Question

Show that for any sets $A$ and $B$,
$A=(A\cap B)\cup (A–B)$ and $A\cup (B–A)=(A\cup B)$.

Answer 1

To show, $A=(A\cap B)\cup (A–B)$
Let $x\in A$
We have to show that, $x\in (A\cap B)\cup (A–B)$
In Case I,
$x\in (A\cap B)$
Then, $x\in (A\cap B)\subset (A\cup B)\cup (A–B)$
In Case II,
$x\notin A\cap B$
$\Rightarrow x\notin B$ or $x\notin A$
$\therefore x\notin B(x\notin A)$
$\therefore x\notin A–B\subset (A\cup B)\cup (A–B)$
$\therefore A\subset (A\cap B)\cup (A–B).......\left(i\right)$
It is clear that,
$A\cap B\subset A$ and $(A–B)\subset A$
$\therefore (A\cap B)\cup (A–B)\subset A........\left(ii\right)$
From (i) and (ii), we obtain
$A=(A\cap B)\cup (A–B)$
To prove: $A\cup (B–A)\subset A\cup B$
Let $x\in A\cup (B–A)$
$\Rightarrow x\in A$ or ($x\in B$ and $x\notin A$)
$\Rightarrow$ ($x\in A$ or $x\in B$) and ($x\in A$ and $x\notin A$)
$\Rightarrow x\in (B\cup A)$
$\therefore A\cup (B–A)\subset (A\cup B).......\left(iii\right)$
Next, we show that, $(A\cup B)\subset A\cup (B–A)$
Let $y\in A\cup B$
$\Rightarrow y\in A$ or $y\in B$
$\Rightarrow$ ($y\in A$ or $y\in B$) and ($y\in A$ and $y\notin A$)
$\Rightarrow y\in A$ or ($y\in B$ and $y\notin A$)
$\Rightarrow y\in A\cup (B–A)$
$\therefore A\cup B\subset A\cup (B–A)..........\left(iv\right)$
Hence, from (iii) and (iv), we obtain $A\cup (B–A)=A\cup B$.