Question

(De Morgan's laws) For any two sets A and B, prove that:
$I.(A\cup B{)}^{\prime }=(A^{\prime }\cap B^{\prime })$

$II.(A\cap B{)}^{\prime }=(A^{\prime }\cup B^{\prime })$

Answer 1

(i) Let x be an arbitary element of $(A\cup B)$' Then,
$xϵ(A\cup B{)}^{\prime }$

$\Rightarrow x\notin A\cup B$

$\Rightarrow x\notin Aandx\notin B$ [note this point]

$\Rightarrow xϵA^{\prime }andxϵB^{\prime }$

$\Rightarrow xϵ(A^{\prime }\cap B^{\prime })$

$\therefore (A\cup B{)}^{\prime }\subseteq (A^{\prime }\cap B^{\prime }).\dots (i)$

Again, let y be an arbitary element of $(A^{\prime }\cap B^{\prime })$. Then,
$yϵ(A^{\prime }\cap B^{\prime })$

$\Rightarrow yϵA^{\prime }andyϵB^{\prime }$

$\Rightarrow y\notin Aandy\notin B$

$\Rightarrow y\notin (A\cup B)$ [note this point]

$\Rightarrow yϵ(A\cup B{)}^{\prime }$

$\therefore (A^{\prime }\cap B^{\prime })\subseteq (A\cup B{)}^{\prime }.\dots (ii)$

From (i) and (ii) we get $A\cup B{)}^{\prime }=(A^{\prime }\cap B^{\prime })$

(ii) Let x be an arbitary element of ${(A\cap B)}^{\prime }$ Then,

$xϵ(A\cap B{)}^{\prime }$
$\Rightarrow x\notin (A\cap B)$

$\Rightarrow x\notin Aorx\notin B$ [note this point]

$\Rightarrow xϵA^{\prime }orxϵB^{\prime }$

$\Rightarrow xϵ(A^{\prime }\cup B^{\prime })$

$\therefore (A\cap B{)}^{\prime }\subseteq (A^{\prime }\cup B^{\prime }).\dots (iii)$

Again, let y be an arbitary element of $(A^{\prime }\cup B^{\prime })$. Then,

$yϵ(A^{\prime }\cup B{)}^{\prime }yϵ(A^{\prime }\cup B^{\prime })\Rightarrow yϵA^{\prime }oryϵB^{\prime }$

$\Rightarrow y\notin Aorx\notin B$

$\Rightarrow y\notin (A\cap B)$ [note this point]

$\Rightarrow yϵ(A\cap B{)}^{\prime }$

$\therefore (A^{\prime }\cup B^{\prime })\subseteq (A\cap B{)}^{\prime }.\dots (iv)$

From (iii) and (iv), we get $(A\cap B{)}^{\prime }=(A^{\prime }\cup B^{\prime })$