From Venn’s diagram, we observe that if we derive union of ( A ∩ B A ∩ B ) to either of the difference sets, then we get the complete individual set.

Figure 2: The difference of two sets is a disjoint set.

Difference of two sets

A = A - B ∪ A ∩ B A = A - B ∪ A ∩ B

and

B = B - A ∪ A ∩ B B = B - A ∪ A ∩ B

The positions of sets about minus operator affect the result. It is clear from the figure above, where “A-B” and “B-A” represent different regions on Venn’s diagram. As such, the difference of sets is not commutative. Let us consider the example used earlier, where :

A = { 1,2,3,4,5,6 } A = { 1,2,3,4,5,6 }

A = { 4,5,6,7,8 } A = { 4,5,6,7,8 }

Then,

⇒ A − B = { 1,2,3 } ⇒ A − B = { 1,2,3 }

and

⇒ B − A = { 7,8 } ⇒ B − A = { 7,8 }

Clearly,

⇒ A − B ≠ B − A ⇒ A − B ≠ B − A

From the Venn’s diagram, we can see that union of two sets is equal to three distinct regions. Alternatively, we can say that the region represented by the union of two sets is equal to the sum of the regions representing three “disjoint” sets (i) difference set A-B (ii) intersection set " A ∩ B A ∩ B " and (iii) difference set B-A.

Figure 3: The difference of two sets is a disjoint set.

Difference of two sets

We use the term “symmetric set” for combining two differences as marked on Venn’s diagram. It is denoted as “ A Δ B A Δ B ”.

A Δ B = A − B ∪ B − A A Δ B = A − B ∪ B − A

Proceeding as before we can read the conditional statement for the complement with the help of two ways arrow as :

x ∈ A ′ ⇔ x ∈ U a n d x ∉ A x ∈ A ′ ⇔ x ∈ U a n d x ∉ A

In terms of minus or difference operation,

A ′ = U − A A ′ = U − A

It is clear from the representation on Venn’s diagram that the universal set comprises of two distinct sets – set A and complement set A’.

⇒ U = A ∪ A ′ ⇒ U = A ∪ A ′

The complement of universal set is empty set. It is so because difference of union set with itself is the empty set (see Venn's diagram).

U ′ = { x : x ∈ U a n d x ∉ U } = φ U ′ = { x : x ∈ U a n d x ∉ U } = φ

The complement of the empty set is universal set. It is so because difference of union set with the empty set is universal set (see Venn's diagram).

φ ′ = { x : x ∈ U a n d x ∉ φ } = U φ ′ = { x : x ∈ U a n d x ∉ φ } = U

The complement of complement set is set itself. The complement set is defined as :

A ′ = U − A A ′ = U − A

Now, complement of complement set is :

⇒ A ′ ′ = U − A ′ ⇒ A ′ ′ = U − A ′

Let us consider the example, where :

U = { 1,2,3,4,5,6,7,8 } U = { 1,2,3,4,5,6,7,8 }

A = { 1,2,3,4,5,6 } A = { 1,2,3,4,5,6 }

Then,

⇒ A ′ = { 1,2,3,4,5,6,7,8 } - { 1,2,3,4,5,6 } = { 7,8 } ⇒ A ′ = { 1,2,3,4,5,6,7,8 } - { 1,2,3,4,5,6 } = { 7,8 }

Again taking complement, we have :

A ′ ′ = { 1,2,3,4,5,6,7,8 } - { 7,8 } = { 1,2,3,4,5,6 } = A A ′ ′ = { 1,2,3,4,5,6,7,8 } - { 7,8 } = { 1,2,3,4,5,6 } = A

The union of a set with its complement is universal set :

A ∪ A ′ = { x : x ∈ U a n d x ∈ A } ∪ { x : x ∈ U a n d x ∈ A } = U A ∪ A ′ = { x : x ∈ U a n d x ∈ A } ∪ { x : x ∈ U a n d x ∈ A } = U

From Venn’s diagram also, we see that universal set consists of set A and component A’.

U = A ∪ A ′ U = A ∪ A ′

The two sets on the right side of the equation are disjoint sets. Hence,

A ∪ A ′ = U A ∪ A ′ = U

There is nothing common between set A and its component A’. Thus, intersection of a set with its complement yields the empty set,

A ∩ A ′ = φ A ∩ A ′ = φ

Here, we shall prove first De-morgan’s law in this section. The second law can be proved in similar fashion. Let us consider an arbitrary element “x” belonging to set ( A ∪ B A ∪ B )’.

x ∈ A ∪ B ′ x ∈ A ∪ B ′

⇒ x ∉ A ∪ B ⇒ x ∉ A ∪ B

Then, by definition of union,

⇒ x ∉ { x : x ∈ A o r x ∈ B } ⇒ x ∉ { x : x ∈ A o r x ∈ B }

Here, “not or” is interpreted same as “and”,

⇒ x ∉ A a n d x ∉ B ⇒ x ∉ A a n d x ∉ B

⇒ x ∈ A ′ a n d x ∈ B ′ ⇒ x ∈ A ′ a n d x ∈ B ′

⇒ x ∈ A ′ ∩ B ′ ⇒ x ∈ A ′ ∩ B ′

But, we had started with ( A ∪ B A ∪ B )’ and used its definition to show that “x” belongs to another set. It means that the other set consists of the elements of the first set – at the least. Thus,

A ∪ B ′ ⊂ A ′ ∩ B ′ A ∪ B ′ ⊂ A ′ ∩ B ′

Similarly, we can start with A ′ ∩ B ′ A ′ ∩ B ′ and reach the conclusion that :

A ′ ∩ B ′ ⊂ A ∪ B ′ A ′ ∩ B ′ ⊂ A ∪ B ′

If sets are subsets of each other, then they are equal. Hence,

A ′ ∩ B ′ = A ∪ B ′ A ′ ∩ B ′ = A ∪ B ′

Problem 1: In the reference of students in a class, the set “B” represents students, who play basketball. The set “F” represents students, who play football. The set “B” and “F” are left and right circles respectively on the Venn's diagram shown below. Identify regions marked 1 to 8 on the Venn’s diagram. Also interpret regions identified by combination U – (6+7).

Figure 7: Interpreting sets

sets

Solution :The meaning of regions market 1 – 8 are as given hereunder :

1 : B-F : It represents the difference of “B” and “F”. It consists of students, who play basketball, but not football.

2 : F-B : It represents the difference of “F” and “B”. It consists of students, who play football, but not basketball.

3 : B ∩ F B ∩ F : It represents the intersection of two sets. It consists of students, who play both basketball and football.

4 : B: It represents the set “B”. It is union of two disjoint sets “B-F” and “ B ∪ F B ∪ F ”. It consists of students, who play basketball.

5 : F: It represents the set “F”. It is union of two disjoint sets “F-B” and “ B ∩ F B ∩ F ”. It consists of students, who play football.

6 : B∪F: It represents the union set of set “B” and “F”. Equivalently, it is union of three disjoint sets “B-F”, “ B ∩ F B ∩ F ” and “F-B”. It consists of students, who play either of two games or both.

7 : ( B ∪ F B ∪ F )’: It represents the component of union set “ B ∪ F B ∪ F ”. It consists of students, who play neither basketball nor football.

8 : B - F ∪ F - B B - F ∪ F - B : It represents union of two disjoint difference sets “B-F” and “F-B”. It consists of students, who play only one game.

The region, identified by U – (6+7), is complement of “ B ∩ F B ∩ F ”. It represents students, who do not play both games, but may play one of them.