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This question already has an answer here:

I'm trying to find an indexed family {An:n∈N} that satisfies:

  1. each An is an infinite subset of the naturals N
  2. the intersection of any two arbitrary sets is empty
  3. the union of all the subsets is is the naturals N
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marked as duplicate by Brian M. Scott elementary-set-theory Jun 8 '16 at 20:28

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  • $\begingroup$ How about even numbers and odd numbers? $\endgroup$ – levap Jun 8 '16 at 20:05
  • $\begingroup$ There must be many possible answers. What exactly was the particular difficulty that prevented you from finding one? $\endgroup$ – David K Jun 8 '16 at 20:13
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Let $f\colon \Bbb N \times \Bbb N \to \Bbb N$ be any bijection, for example $$ f(a,b) = \frac {(a+b)(a+b+1)} 2 + b. $$ Define $$ A_n = \{f(n, k) \mid k \in \Bbb N\}, \quad \text{for $n\in\Bbb N$}. $$ Then the $A_n$ are pairwise disjoint and their union is $\Bbb N$. Every $A_n$ is infinite, because for each $n$, $k \mapsto f(n,k)\colon \Bbb N \to \Bbb N$ is an injection.

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  • $\begingroup$ Thank you, but would each An be an infinite subset in this case? $\endgroup$ – Caerus Jun 8 '16 at 20:22
  • $\begingroup$ Yes, every $A_n$ is infinite if $f$ is a bijection, because for each $n$, $k \mapsto f(n,k)$ is an injection. I'll add that to the answer. $\endgroup$ – BrianO Jun 8 '16 at 20:25
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Given any ordering the natural numbers, just assign every number to an entry of an infinite matrix following the same diagonal path as in the proof that the rational numbers are countable. enter image description here.

Define $A_n$ to be the set of all natural numbers on the $n$-th row. Needless to say that each $A_n$ is infinite.

There are as many ways of performing this task as there are orderings of $\mathbb{N}$.

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