseteq B \cup C \). This implies \( A \subseteq B \) or \( A \subseteq C \).
#### Provided Proof
To prove or disprove the proposition, follow these steps:
1. **Assumption:**
Let \( x \) be any arbitrary object and suppose \( x \in A \).
2. **Implication:**
Since \( A \subseteq B \cup C \), it implies \( x \in B \cup C \).
3. **Definition of Union:**
By the definition of the union, \( x \in B \) or \( x \in C \).
4. **Conclusion:**
Therefore, for any object \( x \), if \( x \in A \), then \( x \in B \) or \( x \in C \).
5. **Subset Definition:**
Hence, by the definition of a subset, we can conclude \( A \subseteq B \) or \( A \subseteq C \).
#### Analysis and Error Identification
The provided "proof" contains a flaw. It incorrectly assumes that if \( x \) belongs to the union \( B \cup C \), then \( x \) necessarily belongs to either \( B \) or \( C \) independently for all elements in \( A \). However, there could be elements of \( A \) that belong to \( B \cup C \) but not entirely to one set \( B \) or \( C \).
#### Counterexample
Consider the following sets:
- \( A = \{1, 2\} \)
- \( B = \{1\} \)
- \( C = \{2\} \)
Here, \( A \subseteq B \cup C \) because every element of \( A \) is in the union of \( B \) and \( C \):
\[ A = \{1, 2\} \text{ and } B \cup C = \{1, 2\} \]
However, \( A \) is not a subset of \( B \) (since \(](https://content.bartleby.com/qna-images/question/4d68e090-7e11-4ab8-a19a-f8ec202a32a2/c65f6e0f-e4bc-4613-8d04-9ad72133d68c/v2clnki_processed.jpeg)
Transcribed Image Text:### Proposition Analysis
**Problem:**
Examine the provided "proof" of the given "proposition." Is this "proposition" true? If not, identify a counterexample.
#### Proposition
Consider sets \( A \), \( B \), and \( C \) such that \( A \subseteq B \cup C \). This implies \( A \subseteq B \) or \( A \subseteq C \).
#### Provided Proof
To prove or disprove the proposition, follow these steps:
1. **Assumption:**
Let \( x \) be any arbitrary object and suppose \( x \in A \).
2. **Implication:**
Since \( A \subseteq B \cup C \), it implies \( x \in B \cup C \).
3. **Definition of Union:**
By the definition of the union, \( x \in B \) or \( x \in C \).
4. **Conclusion:**
Therefore, for any object \( x \), if \( x \in A \), then \( x \in B \) or \( x \in C \).
5. **Subset Definition:**
Hence, by the definition of a subset, we can conclude \( A \subseteq B \) or \( A \subseteq C \).
#### Analysis and Error Identification
The provided "proof" contains a flaw. It incorrectly assumes that if \( x \) belongs to the union \( B \cup C \), then \( x \) necessarily belongs to either \( B \) or \( C \) independently for all elements in \( A \). However, there could be elements of \( A \) that belong to \( B \cup C \) but not entirely to one set \( B \) or \( C \).
#### Counterexample
Consider the following sets:
- \( A = \{1, 2\} \)
- \( B = \{1\} \)
- \( C = \{2\} \)
Here, \( A \subseteq B \cup C \) because every element of \( A \) is in the union of \( B \) and \( C \):
\[ A = \{1, 2\} \text{ and } B \cup C = \{1, 2\} \]
However, \( A \) is not a subset of \( B \) (since \(
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