Infinite collection of infinite sets whose union converges to a defined set?

Question

I was driving around and thinking about the sets that we commonly use in mathematical proofs such as $\mathbb{Z}$ or $\mathbb{Q}$. In many introductory proofs courses, these sets seem to be the building blocks of many definitions (such as what makes a number even or odd) or lemmas. In addition to having their own symbols, this sometimes gives off a perception to many introductory proofs students that these sets, while having infinite elements, are somehow singular and entities$^*$.

$^*$(I'm aware that some of these common sets are subsets of the others and aren't "singular" per se. I was trying to convey a description of their perception among students, including myself, due to their prevalent use which spurred this question to begin with).

It got me thinking, can you take a "reductionist" approach and fragment these common sets into an infinite mutually exclusive infinite sets of which the totality of their elements converges to these "special sets"? For example, to start simple, I was wondering if it was possible to have an infinite collection of infinite sets $n_1, n_2, n_3....n_j, n_{j+1}....$ such that:

1. $n_a\cap n_b=\emptyset$ for $a\neq b$, $\forall$ $a$ and $b$ that are indices to describe the infinite set.

2. $\bigcup^\infty_{i=1} n_i = \mathbb{N}$

Answer 1

Yes, this is possible. Simply take every other number and put it into $n_1$. This means $n_1 = \{0, 2, 4, 6, \ldots\}$. From the remaining numbers in $\mathbb{N}\setminus n_1$, take every other number and put it into $n_1$, yielding $n_2 = \{1, 5, 9, \ldots\}$. From the remaining set $\mathbb{N}\setminus(n_1 \cup n_2)$ take every other number etc..

More formally, you choose $n_i$ to be the set of all numbers whose binary representations ends with exactly $i - 1$ ones.

Answer 2

There are a lot of possibilities in principle. Here's one: take $n_i$ to be the collection of all multiples of $p_i$ which are not multiples of any elements of $\{ p_1,\dots,p_{i-1} \}$, where $p_i$ is the $i$th prime. Then deal with $1$ by putting it in one of them (doesn't matter which).

Answer 3

Here are a couple of related straightforward ones. Each positive integer can be written uniquely in the form $2^k(2m+1)$, where $k,m\in\Bbb N$. For $k\in\Bbb N$ let

$$A_k'=\{2^k(2m+1):m\in\Bbb N\}\;,$$

and for $m\in\Bbb N$ let

$$B_m'=\{2^k(2m+1):k\in\Bbb N\}\;.$$

Now let $A_0=A_0'\cup\{0\}$, $B_0=B_0'\cup\{0\}$, $A_k=A_k'$ for $k>0$, and $B_m=B_m'$ for $m>0$. Then

$$\Bbb N=\bigcup_{k\in\Bbb N}A_k=\bigcup_{m\in\Bbb N}B_m\;,$$

the sets $A_k$ are pairwise disjoint, and the sets $B_k$ are pairwise disjoint.

In words, $A_k$ is the set of positive integers $n$ such that $2^k$ is the highest power of $2$ dividing $n$, with $0$ arbitrarily thrown into $A_0$, and $B_k$ is the set of positive integers $n$ that are a power of $2$ times the odd number $2m+1$, with $0$ arbitrarily thrown into $B_0$. Ignoring $0$:

$$\begin{array}{c|cc} &B_0&B_1&B_2&B_3&B_4&B_5&B_6&\ldots\\ \hline A_0&1&3&5&7&9&11&13&\ldots\\ A_1&2&6&10&14&18&22&26&\ldots\\ A_2&4&12&20&28&36&44&52&\ldots\\ A_3&8&24&40&56&72&88&104&\ldots\\ A_4&16&48&80&112&144&176&208&\ldots\\ A_5&32&96&160&224&288&352&416&\ldots\\ A_6&64&192&320&448&576&704&832&\ldots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots& \end{array}$$

Answer 4

Set-theorists often consider the natural numbers (including zero) and the set of finite ordinals to be equal .The "ordinal zero" is $0=\phi$, the empty set.When $x$ is an ordinal, the ordinal $x+1$ is defined by $x+1=x \cup \{x\}$.So $1=\{0\}$, $2=\{0,1\}$, $3=\{0,1,2\}$ , etc. This might help the students to understand that even the individual numbers are not necessarily "atoms" and so sets of them usually aren't either. [An ordinal is a transitive set for which membership $\in$ is a well-order on $S$, and a set is $S$ transitive iff every member of $ S$ is a subset of $S$.]