Question

(Commultative laws) For any two sets a and B, prove that:
$I.A\cup B=B\cup A$ [Commutative law for union of sets]

$II.A\cap B=B\cap A$ [Commutative law for intersection of sets]

Answer 1

I. Let x be an arbitary element of $A\cup B$. Then,

$x\epsilon A\cup B$$\Rightarrow xϵAorxϵB\Rightarrow xϵBorxϵA\Rightarrow \Rightarrow xϵB\cup A.$

$\therefore A\cup B\subseteq B\cup A.\dots \left(i\right)$

Again, let y be an arbitary elements of $B\cup A$. Then,

$yϵB\cup A$$\Rightarrow yϵBoryϵA\Rightarrow yϵAoryϵB\therefore B\cup A\subseteq A\cup B.\dots \left(ii\right)$

From (i) and (ii), we get $A\cup B=B\cup A$.

II. Let x be an arbitary element of $A\cap B$. Then,

$xϵA\cap B$$\Rightarrow xϵAandxϵB\Rightarrow xϵBandxϵA\Rightarrow xϵB\cap A$
$\therefore A\cap B\subseteq B\cap A.\dots \left(iii\right)$

Again, let y be an arbitray element of $B\cap A$. Then,
$yϵB\cap A$$\Rightarrow yϵBandyϵA\Rightarrow yϵAandyϵB\Rightarrow yϵA\cap B$
$\therefore B\cap A\subset A\cap B\dots \left(iv\right)$

From (iii) and (iv), we get $A\cap B=B\cap A$.