Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 6 - Set Theory - Exercise Set 6.2 - Page 366: 32

Answer

$For\,\,all\,\,sets\,\,A\,\,and\,\,B,\,\,,and\,C\,\,if\,\,A\subseteq B\,\,and\,\,B\cap C=\varnothing \\ then\,\,A\cap C=\varnothing \\ To\,\,prove\,\,that\,\,a\,\,set\,\,A\cap C\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A\cap C\,\,has\,\,no\,\,elements. \\ To\,\, do\,\,this, suppose\,\,A\cap C\,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\ x\in A\cap C\Rightarrow x\in A \,and\,x\in C\\ (by\,\,def.\,\,of\,\,inter\! section) \\ but\,A\subseteq B \\ this\,\,means\,\,if\,x\in A\Rightarrow x\in B\\ \therefore x\in A \,and\,x\in C \Rightarrow x\in B \,and\,x\in C\\ (by\,\,def.\,\,of\,\,inter\! section) \\ x\in B \,and\,x\in C\Rightarrow x\in B\cap C\Rightarrow B\cap C\neq \varnothing \\ (this\,\,is\,\,a\,\,contradiction\,\,because\,B\cap C=\varnothing)\\ so\,\,A\cap C=\varnothing $

Work Step by Step

$For\,\,all\,\,sets\,\,A\,\,and\,\,B,\,\,,and\,C\,\,if\,\,A\subseteq B\,\,and\,\,B\cap C=\varnothing \\ then\,\,A\cap C=\varnothing \\ To\,\,prove\,\,that\,\,a\,\,set\,\,A\cap C\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,A\cap C\,\,has\,\,no\,\,elements. \\ To\,\, do\,\,this, suppose\,\,A\cap C\,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\ x\in A\cap C\Rightarrow x\in A \,and\,x\in C\\ (by\,\,def.\,\,of\,\,inter\! section) \\ but\,A\subseteq B \\ this\,\,means\,\,if\,x\in A\Rightarrow x\in B\\ \therefore x\in A \,and\,x\in C \Rightarrow x\in B \,and\,x\in C\\ (by\,\,def.\,\,of\,\,inter\! section) \\ x\in B \,and\,x\in C\Rightarrow x\in B\cap C\Rightarrow B\cap C\neq \varnothing \\ (this\,\,is\,\,a\,\,contradiction\,\,because\,B\cap C=\varnothing)\\ so\,\,A\cap C=\varnothing $
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