Algebra of Sets

Sets under the operations of union, intersection, and complement satisfy various laws (identities) which are listed in Table 1.

Table: Law of Algebra of Sets

Idempotent Laws	(a) A ∪ A = A	(b) A ∩ A = A
Associative Laws	(a) (A ∪ B) ∪ C = A ∪ (B ∪ C)	(b) (A ∩ B) ∩ C = A ∩ (B ∩ C)
Commutative Laws	(a) A ∪ B = B ∪ A	(b) A ∩ B = B ∩ A
Distributive Laws	(a) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)	(b) A ∩ (B ∪ C) =(A ∩ B) ∪ (A ∩ C)
De Morgan's Laws	(a) (A ∪B)^c=A^c∩ B^c	(b) (A ∩B)^c=A^c∪ B^c
Identity Laws	(a) A ∪ ∅ = A (b) A ∪ U = U	(c) A ∩ U =A (d) A ∩ ∅ = ∅
Complement Laws	(a) A ∪ A^c= U (b) A ∩ A^c= ∅	(c) U^c= ∅ (d) ∅^c = U
Involution Law	(a) (A^c)^c = A

Table 1 shows the law of algebra of sets.

Example 1: Prove Idempotent Laws:

Solution:

Since, B ⊂ A ∪ B, therefore A ⊂ A ∪ A
Let   x ∈ A ∪ A ⇒ x ∈ A  or   x ∈ A ⇒  x ∈ A
∴ A ∪ A ⊂ A
As  A ∪ A ⊂ A and  A ⊂ A ∪ A ⇒ A =A ∪ A. Hence Proved.

Solution:

Since, A ∩ B ⊂ B, therefore A ∩ A ⊂ A
Let x ∈ A ⇒ x ∈ A  and x ∈ A  
⇒ x ∈ A ∩ A         ∴ A ⊂ A ∩ A
As A ∩ A ⊂ A and A ⊂ A ∩ A ⇒ A = A ∩ A. Hence Proved.

Example 2: Prove Associative Laws:

Solution:

Let some x ∈ (A'∪ B) ∪ C
   ⇒  (x ∈ A   or   x ∈ B)    or   x ∈ C
   ⇒   x ∈ A   or   x ∈ B     or  x ∈ C
  ⇒    x ∈ A   or   (x ∈ B    or  x ∈ C)
  ⇒   x ∈ A   or   x ∈ B ∪ C 
  ⇒   x ∈ A ∪ (B ∪ C).
Similarly, if some   x ∈ A ∪ (B ∪ C), then  x ∈ (A ∪ B) ∪ C.
Thus, any 	         x ∈ A ∪ (B ∪ C) ⇔  x ∈ (A ∪ B) ∪ C. Hence Proved.

Solution:

Let some x ∈ A ∩ (B ∩ C) ⇒   x ∈ A and x ∈ B ∩ C 
   ⇒   x ∈ A  and (x ∈ B and x ∈ C)  ⇒   x ∈ A  and x ∈ B and x ∈ C
  ⇒   (x ∈ A  and x ∈ B) and x ∈ C)  ⇒   x ∈ A ∩ B and x ∈ C
  ⇒   x ∈ (A ∩ B) ∩ C.
Similarly, if some   x ∈ A ∩ (B ∩ C), then x ∈ (A ∩ B) ∩ C
Thus, any 	         x ∈ (A ∩ B) ∩ C  ⇔  x ∈ A ∩ (B ∩ C). Hence Proved.

Example3: Prove Commutative Laws

Solution:

To Prove 
      A ∪ B = B ∪ A
      A ∪ B = {x: x ∈ A or x ∈ B}
            = {x: x ∈ B or x ∈ A}   (∵ Order is not preserved in case of sets)
      A ∪ B = B ∪ A. Hence Proved.

Solution:

To Prove 
      A ∩ B = B ∩ A
      A ∩ B = {x: x ∈ A and x ∈ B}
            = {x: x ∈ B and x ∈ A}   (∵ Order is not preserved in case of sets)
      A ∩ B = B ∩ A. Hence Proved.

Example 4: Prove Distributive Laws

Solution:

To Prove 
	     Let x ∈ A ∪ (B ∩ C)  ⇒ x ∈ A or  x ∈ B ∩ C 
      ⇒   (x ∈ A  or x ∈ A) or (x ∈ B and   x ∈ C)
      ⇒   (x ∈ A  or x ∈ B) and (x ∈ A  or x ∈ C)
      ⇒   x ∈ A ∪ B and   x ∈ A ∪ C
      ⇒   x ∈ (A ∪ B) ∩ (A ∪ C)
	  
Therefore, A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C)............(i)
Again, Let y ∈ (A ∪ B)  ∩ (A ∪ C) ⇒   y ∈ A ∪ B and y ∈ A ∪ C
      ⇒   (y ∈ A or y ∈ B) and (y ∈ A or y ∈ C)
      ⇒   (y ∈ A and y ∈ A) or (y ∈ B and y ∈ C)
      ⇒   y ∈ A    or    y ∈ B ∩ C
      ⇒   y ∈ A  ∪ (B ∩ C)
Therefore, (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C)............(ii)

Combining (i) and (ii), we get A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). Hence Proved

Solution:

To Prove 
      Let x ∈ A ∩ (B ∪ C)   ⇒   x ∈ A and x ∈ B ∪ C
	⇒  (x ∈ A and x ∈ A) and (x ∈ B  or x ∈ C)
         ⇒  (x ∈ A and x ∈ B) or  (x ∈ A and x ∈ C)
         ⇒   x ∈ A ∩ B or  x ∈ A ∩ C
         ⇒   x ∈ (A ∩ B) ∪ (A ∪ C)
		 
Therefore, A ∩ (B ∪ C) ⊂ (A ∩ B) ∪ (A ∪ C)............ (i)
Again, Let  y ∈ (A ∩ B) ∪ (A ∪ C) ⇒ y ∈ A ∩ B or y ∈ A ∩ C
	  ⇒  (y ∈ A and y ∈ B) or (y ∈ A and y ∈ C)
	  ⇒  (y ∈ A or y ∈ A) and (y ∈ B or y ∈ C)
	  ⇒ y ∈ A and  y ∈ B ∪ C
           ⇒ y ∈ A ∩ (B ∪ C)
Therefore, (A ∩ B) ∪ (A ∪ C) ⊂ A ∩ (B ∪ C)............ (ii)

Combining (i) and (ii), we get A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∪ C). Hence Proved

Example 5: Prove De Morgan's Laws

(a) (A ∪B)^c=A^c∩ B^c

Solution:

To Prove (A ∪B)^c=A^c∩ B^c
Let x ∈ (A ∪B)^c  ⇒  x ∉  A ∪ B			(∵ a ∈ A ⇔ a ∉ A^c)
           ⇒  x ∉  A and x ∉ B
           ⇒  x ∉  A^c and x ∉ B^c
           ⇒  x ∉  A^c∩ B^c
Therefore,  (A ∪B)^c ⊂ A^c∩ B^c............. (i)
Again, let x ∈ A^c∩ B^c ⇒ x ∈ A^c and x ∈ B^c
            ⇒ x ∉  A and x ∉ B
            ⇒  x ∉  A ∪ B
            ⇒ x ∈ (A ∪B)^c
Therefore, A^c∩ B^c  ⊂ (A ∪B)^c............. (ii)
Combining (i) and (ii), we get A^c∩ B^c =(A ∪B)^c. Hence Proved.

(b) (A ∩B)^c = A^c∪ B^c

Solution:

Let x ∈ (A ∩B)^c ⇒ x ∉  A ∩ B    (∵ a ∈ A ⇔ a ∉ A^c)
           ⇒ x ∉  A or x ∉ B
           ⇒ x ∈ A^c and x ∈ B^c
           ⇒ x ∈ A^c∪ B^c
∴ (A ∩B)^c⊂ (A ∪B)^c.................. (i)
Again, Let x ∈ A^c∪ B^c   ⇒ x ∈ A^c or x ∈ B^c
            ⇒ x ∉  A or x ∉ B
            ⇒ x ∉  A ∩ B
            ⇒ x ∈ (A ∩B)^c
∴ A^c∪ B^c⊂ (A ∩B)^c.................... (ii)
Combining (i) and (ii), we get(A ∩B)^c=A^c∪ B^c. Hence Proved.

Example 6: Prove Identity Laws.

Solution:

To Prove A ∪ ∅ = A
	         Let  x ∈ A ∪ ∅ ⇒ x ∈ A   or  x ∈ ∅
              ⇒ x ∈ A        (∵x ∈ ∅, as ∅ is the null set )
        Therefore, x ∈ A ∪ ∅ ⇒ x ∈ A   
  Hence,     A ∪ ∅ ⊂ A.
We know that A ⊂ A ∪ B for any set B.
 But for B = ∅, we have A ⊂ A ∪ ∅ 
From above, A ⊂ A ∪ ∅ , A ∪ ∅ ⊂ A ⇒ A = A ∪ ∅. Hence Proved.

Solution:

To Prove A ∩ ∅ = ∅
If  x ∈ A, then x ∉  ∅             (∵∅ is a null set)
Therefore, x ∈ A, x ∉  ∅ ⇒ A ∩ ∅ = ∅. Hence Proved.

Solution:

To Prove A ∪ U = U
Every set is a subset of a universal set.
   ∴   A ∪ U ⊆ U
    Also,   U ⊆ A ∪ U
Therefore, A ∪ U = U. Hence Proved.

Solution:

To Prove A ∩ U = A
We know   A ∩ U ⊂ A................. (i)
So we have to show that A ⊂ A ∩ U
Let  x ∈ A ⇒ x ∈ A and x ∈ U        (∵ A ⊂ U so x ∈ A ⇒ x ∈ U )         
   ∴     x ∈ A ⇒ x ∈ A ∩ U
   ∴     A ⊂ A ∩ U................. (ii)
From (i) and (ii), we get A ∩ U = A. Hence Proved.

Example7: Prove Complement Laws

(a) A ∪ A^c= U

Solution:

To Prove A ∪ A^c= U
  Every set is a subset of U
    ∴  A ∪ A^c ⊂ U.................. (i)
We have to show that U ⊆ A ∪ A^c
  Let x ∈ U  ⇒  x ∈ A    or    x ∉  A     
      ⇒  x ∈ A    or   x ∈ A^c    ⇒ x ∈ A ∪ A^c
    ∴ U ⊆ A ∪ A^c................... (ii)
From (i) and (ii), we get A ∪ A^c= U. Hence Proved.

(b) A ∩ A^c=∅

Solution:

As ∅ is the subset of every set
     ∴     ∅ ⊆ A ∩ A^c..................... (i)
We have to show that A ∩ A^c ⊆ ∅
Let x ∈ A ∩ A^c  ⇒ x ∈ A and x ∈  A^c       
     ⇒ x ∈ A  and x ∉  A
      ⇒ x ∈ ∅
    ∴      A ∩ A^c ⊂∅..................... (ii)

From (i) and (ii), we get A∩ A^c=∅. Hence Proved.

(c) U^c= ∅

Solution:

Let x ∈ U^c   ⇔ x ∉ U ⇔ x ∈ ∅
    ∴ U^c= ∅. Hence Proved.     (As U is the Universal Set).

(d) ∅^c = U

Solution:

Let x ∈ ∅^c ⇔ x ∉ ∅  ⇔ x ∈ U       (As ∅ is an empty set)
	∴ ∅^c = U.  Hence Proved.

Example8: Prove Involution Law

(a) (A^c )^c A.

Solution:

Let x ∈ (A^c )^c ⇔ x ∉ A^c⇔  x ∈ a
     ∴ (A^c )^c =A. Hence Proved.

Duality:

The dual E∗ of E is the equation obtained by replacing every occurrence of ∪, ∩, U and ∅ in E by ∩, ∪, ∅, and U, respectively. For example, the dual of

It is noted as the principle of duality, that if any equation E is an identity, then its dual E∗ is also an identity.

Principle of Extension:

According to the Principle of Extension two sets, A and B are the same if and only if they have the same members. We denote equal sets by A=B.

 If A= {1, 3, 5} and B= {3, 1, 5}, then A=B i.e., A and B are equal sets.
 If A= {1, 4, 7} and B= {5, 4, 8}, then A≠ B i.e.., A and B are unequal sets.

Cartesian product of two sets:

The Cartesian Product of two sets P and Q in that order is the set of all ordered pairs whose first member belongs to the set P and second member belong to set Q and is denoted by P x Q, i.e.,

Example: Let P = {a, b, c} and Q = {k, l, m, n}. Determine the Cartesian product of P and Q.

Solution: The Cartesian product of P and Q is