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.


1 Answer 1


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
    Commented Apr 14, 2017 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... $\endgroup$ Commented Apr 14, 2017 at 1:22
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    $\begingroup$ 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? $\endgroup$ Commented Apr 14, 2017 at 2:18
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    $\begingroup$ 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
    Commented Apr 14, 2017 at 10:30
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    $\begingroup$ @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 $B$). $\endgroup$ Commented Apr 15, 2017 at 3:04

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