How to Decompose $\mathbb{N}$ like this? [duplicate]

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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

Solution 1:

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}

Solution 2:

Take any bijection $f: \mathbb N\times\mathbb N \to \mathbb N$. (There are many such bijections, see Wikipedia article Pairing function.)

Define $X_i=f[\mathbb N\times\{i\}]$.