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: 31

Answer

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

Work Step by Step

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