Chapter 6 - Set Theory - Exercise Set 6.3 - Page 373: 44

$a-\,\,\\ an\,element\,argument\,to\,prove\,that\,A-B,B\,\,are\,disjoint \\ means\,\,(A-B)\cap B=\varnothing \\ To\,\,prove\,\,that\,\,a\,\,set\,\,(A-B)\cap B\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,(A-B)\cap B\,\,has\,\,no\,\,elements. \\ To\,\, do\,\,this, suppose\,\,(A-B)\cap B \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\ proof:\\ x\in (A-B)\cap B \Rightarrow x\in (A-B)\,\,and\,\,B \\(def.\,\,of\,intersection)\\ \Rightarrow x\in A \,and\,x\notin B \,and\, x\in B \\ (def.\,\,of\,\,set\,\,dif\! ference)\\ but\,\,x\in B \,and\, x\notin B \\ (this\,is\,a\,\,contradiction)\\ so\,(A-B)\cap B=\varnothing $ $b-\,\,\\ an\,algebraic\,\,\,argument\,\,to\,prove\,that\,A-B,B\,\,are\,disjoint \\ means\,\,(A-B)\cap B=\varnothing \\ =(A\cap B^c)\cap B\\ (set\,\,dif\! ference\,\,law)\\ =A\cap \left ( B^c\cap B \right ) \\ (associative\,\,law)\\ =A\cap \left ( B\cap B^c \right ) \\ (commutative\,\,law\,\,for\,\,\cap)\\ =A\cap \varnothing \\(complement\,law)\\ =A \\(def.of\varnothing )$ $c-\,\,\\ algebraic\,\,\,argument\,is\,easier\,than\,element\,\,\,argument$

$a-\,\,\\ an\,element\,argument\,to\,prove\,that\,A-B,B\,\,are\,disjoint \\ means\,\,(A-B)\cap B=\varnothing \\ To\,\,prove\,\,that\,\,a\,\,set\,\,(A-B)\cap B\,\,is\,\,equal\,\,to\,\,the\,\,empty\,\,set\,\, \varnothing ,\\ prove\,\,that\,\,(A-B)\cap B\,\,has\,\,no\,\,elements. \\ To\,\, do\,\,this, suppose\,\,(A-B)\cap B \,\,has\,\,an\,element\,\,and\,\,derive\,\,a\,\,contradiction \\ proof:\\ x\in (A-B)\cap B \Rightarrow x\in (A-B)\,\,and\,\,B \\(def.\,\,of\,intersection)\\ \Rightarrow x\in A \,and\,x\notin B \,and\, x\in B \\ (def.\,\,of\,\,set\,\,dif\! ference)\\ but\,\,x\in B \,and\, x\notin B \\ (this\,is\,a\,\,contradiction)\\ so\,(A-B)\cap B=\varnothing $ $b-\,\,\\ an\,algebraic\,\,\,argument\,\,to\,prove\,that\,A-B,B\,\,are\,disjoint \\ means\,\,(A-B)\cap B=\varnothing \\ =(A\cap B^c)\cap B\\ (set\,\,dif\! ference\,\,law)\\ =A\cap \left ( B^c\cap B \right ) \\ (associative\,\,law)\\ =A\cap \left ( B\cap B^c \right ) \\ (commutative\,\,law\,\,for\,\,\cap)\\ =A\cap \varnothing \\(complement\,law)\\ =A \\(def.of\varnothing )$ $c-\,\,\\ algebraic\,\,\,argument\,is\,easier\,than\,element\,\,\,argument$