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Problem 2: Prove or disprove using an element-based argument: Given a universal set U, for all sets A, B, C CU , if AC B then BCUCC A¢UC.

Problem 2: Prove or disprove using an element-based argument: Given a universal set U, for all sets A, B, C CU , if AC B then BCUCC A¢UC.

Advanced Engineering Mathematics
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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**Problem 2:** **Objective:** Prove or disprove using an element-based argument: **Given** a universal set \( U \), for all sets \( A, B, C \subseteq U \), if \( A \subseteq B \) then \( B^c \cup C \subseteq A^c \cup C \). **Explanation:** This problem involves set theory and logical reasoning. The goal is to either prove or disprove the statement by considering elements and their relationships within the sets provided. **Key Points to Consider:** 1. **Universal Set \( U \):** This is the set that contains all possible elements under consideration. 2. **Subsets:** \( A, B, \) and \( C \) are subsets of \( U \). This means each element in these sets is also an element of \( U \). 3. **Subset Condition \( A \subseteq B \):** This indicates that every element of set \( A \) is also an element of set \( B \). 4. **Complement Sets:** - \( B^c \) is the complement of set \( B \), consisting of all elements in \( U \) that are not in \( B \). - \( A^c \) is the complement of set \( A \), consisting of all elements in \( U \) that are not in \( A \). 5. **Unions:** - \( B^c \cup C \) represents the set of elements that are either in \( B^c \) or in \( C \) (or in both). - \( A^c \cup C \) represents the set of elements that are either in \( A^c \) or in \( C \) (or in both). **Approach:** To prove the statement, one must show that any element in \( B^c \cup C \) is also in \( A^c \cup C \), assuming \( A \subseteq B \). Conversely, to disprove, one would need to find at least one instance where an element in \( B^c \cup C \) is not in \( A^c \cup C \). This exercise enhances understanding of the relationships between set theory operations such as union and complement within a universal set framework.
Transcribed Image Text:**Problem 2:** **Objective:** Prove or disprove using an element-based argument: **Given** a universal set \( U \), for all sets \( A, B, C \subseteq U \), if \( A \subseteq B \) then \( B^c \cup C \subseteq A^c \cup C \). **Explanation:** This problem involves set theory and logical reasoning. The goal is to either prove or disprove the statement by considering elements and their relationships within the sets provided. **Key Points to Consider:** 1. **Universal Set \( U \):** This is the set that contains all possible elements under consideration. 2. **Subsets:** \( A, B, \) and \( C \) are subsets of \( U \). This means each element in these sets is also an element of \( U \). 3. **Subset Condition \( A \subseteq B \):** This indicates that every element of set \( A \) is also an element of set \( B \). 4. **Complement Sets:** - \( B^c \) is the complement of set \( B \), consisting of all elements in \( U \) that are not in \( B \). - \( A^c \) is the complement of set \( A \), consisting of all elements in \( U \) that are not in \( A \). 5. **Unions:** - \( B^c \cup C \) represents the set of elements that are either in \( B^c \) or in \( C \) (or in both). - \( A^c \cup C \) represents the set of elements that are either in \( A^c \) or in \( C \) (or in both). **Approach:** To prove the statement, one must show that any element in \( B^c \cup C \) is also in \( A^c \cup C \), assuming \( A \subseteq B \). Conversely, to disprove, one would need to find at least one instance where an element in \( B^c \cup C \) is not in \( A^c \cup C \). This exercise enhances understanding of the relationships between set theory operations such as union and complement within a universal set framework.
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