Today is our topic of discussion Finite and Infinite Sets .
Finite and Infinite Sets
Counting the elements of the set A = {15, 16, 17, 18, 19, 20, 21, 22}, we see that A has 8 elements. This counting is completed by establishing a one-one correspondence of the set A with the set B = {1, 2, 3, 4, 5, 6, 7, 8}, as shown below:
Definition 3 (Finite and Infinite Sets). A set whose elements can be fixed by counting, is called a finite set. If any set A is not a finite set, then it is called infinite set.
1) Empty set is a finite set, whose number of elements is 0.
2) If any set A and Jm = {1,2,3,…,m} are equivalent, where m Є N, then A is a finite set and the number of elements of A is m.
3) If A is a finite set, the number of elements of A is denoted by n(A).
Note:
1) J₁ = {1}, J₂ = {1,2}, J3 = {1, 2, 3} is the finite subset of N and ∩(J₁) = 1, ∩(J2) = 2, n(J3) = 3 etc. In real sense, Jm~ Jm and n(Jm) = m.
2) Only the number of elments in a finite set can be fixed. So notation n(A) implies that A is a finite set.
3) If A and B are equivalent sets and if one of them is finite, then the other set must be finite and n(A) = n(B) will hold.
Proposition 6. If A is a finite set and B is a proper subset of A, then B will be a finite set and there will be n(B) < n(A).
Proposition 7. Set A is infinite if and only if there exists A and a proper subset equivalent to A.
Number of Elements of Finite Sets
Elements of finite set A is denoted by n(A) and how to determine n(A) is explained. Now, suppose n(A) = p > 0, n(B) = q> 0 where AПB = Ø.
So, n(A∪B) = p + q = n(A) + n(B). From this we can say the following proposition.
Proposition 8. If A and B are disjoint finite set to each other, then n(AUB) = n(A) + n(B).
Expanding the proposition, we can say, n(A∪B∪C) = n(A) + n(B) + n(C).
Similarly, n(A∪B∪UC∪D) = n(A) + n(B) +n(C) +n(D) etc,
where A, B, C, D sets are disjoint finite set to one another.
Proposition 9. For any finite set A and B, n(AUB) = n(A)+n(B)−n(ANB).
Proof: Here, A\B, An B and B \ A are disjoint sets to each other [See Venn Diagram].
So, A = (A\ B) ∪ (A∩B) and B = (B \ A) U (A∩B)
Therefore, AUB = (A\ B) u (An B) U (B\ A)
n(A) = n(A \ B) + n(A∩B) (1)
n(B) = n(B \ A) + n(A ∩ B) · · · · · (2)
n(AUB) = n(A\ B) + n(An B) + n(B \ A) · · · · (3)
So, from (1) we get, n(A \ B) = n(A) – n(ANB) and from (2) we get, n(B \ A) = n(B) – n(AB)
Now, putting n(A \ B) and n(B \ A) in (3), we get n(AUB) = n(A) – n(AП B) +n(B) — n(Aп B) +n(ANB) –
n(AUB) = n(A) +n(B) − n(A^ B)
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