Properties of Union and Intersection (Commutative, Associative, and Distributive) w. r. t. Union and Intersection

Properties of Union and Intersection

a)  Commutative Property of Union

For any two sets $A$ and $B$
$A\cup B$ $=B$ $\cup A$

Proof:

Let  $x$ $\epsilon A$ $\cup B$

$x$ $\epsilon A$  or  $x$ $\epsilon B$        (By the definition of the union sets)

$x$ $\epsilon B$  or  $x$ $\epsilon A$

$x$ $\epsilon B$ $\cup A$

$A$ $\cup B$ $\subseteq B$ $\cup A$     $(i)$

Now let
$y$ $\epsilon B\cup A$
$y$ $\epsilon B$    or    $y$ $\epsilon A$

$y$ $\epsilon A$    or    $y$ $\epsilon B$

$y$ $\epsilon A$ $\cup B$
$B$ $\cup A$ $\subseteq A$ $\cup B$   $(ii)$

From $(i)$ and $(ii)$, we have
$B$ $\cup A$ $= A$ $\cup B$   (by the definition of equal sets)

b)    Commutative property of intersection

For any two sets $A$ and $B$       
$A\cap B$ $= B\cap A$Proof:
Let   $x$ $\epsilon A\cap B$
$x$ $\epsilon A$  and  $x\epsilon B$    (By the definition of intersection sets)
$x\epsilon B$  and  $x\epsilon A$
$x$ $\epsilon B$ $\cap A$
$A$ $\cap B$ $\subseteq B$ $\cap A$       $(i)$
Now let 
$y$ $\epsilon B$ $\cap A$

$y$ $\epsilon B$   and  $y$ $\epsilon A$
$y$ $\epsilon A$    and    $y$ $\epsilon B$
$y$ $\epsilon A$ $\cap B$
$B$ $\cap A\subseteq A$ $\cap B$      $(ii)$

From $(i)$ and $(ii)$, we have
$A\cap B$ $=B\cap A$             (By the definition of equal sets)

c)    Associative property of the union

For any three sets $A$, $B$, and $C$

$A$ $\cup (B\cup C)$ $=(A\cup B)\cup C$

$y$ $\epsilon(A\cup B)$ $\cup C$

$y$ $\epsilon(A\cup B)$  or  $y$ $\epsilon C$

$y$ $\epsilon A$  or  $y\epsilon B$  or  $y\epsilon C$

$y$ $\epsilon A$  or  $y\epsilon (B\cup C)$

$y$ $\epsilon A$ $\cup (B\cup C)$

$y$ $\epsilon (A\cup B)$ $\cup C$ v\subseteq A$ $\cup (B\cup C)$        $(ii)$

Now let
$A$ $\cap (B\cap C)$ $=(A\cap B)$ $\cap C$

$x$ $\epsilon A$  or  $x$ $\epsilon (B\cap C)$

$x\epsilon A$  or  $(x\epsilon B$ and $x\epsilon C)$

$(x\epsilon A$ or $x\epsilon B)$    and    $(x\epsilon A$ or $x\epsilon C)$
$x\epsilon (A\cup B)$    and    $x\epsilon (A\cup C)$
$x\epsilon (A\cup B)$ $\cap (B\cup C)$

$A\cup (B\cap C)$ $\subseteq (A\cup B)$ $\cap (B\cup C)$        $(i)$

$y$ $\epsilon (A\cup B)$ $\cap (A\cup C)$

$y$ $\epsilon (A\cup B)$  and  $y$ $\epsilon (A\cup C)$

$y$ $\epsilon A$  or  $y\epsilon B$  and  $y\epsilon A$  or  $y\epsilon C$

$y$ $\epsilon A$  or  $(y$ $\epsilon B$ and $y$ $\epsilon C)$

$y$ $\epsilon A$  or  $y\epsilon (B\cap C)$

$y$ $\epsilon A$ $\cup (B\cap C)$

$(A\cup B)$ $\cap (A\cup C)$$\subseteq A$ $\cup (B\cap C)$        $(ii)$

Proof:
Let:  $x\epsilon A$ $\cup (B\cup C)$

$x\epsilon A$  or  $x$ $\epsilon (B\cup C)$

$x\epsilon A$  or  $x$ $\epsilon B$  or   $x$ $\epsilon C$

$x\epsilon (A\cup B)$  or  $x$ $\epsilon C$

$x\epsilon (A\cup B)$ $\cup C$

$A\cup (B\cup C)$ $\subseteq (A\cup B)\cup C$        $(i)$

From $(i)$ and $(ii)$ we have
$A$ $\cup (B\cup C)$ $=(A\cup B)$ $\cup C$

d)   Associative property of Intersection
       For any three sets $A$, $B$, and $C$
       $A$ $\cap (B\cap C)$ $=(A\cap B)\cap C$

Proof:
Let $x$ $\epsilon A$ $\cap (B\cap C)$

      $x$ $\epsilon A$    and    $x$ $\epsilon (B\cap C)$

      $x\epsilon A$    and    $x\epsilon B$    and    $x\epsilon C$
      $x\epsilon (A\cap B)$    and    $x\epsilon C$
      $x\epsilon (A\cap B)\cap C$
      $A\cap (B\cap C)$ $\cap C $ $\subseteq (A\cap B)$ $\cap C$        $(i)$

Now let
       $y\epsilon (A\cap B)$ $\cap C$

       $y\epsilon (A\cap B)$   and   $y\epsilon C$
       $y\epsilon A$   and   $y\epsilon B$    and    $y\epsilon C$
       $y\epsilon A$    and    $y\epsilon (B\cap C)$
       $y\epsilon A$ $\cap (B\cap C)$

       $(A\cap B)$ $\cap C$ $\subseteq A$ $\cap (B\cap C)$        $(ii)$

From $(i)$ and $(ii)$ we have

e)    Distributive property of the union over the Intersection
        For any three sets $A$, $B$, and $C$
        $A$ $\cup (B\cap C)$ $=(A\cup B)$ $\cap (A\cup C)$

Proof:
Let  $x\epsilon A$ $\cup (B\cap C)$

Now let
From $(i)$ and $(ii)$ we have
$A\cup (B\cap C)$ $=(A\cup B)\cap (A\cup C)$

Similarly, we can prove
Distributive property of intersection over union
For any three sets
$A$ $\cap (B\cup C)$ $=(A\cap B)\cup (A\cap C)$