How to Decompose $\mathbb{N}$ like this? 
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Partitioning an infinite set
Partition of N into infinite number of infinite disjoint sets? 

Is it possible to find a family of sets $X_{i}$, $i\in\mathbb{N}$, such that:


*

*$\forall i$, $X_i$ is infinite, 

*$X_i\cap X_j=\emptyset$ for $i\neq j$, 

*$\mathbb{N}=\bigcup_{i=1}^{\infty}X_i$
Maybe it is an easy question, but I'm curious about the answer and I couldnt figure out any solution.
Thanks
 A: This is much like the answer by GEdgar, but takes into account the fact that $0\in\Bbb N$. Define $X_i$ to be the set of numbers whose representation ends with exactly $i$ digits $1$ (so in particular $X_0$ is the set of numbers that do not end with a digit $1$; it is obviously infinite (and $0\in X_0$; this is why I didn't take digits $0$), and you can get $X_i$ from $X_0$ by adding $i$ digits $1$ to the end of each element). I was originally thinking of binary representation, but it actually works for any base, in particular for base $10$.
A: Take any bijection $f: \mathbb N\times\mathbb N \to \mathbb N$. (There are many such bijections, see Wikipedia article Pairing function.)
Define $X_i=f[\mathbb N\times\{i\}]$.
A: An explicit one.  $\mathbb N = \{1,2,3,4, \dots\}$.
For each $i$, let $A_i$ be the set of odd multiples of $2^{i-1}$.
\begin{align}
A_1 &= \{1,3,5,7,\dots\}
\\
A_2 &= \{2,6,10,14,\dots\}
\\
A_3&=\{4,12,20,28,\dots\}
\\
&\dots
\end{align}
A: Let $p_n$ be the $n$-th prime number. That is $p_1=2; p_2=3; p_3=5; p_4=7$ and so on.
For $n>0$ let $X_n=\{(p_n)^k\mid k\in\mathbb N\setminus\{0\}\}$. 
For $X_0$ take all the rest of the numbers available, namely $k\in X_0$ if and only if $k$ can be divided by two distinct prime numbers, or if $k=1$.
