If the sets $A$ and $B$ are defined as $A = \{ (x,\,y):y = {e^x},\,x \in R\} $; $B = \{ (x,\,y):y = x,\,x \in R\} ,$ then
$B \subseteq A$
$A \subseteq B$
$A \cap B = \phi $
$A \cup B = A$
If $A, B$ and $C$ are non-empty sets, then $(A -B) \cup (B -A)$ equals
If $A =$ [$x:x$ is a multiple of $3$] and $B =$ [$x:x$ is a multiple of $5$], then $A -B$ is ($\bar A$ means complement of $A$)
If $A=\{3,5,7,9,11\}, B=\{7,9,11,13\}, C=\{11,13,15\}$ and $D=\{15,17\} ;$ find
$A \cap B$
Show that $A \cap B=A \cap C$ need not imply $B = C$
If the sets $A$ and $B$ are defined as $A = \{ (x,\,y):y = {1 \over x},\,0 \ne x \in R\} $ $B = \{ (x,y):y = - x,x \in R\} $, then