Partition of N into infinite number of infinite disjoint sets? [duplicate]

Solution 1:

Maybe we can start slowly, by doing a decomposition of $\mathbb{N}$ into $2$ disjoint sets, say the odds and the evens.

Let's now go for a decomposition into $3$ disjoint sets. Leave the odds alone, and decompose the evens into those divisible by $2$ but no higher power of $2$, and those divisible by $4$. To put it another way, we are using the odds, twice the odds, and the rest.

Continue, and let's introduce some notation. Let $W_0$ be the set of odd positive integers. Let $W_1$ be the set of positive integers which are $2^1$ times an odd number. Let $W_2$ be the set of positive integers which are $2^2$ times an odd number. In general let $W_n$ be the set of integers which are $2^n$ times an odd number.

It is clear that the $W_k$ are all infinite, pairwise disjoint, and that their union is all of $\mathbb{N}$.

Solution 2:

Sure. For example, let $A_n$ be the natural numbers with exactly $n$ ones in their binary expansion.

Alternately, pick your favorite way of decomposing $\mathbb{N}$ into two disjoint infinite subsets $A, B$, and pick a bijection $f : B \to \mathbb{N}$. Then $f(B)$ can be decomposed into two disjoint subsets $A, B$, hence $B$ can be decomposed into two disjoint subsets $f^{-1}(A), f^{-1}(B)$. Rinse and repeat. This argument is fairly general and works for any infinite set which admits a decomposition into two disjoint subsets of the same cardinality as it (which under the Axiom of Choice is all of them).