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A friend thought of this problem and I found it interesting to think about so I want to share it with you. I am intrigued how you will solve the problem.

Find countably infinite many sets $U_i$ such that $$\mathbb{N}=\cup_{i=1}^{\infty}U_i,$$ with $U_i$ disjunct and countably infinite.

the.polo
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9 Answers9

10

Let $U_i=\{2^in\colon n\text{ is an odd natural number}\}$.

Harald Hanche-Olsen
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For each $i$, let $U_i$ be the set of integers whose decimal expansion contains exactly $i$ ones. For example, $2779\in U_0$, $114567\in U_2$. Then the sets are disjoint and cover $\mathbb N$.

Or for each $i$, let $U_i$ be the set of integers with exactly $i$ distinct prime factors. (For completion, put $1\in U_1$). Then since the set of primes is infinite, each $U_i$ is infinite, and the $U_i$'s are distinct and cover $\mathbb N$ by the fundamental theorem of arithmetic.

Mathmo123
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$$U_i = \{n \mid \text{the $i$th prime number is the smallest number in $n$'s prime factorization}\}$$

The $U_i$'s are:

  • Each infinite, since if $p(i)$ is the $i$th prime number, the sequence $p(i)^1, p(i)^2 \dots$ is infinite and in $U_i$.
  • Of infinite number, since the primes are infinite.
  • Disjoint, since no two numbers can both be the smallest in a set.
Eli Rose
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For each $i \geq 1$, let $U_i'$ denote the set of all natural numbers divisible by exactly $i$ prime numbers, counting multiplicity. Set $U_1 = U_1'\cup\{1\}$ and $U_i = U_i'$ for i $\geq 2$ (or do whatever else you'd like with that one straggler $1$).

Then $\mathbb{N} = \cup_{i = 1}^\infty U_i$, and each $U_i$ is infinite.

Dan
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Let $p_1,p_2,...$ enumerate the primes. Let $$U_n = \{ p_n^m : m \in \mathbb{N}\}.$$

Then $\mathbb{N} = \left(\mathbb{N} \setminus \cup_{n=1}^\infty U_n\right) \cup \left(\cup_{n=1}^\infty U_n\right).$ If we call $U_0 := \mathbb{N} \setminus \cup_{n=1}^\infty U_n$ then $\mathbb{N} = \cup_{n=0}^\infty U_n$.

Finally $U_0$ is infinite since, $2p_n \not \in U_m$ for all $m$ unless $p_n = 2$.

Joel
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$$U_n=\{x\ |\ x-\lfloor\sqrt x\rfloor^2=n\}$$ Graph of $x-\lfloor\sqrt x\rfloor^2$ for integer $x$: x-floor(sqrt(x))^2

Akiva Weinberger
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Here is anpother one. Let $\{p_1, p_2, p_3, \dots\}$ be the prime numbers. Let $$ U_1 = \{n\in \mathbb{N} : p_1 \mid n \text{ and } p_i \nmid n\text{ for }i\geq 2 \} \cup \{1\} $$ and for $i\geq 2$ $$ U_i = \{n\in \mathbb{N} : p_j \mid n\text{ exactly for }j\leq i \}. $$ For example $$\begin{align} U_1 &= \{2^i: i\geq 1\}\cup \{1\} \\ U_2 &= \{2^i3^j: i,j \geq 1\} \\ U_3 &= \{2^i3^j5^k: i,j,k \geq 1\}. \end{align} $$ Then $$ \mathbb{N} = \bigcup_{i=2}^{\infty} U_i $$ (So the union starts at $2$, but I am sure you can modify this if you want the union to start at $1$.)

Thomas
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Tons of approaches exist. Here's another one: let $U_1=\{ 1 \}$ and recursively define

$$U_{n+1} = \left \{ k \in \mathbb{N} \setminus \bigcup_{j=1}^n U_j : k \text{ is divisible by } p_n \right \}$$

where $p_n$ is the $n$th prime number. So $U_2$ are the even numbers, $U_3$ are the multiples of $3$ which are not multiples of $2$, etc. As in Dan's answer you can throw $1$ into whichever of the others you like, if you want all of the $U$'s to be infinite.

Ian
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I will assume that $\mathbb{N}$ does not include $0$.

Let $f: \mathbb{N}\times \mathbb{N}\to \mathbb{N}$ be any pairing function, that is, a one-to-one onto map from $\mathbb{N}\times \mathbb{N}$ to $\mathbb{N}$.

For any natural number $k$, let $U_k$ be the set of all $f(k,y)$, where $y$ ranges over $\mathbb{N}$. Then the sets $U_k$ do the job.

In particular, one can use the standard Cantor Pairing Function $$f(x,y)=\frac{(x+y-2)(x+y-1)}{2}+y.$$

André Nicolas
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