Consider the following statement. Assume that all sets are subsets of a universal set U. For all sets A and B, if A C B then BC S AC. Use an element argument to construct a proof for the statement by putting selected sentences from the following scrambled list in the correct order. If x were in A, then x would have to be in B by definition of subset. But x € B, and so x € A. Therefore, by definition of complement x E AC, and thus, by definition of subset, BC C AC. By definition of complement, x € B. Suppose A and B are any sets such that A C B, and suppose x E BC. Hence, x € A, because A N B = Ø. Suppose A and B are any sets such that A S B, and suppose x E B.
Consider the following statement. Assume that all sets are subsets of a universal set U. For all sets A and B, if A C B then BC S AC. Use an element argument to construct a proof for the statement by putting selected sentences from the following scrambled list in the correct order. If x were in A, then x would have to be in B by definition of subset. But x € B, and so x € A. Therefore, by definition of complement x E AC, and thus, by definition of subset, BC C AC. By definition of complement, x € B. Suppose A and B are any sets such that A C B, and suppose x E BC. Hence, x € A, because A N B = Ø. Suppose A and B are any sets such that A S B, and suppose x E 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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Transcribed Image Text:### Understanding Set Theory: Proof involving Complements and Subsets
**Statement:**
Consider the following statement. Assume that all sets are subsets of a universal set \( U \).
For all sets \( A \) and \( B \), if \( A \subseteq B \) then \( B^C \subseteq A^C \).
**Objective:**
Use an element argument to construct a proof for the statement by putting selected sentences from the following scrambled list in the correct order.
**Proof Steps:**
1. **Assumption and Goal:**
- Suppose \( A \) and \( B \) are any sets such that \( A \subseteq B \), and suppose \( x \in B^C \).
2. **Complement Definition:**
- By definition of complement, \( x \notin B \).
3. **Subset Implication:**
- If \( x \) were in \( A \), then \( x \) would have to be in \( B \) by definition of subset. But \( x \notin B \), and so \( x \notin A \).
4. **Complement Membership:**
- Therefore, by definition of complement \( x \in A^C \), and thus, by definition of subset, \( B^C \subseteq A^C \).
5. **Conclusion:**
- Hence, \( x \notin A \), because \( A \cap B = \emptyset \).
This proof demonstrates that if one set \( A \) is a subset of another set \( B \), then the complement of \( B \) is a subset of the complement of \( A \), using the concept of set complements and the relationships between subsets.
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