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$$\mathbb N =\bigcup_{j\in \mathbb N}\Delta_j $$ where each $\Delta_j$ is an infinite subset of $\mathbb N$ and $\Delta_j\cap \Delta_i=\Phi \ for\ i\neq j.$

Now what I need is a few examples of such decompositions. The only one I can think of now is the collection of the odd numbers and the even numbers. That satisfies it.

Another possibility I was considering was like this ::

$$\Delta_1=2\mathbb N\\ \Delta_2=3\mathbb N \backslash \Delta_1\\ \Delta_3=5\mathbb N\backslash (\Delta_1\cup\Delta_2)\\.\\.\\.\\.\\.\\.\\so\ \ on.$$ The technique here is for any arbitrary $k$ , $\Delta_k=p\mathbb N\backslash \left(\bigcup_{i=1}^{k-1}\Delta_i\right)$. Clearly I can see all these sets are mutually disjoint and also infinite since there are infinitely many prime numbers.

So , are there any other construction of this kind possible $?$ If so please let me know . Also , if there is any fault in my above construction point it out .

Thank you.

Peter Taylor
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user118494
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    Not an answer to your question, but you may find this interesting: clearly I can only decompose $\mathbb{N}$ into countably many disjoint pieces. However, what if I weaken that requirement from "disjoint" to just "almost disjoint" - that is, I demand that $A\cap B$ be finite (not necessarily empty) for distinct $A, B$ in my collection? It turns out this has a huge effect - we can now get an uncountable almost disjoint family! See e.g. http://math.stackexchange.com/questions/1617178/some-kind-of-bijection-between-mathbb-r-and-mathbb-n. – Noah Schweber Apr 17 '16 at 05:21
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    You appear to have forgotten $1$ in your decomposition, as it is not a multiple of any prime. Also, odds and evens doesn't satisfy since there are only two sets in that case and you need infinitely many. – Adam Hughes Apr 17 '16 at 05:23
  • @AdamHughes : So , my decomposition is totally wrong or is there some modification possible to make it work. I'm guessing I could choose $\Delta_1={1}\cup 2\mathbb N.$ That would include $1$ and solve the problem . Right $?$ Or is there still some problem pls notify. – user118494 Apr 17 '16 at 10:16
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    @user118494 yeah, it's very easy to fix, that was just a minor quibble. Put $1$ into the first one and you're golden. – Adam Hughes Apr 17 '16 at 15:16
  • See also http://math.stackexchange.com/questions/12629/partitioning-an-infinite-set adn http://math.stackexchange.com/questions/1750432/equivalence-relation-on-mathbbn-with-infinitely-many-equivalence-classes – Martin Sleziak Apr 20 '16 at 02:55

1 Answers1

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For every $j\ge 1$, let $\Delta_j$ consist of all natural numbers of the form $2^{j-1} m$, where $m$ is odd.

Remarks: $1.$ This can be used to produce a simple bijection between $\mathbb{N}$ and $\mathbb{N}\times \mathbb{N}$.

$2.$ You can get an interesting example of a different character by using the Cantor Pairing Function. Note that the linked article includes $0$ among the natural numbers, so you will have to modify it a little if you want $\mathbb{N}$ to exclude $0$.

André Nicolas
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