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

