Question

Show that for any set A and B,

(i) $A=(A\cap B)\cup (A-B)$

(ii) $A\cup (B-A)=A\cup B$

Answer 1

(i) To show :

$A=(A\cap B)\cup (A-B)$

Let $xϵA$

Then either $xϵ(A-B)orxϵ(A\cap B)$ i.e.

$xϵ(A-B)\cup (A\cap B)$

$\Rightarrow A\subset (A-B)\cup (A\cap B)$

Conversely,

Let , $xϵ(A\cap B)\cup (A-B)$

$\Rightarrow xϵ(A\cap B)orxϵ(A-B)$

$\Rightarrow xϵAandxϵBorxϵAandx\text{/}ϵB$

$\Rightarrow xϵA$ [$\because xϵBandx\text{/}ϵB$ are contrary to each other ]

$\therefore (A\cap B)\cup (A-B)\subset A$

Hence, A = $(A\cap B)\cup (A-B)$ proved.

(ii) To show : $A\cup (B-A)=A\cup B$

Let , $xϵA\cup (B-A)$

$\Rightarrow xϵAorxϵ(B-A)$

$\Rightarrow xϵAorxϵBandx\text{/}ϵA$

$\Rightarrow xϵB$

$\Rightarrow xϵA\cup B$ [$\because B\subset (A\cup B)$]

This is true for all $xϵA\cup (B-A)$

$A\cup (B-A)\subset (A\cup B)$

conversely,

Let , $xϵA\cup B$

$xϵAorxϵB$

Case I :

Let, $xϵA$

$\Rightarrow xϵA\cup (B-A)$ [$\because A\subset A\cup (B-A)$]

$\Rightarrow A\cup B\subset A\cup (B-A)$

Case II:

Let, $xϵB$

$\Rightarrow xϵB-A$

$\Rightarrow xϵA\cup (B-A)$

[ $\because (B-A)\subset A\cup (B-A)$ ]

$\Rightarrow A\cup B\subset A\cup (B-A)$ in both cases.

Hence, $A\cup (B-A)=A\cup B$ proved.