Question

For any two sets of A and B, prove that :

(i) $A^{\prime }\cup B=U\Rightarrow A\subset B$

(ii) $B^{\prime }\subset A^{\prime }\Rightarrow A\subset B$

Answer 1

(i) We have $B^{\prime }\subset A^{\prime }$

To show : $A\subset B$

Let, $xϵA$

$\Rightarrow x\text{/}ϵA^{\prime }$ [$\because A\cap A^{\prime }=\varphi$]

$\Rightarrow x\text{/}ϵB^{\prime }$ [$\because B^{\prime }\subset A^{\prime }$]

$\Rightarrow x\text{/}ϵB$ [$\because B\cap B^{\prime }=\varphi$]

Thus, $xϵA\Rightarrow xϵB$

This is true for all $xϵA$

$\therefore A\subset B$

(ii) We have $A\cup B=\cup$, the universal set

To show : $A\subset B$

Let , $xϵA$

$\Rightarrow x\text{/}ϵA^{\prime }$ [$\because A\cap A^{\prime }=\varphi$]

$\because xϵAandA\subset \cup$

$\Rightarrow xϵ(A^{\prime }\cup B)$ [$\therefore \cup =A^{\prime }\cup B$]

$\Rightarrow xϵA^{\prime }orxϵB$

But, $x\text{/}ϵA^{\prime },$

$\therefore$ $xϵB$

Thus, $xϵA\Rightarrow xϵB$

This is true for all $xϵA$

$\therefore A\subset B$