Question

Produce an infinite collection of sets A1,A2,A3, . . . with the property that every Ai has an infinite number of elements, Ai ∩ Aj = ∅ for all i = j, and [infinity] i=1 Ai = N.

Answer 1

Answer:

Produce an infinite collection of sets A1,A2,A3, . . . with the property that every Ai has an infinite number of elements, Ai ∩ Aj = ∅ for all i = j, and [infinity] i=1 Ai = N.

Explanation:

Solution

For n ∈ N,

define  A_n = {2 ^n−1  ,(3)(2n−1 ),(5)(2^n−1 ),(7)(2^n−1 ), . . .}

I.e. A_n is all odd multiples of 2^n−1 . We must show that these sets satisfy the desired properties.

• (Infinite Number of Elements).

It is clear that the set A_n = {2 ^n−1 ,(3)(2^n−1 )(5)(2^n−1 ),(7)(2^n−1 ), . . .}  has infinitely many elements.

• (Disjoint).

Given A_n and A_m with n ≠ m, we can assume, without loss of generality, that n < m. Suppose  that there existed some x ∈ A_n ∩ A_m. Then by definition of these sets, there exists some odd numbers k  and l such that x = 2^n−1 . k = 2^m−1  . l.

However since n < m, we have that n ≤ m − 1, and therefore we  can write 2^m−1 = (2^n )(2 i ) with i ≥ 0. Hence we have 2^n−1 . k = 2^n. 2 ^i. l  

Dividing both sides by 2^n−1 yields  k = (2)(2^i ) .l, which contradicts the assumption that k is odd. Therefore A_n ∩ A_m = ∅.

• (Union is N).

We want to show that  [infinity] i=1 A_n = N.

(⊆). Since each A_n is a subset of N, the union of these sets is a subset of N as well.

(⊇).Given any x ∈ N, we can write x = 2^n−1 . k for some n ∈ N where k is odd. Then x ∈ A_n, as  desired.