You have landed on the right page to learn about Operations on Sets. In our daily lives, we often deal with collecting objects like books, stamps, coins, etc. Set theory is a mathematical way of representing a collection of objects. A set is a well-defined collection of objects.

Here a well-defined compilation of objects means that given a specific object, it must be possible for us to decide whether the object is an element of the given collection or not. The objects of a set are called its representatives or elements. Continue reading to know more.

Define Operations on Sets

The set operations are carried out on two or more sets to obtain a mixture of elements, as per the operation performed on them. 

There are three fundamental operations for constructing new sets from given sets.

1. Union of sets \({\rm{(U)}}\)
2. Intersection of sets \(( \cap )\)
3. Difference of sets \({\rm{( – )}}\)

Operations on Sets Notes

Union of Two Sets: The union of two sets \(A\) and \(B\) is the set of elements either in \(A\) or in \(B\) or both. It is indicated by \(A∪B\) and read as \(A\) union \(B\).

In symbol, \(A \cup B = \left\{ {x:x \in A\,{\rm{or}}\,x \in B} \right\}\)

For example,

If \(P = \left\{ {{\rm{Asia,}}\,{\rm{Africa,}}\,{\rm{Antarctica}},\,{\rm{Australia}}} \right\}\) and

\(Q = \left\{ {{\rm{Europe,}}\,{\rm{North}}\,{\rm{America,}}\,{\rm{South}}\,{\rm{America}}} \right\}\) then the union set of \(P\) and \(Q\) is

\(P \cup Q = \left\{ {{\rm{Asia,}}\,{\rm{Africa,}}\,{\rm{Antarctica,}}\,{\rm{Australia,}}\,{\rm{Europe,}}\,{\rm{North}}\,{\rm{America,}}\,{\rm{South}}\,{\rm{America}}} \right\}\) respectively .

Intersection of Two Sets: The intersection of two sets \(A\) and \(B\) the set containing all those common elements to both the sets \(A\) and \(B\). It is indicated by \(A∩B\) and read as \(A\) intersection \(B\).

In symbol, \(A \cap B = \left\{ {x:x \in A\,{\rm{and}}\,x \in B} \right\}\)

For example, 

If \(A = \left\{ {1,\,2,\,6} \right\},\,B = \left\{ {2,\,3,\,4} \right\}\) then \(A \cap B = \left\{ 2 \right\}\) because \(2\) is the common element of the sets \(A\) and \(B\).

Disjoint Sets: Two sets \(P\) and \(Q\) are called disjoint sets if they do not possess any common element. The intersection of two disjoint sets is the empty set. Sets \(P\) and \(Q\) are disjoint sets if \(P∩Q=∅\).

For example: If \(P = \left\{ {a,\,b,\,c,\,d} \right\}\) and \(Q = \left\{ {p,\,q,\,r,\,s,\,t} \right\}\) then \(P∩Q=∅\). Hence, \(P\) and \(Q\) are disjoint sets.

Difference of Two Sets: Let \(P\) and \(Q\) be two sets; the difference of sets \(P\) and \(Q\) is the set of all elements which are in \(P\), but not in \(Q\). It is specified by \(P-Q\) or \(P\backslash Q\) and read as \(P\) difference \(Q\).

In symbol, \(P – Q = \left\{ {x:x \in P\,{\rm{and}}\,x \notin Q} \right\}\) and \(Q – P = \left\{ {x:x \in Q\,{\rm{and}}\,x \notin P} \right\}\)

For example, 

If \(P = \left\{ { – 3,\, – 2,\,1,\,4} \right\}\) and \(Q = \left\{ {0,\,1,\,2,\,4} \right\},\) find \(P-Q\) and \(Q-P.\)

Answer:

\(P – Q = \left\{ { – 3,\, – 2,\,1,\,4} \right\} – \left\{ {0,\,1,\,2,\,4} \right\} = \left\{ { – 3,\, – 2} \right\}\)

\(Q – P = \left\{ {0,\,1,\,2,\,4} \right\} – \left\{ { – 3,\, – 2,\,1,\,4} \right\} = \left\{ {0,\,2} \right\}\)

Symmetric Difference of Sets: The symmetric difference of two sets \(P\) and \(Q\) is the set \((P – Q) \cup (Q – P)\). It is indicated by \(P\Delta Q\). 

\(P\Delta Q = \{ x:x \in P – Q\) or \(Q – P\} \)

For example,

If \(P = \left\{ {6,\,7,\,8,\,9} \right\}\) and \(Q = \left\{ {8,\,10,\,12} \right\},\) find \(P\Delta Q.\)

Answer:

\(P – Q = \left\{ {6,\,7,\,8,\,9} \right\} – \left\{ {8,\,10,\,12} \right\} = \left\{ {6,\,7,\,9} \right\}\)

\(Q – P = \left\{ {8,\,10,\,12} \right\} – \left\{ {6,\,7,\,8,\,9} \right\} = \left\{ {10,\,12} \right\}\)

\(P\Delta Q = \left( {P – Q} \right) \cup \left( {Q – P} \right) = \left\{ {6,\,7,\,9} \right\} \cup \left\{ {10,\,12} \right\}\)

\(P\Delta Q = \left\{ {6,\,7,\,9,\,10,\,12} \right\}\)

Complement of a Set: The complement of a set \(P\) is the set of all the elements of the universal set which are not in \(P\).

The symbol for the complement of \(P\) is \(\left( {P’} \right).\)

Operations on Sets Properties

It is interesting 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 are explored.

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

Commutative Property

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

Let \(P = \left\{ {3,\,6,\,8,\,10} \right\}\) and \(Q = \left\{ {2,\,6,\,8,\,12} \right\}\) be two sets.

Then, \(P \cup Q = \left\{ {2,\,3,\,6,\,8,\,10,\,12} \right\}\) and \(Q \cup P = \left\{ {2,\,3,\,6,\,8,\,10,\,12} \right\}\)

From the above, we see that \(P \cup Q = Q \cup P.\)

This is referred to as the Commutative property of union of sets.

Then, \(P \cap Q = \left\{ {6,\,8} \right\}\) and \(Q \cap P = \left\{ {6,\,8} \right\}\)

From the above, we see that \(P \cap Q = Q \cap P.\)

This is referred to as the Commutative property of intersection of sets.

Commutative property: For any two sets \(P\) and \(Q\)

1. \(P∪Q=Q∪P\)
2. \(P∩Q=Q∩P\)

Associative Property

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

Let \(A = \left\{ { – 1,\,0,\,1,\,2} \right\},\,B = \left\{ { – 3,\,0,\,2,\,3} \right\}\) and \(C = \left\{ {0,\,1,\,3,\,4} \right\}\) be three sets. 

Now, \(B \cup C = \left\{ { – 3,\,0,\,1,\,2,\,3,\,4} \right\},\,A \cup \left( {B \cup C} \right) = \left\{ { – 1,\,0,\,1,\,2} \right\} \cup \left\{ { – 3,\,0,\,1,\,2,\,3,\,4} \right\}\)

\( = ( – 3,\, – 1,\,0,\,1,\,2,\,3,\,4\} \) —-(i)

Then, \(A \cup B = ( – 3,\, – 1,\,0,\,1,\,2,\,3,\,4\} ,\,\left( {A \cup B} \right) \cup C = \left\{ { – 3,\, – 1,\,0,\,1,\,2,\,3} \right\} \cup \left\{ {0,\,1,\,3,\,4} \right\}\)

\( = \left\{ { – 3,\, – 1,\,0,\,1,\,2,\,3,\,4} \right\}\) —-(ii)

From (i) and (ii), \(A \cup \left( {B \cup C} \right) = \left( {A \cup B} \right) \cup C\)

This represents the associative property of union among sets \(A, B\) and \(C.\) 

Now, \(B \cap C = \left\{ {0,\,3} \right\}\)

\(A \cap \left( {B \cap C} \right) = \left\{ { – 1,\,0,\,1,\,2} \right\} \cap \left\{ {0,\,3} \right\} = \left\{ 0 \right\}\)—-(iii)

Then, \(A \cap B = \left\{ {0,\,2} \right\},\,\left( {A \cap B} \right) \cap C = \left\{ {0,\,2} \right\} \cap \left\{ {0,\,1,\,3,\,4} \right\} = \left\{ 0 \right\}\)—-(iv) 

From (iii) and (iv), \(A∩B∩C=A∩B∩C\). 

This represents the associative property of intersection among sets \(A, B\), and \(C\).

Associative property: For some three sets \(A, B\) and \(C\) 

1. \(A∪(B∪C)=(A∪B)∪C\)
2. \(A∩(B∩C)=(A∩B)∩C\)

Distributive Property

We have studied the 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 some three sets \(A, B\) and \(C\)

1. \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) (Intersection over union)
2. \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\) (Union over intersection)

De Morgan’s Laws

Augustus De Morgan \((1806-1871)\) was a British mathematician. He was born on \(27th\) June, \(1806\) in Madurai, Tamilnadu, India. His father was sent to India by the East India Company. When he was seven months old, his family shifted back to England. De Morgan was skilled at Trinity College, Cambridge, London. He formulated laws for set difference and complementation. These are called De Morgan’s laws.

De Morgan’s Laws for Set Difference

These laws relate to the set operations union, intersection and set difference.

For any three sets \(A, B\) and \(C\)

1. \(A – (B \cup C) = (A – B) \cap (A – C)\)

2. \(A – (B \cap C) = (A – B) \cup (A – C)\)

De Morgan’s Laws for Complementation

These laws relate to the set operations union and intersection.

Let \(‘U’\) be the universal set that contains finite sets \(A\) and \(B.\) Then

1. \({(A \cup B)^\prime } = {A^\prime } \cap {B^\prime }\)

2. \({(A \cap B)^\prime } = {A^\prime } \cup {B^\prime }\)

Cardinality of Sets

If \(A\) and \(B\) are two finite sets, then:

1. \(n(A \cup B) = n(A) + n(B) – n(A \cap B)\)

2. \(n(A – B) = n(A) – n(A \cap B)\)

3. \(n(B – A) = n(B) – n(A \cap B)\)

4. \(n\left( {{A^\prime }} \right) = n(U) – n(A)\)

Note:

From the above results, we may get:

1. \(n(A \cap B) = n(A) + n(B) – n(A \cup B)\)

2. \(n(U) = n(A) + n(A’)\)

3. If \(A\) and \(B\) are disjoint sets then,\(n(A \cup B) = n(A) + n(B)\)

Venn Diagrams

Venn diagrams were developed by the mathematician Jhon Venn. In these diagrams, the universal set \((U)\) is represented by a rectangle and the sets within are represented by circles.

The union of two sets can be represented by Venn diagram as given below:

Intersection of two sets can be represented by a Venn diagram as given below:

Venn diagram for set difference:

Venn diagram for symmetric difference of sets:

\(A\Delta B = (A – B) \cup (B – A)\)

Venn diagram for complement of a set:

Solved Examples: Operations on Sets

Q.1. If \(A = \left\{ {1,\,2,\,3,\,4} \right\}\) and \(U = \left\{ {{\rm{natural}}\,{\rm{numbers}}\,{\rm{less}}\,{\rm{than}}\,10} \right\},\) then find \(A’.\)
Ans:
Given: \(A = \left\{ {1,\,2,\,3,\,4} \right\}\)
\(U = \left\{ {1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9} \right\}\)
Now,\(A’ = \left\{ {5,\,6,\,7,\,8,\,9} \right\}\)
Hence, the complement of the set \(A\) is \(\left\{ {5,\,6,\,7,\,8,\,9} \right\}.\)

Q.2. Let \(A = \left\{ {2,\,3,\,5,\,7,\,9} \right\}\) and \(B = \left\{ {7,\,9,\,11,\,13} \right\}.\) Verify that \(n(A – B) = n(A) – n(A \cap B).\)
Ans:
Given: \(A = \left\{ {2,\,3,\,5,\,7,\,9} \right\}\) and \(B = \left\{ {7,\,9,\,11,\,13} \right\}\)
We need to verify \(n(A – B) = n(A) – n(A \cap B)\)
Now, \(\left( {A – B} \right) = \left\{ {2,\,3,\,5} \right\}\)
\(n(A – B) = 3\)………(i)
\(\left( {A \cap B} \right) = \left\{ {7,\,9} \right\}\)
\(n(A \cap B) = 2\) and \(n(A) = 5\)
So, \(n(A) – n(A \cap B) = 5 – 2 = 3\)
From equation (i) and (ii), we get \(n(A – B) = n(A) – n(A \cap B).\)

Q.3. From the Venn diagram, verify that \(n(A \cup B) = n(A) + n(B) – n(A \cap B)\).
Ans:

From the Venn diagram,
\(A = \left\{ {5,\,10,\,15,\,20} \right\},\,B = \left\{ {10,\,20,\,30,\,40,\,50} \right\}\)
Then \(A \cup B = \left\{ {5,\,15,\,10,\,20,\,30,\,40,\,50} \right\}\) and \(A \cap B = \left\{ {10,\,20} \right\}\)
\(n(A) = 4,n(B) = 5,n(A \cup B) = 7,n(A \cap B) = 2\)
\(n(A \cup B) = 7\)—–(i)
\(n(A) + n(B) – n(A \cap B) = 4 + 5 – 2 = 7\)——-(ii)
From equation (i) and (ii), \(n(A \cup B) = n(A) + n(B) – n(A \cap B)\)

Q.4. If \(A = \left\{ {6,\,7,\,8,\,9} \right\}\) and \(B = \left\{ {8,\,10,\,12} \right\},\) Find \(A\Delta B\)
Ans:
\(A – B = \left\{ {6,\,7,\,9} \right\}\)
\(B – A = \left\{ {10,\,12} \right\}\)
\(A\Delta B = \left( {A – B} \right) \cup \left( {B – A} \right) = \left\{ {6,\,7,\,9} \right\} \cup \left\{ {10,\,12} \right\}\)
\(A\Delta B = \left\{ {6,\,7,\,9,\,10,\,12} \right\}\)
Therefore, \(A\Delta B = \left\{ {6,\,7,\,9,\,10,\,12} \right\}\)

Q.5. If \(A = \left\{ {b,\,e,\,g,\,h} \right\}\) and \(B = \left\{ {c,\,e,\,g,\,h} \right\},\) then verify the commutative property of union of sets and intersection of sets.
Ans:
Given: \(A = \left\{ {b,\,e,\,g,\,h} \right\}\) and \(B = \left\{ {c,\,e,\,g,\,h} \right\}\)
To verify commutative property of union of sets:
\(A \cup B = \left\{ {b,\,c,\,e,\,g,\,h} \right\}\)—-(i)
\(B \cup A = \left\{ {b,\,c,\,e,\,g,\,h} \right\}\)—–(ii)
From (i) and (ii) we have \(A \cup B = B \cup A\)
It is verified that union of sets is commutative.
\(A \cap B = \left\{ {e,\,g} \right\}\)
\(B \cup A = \left\{ {e,\,g} \right\}\)
From (iii) and (iv) we have \(A \cap B = B \cap A\)
It is verified that intersection of sets is commutative.

Summary

In this article, we learnt about the definition of operations on sets, properties of set operations, De Morgan’s laws, the cardinality of sets, Venn diagrams, solved examples on sets and frequently asked questions on operations on sets.

This article’s learning outcome is that set theory is a mathematical abstract concerned with the grouping of sets of numbers that have commonality. For example, all even numbers make up a set, and all odd numbers comprise a set. 

Frequently Asked Questions

Q.1. What are the basic operations on sets?
Ans: There are three basic operations for constructing new sets from given sets:
1. Union of sets \({\rm{(}} \cup {\rm{)}}\)
2. Intersection of sets \(\left( \cap \right)\)
3. Difference of sets \(\left( – \right)\)

Q.2. What are the \(4\) operations of sets?
Ans: The four operations on sets are
1. Union of Two Sets
2. The intersection of Two Sets
3. Difference of Two Sets
4. Complement of a Set

Q.3. What are De Morgan’s Laws in operation on sets?
Ans: De Morgan’s Laws for Set Difference
These laws relate to the set operations union, intersection and set difference.
For any three sets \(A,B\) and \(C\)
\(A – (B \cup C) = (A – B) \cap (A – C)\)
\(A – (B \cap C) = (A – B) \cup (A – C)\)
De Morgan’s Laws for Complementation
These laws relate to the set operations union, intersection and set difference.
Let ‘U’ be the universal set containing finite sets \(A\) and \(B\). Then:
\({(A \cup B)^\prime } = {A^\prime } \cap {B^\prime }\)
\({(A \cap B)^\prime } = {A^\prime } \cup {B^\prime }\)

Q.4. What is the complement of sets in operation on sets?
Ans: The complement of a set \(A\) is the set of all the elements of the universal set which are not in \(A\).
The symbol for the complement of \(A\) is \(A’.\)

Q.5. What are the \(3\) properties of set operations?
Ans: The properties of set operations on union and intersection are Commutative Property, Associative Property, and Distributive Property.

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