Question
Show that for any sets A and B,$\text{A}=(\text{A}\cap\text{B})\cap(\text{A - B})$
✓
Answer
We know that $(\text{A}\cap\text{B})\subset\text{A and (A - B)}\subset\text{A}$
$\Rightarrow(\text{A}\cap\text{B})\cap(\text{A - B})\subset\text{A}......(\text{i})$
Let and $\text{x}\in(\text{A}\cap\text{B})\cap(\text{A - B})$
$\Rightarrow\text{x}\in(\text{A}\cap\text{B})\text{ and x}\in(\text{A}-\text{B})$
$\Rightarrow\text{x}\in\text{A and}\text{ x}\in\text{B and }\text{x}\in\text{A and }\text{x}\not\in\text{B}$
$\Rightarrow\text{x}\in\text{A and }\text{x}\in\text{A}$ [$\because\text{ x}\in\text{A}\text{ x}\not\in\text{A}$ are not possible simutaneusly]
$\Rightarrow\text{x}\in\text{A}$
$\therefore(\text{A}\cap\text{B})\cap\text{(A - B)}\subset\text{A}..........\text{(ii)}$
From (i) and (ii), we get
$\text{A}=(\text{A}\cap\text{B})\cap(\text{A - B}).$
$\Rightarrow(\text{A}\cap\text{B})\cap(\text{A - B})\subset\text{A}......(\text{i})$
Let and $\text{x}\in(\text{A}\cap\text{B})\cap(\text{A - B})$
$\Rightarrow\text{x}\in(\text{A}\cap\text{B})\text{ and x}\in(\text{A}-\text{B})$
$\Rightarrow\text{x}\in\text{A and}\text{ x}\in\text{B and }\text{x}\in\text{A and }\text{x}\not\in\text{B}$
$\Rightarrow\text{x}\in\text{A and }\text{x}\in\text{A}$ [$\because\text{ x}\in\text{A}\text{ x}\not\in\text{A}$ are not possible simutaneusly]
$\Rightarrow\text{x}\in\text{A}$
$\therefore(\text{A}\cap\text{B})\cap\text{(A - B)}\subset\text{A}..........\text{(ii)}$
From (i) and (ii), we get
$\text{A}=(\text{A}\cap\text{B})\cap(\text{A - B}).$
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