Properties of Set Operations

Properties of Set Operations

It is an interesting investigation to find out if operations among sets (like union, intersection, etc) follow mathematical properties such as Commutativity, Associativity, etc., We have seen numbers having many of these properties; whether sets also possess these, is to be explored.

We first take up the properties of set operations on union and intersection.

1. Commutative Property

In set language, commutative situations can be seen when we perform operations. For example, we can look into the Union (and Intersection) of sets to find out if the operation is commutative.

Note

For any set A,

• A ∪ A = A and A ∩  A = A [Idempotent Laws].

•  A ∪ Ï• = A and A ∩ U = A [Identity Laws].

Let A = {2, 3, 8,10} and B = {1, 3,10, 13} be two sets.

Then, A ∪ B = {1, 2, 3, 8,10, 13} and

B ∪ A = {1,2, 3, 8,10,13}

From the above, we see that A ∪ B = B ∪ A.

This is called Commutative property of union of sets.

Now, A ∩ B = {3, 10} and B ∩ A = {3, 10}. Then, we see that A ∩ B = B ∩ A .

This is called Commutative property of intersection of sets.

Commutative property: For any two sets A and B

(i) A ∪ B =B ∪ A

(ii) A ∩ B =B ∩ A

Example 1.19

If A = {b, e, f ,g} and B = {c, e, g,h}, then verify the commutative property of (i) union of sets (ii) intersection of sets.

Solution

Given, A = { b, e, f ,g } and B = { c, e, g,h }

(i) A ∪ B = { b, c,e, f ,g , h }   ... (1)

 B ∪ A = { b, c, e, f, g, h }  ... (2)

From (1) and (2) we have A ∪ B = B ∪ A

It is verified that union of sets is commutative.

(ii) A∩B = { e, g } ... (3)

 B∩A= { e, g } ... (4)

From (3) and (4) we get, A ∩ B = B ∩ A

It is verified that intersection of sets is commutative.

Thinking Corner

Given, P = { l, n , p } and P ∪ Q  = { j ,l ,m , n,o , p}. If P and Q are disjoint sets, then what will be Q and P ∩Q ?

Note

Recall that subtraction on numbers is not commutative. Is set difference commutative? We expect that the set difference is not commutative as well. For instance, consider  A = { a , b, c }, B = { b, c, d }. A−B = { a }, B−A= { d } ; we see that A − B ≠ B − A.

2. Associative Property

Now, we perform operations on union and intersection for three sets.

Let A = {− 1, 0, 1, 2}, B = {−3, 0, 2, 3} and C = {0, 1, 3, 4} be three sets.

Now, B ∪ C = {-3, 0,1,2, 3, 4}

 A∪(B ∪C) = {−1, 0,1,2} ∪{−3, 0,1,2, 3, 4}

 = {-3,-1, 0,1,2, 3, 4}    ... (1)

Then, A ∪ B = {-3,-1, 0,1,2, 3}

 (A∪B)∪C = {−3,−1, 0,1,2, 3} ∪{0,1, 3, 4}

 = {-3,-1, 0,1,2, 3, 4} ... (2)

From (1) and (2), A ∪ (B ∪C ) = (A ∪ B ) ∪C .

This is associative property of union among setsA, B, and C.

Now, B ∩C = {0, 3}

 A∩(B ∩C) = {−1, 0,1,2} ∩{0, 3}

 = {0} ... (3)

Then, A ∩ B = {0,2}

 (A∩B)∩C = {0,2} ∩ {0,1, 3, 4}

 = {0} ... (4)

From (3) and (4), A ∩ (B ∩C ) = (A ∩ B ) ∩C .

This is associative property of intersection among setsA, B and C.

Associative property: For any three sets A, B and C

(i) A ∪ (B ∪C ) = (A ∪ B ) ∪C

(ii) A ∩ (B ∩C ) = (A ∩ B ) ∩C

Example 1.20

If  then verify A ∩ (B ∩C ) = (A ∩ B ) ∩C .

Solution

From (1) and (2), it is verified that

(A∩B)∩C = A∩(B ∩C)

Note

The set difference in general is not associative

that is, (A–B)–C ≠A–(B–C).

But, if the sets A, B and C are mutually disjoint then the set difference is associative

that is, (A–B)–C=A–(B–C).

3. Distributive Property

In lower classes, we have studied distributive property of multiplication over addition on numbers. That is, a × (b +c ) = (a ×b ) + (a ×c). In the same way we can define distributive properties on sets.

Distributive property: For any three sets A, B and C

(i) A ∩ (B ∪C ) = (A ∩ B ) ∪ (A ∩C) [Intersection over union]

(ii) A ∪ (B ∩C ) = (A ∪ B ) ∩ (A ∪C) [Union over intersection]

Example 1.21

If A = {0, 2, 4, 6, 8}, B = {x : x is a prime number and x < 11} and C= {x : x ∈ N and 5 ≤ x < 9 } then verify A ∪ (B ∩C ) = (A ∪ B ) ∩ (A ∪C) .

Solution

Given    A = { 0,2,4,6,8 } , B = { 2,3,5,7 } and C = { 5,6,7,8 }

First, we find B ∩C = {5,7 }, A ∪(B ∩C) = {0,2,4,5, 6,7,8} ... (1)

Next, A ∪ B = { 0,2,3,4,5,6,7,8 } , A∪C= { 0,2,4,5,6,7,8 }

Then, (A ∪ B) ∩ (A ∪C) = { 0,2,4,5,6,7,8 } ... (2)

From (1) and (2), it is verified that A ∪ (B ∩C ) = (A ∪ B ) ∩ (A ∪C).

Example 1.22

Verify A ∩ (B ∪C ) = (A ∩ B ) ∪ (A ∩C) using Venn diagrams.

Solution

From (1) and (2), A ∩ (B ∪C ) = (A ∩ B ) ∪ (A ∩C) is verified.