Let $A,B$ be two sets such that

$A\cup B = \mathbb{N}, A\cap B = \emptyset$ with $|A| = |B| = \infty$

Then for every $r \in \mathbb{R}>0$, exists sequences such that

$a_n\in A$, $b_n \in B$

such that

$\lim_{n\to \infty}\frac{a_n}{b_n} = r$

I do not know how to construct this two sequences. My ideas was using the density of the rationals in the reals and construct a sequence or using the cantor series but I can not get anything.


First it suffices to show the case of $r\geq 1$. Then it suffices to show that for any $r\geq 1$, $$\inf_{a\in A}\inf_{b\in B}|\frac{a}{b}-r|=0.$$ Suppose not. Then there exists $r>c>0$ such that for all $a\in A,b\in B$, $|\frac{a}{b}-r|>c$, which implies $a\notin[b(r-c),b(r+c)]$ for all $a\in A$. This implies $[b(r-c),b(r+c)]\cap \mathbb{Z}\subset B$ for all $b\in B$. This further implies that for all $n\in\mathbb{N}$, $[b(r-c)^n,b(r+c)^n]\cap\mathbb{Z}\subset B$.

Now for $n$ large enough, $(r+c)^n>(r-c)^{n+1}$, thus the fact that $[b(r-c)^n,b(r+c)^n]\cap\mathbb{Z}\subset B$ for all $n\in\mathbb{N}$ implies that $B$ contains all numbers sufficiently large, which contradicts $A\cap B=\emptyset$ and $|A|=\infty$. Thus the result is proved.

$\bf{Edit}$: The above argument has an error in the following statement:

"...$[b(r-c),b(r+c)]\cap \mathbb{Z}\subset B$ for all $b\in B$. This further implies that for all $n\in\mathbb{N}$, $[b(r-c)^n,b(r+c)^n]\cap\mathbb{Z}\subset B$."

Here is a modification. We have $[b(r-c),b(r+c)]\cap \mathbb{Z}\subset B$ for all $b\in B$. This implies that $[(b(r-c)+1)(r-c),(b(r+c)-1)(r+c)]\cap\mathbb{Z}\subset B$." Inductively, we get $[b(r-c)^n+(r-c)^{n-1}+\cdots+(r-c), b(r+c)^{n}-(r+c)^{n-1}-\cdots-(r+c)] \cap\mathbb{Z}\subset B$ for all $n\in\mathbb{N}$.

But note that $b(r-c)^{n+1}+(r-c)^{n}+\cdots+(r-c)<b(r+c)^{n}-(r+c)^{n-1}-\cdots-(r+c)$ for $n$ large enough, say $n\geq N$. Thus the intervals $[b(r-c)^n+(r-c)^{n-1}+\cdots+(r-c), b(r+c)^{n}-(r+c)^{n-1}-\cdots-(r+c)]$ for $n\geq N$ connect to form $[b(r-c)^N+(r-c)^{N-1}+\cdots+(r-c), \infty)$, and $B\supset[b(r-c)^N+(r-c)^{N-1}+\cdots+(r-c), \infty)\cap \mathbb{N}$.

  • 1
    $\begingroup$ How do you prove " $\forall n\in\mathbb N$, $[b(r-c)^n, b(r+c)^n]\cap\mathbb Z\subset B$ " ? If you don't intersect with $\mathbb Z$ I see the induction, but with the intersection you're forced to work with $\lceil b(r-c)\rceil$ and $\lfloor b(r+c)\rfloor$, and I don't see how you get rid of the floor/ceil... $\endgroup$
    – N.Bach
    Apr 14 '17 at 1:06
  • 1
    $\begingroup$ @N.Bach Good point. But it still works under a slight modification. Now we need to integrate this inductive argument with the fact that $(r+c)^n>(r-c)^{n+1}$ for $n$ large enough. Editing... $\endgroup$ Apr 14 '17 at 1:22
  • 1
    $\begingroup$ The key is "This further implies that for all $n\in\mathbb{N}$, $[b(r-c)^n,b(r+c)^n]\cap\mathbb{Z}\subset B$." I don't find this immediately obvious.How is this proved? $\endgroup$ Apr 14 '17 at 2:18
  • 1
    $\begingroup$ Read the edit, looks good, but I have another problem now... Why can we ignore the values $0<r<1$? At first I thought we could just take the sequences for $\frac 1 r$ and swap $(a_n)$ and $(b_n)$, but after re-reading the problem we're not allowed to do that. $\endgroup$
    – N.Bach
    Apr 14 '17 at 10:30
  • 3
    $\begingroup$ @N.Bach In fact, after we proved the case of $r\geq 1$, for the case of $0<r<1$, we may swap the sets $A$ and $B$ (not swap the sequences $(a_n)$ and $(b_n)$ for the fixed sets $A$ and $B$). $\endgroup$ Apr 15 '17 at 3:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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