Finding an infinite union of disjoint infinite sets to add up to natural numbers

Question

I want to prove the following statement:

Produce an infinite collection of sets $A_1, A_2, A_3, ...$ with the property that every $A_i$ has an infinite number of elements, $A_i ∩ A_j = ∅$ for all $i \ne j$, and $\cup^{∞}_{i=1} A_i = \mathbb{N}$.

My proof is a bit informal and non-rigorious due to me not exercising in writing a rigorous proof for a long time, so I would like to get comment on how to prove the following informal thought - if it can lead to correct proof at all.

Suggestion: Let $n_1, n_2, n_3, ...$ be a sequence of all prime numbers. Let $A_1$ be $\{ \forall k_1 \in \mathbb{N}: n_1^{k_1} \}$, let $A_2$ be $\{ \forall k_1, k_2 \in \mathbb{N}: n_1^{k_1} \cdot n_2^{k_2} \}$, and so on. All of those sets are obviously infinite. Since all natural numbers are either prime numbers, their powers, compositions of prime numbers or compositions of their powers, every natural number will be found in one of those sets. On the other hand, you can always find a natural number not in any of those sets $A_1, A_2, ..., A_m$, if it's prime decomposition does not contain prime number $n_m$.

Answer 1

Formally, your proof can be written as follows.

Let $(p_i)_{i\in\mathbb N}$ be the sequence of all prime numbers, in increasing order (i.e. $p_1=2, p_2=3, p_3=5$ etc.)

For $i\in\mathbb N, i\ge 2$, define $A_i=\{n\in\mathbb N: (p_i\mid n) \land (\forall j>i)(p_j\not\mid n)\}$. In other words, $A_i$ consists of all natural numbers whose largest prime factor is $p_i$.

Also, define $A_1=\{1,2,2^2,2^3,\ldots\}$. Obviously, $A_1$ is the only set containing $1$.

One can see easily that:

• All $A_i$ are infinite. Namely, $p_i, p_i^2, p_i^3, \ldots\in A_i$.
• $A_i\cap A_j=\emptyset$ for $i\ne j$. Namely, if $n\in A_i\cap A_j$, then either $n=1$ and so $i=j=1$ or $n$ has a largest prime factor $p_k$, so $i=k=j$.
• $\bigcup_{i=1}^\infty A_i=\mathbb N$. Namely, $1\in A_1$ and if $n\ge 2$, take the largest prime factor $p_i$ of $n$, and it follows that $n\in A_i$.

---

Having said all that, here is potentially a simpler solution: $A_i$ is the set of the natural numbers having exactly $i-1$ zeros at the end:

• $A_1=\{1,2,\ldots,9,11,12,\ldots,19,21,22,\ldots,29,\ldots\}$
• $A_2=\{10,20,\ldots,90,110,120,\ldots,190,\ldots\}$
• $A_3=\{100,200,\ldots,900,1100,1200,\ldots,1900,\ldots\}$
• etc.

Answer 2

You have to be a little more careful with your proof-writing. What you're saying in English is that each $A_i$ is a product of powers of the first $i$ primes. The problem is that, as you've written it, it's not true that $A_i \cap A_j = \varnothing$. It's easy to fix the problem, though; just require in each $A_i$ that $k_i \geq 1$.

Also, as noted above, you need to place $1$ (and possibly $0$) in one of your sets, which you can do arbitrarily.

Answer 3

Let $ A_{1}= \{ 1 \} \cup P$ where $P$ is the set of primes, and Let $A_k = \{n | \Omega(n)=k\}$ where $\Omega(n)$ is the # number of prime divisors of $n$ and $k\geq2$.

Surely $A_k $ has infinitely many numbers for any $k\geq 1$

And that $A_i\cap A_j = \emptyset$ because you can't have a number $n\geq 2$ with different number of prime divisors !!