2
$\begingroup$

This question already has an answer here:

Prove that there exists a countably infinite set $A \subseteq $ $P(\mathbb{N})$ that satisfies all of the following conditions:

$i)$ $X \cap Y = \emptyset$ for all $X, Y \in A$ such that $X \neq Y$
$ii)$ $\mathbb{N} = \bigcup A $
$iii)$ Every element in $A$ is countably infinite

For all the sets I tried, the three conditions were satisfied but $A$ was not countably infinite, so now I'm stuck. Any help would be greatly appreciated.

$\endgroup$

marked as duplicate by Hanul Jeon, Asaf Karagila, David, Claude Leibovici, mau May 8 '14 at 8:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ If $A$ was not countably infinite, condition ii) would fail. $\endgroup$ – Umberto P. May 8 '14 at 4:26
  • $\begingroup$ What do you know about partitions? $\endgroup$ – Asaf Karagila May 8 '14 at 4:27
  • 1
    $\begingroup$ A simple example is $A=\{A_0,A_1,A_2,\ldots\}$, where $A_i$ is the set of natural numbers divisible by $2^i$ but not $2^{i+1}$. $\endgroup$ – mjqxxxx May 8 '14 at 4:29
  • $\begingroup$ I'd suggest looking at sets made up of multiples of primes that are not divisible by the primes before. $A_2={1,2,4,6,...}, A_3={3,9,15,21,...}$. And so on. $\endgroup$ – Steven F May 8 '14 at 4:32
  • $\begingroup$ @UmbertoP. It could be that ii) holds and i) fails. $\endgroup$ – bof May 8 '14 at 4:38
3
$\begingroup$

Define the set $$ A_1=\{1\}\cup\{n\in\mathbb{N}:n\;\mathrm{is\; product\; of\; two\; or \; more\; distinct \; primes}\} $$ And for each prime number $p$, let $A_p\{p^n:n\in\mathbb{N}$. Finally, let $A$ be the collection of this sets. It is not hard to see that this $A$ is the desires partition of $\mathbb{N}$.

$\endgroup$
2
$\begingroup$

Let $f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ be a bijection.

Define

$$A_n=\{ f(n,m) | m \in \mathbb{N} \} \,.$$

$\endgroup$
1
$\begingroup$

Let each set contain integers that are the product of $k$ primes. Add the value 0 and 1 into the first set.

It is obvious that all the conditions are satisfied.

$\endgroup$
  • $\begingroup$ Except $0$ and $1$ are in their own classes. :-) $\endgroup$ – Asaf Karagila May 8 '14 at 5:56
  • $\begingroup$ @asaf lol thanks. $\endgroup$ – Calvin Lin May 8 '14 at 12:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.