Relationship in Sets using Venn Diagram

The relationship in sets using Venn diagram are discussed below:

• The union of two sets can be represented by Venn diagrams by the shaded region, representing A ∪ B.

$A ∪ B when A ⊂ B$

A ∪ B when A ⊂ B

$A ∪ B when neither A ⊂ B nor B ⊂ A$

A ∪ B when neither A ⊂ B nor B ⊂ A

$A ∪ B when A and B are Disjoint Sets$

A ∪ B when A and B are disjoint sets

• The intersection of two sets can be represented by Venn diagram, with the shaded region representing A ∩ B.

$A ∩ B when A ⊂ B, i.e., A ∩ B = A$

A ∩ B when A ⊂ B, i.e., A ∩ B = A

$A ∩ B when neither A ⊂ B nor B ⊂ A$

A ∩ B when neither A ⊂ B nor B ⊂ A

$A ∩ B = ϕ No shaded Part$

A ∩ B = ϕ No shaded part

• The difference of two sets can be represented by Venn diagrams, with the shaded region representing A - B.

$A – B when B ⊂ A$

A – B when B ⊂ A

$A – B when neither A ⊂ B nor B ⊂ A$

A – B when neither A ⊂ B nor B ⊂ A

$A – B when A and B are Disjoint Sets$

A – B when A and B are disjoint sets.

Here A – B = A

$A – B when A ⊂ B$

A – B when A ⊂ B

Here A – B = ϕ

Relationship between the three Sets using Venn Diagram

• If ξ represents the universal set and A, B, C are the three subsets of the universal sets. Here, all the three sets are overlapping sets.

Let us learn to represent various operations on these sets.

$A ∪ B ∪ C$

A ∪ B ∪ C

$A ∩ B ∩ C$

A ∩ B ∩ C

$A ∪ (B ∩ C)$

A ∪ (B ∩ C)

$A ∩ (B ∪ C)$

A ∩ (B ∪ C)

Some important results on number of elements in sets and their use in practical problems.

Now, we shall learn the utility of set theory in practical problems.

If A is a finite set, then the number of elements in A is denoted by n(A).

Relationship in Sets using Venn Diagram
Let A and B be two finite sets, then two cases arise:

$A and B be Two Finite Sets$

Case 1:

A and B are disjoint.

Here, we observe that there is no common element in A and B.

Therefore, n(A ∪ B) = n(A) + n(B)

$A and B are not Disjoint Sets$

Case 2:

When A and B are not disjoint, we have from the figure

(i) n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

(ii) n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B)

(iii) n(A) = n(A - B) + n(A ∩ B)

(iv) n(B) = n(B - A) + n(A ∩ B)

$Sets A – B$

A – B

$Sets B – A$

B – A

$A ∩ B Sets$

A ∩ B

Let A, B, C be any three finite sets, then

n(A ∪ B ∪ C) = n[(A ∪ B) ∪ C]

                = n(A ∪ B) + n(C) - n[(A ∪ B) ∩ C]

                = [n(A) + n(B) - n(A ∩ B)] + n(C) - n [(A ∩ C) ∪ (B ∩ C)]

                = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)

         [Since, (A ∩ C) ∩ (B ∩ C) = A ∩ B ∩ C]

Therefore, n(A ∪B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)