Chapter 6 - Set Theory - Exercise Set 6.3 - Page 372: 8

$For\,\,all\,\,sets\,\,A,\,B,\,and\,\,C\,\\ if\,\,A^{c} \subseteq B\,\,then\,\,A \cup B = U \\ this\,\,is\,\,true:\\ A \cup B = U\\ talking\,\,complement\,\,we\,get\,: (A \cup B)^c = U^c \\ if\,\,we\,\,prove\,\,this\,(A \cup B)^c = U^c\,to\,be\,true\,\,\\then\,A \cup B = U\,\,is\,true \\ (A \cup B)^c = U^c \\ L.H.S:\\ (A \cup B)^c=A^{c}\cap B^c (by\,De\,Morgan\,law)\\ \because A^c\subseteq B \Rightarrow\therefore A^{c}\cap B^c=B\cap B^c=\varnothing \\ L.H.S=\varnothing \\ R.H.S:\\ U^c=\varnothing (by\,def\,of\,universal\,set)\\ L.H.S=R.H.S=\varnothing\\ \because (A \cup B)^c = U^c \\ \therefore (A \cup B) = U (if\,A^c\subseteq B) $

$For\,\,all\,\,sets\,\,A,\,B,\,and\,\,C\,\\ if\,\,A^{c} \subseteq B\,\,then\,\,A \cup B = U \\ this\,\,is\,\,true:\\ A \cup B = U\\ talking\,\,complement\,\,we\,get\,: (A \cup B)^c = U^c \\ if\,\,we\,\,prove\,\,this\,(A \cup B)^c = U^c\,to\,be\,true\,\,\\then\,A \cup B = U\,\,is\,true \\ (A \cup B)^c = U^c \\ L.H.S:\\ (A \cup B)^c=A^{c}\cap B^c (by\,De\,Morgan\,law)\\ \because A^c\subseteq B \Rightarrow\therefore A^{c}\cap B^c=B\cap B^c=\varnothing \\ L.H.S=\varnothing \\ R.H.S:\\ U^c=\varnothing (by\,def\,of\,universal\,set)\\ L.H.S=R.H.S=\varnothing\\ \because (A \cup B)^c = U^c \\ \therefore (A \cup B) = U (if\,A^c\subseteq B) $