Pairwise disjoint sets of the form $\{\lfloor n\alpha \rfloor : n \in \mathbb{N}\}.$

Question

Let $S_{\alpha} = \{\lfloor n\alpha \rfloor : n \in \mathbb{N}\}.$ I am working on a problem which asks to show that $\mathbb{N}$ cannot be partitioned as the pairwise disjoint union of $S_{\alpha}, S_{\beta}, S_{\gamma}$ for some $\alpha, \beta, \gamma.$ I am familiar with Beatty sequences, but this problem is slightly different.

To start attacking this problem, I first wish to place conditions under which $S_{\alpha}, S_{\beta}$ are disjoint. Let $S = \frac{1}{\alpha}+\frac{1}{\beta}.$ If disjointness is equivalent to $S = 1$ and $\alpha, \beta$ being irrational, we would be done by Beatty's Theorem. I have shown $S_{\alpha}, S_{\beta}$ intersect if $S>1.$ I have also shown they intersect if one of $\alpha, \beta$ is rational. But what if $S<1$? It seems counterintuitive that making the sequences less dense would still allow them to intersect, no matter how small $S$ is. But whenever I test $2$ sequences on Desmos with various small values of $S,$ they always overlap. This is my only justification for why the result must be true.

Here is my attempt at finding an intersection: $\alpha, \beta$ being irrational and $\lfloor \alpha n \rfloor = \lfloor \beta m \rfloor = k$ is equivalent to $\frac{k}{\alpha} < n < \frac{k+1}{\alpha}, \frac{k}{\beta} < m < \frac{k+1}{\beta}.$ If we add the $2$ equations, we get $kS < m+n < (k+1)S.$ Since $S < 1,$ we can certainly find integers $m,n, k$ to satisfy this. But we cannot proceed backwards because only $2$ inequalities have been satisfied while we need $4.$ How should I proceed? Any hints?

Update: I have solved the original problem with a different method, but I'm still curious about the answer to this question.

For a subset $A$ of $\mathbb{N}$ if $d(A)=\lim_{n\rightarrow}\frac{|A\cap[1,n]|}{n}$ exists then $d(A)+d(\mathbb{N}\setminus A)=1$. Now you can show that $d(S_{\alpha})=\frac{1}{\alpha}$ and hence if $S_{\alpha},S_{\beta}$ partition $\mathbb{N}$ we should have $\frac{1}{\alpha}+\frac{1}{\beta}=1$. $d(A)$ is known as natural density of $A$. For details see the link https://en.wikipedia.org/wiki/Natural_density — ShBh, Jun 21 '20 at 15:44
@ShubhrajitBhattacharya $S_{\alpha}, S_{\beta}$ do not partition $\mathbb{N}.$ I am investigating conditions under which $S_{\alpha}, S_{\beta}$ are merely disjoint. — Display name, Jun 21 '20 at 15:49
Okay. But I think you can use the concept of natural density. — ShBh, Jun 21 '20 at 15:51
@ShubhrajitBhattacharya Look up Beatty sequences to not be at a disadvantage here. — Jyrki Lahtonen, Jun 21 '20 at 15:53
@ShubhrajitBhattacharya I already used it to show that they are not disjoint if $\frac{1}{\alpha} + \frac{1}{\beta} > 1.$ But it does not help at all in the case $S<1.$ — Display name, Jun 21 '20 at 15:53
If $\alpha$ and $\beta$ are a Beatty pair, then surely $S_\alpha$ and $S_{2\beta}$ are distinct. Not that it would help you in finding a $\gamma$ :-( — Jyrki Lahtonen, Jun 21 '20 at 15:56
Or generalizing that trivial observation: If $(\alpha,\beta)$ is a Beatty pair, then $S_{n\alpha}\cap S_{m\beta}=\emptyset$ for all pairs $m,n>0$. — Jyrki Lahtonen, Jun 21 '20 at 16:49

Jyrki Lahtonen · Accepted Answer · 2020-06-21T16:34:19.073

I learned about Beatty sequences from Joe Roberts' excellent book Elementary number theory - a problem oriented approach. There he cites a theorem due to Skolem (1957).

There do not exist positive irrational numbers $\alpha,\beta,\gamma$ such that $S_\alpha$, $S_\beta$ and $S_\gamma$ would be pairwise disjoint.

An earlier (1927) result by Uspensky states that it is impossible to find $\alpha,\beta,\gamma$ such that $S_\alpha$, $S_\beta$, $S_\gamma$ would form a partition on $\Bbb{Z}_{>0}$. I'm fairly sure that the link on WP on Beatty sequences is exactly the reference.

Roberts has listed both these results as exercises (he builds up to the proofs by sequences of exercises). I'm not sure I have the time to get to the heart of the matter :-( Anyway, here is the sequence of exercises leading to Uspensky's theorem. If I got it right, the approach is due to R.L. Graham (1963)

Assume $\alpha<\beta<\gamma$. Let $m$ be the smallest natural number not in $S_\alpha$.

$\alpha=1+\delta$ where $0<\delta<1$.
$S_\alpha$ does not miss any pair of consecutive integers.
$(m-1)\delta<1\le m\delta$.
$m$ is the first element of $S_\beta$ and $\beta=m+\epsilon, 0\le\epsilon<1$.
If $x$ is a positive integer $\notin S_\alpha$, then the next integer $\notin S_\alpha$ is either $x+m$ or $x+m+1$.
The next element in $S_\beta$ after $\lfloor n\beta\rfloor$ is either $\lfloor n\beta\rfloor+m$ or $\lfloor n\beta\rfloor+m+1$.
The $k$th positive integer missing from $S_\alpha$ is the $k$th element in $S_\beta$. The claim follows.

In terms of solving the original problem, this doesn't help since the problem was surely not meant to be solved by appealing to a theorem stronger than it. In terms of determining when $S_{\alpha}, S_{\beta}$ are disjoint, this doesn't help either. — Display name, Jun 21 '20 at 16:11
@Displayname I realize that this is not very useful as an answer for the reason you explained. If everything else fails, you can check out Roberts' book. A free download seems to be available. Or check out the reference to Uspensky's paper in WP. — Jyrki Lahtonen, Jun 21 '20 at 16:15
@Displayname The approach seems to be that when $S_\alpha$, $\alpha\in(1,2)$ and $S_\beta$ do not intersect, and $S_\beta$ contains the smallest integer missing from $S_\alpha$, then they must partition $\Bbb{Z}{>0}$ already - leaving no room for $S\gamma$. — Jyrki Lahtonen, Jun 21 '20 at 16:42

Pairwise disjoint sets of the form $\{\lfloor n\alpha \rfloor : n \in \mathbb{N}\}.$

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