Chapter 5: Problem 5
Prove Theorem 5.25, Part (5): \(A \times(B-C)=(A \times B)-(A \times C)\).
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 5
Prove Theorem 5.25, Part (5): \(A \times(B-C)=(A \times B)-(A \times C)\).
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
Let \(S, T, X,\) and \(Y\) be subsets of some universal set. Assume that (i) \(S \cup T \subseteq X \cup Y\) (ii) \(S \cap T=\emptyset\); and (iii) \(X \subseteq S\) (a) Using assumption (i), what conclusion(s) can be made if it is known that \(a \in T ?\) (b) Using assumption (ii), what conclusion(s) can be made if it is known that \(a \in T ?\) (c) Using all three assumptions, either prove that \(T \subseteq Y\) or explain why it is not possible to do so.
For each natural number \(n,\) let \(A_{n}=\\{x \in \mathbb{R} \mid n-1
Let \(\Lambda\) and \(\Gamma\) be nonempty indexing sets. (Note: The letter \(\Gamma\) is the uppercase Greek letter gamma.) Also, let \(\mathcal{A}=\left\\{A_{\alpha} \mid \alpha \in \Lambda\right\\}\) and \(\mathcal{B}=\left\\{B_{\beta} \mid \beta \in \Gamma\right\\}\) be indexed families of sets. Use the distributive laws in Exercise (5) to: (a) Write \(\left(\bigcup_{\alpha \in \Lambda} A_{\alpha}\right) \cap\left(\bigcup_{\beta \in \Gamma} B_{\beta}\right)\) as a union of intersections of two sets. (b) Write \(\left(\bigcap_{\alpha \in \Lambda} A_{\alpha}\right) \cup\left(\bigcap_{\beta \in \Gamma} B_{\beta}\right)\) as an intersection of unions of two sets.
For each natural number \(n,\) let \(A_{n}=\\{k \in \mathbb{N} \mid k \geq n\\} .\) Assuming the universal set is \(\mathbb{N}\), use the roster method or set builder notation to specify each of the following sets: (a) \(\bigcap_{j=1}^{5} A_{j}\) (e) \(\bigcup_{j=1}^{5} A_{j}\) (b) \(\left(\bigcap_{j=1}^{5} A_{j}\right)^{c}\) $$ \text { (f) }\left(\bigcup_{j=1}^{5} A_{j}\right)^{c} $$ (c) \(\bigcap_{j=1}^{5} A_{j}^{c}\) (g) \(\bigcap_{j \in \mathbb{N}} A_{j}\) (d) \(\bigcup_{j=1}^{5} A_{j}^{c}\) (h) \(\bigcup_{j \in \mathbb{N}} A_{j}\)
See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) Let \(A, B,\) and \(C\) be subsets of some universal set. If \(A \nsubseteq B\) and \(B \nsubseteq C,\) then \(A \nsubseteq C\) Proof. We assume that \(A, B,\) and \(C\) are subsets of some universal set and that \(A \nsubseteq B\) and \(B \nsubseteq C\). This means that there exists an element \(x\) in \(A\) that is not in \(B\) and there exists an element \(x\) that is in \(B\) and not in \(C\). Therefore, \(x \in A\) and \(x \notin C,\) and we have proved that \(A \nsubseteq C\). Let \(A, B,\) and \(C\) be subsets of some universal set. If \(A \cap B=A \cap C\) then \(B=C\) Proof. We assume that \(A \cap B=A \cap C\) and will prove that \(B=C\). We will first prove that \(B \subseteq C\). So let \(x \in B\). If \(x \in A\), then \(x \in A \cap B\), and hence, \(x \in A \cap C\). From this we can conclude that \(x \in C .\) If \(x \notin A,\) then \(x \notin A \cap B\), and hence, \(x \notin A \cap C\). However, since \(x \notin A\), we may conclude that \(x \in C .\) Therefore, \(B \subseteq C\). (b) Let \(A, B,\) and \(C\) be subsets of some universal set. If \(A \cap B=A \cap C\), then \(B=C\) \mathrm{\\{} \text { Proof } . We assume that \(A \cap B=A \cap C\) and will prove that \(B=C\). We will first prove that \(B \subseteq C\) So let \(x \in B .\) If \(x \in A,\) then \(x \in A \cap B,\) and hence, \(x \in A \cap C\) From this we can conclude that \(x \in C .\) If \(x \notin A,\) then \(x \notin A \cap B\), and hence, \(x \notin A \cap C .\) However, since \(x \notin A,\) we may conclude that \(x \in C .\) Therefore, \(B \subseteq C\) The proof that \(C \subseteq B\) may be done in a similar manner. Hence, \(B=\) \(C\). (c) Let \(A, B,\) and \(C\) be subsets of some universal set. If \(A \nsubseteq B\) and \(B \subseteq C,\) then \(A \nsubseteq C\). Proof. Assume that \(A \nsubseteq B\) and \(B \subseteq C\). Since \(A \nsubseteq B\), there exists an element \(x\) such that \(x \in A\) and \(x \notin B\). Since \(B \subseteq C,\) we may conclude that \(x \notin C\). Hence, \(x \in A\) and \(x \notin C,\) and we have proved that \(A \nsubseteq C\).
What do you think about this solution?
We value your feedback to improve our textbook solutions.