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Possible to construct a disjoint sequence of infinite subsets of $\mathbb{N}$ whose union is not $\mathbb{N}$?

I am thinking this must be possible.

I've considered trying to define each sequence as the range of some function. I have to find a way to hop over natural numbers in some way to leave infinitely many left for each subsequent subset of $\mathbb{N}$. Thinking about just leaving out $1$ so then the union isn't $\mathbb{N}$. I haven't tried to do anything like this before. Any possible hints on how this can be done and I can think a little more to see if I can come up with something? If not I may just ask for an answer heh.

Thanks much for your time and input I appreciate it very much.

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marked as duplicate by Martin Sleziak, Arctic Char, arjafi Sep 7 '15 at 20:00

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.

up vote 9 down vote accepted

Consider $A_n = \{ p_n^k : k \geq 1 \}$ where $p_n$ is the $n$th prime number.

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Sure. For $k\in\Bbb N$ let $A_k=\{(4k+1)2^n:n\in\Bbb N\}$. It’s easy to check that the sets $A_k$ are infinite and pairwise disjoint, and their union contains no number of the form $(4k+3)2^n$.

Or let $\{A_k:k\in\Bbb N\}$ be any partition of $\Bbb N$ into infinite sets. For $k\in\Bbb N$ let $B_k=\{2n:n\in A_k\}$. Then $\{B_k:k\in\Bbb N\}$ is a partition of the set of non-negative even integers into infinite sets, whose union therefore contains none of the odd integers.

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An all-too-easy answer: choose your favorite way of mapping $\mathbb{N}$ to a subset of $\mathbb{N}$ (say, the even numbers). Now, take a partition of $\mathbb{N}$ into disjoint infinite subsets (e.g., via the inverse of the pairing function), and apply that map to each of the subsets.

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You can just do arithmetic progressions:
Choose distinct $x,y \in \mathbb{N}$

Define $$b: |x-y|+1 $$ $$ O = \bigcup_{x \in \mathbb{N}, x \ne 1} \{x,x+2b,x+3b,\ldots\} $$ Now just ensure that you omit at least one progression from your collection, (by the $x\ne 1$ condition), and their disjoint, infinite union will not be $\mathbb{N}$.

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