Chapter 5: Problem 9
For all sets \(A, B\), and \(C\), $$ (A-B) \cap(C-B)=(A \cap C)-B $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 5: Problem 9
For all sets \(A, B\), and \(C\), $$ (A-B) \cap(C-B)=(A \cap C)-B $$
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Use mathematical induction and the following definitions to prove each statement in 35-37. If \(n\) is an integer with \(n \geq 3\) and if \(C_{1}, C_{2}, C_{3}, \ldots, C_{n}\) are any sets, \(C_{1} \cup C_{2} \cup C_{3} \cup \cdots \cup C_{n}=\left(C_{1} \cup C_{2} \cup C_{3} \cup \cdots \cup C_{n-1}\right) \cup C_{n}\), and \(C_{1} \cap C_{2} \cap C_{3} \cap \cdots \cap C_{n}=\left(C_{1} \cap C_{2} \cap C_{3} \cap \cdots \cap C_{n-1}\right) \cap C_{n} .\) (More rigorous versions of the definitions are given in Section 8.4.) Generalized Distributive Law for Sets: For any integer \(n \geq 1\), if \(A\) and \(B_{1}, B_{2}, B_{3}, \ldots, B_{n}\) are any sets, then $$ \begin{aligned} \left(A \cap B_{1}\right) \cup\left(A \cap B_{2}\right) \cup & \cdots \cup\left(A \cap B_{n}\right) \\ &=A \cap\left(B_{1} \cup B_{2} \cup B_{3} \cup \cdots \cup B_{n}\right) \end{aligned} $$
For all sets \(A\) and \(B\), $$ (A-B) \cup(B-A)=(A \cup B)-(A \cap B) $$
The following is a proof that for all sets \(A, B\), and \(C\), if \(A \subseteq B\) and \(B \subseteq C\), then \(A \subseteq C\). Fill in the blanks. Proof: Suppose \(A, B\), and \(C\) are sets and \(A \subseteq B\) and \(B \subseteq C\). To show that \(A \subseteq C\), we must show that every element in \(\frac{(1)}{}\) is in \((2)\). But given any element in \(A\), that element is in \(\underline{(3)}\) (because \(A \subseteq B\) ), and so that element is also in \(\stackrel{(4)}{\text { (because }}(5)\) ). Hence \(A \subseteq C\).
Some English adjectives are descriptive of themselves (for instance, the word polysyllabic is polysyllabic) whereas others are not (for instance, the word monosyllabic is not monosyllabic). The word heterological refers to an adjective that does not describe itself. Is heterological heterological? Explain your answer.
Prove each statement that is true and find a counterexample for each statement that is false. Assume all sets are subsets of a universal set \(U\). For all sets \(A\) and \(B\), if \(A \cap B=\emptyset\) then \(A \times B=\emptyset\).
What do you think about this solution?
We value your feedback to improve our textbook solutions.