TITLE: How to Decompose $\mathbb{N}$ like this?
QUESTION [8 upvotes]: Possible Duplicate:
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

REPLY [5 votes]: 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 $10$.