In the previous section, we saw some interesting relations between two sets. In this section, we will see such relations between three sets. We will see them in the form of solved examples.
Solved example 1.64
Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
Solution:
• We will solve this problem in two parts (a) and (b).
♦ In part (a), we will prove that B ⊂ C
♦ In part (b), we will prove that C ⊂ B
Part (a):
1. Let x be an element of B
• Using symbols, we write this as: x ∈ B
2. Then x will be an element of A ∪ B also.
• Using symbols, we write this as: x ∈ B ⇒ x ∈ (A ∪ B)
3. Given that, A ∪ B = A ∪ C
• So x must be an element of A ∪ C also.
• Using symbols, we write this as:
x ∈ (A ∪ B) ⇒ x ∈ (A ∪ C)
4. If x is an element of A ∪ C, it can be an element of A or C
• Using symbols, we write this as:
x ∈ (A ∪ C) ⇒ x ∈ A or x ∈ C
5. Let us assume that, x is an element of A
• In (1), we have already said that x is an element of B
♦ So x is an element of both A and B
♦ In that case, x will be an element of A ∩ B
• Using symbols, we write this as: x ∈ (A ∩ B)
6. Given that, A ∩ B = A ∩ C
♦ So x must be an element of A ∩ C also.
• Using symbols, we write this as:
x ∈ (A ∩ B) ⇒ x ∈ (A ∩ C)
7. If x is an element of A ∩ C, it must be an element of both A and C
• Using symbols, we write this as:
x ∈ (A ∩ C) ⇒ x ∈ A and x ∈ C
• That means, if an element x which belongs to B, is present in A also, it will be present in C also
8. In step (5), instead of assuming A, let us assume that, x belongs to C
• Then we can write:
The element x, which belongs to B, is present in C also
9. So whatever be the assumption that we make in (5), the element x, which belongs to B, will be present in C
◼ Thus we get: B ⊂ C
Part (b):
1. Let y be an element of C
• Using symbols, we write this as: y ∈ C
2. Then y will be an element of A ∪ C also.
• Using symbols, we write this as: y ∈ C ⇒ y ∈ (A ∪ C)
3. Given that, A ∪ B = A ∪ C
• So x must be an element of A ∪ B also.
• Using symbols, we write this as:
y ∈ (A ∪ C) ⇒ y ∈ (A ∪ B)
4. If y is an element of A ∪ B, it can be an element of A or B
• Using symbols, we write this as:
y ∈ (A ∪ B) ⇒ y ∈ A or y ∈ B
5. Let us assume that, y is an element of A
• In (1), we have already said that y is an element of C
♦ So y is an element of both A and C
♦ In that case, y will be an element of A ∩ C
• Using symbols, we write this as: y ∈ (A ∩ C)
6. Given that, A ∩ B = A ∩ C
♦ So y must be an element of A ∩ B also.
• Using symbols, we write this as:
y ∈ (A ∩ C) ⇒ y ∈ (A ∩ B)
7. If y is an element of A ∩ B, it must be an element of both A and B
• Using symbols, we write this as:
y ∈ (A ∩ B) ⇒ y ∈ A and y ∈ B
• That means, if an element y which belongs to C, is present in A also, it will be present in B also
8. In step (5), instead of assuming A, let us assume that, y belongs to B
• Then we can write:
The element y, which belongs to C, is present in B also
9. So whatever be the assumption that we make in (5), the element y, which belongs to C, will be present in B
◼ Thus we get: C ⊂ B
◼ In part (a), we proved: B ⊂ C
◼ In part (b), we proved: C ⊂ B
◼ So we get: B = C
Solved example 1.65
Show that if A ⊂ B, then C – B ⊂ C – A
Solution:
1. Let x be an element of C - B
• Using symbols, we write this as: x ∈ (C - B)
2. C - B will not contain any element of B
• So we can write:
♦ x will be an element of C.
♦ But x will not be an element of B
• Using symbols, we write this as:
x ∈ (C - B) ⇒ x ∈ C and x ∉ B
3. Given that A is a subset of B
• So all elements of A will be present in B
• In (2), we saw that x is not present in B
♦ If x is not present in B, x will not be present in A either.
4. If x is not present in A,
♦ x will not be deleted from C when (C - A) is formed
• That means, x will be present in (C - A)
5. So we can write:
x, which is an element of C - B, is an element of C - A also
◼ Thus we get: C - B ⊂ C - A
Solved example 1.66
Show that A ∩ B = A ∩ C need not imply B = C.
Solution:
We can show this using an example. It can be written in 3 steps:
1. Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {3, 4, 7, 8}
2. We get: A ∩ B = {3, 4} and A ∩ C = {3, 4}
3. Here A ∩ B = A ∩ C but A ≠ C
Solved example 1.67
Let A and B be sets. If A ∩ X = B ∩ X = ɸ and A ∪ X = B ∪ X for some set X, show that A = B.
(Hints A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use Distributive law)
Solution:
• We will solve this problem in two parts (a) and (b)
Part (a):
1. We have: A = A ∩ (A ∪ X)
• (A ∪ X) can be replaced by (B ∪ X) because, it is given that they are equal.
• So we get: A = A ∩ (B ∪ X)
2. The RHS can be expanded using the distributive law of intersection. We get:
A = (A ∩ B) ∪ (A ∩ X)
⇒ A = (A ∩ B) ∪ ɸ [∵ (A ∩ X) = ɸ]
⇒ A = (A ∩ B)
Part (b):
1. We have: B = B ∩ (B ∪ X)
• (B ∪ X) can be replaced by (A ∪ X) because, it is given that they are equal.
• So we get: B = B ∩ (A ∪ X)
2. The RHS can be expanded using the distributive law of intersection. We get:
B = (B ∩ A) ∪ (B ∩ X)
⇒ B = (B ∩ A) ∪ ɸ [∵ (B ∩ X) = ɸ]
⇒ B = (A ∩ B)
• From part (a), we have A = (A ∩ B)
• From part (b), we have B = (A ∩ B)
◼ So we can write: A = B
Solved example 1.68
Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = ɸ.
Solution:
1. A ∩ B should be non-empty. That means, there must be at least one common element in A and B.
• So let A = {1, 2} and B = {2, 3}
2. B ∩ C should be non-empty. That means, there must be at least one common element in B and C.
3. A ∩ C should be non-empty. That means, there must be at least one common element in A and C.
• So let C = {1, 3}
4. So we get:
♦ A ∩ B = {2}
♦ B ∩ C = {3}
♦ A ∩ C = {1}
• They are all non-empty sets
5. Now, there is not even a single element which is common in all three sets.
◼ So we get: A ∩ B ∩ C = ɸ.
In
the next
section, we will see the practical problems which involves three sets.
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