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Application of Sets

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Application of Sets - Lesson Summary

Results based on intersection, union and difference of two sets.
1) n(A ∪ B) = n(A) + n(B)

A ∩ B = ∅


Example: Consider two sets A and B.

A = {1,2,3,4} B = {5,6,7,8,9}

A ∩ B = ∅

n(A) = 4

n(B) = 5

A ∪ B = {1,2,3,4,5,6,7,8,9}

n(A ∪ B) = 9

n(A ∪ B) = n(A) + n(B)

⇒ 9 = 4 + 5 = 9

Therefore, the relation is verified.

2) n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

A and B are finite sets.
Verification of the relation with the help of a Venn diagram:
It can be observed that,

(A - B) ∩ (A ∩ B) ∩ (B - A) = ∅

(A - B) ∩ (A ∩ B) ∪ (B - A) = A ∪ B

n(A ∪ B) = n(A - B) + n(A ∩ B) + n(B - A)

n(A ∪ B) = n(A - B) + n(A ∩ B) + n(B - A) + n(A ∩ B) - n(A ∩ B)

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Hence, the relation is verified.

3) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C)

n(A ∪ D) = n(A) + n(D) - n(A ∩ D)

n(A ∪ B ∪ C) = n(A) + n(B ∪ C) - n(A ∩ (B ∪ C))

n(A ∪ B ∪ C) = n(A) + n(B ∪ C) - n((A ∩ B) ∪ (A ∩ C))

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(B ∩ C) - n(A ∩ B) - n(A ∩ C) + n(A ∩ B ∩ C) .

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