Question

If $A\cap B^{\prime }=\Phi ,$ show that $A=A\cap B$ and hence show that $A\subseteq B.$

Answer 1

Given, $A\cap B^{\prime }=\Phi$

Then, $A=A\cap U,$ where U is the universal set.

= $A\cap (B\cup B^{\prime })$ $[\because B\cup B^{\prime }=U]$

= $(A\cap B)\cup (A\cap B^{\prime })$ [by distributive law]

= $A\cap B\cup \Phi$ $[\because A\cap B^{\prime }=\Phi ]$

= $A\cap B$

But $A\cap B=A,$ it means $A\subseteq B.$

Hence proved.