Chapter 1.10 - Some Interesting Relations Between Two Sets

In the previous section, we saw some practical problems involving sets. In this section, we will see some interesting relations between sets. We will see them in the form of solved examples.

Solved example 1.56
Show that A ∪ B = A ∩ B implies A = B
Solution:
• We will solve this problem in two parts: Part (a) and Part (b).
    ♦ In part (a), we prove that A ⊂ B
    ♦ In part (b), we prove that B ⊂ A
Part (a):
1. Let a be an element of A.
    ♦ Using symbols, we write this as: a ∈ A
2. Since a is an element of A, the same a will be an element of A ∪ B also.
    ♦ Using symbols, we write this as: a ∈ A ∪ B
3. Given that A ∪ B = A ∩ B
    ♦ So a will be an element of A ∩ B also.
    ♦ Using symbols, we write this as: a ∈ A ∩ B
4. A ∩ B is the set which contains only the common elements of A and B.
    ♦ So, if a is an element of A ∩ B, the same a will be an element of B also.
    ♦ Using symbols, we write this as: a ∈ B
5. Comparing two statements:
    ♦ In (1), we wrote that, a is an element of A.
    ♦ In (4), we saw that, the same a is an element of B also.
• That means, every element in A is an element of B also.
    ♦ Using symbols, we write this as: A ⊂ B

Part (b):
1. Let b be an element of B.
    ♦ Using symbols, we write this as: b ∈ B
2. Since b is an element of B, the same b will be an element of A ∪ B also.
    ♦ Using symbols, we write this as: b ∈ A ∪ B
3. Given that A ∪ B = A ∩ B
    ♦ So b will be an element of A ∩ B also.
    ♦ Using symbols, we write this as: b ∈ A ∩ B
4. A ∩ B is the set which contains only the common elements of A and B.
    ♦ So, if b is an element of A ∩ B, the same b will be an element of A also.
    ♦ Using symbols, we write this as: b ∈ A
5. Comparing two statements:
    ♦ In (1), we wrote that, b is an element of B.
    ♦ In (4), we saw that, the same a is an element of A also.
• That means, every element in B is an element of A also.
    ♦ Using symbols, we write this as: B ⊂ A

◼ In part (a), we proved that, A ⊂ B
◼ In part (b), we proved that, B ⊂ A
◼ This is possible only if the two sets are the same. So we can write: A = B

Solved example 1.57
In each of the following, determine whether the statement is true or false. If it is
true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B , then x ∈ B
(ii) If A ⊂ B and B ∈ C , then A ∈ C
(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
(iv) If A ⊄ B and B ⊄ C , then A ⊄ C
(v) If x ∈ A and A ⊄ B , then x ∈ B
(vi) If A ⊂ B and x ∉ B , then x ∉ A
Solution:
(i) If x ∈ A and A ∈ B , then x ∈ B
This is false. Let us see an example:
• A = {1, 2}, B = {3, {1, 2}, 4}
    ♦ Here A is an element of B.
    ♦ But '1' and '2' are not elements of B.

(ii) If A ⊂ B and B ∈ C , then A ∈ C
This is false. Let us see an example:
A = {1, 2}, B = {1, 2, 3}, C = {4, {1, 2, 3}, 5}
    ♦ Here B is an element of C.
    ♦ But {1, 2} is not an element of C.

(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
This is true. Proof can be written in 3 steps:
1. Let a be an element of A.
    ♦ Using symbols, we write this as: a ∈ A
2. Given that A ⊂ B
    ♦ So all elements of A are present in B.
    ♦ So a ∈ B
3. Given that B ⊂ C
    ♦ So all elements of B are present in C.
    ♦ So a ∈ C
• Thus we get A ⊂ C
◼ The Venn diagram in fig.1.26(a) also proves this.
    ♦ The circle B encloses circle A. This is because A ⊂ B
    ♦ The circle C encloses circle B. This is because B ⊂ A
    ♦ So the circle C automatically encloses circle A, indicating A ⊂ C

Fig.1.26

(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
This is false. Let us see an example:
• A = {1, 2}, B = {2, 3, 4}, C = {1, 2, 3, 5}
    ♦ Here A ⊄ B and B ⊄ C
    ♦ But A ⊂ C
◼ The Venn diagram in fig.1.26(b) also proves that this is false.
   ♦ The circle B does not enclose circle A because A ⊄ B
   ♦ The circle C does not enclose circle B because B ⊄ C
   ♦ But the circle C can enclose circle A because A ⊂ C

(v) If x ∈ A and A ⊄ B , then x ∈ B
This is false. It can be written in steps:
1. Given that x is an element of A.
2. Given that A is not a subset of B. So all elements of A are not present in B.
3. So x need not be present in B.
Let us see an example:
• A = {1, 2}, B = {2, 3, 4}
    ♦ Here A ⊄ B
    ♦ Also, '1' is not present in B.

(vi) If A ⊂ B and x ∉ B , then x ∉ A.
This is true. It can be proved in 2 steps:
1. Given that A is a subset of B. So all elements of A will be present in B.
2. So if x is not present in B, it cannot be present in A.

Solved example 1.58
Show that the following four conditions are equivalent :
(i) A ⊂ B  (ii) A – B = ɸ  (iii) A ∪ B = B  (iv) A ∩ B = A
Solution:
Part (a):
• In this part, we prove that: A ⊂ B  ⇒ A – B = ɸ
   ♦ It can be proved in 4 steps:
1. To find A - B, we first discard all elements of B.
2. Then we discard elements which are common in both A and B.
3. But when we discard the elements in B, we automatically discard the following two items:
   ♦ Common elements mentioned in (2).
   ♦ All elements of A.
4. So after the operation A - B, none of the elements of either A or B will remain.
• So we can write:
If A ⊂ B, the difference A - B will be a null set.
• In other words, A ⊂ B  ⇒ A – B = ɸ

Part (b):
• In this part, we prove that: A ⊂ B ⇒ A ∪ B = B
   ♦ It can be proved in 4 steps:
1. To find A ∪ B, we write all elements of A and B together, but repeating elements will be written only once.
2. Here, all elements of A will be repeating. So none of the elements of A need to be considered.
3. All we need to do is: Write the elements of B.
4. So we can write: A ⊂ B ⇒ A ∪ B = B

Part (c):
In this part, we prove that: A ⊂ B ⇒ A ∩ B = A
   ♦ It can be proved in 4 steps:
1. To find A ∩ B, we write all elements which are common to both A and B.
2. Here, all elements of A are the common elements. In fact, they are the only common elements.
3. All we need to do is: Write the elements of A.
4. So we can write: A ⊂ B ⇒ A ∩ B = A

◼ Thus we see that, all given conditions are equivalent.

Solved example 1.59
Assume that P(A) = P(B). Show that A = B
Solution:
• We will solve this problem in two parts: Part (a) and Part (b).
    ♦ In part (a), we prove that A ⊂ B
    ♦ In part (b), we prove that B ⊂ A
Part (a):
1. P(A) will contain all subsets of A.
• So P(A) will contain A.
• Using symbols, we write this as: A ∈ P(A)
2. If A is an element of P(A), the same A will be an element of P(B) also.
   ♦ Because, it is given that P(A) = P(B)
• Using symbols, we write this as:
A ∈ P(A) ⇒ A ∈ P(B) [∵ P(A) = P(B)]
3. If A is an element of P(B), A will be a subset of B.
• Using symbols, we write this as: A ∈ P(B) ⇒ A ⊂ B
◼ Thus we proved that A ⊂ B

Part (b):
1. P(B) will contain all subsets of B.
• So P(B) will contain B.
• Using symbols, we write this as: B ∈ P(B)
2. If B is an element of P(B), the same B will be an element of P(A) also.
   ♦ Because, it is given that P(A) = P(B)
• Using symbols, we write this as:
B ∈ P(B) ⇒ B ∈ P(A) [∵ P(A) = P(B)]
3. If B is an element of P(A), B will be a subset of A.
• Using symbols, we write this as: B ∈ P(A) ⇒ B ⊂ A
◼ Thus we proved that B ⊂ A

• So we get: A ⊂ B and B ⊂ A
   ♦ This is possible only if A and B are equal.
◼  That means, A = B

Solved example 1.60
Is it true that for any sets A and B, P(A) ∪ P(B) = P(A ∪ B)? Justify your answer.
Solution:
This is false. Let us see an example. It can be written in steps:
1. Let A = {1, 2} and B = {2, 3}
• Then the power sets are:
   ♦ P(A) = {{1}, {2}, {1, 2}, ɸ}
   ♦ P(B) = {{2}, {3}, {2, 3}, ɸ}
2. So P(A) ∪ P(B) will be: {{1}, {2}, {3}, {1, 2}, {2, 3}, ɸ}
3. (A ∪ B) will be: {1, 2, 3}
• Then the power set of (A ∪ B) will be:
P(A ∪ B) = {{1}, {2}, {3}, {1, 2, 3}, {1, 2}, {2, 3}, {1, 3}, ɸ}
4. We see that, result in (2) is different from result in (3).

Solved example 1.61
Show that for any sets A and B,
A - B = A ∩ B'
Solution:
1. Consider the LHS: A - B
   ♦ To find A - B, we discard all elements of B.
   ♦ Then we discard those elements of A which are present in B also.
• The result will be the left side crescent in fig.1.27(b) below:

Fig.1.27

(The dotted B circle is imaginary. It does not come in the result. It is shown only to indicate relative positions)
2. Consider the RHS A ∩ B'
• B' is shown in fig.1.28(a) below. It is the red shaded portion.

Fig.1.28

• We want the intersection of two regions:
   ♦ Region B'
   ♦ Region A
3. The resulting intersection is the region,
   ♦ Where there is shading from both regions B' and A.
• So the result will be as in fig.1.28(b)
4. We see that:
Fig.1.27(b) is same as fig.1.28(b)
◼ So we can write: A - B = A ∩ B'
◼ Similarly, we can prove that: B - A = B ∩ A'

Solved example 1.62
Show that for any sets A and B,
(a) A = (A ∩ B) ∪ (A – B)
(b) A ∪ (B – A) = (A ∪ B)
Solution:
Part (a):
1. We have to prove that A = (A ∩ B) ∪ (A – B)
• Consider (A - B) in the RHS
• From the results in the previous solved example 1.61, we can write (A ∩ B') in the place of (A - B)
2. So the RHS becomes:
(A ∩ B) ∪ (A ∩ B')
3. Now recall the distributive law of intersection (Fig.1.14 of section 1.6):
X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z)
• The RHS in this law is similar to the result written in (2)
• So the result in (2) becomes: A ∩ (B ∪ B')
4. So the question becomes:
Prove: A = A ∩ (B ∪ B')
5. Consider (B ∪ B') in the RHS
• (B ∪ B') is U
• So the RHS in (4) becomes: A ∩ U
6. But A ∩ U is A
• So LHS and RHS are equal.

Part (b):
1. We have to prove that A ∪ (B – A) = (A ∪ B)
• Consider (B - A) in the LHS.
• From the results in the previous solved example 1.61, we can write (B ∩ A') in the place of (B - A)
2. So the LHS becomes:
A ∪ (B ∩ A')
3. Now recall the distributive law of union:
X ∪ (Y ∩ Z) = (X ∪ Y) ∩ (X ∪ Z)
• The LHS in this law is similar to the result written in (2)
• So the result in (2) becomes: (A ∪ B) ∩ (A ∪ A')
4. So the question becomes:
Prove: (A ∪ B) ∩ (A ∪ A') = (A ∪ B)
5. Consider (A ∪ A') in the RHS
• (A ∪ A') is U
• So the LHS in (4) becomes: (A ∪ B) ∩ U
6. But (A ∪ B) ∩ U is (A ∪ B)
• So LHS and RHS are equal.

Alternate method using Venn diagrams:
1. Fig.1.29(a) below shows our familiar 'Venn diagram of two sets'.

Fig.1.29

2. Fig.1.29(b) shows the three portions separated from each other. This is also familiar to us.
◼ It is clear that, the union of
    ♦ left crescent portion
    ♦ and the middle portion
    ♦ will give set A 
• This proves part (a): A = (A ∩ B) ∪ (A – B)
3. From part (a), we have: A = (A ∩ B) ∪ (A – B)
• So A ∪ (B – A) = (A ∪ B) is same as: (A ∩ B) ∪ (A – B) ∪ (B – A)
    ♦ This is the union of the three portions in fig.1.29(b)
• The union of those three portions will give (A ∪ B)
• This proves part (b): A ∪ (B – A) = (A ∪ B)   

Solved example 1.63
Using properties of sets, show that
(a) A ∪ ( A ∩ B ) = A
(b) A ∩ ( A ∪ B ) = A.
Solution:
Part (a):
1. We have to prove that A ∪ ( A ∩ B ) = A
• Consider the LHS. Recall the distributive law of union:
X ∪ (Y ∩ Z) = (X ∪ Y) ∩ (X ∪ Z)
2. Applying this law, the LHS becomes:
(A ∪ A) ∩ (A ∪ B)
3. But (A ∪ A) is A
• So the LHS becomes:
A ∩ (A∪B)
4. But A ∩ (A∪B) is A
• So the LHS becomes A
5. Thus we get LHS = RHS

Part (b):
1. We have to prove that A ∩ ( A ∪ B ) = A
• Consider the LHS. Recall the distributive law of intersection (Fig.1.14 of section 1.6):
X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z)
2. Applying this law, the LHS becomes:
(A ∩ A) ∪ (A ∩ B)
3. But (A ∩ A) is A
• So the LHS becomes:
A ∪ (A ∩ B)
4. But in part (a), we proved that A ∪ (A ∩ B) is A.
• So the LHS becomes A.
5. Thus we get LHS = RHS

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In the next section, we will see some relations between three sets.

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Chapter 1.7 - Difference of Sets

In the previous section, we saw intersection of sets. In this section, we will see difference of sets.

Difference of sets

• This can be explained in 9 steps:
1. Let A and B be any two sets.
• We perform an operation called ‘difference of sets’ between A and B.
2. The set which is formed as a result of that operation,
    ♦ will contain only those elements of A which do not belong to B.
• This can be explained further as follows:
    ♦ The resulting set can contain elements of A.
    ♦ The resulting set can not contain any element of B.
    ♦ The resulting set can not contain any element which is common to A and B.
3. The symbol ‘－’ is used to represent difference.
• So the difference of A and B can be represented as A － B.
    ♦ It is read as: A minus B.
4. Let us see some examples:
A = {2, 4, 6, 8, 10} and B = {2, 4, 6, 12, 14}. Find A － B.
Solution:
• A － B can contain elements of A.
• A － B can not contain any element of B.
    ♦ So we can right away discard set B.
• Also, A － B cannot contain any element which is common to both A and B.
    ♦ So we can discard 2, 4 and 6 from A.
• Thus we get: A － B = {8, 10}
• We can represent this difference using Venn diagrams also.
    ♦ Fig.1.16(a) below shows A and B before the difference.
    ♦ Fig.1.16(b) shows the new set which is A － B.

Fig.1.16

5. In the above example, find B － A
Solution:
• B － A can contain elements of B.
• B － A cannot contain any element of A.
    ♦ So we can right away discard set A.
• Also, B － A cannot contain any element which is common to both B and A.
    ♦ So we can discard 2, 4 and 6 from B.
• Thus we get: B － A = {12, 14}
• We can represent this difference using Venn diagrams also.
    ♦ Fig.1.17(a) below shows A and B before the difference.
    ♦ Fig.1.17(b) shows the new set which is B － A.

Fig.1.17

6. The above two examples show that:
    ♦ A － B
    ♦ is not equal to
    ♦ B － A
• So we must be careful about the order while specifying the difference.
• We must clearly examine the two sets:
    ♦ The set on the left side of '－' sign.
    ♦ The set on the right side of '－' sign.
7. Another example:
Let V = {a, e, i, o, u} and B = {a, i, k, u}. Show that V － B ≠ B － V 
Solution:
(i) V － B can contain elements of V.
• V － B cannot contain any element of B.
    ♦ So we can right away discard set B.
• Also, V － B cannot contain any element which is common to both V and B.
    ♦ So we can discard a, i and u from V.
• Thus we get: V － B = {e, o}
(ii) B － V can contain only elements of B.
• B － V cannot contain any element of V.
    ♦ So we can right away discard set V.
• Also, B － V cannot contain any element which is common to both V and B.
    ♦ So we can discard a, i and u from B.
• Thus we get: V － B = {k}
(iii) Thus we see that: V － B ≠ B － V
8. Thus we can write the definition:

Definition 8: The difference of two sets A and B is the set C which consists of all those elements which belong to A but not to B.

9. Let us see the relation between intersection and difference. It can be written in 4 steps:
(i) Consider two sets A and B.
• We can performs the operations of intersection and difference on the two sets.
(ii) As a result of those operations, we get three different sets:
A － B, B － A and A ∩ B
(Recall that, A ∩ B is same as B ∩ A. So the operation of intersection will give only one set)
(iii) Fig.1.18(a) below shows the two sets A and B before the operations.
• Fig.b shows the three resulting sets.

Fig.1.18

(iv) We see that, the three resulting sets do not have any overlapping parts.
◼ Thus we can write:
A － B, B － A and A ∩ B are mutually disjoint. They will never have any common elements.

Now we will see some solved examples:
Solved example 1.26
Find the union of each of the following pairs of sets :
(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = [ a, e, i, o, u} B = {a, b, c}
(iii) A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}
(iv) A = {x : x is a natural number and 1 < x ≤ 6 }
B = {x : x is a natural number and 6 < x < 10 }
(v) A = {1, 2, 3}, B = ɸ
Solution:
(i) X ∪ Y = {1, 2, 3, 5}
(ii) A ∪ B = {a, b, c, e, i, o, u}
(iii) In roster form, A = {3, 6, 9, 12, . . .}
• In roster form, B = {1, 2, 3, 4, 5}
• So A ∪ B = {x : x = 1, 2, 4, 5 or a multiple of 3}
(iv) In roster form, A = {2, 3, 4, 5, 6}
• In roster form, B = {7, 8, 9}
• So A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
(v) A ∪ B = {1, 2, 3}

Solved example 1.27
Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?
Solution:
• All elements of A are present in B. So A ⊂ B
• A ∪ B = {a, b, c} = B

Solved example 1.28
If A and B are two sets such that A ⊂ B, then what is A ∪ B
Solution:
If A is a subset of B, then A ∪ B will be B

Solved example 1.29
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 } and D = { 7, 8, 9, 10 }; find
(i) A ∪ B  (ii) A ∪ C  (iii) B ∪ C  (iv) B ∪ D  (v) A ∪ B ∪ C  (vi) A ∪ B ∪ D  (vii) B ∪ C ∪ D
Solution:
(i) A ∪ B = {1, 2, 3, 4, 5, 6}
(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(iii) B ∪ C = {3, 4, 5, 6, 7, 8}
(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}
(v) A ∪ B = {1, 2, 3, 4, 5, 6}
   ♦ So A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(vi) A ∪ B = {1, 2, 3, 4, 5, 6}
   ♦ So A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C = {3, 4, 5, 6, 7, 8}
   ♦ So B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

Solved example 1.30
Find the intersection of each pair of sets of example 1.26 above.
Solution:
(i) X ∩ Y = {1, 3}
(ii) A ∩ B = {a}
(iii) In roster form, A = {3, 6, 9, 12, . . .}
• In roster form, B = {1, 2, 3, 4, 5}
• So A ∩ B = {3}
(iv) In roster form, A = {2, 3, 4, 5, 6}
• In roster form, B = {7, 8, 9}
• So A ∩ B = ɸ
(v) A ∩ B = ɸ {1, 2, 3}

Solved example 1.31
If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find
(i) A ∩ B  (ii) B ∩ C  (iii) A ∩ C ∩ D  (iv) A ∩ C  (v) B ∩ D  (vi) A ∩ (B ∪ C) (vii) A ∩ D
(viii) A ∩ (B ∪ D)  (ix) ( A ∩ B ) ∩ ( B ∪ C )  (x) (A ∪ D) ∩ (B ∪ C)
Solution:
(i) A ∩ B = {7, 9, 11}
(ii) B ∩ C = {11, 13}
(iii) A ∩ C = {11}
   ♦ So A ∩ C ∩ D = ɸ
(iv) A ∩ C = {11}
(v) B ∩ D = ɸ
(vi) B ∪ C = {7, 9, 11, 13, 15}
   ♦ So A ∩ (B ∪ C) = {7, 9, 11}
(vii) A ∩ D = ɸ
(viii) B ∪ D = {7, 9, 11, 13, 15, 17}
   ♦ So A ∩ (B ∪ D) = {7, 9, 11}
(ix) A ∩ B = {7, 9, 11}
• B ∪ C = {7, 9, 11, 13, 15}
   ♦ So ( A ∩ B ) ∩ ( B ∪ C )   = {7, 9, 11}
(x) A ∪ D = {3, 5, 7, 9, 11, 15, 17}
• B ∪ C = {7, 9, 11, 13, 15}
   ♦ So ( A ∩ B ) ∩ ( B ∪ C ) = {7, 9, 11, 15}

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The link below gives some solved examples:

Solved examples 1.32 to 1.37

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In the next section, we will see complement of a set.

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Chapter 1.3 - Subsets

In the previous section, we saw the empty set, finite set and equal sets. In this section, we will see subsets

Some basics about subsets can be written in 14 steps:
1. Consider the following two sets:
X = {x : x is a student in the school}
Y = {y : y is a student in class 11}
2. It is clear that, all elements in Y will be present in X.
• We say that: Y is a subset of X.
   ♦ Using symbols, this is written as: Y ⊂ X
   ♦ The symbol ‘⊂’ stands for ‘is a subset of’ or ‘is contained in’.
3. We can write the definition:

Definition 4 A set A is said to be a subset of set B if every element of A is contained in B.

4. We already know that:
If a is an element of A, we can write a ∈ A.
• But since A is a subset of B, the element a will be present in B also. So we can write: a ∈ B
5. That means,
If a ∈ A implies a ∈ B then, A ⊂ B.
• The word implies can be replaced by the symbol ‘⇒ ’. So we can write:
If a ∈ A ⇒ a ∈ B then, A ⊂ B.
6. There are two situations where A can be a subset of B
(i) Every element in A is present in B. Further, B has some extra elements also.
(ii) Every element in A is present in B. There are no extra elements in B.
   ♦ This means B has the exact same elements as A.
7. Let us consider the second situation carefully:
• If B has the exact same elements as A, we can write A ⊂ B alright.
   ♦ But we can write B ⊂ A also.
• That means, if two sets A and B have the exact same elements, we can write both:
A ⊂ B and B ⊂ A.
8. Recall that, if two sets have the exact same elements, they are equal sets.
• So we can write:
If A ⊂ B and B ⊂ A, the sets A and B are equal.
• Using symbols, we can write:
A ⊂ B and B ⊂ A ⇒ A = B
9. Conversely, we can write:
If two sets A and B are equal, then A ⊂ B and B ⊂ C.
10. A theorem and it’s converse can be written using the symbol ‘⇔’.
• This symbol indicates two way implication. So we get: A ⊂ B and B ⊂ A ⇔ A = B
   ♦ The forward direction indicates what we wrote in (8).
   ♦ The reverse direction indicates what we wrote in (9).
11. What we saw in steps (7) to (10) is the analysis of the situation in 6(ii).
• What if the situation is as in 6(i) ?
   ♦ In such a situation, we say two things:
         ✰ A is a proper subset of B.
         ✰ B is a superset of A.
12. Let us see some examples:
Example 1:
• Consider the following two sets:
A = {x : x is a divisor of 56}
B = {x : x is prime divisor of 56}
• Obviously, all elements in B will be present in A.
   ♦ So we can write: B ⊂ A
Example 2:
• Consider the following two sets:
A = {1, 3, 5}
B = {x : x is an odd number less than 6}
• The set B in roster form is {1, 3, 5}.
• So we can write:
   ♦ All elements of A are present in B.
         ✰ So A ⊂ B
   ♦ All elements of B are present in A.
         ✰ So B ⊂ A
• Since A ⊂ B and B ⊂ A, we can write A = B.
Example 3:
• Consider the following two sets:
A = {a, l, o, y}
B = {a, e, i, o, u}
• We see that, A is not a subset of B.
   ♦ We write: A ⊄ B
• We see that, B is not a subset of A.
   ♦ We write: B ⊄ A
13. From the above steps, it is clear that, every set A is a subset of itself
• Also, since the empty set ɸ has no elements, ɸ is a subset of every set
14. If a set A has only one element, we call it a singleton set
• For example, {a} is a singleton set

Solved example 1.18
Consider the sets
ɸ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}.
Insert the symbol ⊂ or ⊄ between each of the following pair of sets:
(i) ɸ . . . B  (ii) A . . . B  (iii) A . . . C  (iv) B . . . C
Solution:
(i) Since is a subset of every set, we can write: ɸ ⊂ B
(ii) 3 is not present in B. So we write: A ⊄ B
(iii) Both 1 and 3 are present in C. So we write: A ⊂ C
(iv) All three elements 1, 5 and 9 are present in C. So we write B ⊂ C

Solved example 1.19
Let A, B and C be three sets. If A ∈ B and B ⊂ C, is it true that A ⊂ C?. If not, give an example.
Solution:
1. Let A = {1, 2}
• Given that, set A is an element of B. So we can write: B = {3, {1, 2}, 4}
• Given that B is a subset of C. So we can write: C = {3, {1, 2}, 4, 5}
2. Here we are inclined to think that, A is a subset of C
• But if we write a subset of C, it will be: {{1, 2}}
   ♦ {{1, 2}} is different from {1, 2}
   ♦ That is: {{1, 2}} is different from A.
• So the answer is: No, A cannot be a subset of C.
3. Another way to analyze this situation can be written in 3 steps:
(i) If A is to be a subset of C, all elements of A should be present in C as separate elements.
(ii) But in C, the elements 1 and 2 are present together as a set. Not as separate elements.
(iii) So the answer is: No, A cannot be a subset of C.

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In the next section, we will see subsets of the set R

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Chapter 1.2 - Empty, Finite and Equal Sets

In the previous section, we saw the roster form and set-builder form. In this section, we will see empty set, finite set and equal sets.

The empty set can be described in 6 steps:
1. Consider the set:
A = {x : x is a student of Class XI presently studying in a school}
• Let us analyze this set:
We can write the names of all students of Class XI in that school. So set A will contain a definite number of names.
2.Consider another set:
B = {x : x is a student of both Class X and Class XI presently studying in that school}
• Let us analyze this set:
We will not find any student who studies in both Class X and Class XI in the same school. So set B will not contain any element.

Definition 1: • A set which does not contain any element is called the empty set    ♦ Another name for empty set is null set    ♦ Yet another name for empty set is void set

3. In our present case,
   ♦ A is not an empty set.
   ♦ B is an empty set.
4. Representing empty set:  
   ♦ Empty set is denoted by the symbol { }
   ♦ Another symbol for empty set is Ø
5. In our present case, we can write:
   ♦ A ≠ { }
   ♦ B = { }
6. Let us see a few more examples:
(i) Let A = {x : 1 < x < 2, x is a natural number}.
• Here x should be a natural number. Also, it must lie between 1 and 2.
• There is no such natural number. So the set will not contain any elements.
• In other words, A is an empty set.

(ii) B = {x : x² – 2 = 0 and x is rational number}.
• Here, the only solution of (x² – 2 = 0) is +√2 and -√2
• Both +√2 and -√2 are irrational numbers. So B is an empty set.

(iii) C = {x : x is an even prime number greater than 2}.
• Here, x should be:
    ♦ A prime number.
    ♦ An even number.
    ♦ Greater than 2.
• All prime numbers greater than 2 are odd numbers. So C is an empty set.

(iv) D = { x : x² = 4, x is odd}.
Here, the only solution of (x² = 4) is +2 and -2
• Both +2 and -2 are even numbers. So D is an empty set.

Solved example 1.10
Which of the following are examples of the null set
(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) { x : x is a natural numbers, x < 5 and x > 7}
(iv) { y : y is a point common to any two parallel lines}
Solution:
(i) None of the odd natural numbers are divisible by 2
• So it is a null set.
(ii) All prime numbers except 2 are odd numbers.
• That means, there is one even prime number, which is '2'. So this is not a null set.
(iii) No natural number can be less than 5 and greater than 7 at the same time. So it is a null set.
(iv) Two parallel lines will never have a common point. So it is a null set.

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Finite and Infinite sets

This can be explained in 7 steps
1. Consider the following set:
A = {2, 4, 6, 8}
• This set A has four elements. We write: n(A) = 4
   ♦ 4 is a finite natural number.
   ♦ So n(A) is a finite natural number.
2. Consider the following set:
V = {a, e, i, o, u}
• This set V has five elements. We write: n(V) = 5
   ♦ 5 is a finite natural number.
   ♦ So n(V) is a finite natural number.
3. Consider the following set:
C = {x : x is a citizen of India}
• This set will contain the names of all citizen of India. The elements are names.
   ♦ The ‘number of elements’ will be very large. But it will be a finite natural number.
   ♦ That is., n(C) will be a very large but finite natural number.
4. Consider the following set:
D = {x : x is a positive even number}
• This set will contain the numbers 2, 4, 6, 8, . . . The elements are even numbers.
   ♦ The ‘number of elements’ will be very very large.
   ♦ We will not be able to count them.
   ♦ That is., n(D) will be infinite.
5. We can now write the definition

Definition 2 • The set A is said to be a finite set if n(A) is a finite number.    ♦ The set A is said to be a finite set even if it is an empty set. • The set A is said to be an infinite set if n(A) is not a finite number.

6. Let us see some more examples:
(i) Consider the set:
D = {x : x is a day of the week}
• Then D = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
• We get: n(D) = 7
   ♦ '7' is a finite natural number. So D is a finite set.
(ii) Consider the set:
S = {x : x² – 16 = 0}.
• Here, x can take two values only. They are 4 and -4. So S = {4, -4}
• We get: n(S) = 2
   ♦ '2' is a finite natural number. So S is a finite set.
(iii) Consider the set:
G = {x : x is a point on a line}.
• There will be infinite number of points on a line.
• So G will contain infinite number of elements.
   ♦ That means, n(G) is not a finite number.
   ♦ So G is an infinite set.
7. We know that, in roster form, we write all the elements and enclose them within braces.
• But if the set is an infinite set, we will not be able to write all the elements.
• If the set is infinite, we write a few elements followed by three dots.
• Those few elements that are written, should clearly indicate the structure of the set.
• For example,
   ♦ If we write {1, 3, 5, 9, . . .}, it clearly indicates odd numbers.
   ♦ If we write {2, 2², 2³, 2⁴, . . .}, it clearly indicates the powers of 2.
• Some times, the three dots may precede the elements.
• For example,
   ♦ If we write { . . . , -3, -2, -1, 0, 1, 2, 3, . . .}, it clearly indicates integers.
◼ This method of providing three dots, cannot be used to indicate real numbers. This is because, real numbers do not follow any particular pattern. For example, the distance (on the number line) between √2 and √3 is not the same as that between √3 and √4.

Solved example 1.11
Which of the following sets are finite or infinite
(i) The set of months of a year
(ii) {1, 2, 3, . . .}
(iii) {1, 2, 3, . . .99, 100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99
Solution:
(i) The set will contain the names of the 12 months. The number of elements is 12. So it is a finite set.
(ii) This set has the definite pattern of natural numbers. But the pattern is followed by 3 dots. The dots indicate that, all the natural numbers must be included in the set. So it is an infinite set.
(iii) This set has the definite pattern of natural numbers. The pattern is followed by 3 dots. The dots usually indicate that, the set is infinite. But here, the last two elements are also given. It is clear that, there will be exactly 100 elements in the set. So it is a finite set.
(iv) There are infinite number of positive integers which are greater than 100. So it is an infinite set.
(v) There is a finite number of prime numbers which are less than 99. So it is a finite set.

Solved example 1.12
State whether each of the following set is finite or infinite:
(i) The set of lines which are parallel to the x-axis
(ii) The set of letters in the English alphabet
(iii) The set of numbers which are multiple of 5
(iv) The set of animals living on the earth
(v) The set of circles passing through the origin (0,0)
Solution:
(i) There are infinite number of lines which are parallel to the x-axis. So it is an infinite set.
(ii) There is a finite number of letters in the English alphabet. So it is a finite set.
(iii) There are infinite numbers which are multiple of 5. So it is an infinite set.
(iv) There is a finite number of animals living on the earth. So it is a finite set.
(iii) There are infinite number of circles passing through the origin (0,0). So it is an infinite set.

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Equal Sets

This can be explained in 5 steps:
1. Suppose that, we have two sets A and B. We will be able to say that the two sets are equal, if two conditions are satisfied:
(i) All elements in A are present in B.
(ii) All elements in B are present in A.
2. But if the above two conditions are satisfied, it is obvious that, the two sets will be having the same elements.
3. So we can write the definition:

Definition 3 • Two sets A and B are said to be equal if they have exactly the same elements.    ♦ The equality is expressed as A = B • Two sets A and B are said to be unequal if they do not have exactly the same elements.    ♦ The inequality is expressed as A ≠ B

4. Let us see some examples:
Example (i):
• Let A = {2, 2¹⁄₂, 3, 3¹⁄₂} and B = {2, 3, 2¹⁄₂, 3¹⁄₂}
• We see that,
   ♦ All elements in A are present in B.
   ♦ All elements in B are present in A.
• A and B have exactly the same elements. They are equal sets.
   ♦ We write: A = B

Example (ii):
• Consider the following set:
A = {x : x is a prime number less than 6}
• Consider another set:
B = {x : x prime factor of 30}
• In roster form, the above sets are: A = {2, 3, 5} and B = {2, 3, 5}
We see that,
   ♦ All elements in A are present in B.
   ♦ All elements in B are present in A.
• A and B have exactly the same elements. They are equal sets.
   ♦ We write: A = B
5. An interesting note:
• Consider the sets: A = {1, 2, 3} and B= {2, 2, 1, 3, 3}
• We see that:
   ♦ All elements in A are present in B.
   ♦ All elements in B are present in A.
   ♦ So A = B
◼ That means, it not necessary to write the repeating elements in B more than once. That is the reason why in general, we avoid writing repeating elements more than once.

Solved example 1.13
Find the pairs of equal sets, if any, give reasons:
A = {0},
B = {x : x > 15 and x < 5},
C = {x : x – 5 = 0 },
D = {x: x² = 25},
E = {x : x is an integral positive root of the equation x² – 2x –15 = 0}.
Solution:
1. Let us convert those last 5 sets which are in set-builder form into roster form:
• We have: B = {x : x > 15 and x < 5}
   ♦ Obviously, no number can be less than 5 and greater than 15 at the same time.
   ♦ It is a null set.
   ♦ We can write: B = {}
• We have: C = {x : x – 5 = 0 }
   ♦ We can write: C = {5}
• We have: D = {x: x² = 25}
   ♦ Solving (x² = 25), we can write: D = {5, -5}
• We have: E = {x : x is an integral positive root of the equation x² – 2x –15 = 0}
   ♦ Solving (x² – 2x –15 = 0), the solution is 5 and -3
   ♦ But the set must contain only integral positive roots
   ♦ So we can write: E = {5}
2. So the sets are:
A = {0}, B = {}, C = {5}, D = {5, -5} and E = {5}
3. Let us compare each set with the others
(i) Set A:
• The element '0' is not present in any other set. So we write:
A ≠ B, A ≠ C, A ≠ D, A ≠ E
(ii) Set B:
• None of the other sets is a null set. So we write:
B ≠ A, B ≠ C, B ≠ D, B ≠ E
(iii) Set C:
• The element '5' is present only in E. The other sets cannot be equal to C. So we write:
C ≠ A, C ≠ B, C ≠ D
• Since '5' is present in 'E', we must check further. We see that, E has only '5'
• So we can write:
   ♦ All elements in C are present in E
   ♦ All elements in E are present in C
   ♦ Thus we get: C = E
(iv) Set D:
• D has '5' and '-5'. No other sets have both these elements. So we write:
D ≠ A, D ≠ B, A ≠ C, D ≠ E
(v) Set E
   ♦ We have already seen that, A, B and D are not equal to E
   ♦ We have also seen that, C = E
(vi) So the only equal pair is: C and E

Solved example 1.14
Which of the following pairs of sets are equal ? Justify your answer.
(i) X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”.
(ii) A = {n : n ∈ Z and n 2 ≤ 4} and B = {x : x ∈ R and x 2 – 3x + 2 = 0}.
Solution:
(i) We have: X = {A, L, O, Y} and B = {L, O, Y, A}.
   ♦ All elements in X are present in B
   ♦ All elements in B are present in X
• So we write: X = B
(ii) We have: A = {–2, –1, 0, 1, 2} and B = {1, 2}.
• We see that, '0' is present only in A
   ♦ All elements in A are not present in B
   ♦ All elements in B are present in A
• Since both conditions are not satisfied, we can write: A ≠ B

Solved example 1.15
In the following, state whether A = B or not:
(i) A = { a, b, c, d }, B = { d, c, b, a }
(ii) A = { 4, 8, 12, 16 }, B = { 8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}, B = { x : x is positive even integer and x ≤ 10}
(iv) A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . }
Solution:
(i) A = { a, b, c, d }, B = { d, c, b, a }
   ♦ All elements in A are present in B
   ♦ All elements in B are present in A
• So we write: A = B
(ii) A = { 4, 8, 12, 16 }, B = { 8, 4, 16, 18}
   ♦ All elements in A are not present in B
   ♦ All elements in B are not present in A
• So we write: A ≠ B
(iii) A = {2, 4, 6, 8, 10}, B = { x : x is positive even integer and x ≤ 10}
• Set B in roster form is: {2, 4, 6, 8, 10}
   ♦ All elements in A are present in B
   ♦ All elements in B are present in A
• So we write: A = B
(iv) A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . }
• Set A in roster form is: {10, 20, 30, 40, . . .}
   ♦ All elements in A are present in B
   ♦ All elements in B are not present in A
• So we write: A ≠ B

Solved example 1.16
Are the following pair of sets equal ? Give reasons.
(i) A = {2, 3}, B = {x : x is solution of x2 + 5x + 6 = 0}
(ii) A = { x : x is a letter in the word FOLLOW}, B = { y : y is a letter in the word WOLF}
Solution:
(i) A = {2, 3}, B = {x : x is solution of x² + 5x + 6 = 0}
• Set B in roster form is: {-2, -3}
   ♦ All elements in A are not present in B
   ♦ All elements in B are not present in A
• So we write: A ≠ B
(ii) A = { x : x is a letter in the word FOLLOW}, B = { y : y is a letter in the word WOLF}
• Set A in roster form is: {F, O, L, W}
• Set B in roster form is: {W, O, L, F}
   ♦ All elements in A are present in B
   ♦ All elements in B are present in A
• So we write: A = B

Solved example 1.17
From the sets given below, select equal sets :
A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2},
E = {–1, 1}, F = { 0, a}, G = {1, –1}, H = { 0, 1}
Solution:
(i) Consider set A
• Take the first element '2'
   ♦ C, E, F, G and H can be ruled out because, '2' is not present in them
   ♦ The remaining sets are B and D
• Take the second element '4'
   ♦ This element is present in B and D
• Take the third element 8
   ♦ B and D can be ruled out because, '8' is not present in them
• So A is not equal to any other set

(ii) Consider set B
• Take the first element '1'
   ♦ A, C and F can be ruled out because, '1' is not present in them
   ♦ The remaining sets are D, E, G and H
• Take the second element '2'
   ♦ E, G and H can be ruled out because, '2' is not present in them
   ♦ The remaining set is D
• Take the third element '3'
   ♦ This element is present in D
• Take the fourth element '4'
   ♦ This element is present in D
   ♦ All the elements in B are present in D
• Comparing B and D, we see that:
   ♦ All the elements in B are present in D
   ♦ All the elements in D are present in B
◼ So B and D are equal sets

(iii) Consider set C
• A and B can be ruled out because, we already compared them with C
• Take the first element '4'
   ♦ E, F, G and H can be ruled out because, '4' is not present in them
   ♦ The remaining set is D
• Take the second element '8'
   ♦ D can be ruled out because, '8' is not present in it
• So C is not equal to any other set

(iv) Consider set D
• We already saw that,
   ♦ D = B
   ♦ B is not equal to any other set
• So D cannot be equal to any set other than B

(v) Consider set E
• A, B, C and D can be ruled out because, we already compared them with E
• Take the first element '-1'
   ♦ F and H can be ruled out because, '-1' is not present in them
   ♦ The remaining set is G
• Take the second element '1'
   ♦ This element is present in G
   ♦ All the elements in E are present in G
• Comparing E and G, we see that:
   ♦ All the elements in E are present in G
   ♦ All the elements in G are present in E
◼  So E and G are equal sets

(vi) Consider set F
• A, B, C, D and E can be ruled out because, we already compared them with F
• Take the second element 'a'
   ♦ G and H can be ruled out because, 'A' is not present in them
   ♦ There are no remaining sets to compare with F
• So F is not equal to any other set

(vii) Consider set G
• We already saw that,
   ♦ E = G
   ♦ E is not equal to any other set
• So G cannot be equal to any set other than E

(viii) Consider set H
• All other sets can be ruled out because, we already compared them with H
• So H is not equal to any other set

◼  Thus we get the final answer:
• The equal sets are:
   ♦ From (ii): B = D
   ♦ From (v): E = G

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In the next section, we will see subsets

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