Last modified on July 12th, 2024

chapter outline

 

Properties of Sets

The properties of real numbers apply to sets as well. Here are the six basic properties of sets involving sets A, B, and C.

Commutative

Intersection and union of sets hold the commutative law.

  • A ∪ B = B ∪ A
  • A ∩ B = B ∩ A

Associative 

Intersection and union of sets hold the associative law.

  • (A ∪ B) ∪ C = A ∪ (B ∪ C)
  • (A ∩ B) ∩ C = A ∩ (B ∩ C)

Distributive

Intersection and union of sets hold the distributive property.

  • A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
  • A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Identity

  • A ∪ ɸ = A
  • A ∩ ∪ = A

Complement

  • A ∪ A’ = ∪

Idempotent 

  • A ∪ A = A
  • A ∩ A = A
Properties of Sets

Others

Apart from the six basic properties, some more properties are used in set theory.

De Morgan’s Laws

For the Union of Sets

It states that the complement of the union of two sets is the intersection of their complements. 

If A and B are two sets, then 

(A ∪ B)’ = A’ ∩ B’

For the Intersection of Sets

It states that the complement of the intersection of two sets is the union of their complements. 

If A and B are two sets, then

(A ∩ B)’ = A’ ∪ B’

The Principle of Duality

The properties mentioned above come in pairs of identities, where each property can be transformed into its counterpart by swapping the union (∪) with the intersection (∩) and the empty set (ɸ) with the universal set (U). This concept is known as the principle of duality.

For Two Finite Sets

If A and B are any two finite sets, then

  • (A – B) ∩ B = ɸ
  • B – A = B ∩ A’
  • A ⊆ B ⇔ B’ ⊆ A’
  • (A – B) ∪ B = A U B
  • A – B = A ∩ B’
  • A – B = A ⇔ A ∩ B = ɸ
  • (A – B) ∪ (B – A) = (A ∪ B) – (B ∩ A)

For Three Finite Sets

If A, B, and C are any three finite sets, then

  • A – (B ∩ C) = (A – B) ∪ (A – C)
  • A – (B ∪ C) = (A – B) ∩ (A – C)
  • A ∩ (B – C) = (A ∩ B) – (A ∩ C)
  • A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Solved Examples

Find A ∪ B if A = {2, 5, 7, 12, 19} and B = {7, 12, 17, 22}, and verify that the union set follows the commutative property.

Solution:

Given A = {2, 5, 7, 12, 19} and B = {7, 12, 17, 22}
Thus, A ∪ B = {2, 5, 7, 12, 19} ∪ {7, 12, 17, 22} = {2, 5, 7, 12, 17, 19, 22}
Now, B ∪ A = {7, 12, 17, 22} ∪ {2, 5, 7, 12, 19} = {2, 5, 7, 12, 17, 19, 22}
A ∪ B = B ∪ A = {2, 5, 7, 12, 17, 19, 22}
Hence, the commutative property is verified.

Find  A ∩ B if A = {2, 5, 7, 12, 19}, B = {7, 12, 17, 22}, and C = {5, 7, 35}, and verify that the intersection set follows the associative property.

Solution:

Given A = {2, 5, 7, 12, 19}, B = {7, 12, 17, 22}, and C = {5, 7, 35}
Here, A ∩ B = {2, 5, 7, 12, 19} ∩ {7, 12, 17, 22} = {7, 12}
(A ∩ B) ∩ C = {7, 12} ∩ {5, 7, 35} = {7}
Thus, the intersection of sets A, B, and C is {7}
Now, B ∩ C = {7, 12, 17, 22} ∩ {5, 7, 35} = {7}
A ∩ (B ∩ C) = {2, 5, 7, 12, 19} ∩ {7} = {7}
(A ∩ B) ∩ C = A ∩ (B ∩ C) = {7}
Hence, the associative property is verified.

Verify the distributive property A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) when A = {even prime number}, B = {8, 10, 12, 14}, and C = {x2 | 2 < x < 6 and x is a natural number}

Solution:

Given sets A = {even prime number}, B = {4, 8, 10, 12, 14}, and C = {x2 | 1 < x < 6 and x is a natural number}
⇒ A = {2}, B = {4, 8, 10, 12, 14}, and C = {4, 9, 16, 25}
Here, B ∩ C = {4, 8, 10, 12, 14} ∩ {4, 9, 16, 25} = {4}
A ∪ B = {2} ∪ {4, 8, 10, 12, 14} = {2, 4, 8, 10, 12, 14}
A ∪ C = {2} ∪ {4, 9, 16, 25} = {2, 4, 9, 16, 25}
Now, A ∪ (B ∩ C) = {2} ∪ {4} = {2, 4} and
(A ∪ B) ∩ (A ∪ C) = {2, 4, 8, 10, 12, 14} ∩ {2, 4, 9, 16, 25} = {2, 4}
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) = {2, 4}
Hence, the distributive property is verified.

If A = {3, 4, 12, 48}, find its union with an empty set.

Solution:

Given A = {3, 4, 12, 48}
Let the empty set be B = { }
Now, A ∪ B = {3, 4, 12, 48} ∪ { } = {3, 4, 12, 48} = A
Thus, the union of set A with the empty set gives the set A itself.

Verify if A = {a, e, i, o, u} and A’ = {b, c, d, f, g, h, p q, v} are complement sets when the universal set is U = {a, b, c, d, e, f, g, h, i, o, p, q, u, v}

Solution:

Given A = {a, e, i, o, u}, A’ = {b, c, d, f, g, h, p q, v}, and U = {a, b, c, d, e, f, g, h, i, o, p, q, u, v}
Using the complement property of sets, we get
A ∪ A’ = {a, e, i, o, u} ∪ {b, c, d, f, g, h, p q, v} 
⇒ A ∪ A’ = {a, b, c, d, e, f, g, h, i, o, p, q, u, v}
⇒ A ∪ A’ = U, the complement property is verified.
Thus, sets A and A’ are complement sets.

Last modified on July 12th, 2024