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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.


marked as duplicate by Asaf Karagila elementary-set-theory Nov 14 '17 at 9:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ 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. math.stackexchange.com/questions/1617178/…. $\endgroup$ – Noah Schweber Apr 17 '16 at 5:21
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    $\begingroup$ 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. $\endgroup$ – Adam Hughes Apr 17 '16 at 5:23
  • $\begingroup$ @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. $\endgroup$ – user118494 Apr 17 '16 at 10:16
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    $\begingroup$ @user118494 yeah, it's very easy to fix, that was just a minor quibble. Put $1$ into the first one and you're golden. $\endgroup$ – Adam Hughes Apr 17 '16 at 15:16
  • $\begingroup$ See also math.stackexchange.com/questions/12629/… adn math.stackexchange.com/questions/1750432/… $\endgroup$ – Martin Sleziak Apr 20 '16 at 2:55

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 $0$.

  • $\begingroup$ For a $\Delta_j$ , when $j$ is fixed, and $m$ varies over the odd numbers ? $\endgroup$ – user118494 Apr 20 '16 at 15:45
  • $\begingroup$ @user118494: If I understand your comment, then yes. For example $\Delta_3$ consists of the numbers $4,12,20,28,36,44,\dots$. $\endgroup$ – André Nicolas Apr 20 '16 at 15:59
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Apr 21 '16 at 6:35

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