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Let $\alpha$ and $\beta$ be positive real numbers such that $\alpha/\beta$ is irrational. Then the following holds

Theorem 1. For all $x\in \R$ and $\varepsilon>0$, there exist integers $k$ and $\ell$ such that $|k\alpha+\ell\beta-x|< \varepsilon$.

I have provided an elementary proof of the above fact at the end. But I suspect that there might be an ergodic theoretic way to show this. For example, a similar looking problem is the following:

Given an irrational number $\alpha$ and a real number $x$, then for all $\varepsilon>0$ there is an integer $n$ such that $x-n\alpha$ is $\varepsilon$-close to an integer.

This can be solved again using an elementary argument or by using ergodic theory by noticing that the problem is equivalent to proving that the irrational circle rotation has dense orbits, which in turn is equivalent to the ergodicity of the rotation. One can use Fourier analysis to do this and we do not need to invoke the pigeon-hole principle.

Another reason I think ergodic theory might help is that the proof below uses only the condition that that $(\mathbb Z\alpha+\mathbb Z\beta)\cap\mathbb Z=\set0$, a condition implied by the irrationality of $\alpha/\beta$. This in turn is equivalent to the ergodicity of the map $R_\alpha\times R_\beta:S^1\times S^1\to S^1\times S^1$, where $R_\alpha$ and $R_\beta$ are rotations of the circle by angles $\alpha$ and $\beta$ respectively (see this MSE post).

So does anyone see a way tp approach this using ergodic theory?

Proof of Theorem 1. It suffices to show this when $x=0$. Let $M$ be large enough so that $\alpha, \beta\leq M$. Fix $n>0$ such that $2/n< \varepsilon$. Consider the set $S=\set{k\alpha+\ell\beta:\ 0\leq k, \ell\leq nM}$. The set $S$ houses $(nM+1)^2$ distinct numbers (we use the irrationality of $\alpha/\beta$ for distinctness). Also, no member of $S$ is larger than $2nM^2$. Therefore by the pigeon-hole principle we have that at least $n/2+1$ many of the members of $S$ are lying between $r$ and $r+1$ for some integer $r$. Partition the interval $[r, r+1]$ into $n/2$ equal parts each of length $2/n$. Again, by the pigeon-hole principal, some two of these members are lying in a single such partition. We conclude that there exist integers $k_1, \ell_1, k_2, \ell_2$ with $(k_1, \ell_1) \neq (k_2, \ell_2)$ such that $|(k_1-k_2)\alpha+ (\ell_1-\ell_2)\beta|< 2/n< \varepsilon$. So we have found integers $k, \ell$, namely $k=k_1-k_2$ and $\ell=\ell_1-\ell_2$ such that $|k\alpha+\ell\beta|< \varepsilon$ and we are done. $\blacksquare$

caffeinemachine
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  • Can't help you with the ergodic theory, but I think there's a simpler non-eergodic proof of Theorem 1. $\alpha/\beta$ irrational implies at least one of $\alpha,\beta$ is irrational. Without loss of generality, we assume $\alpha$ is irrational. Then taking $\ell=0$ reduces to the one-variable case, already solved. – Gerry Myerson Oct 18 '18 at 01:16
  • It seems that your claim is the following: Is $\alpha$ is irrational, then for any given $x\in \mathbb R$, and any given $\epsilon>0$, we have an integer $k$ such that $|k\alpha-x|<\epsilon$. This is clearly false since $\mathbb Z\alpha$ is a discrete subgroup of $\mathbb R$. Perhaps you meant something else. – caffeinemachine Oct 18 '18 at 04:04
  • Sorry, thought we were working mod one, just trying to get close to an integer. – Gerry Myerson Oct 18 '18 at 05:12
  • Trying again – I do think Theorem 1 follows from Dirichlet's Theorem on Diophantine Approximation (which in turn follows from Dirichlet's Pigeonhole Principle, so this may not be too far from what you've already done). Given $\alpha,\beta,x$, let $\gamma=(\alpha/\beta)-(x/(k\beta))$ for some integer $k$. Choosing $k$ large enough, we may assume $\gamma$ is irrational. By Dirichlet there exist arbitrarily large integers $\ell,k$ such that $|\gamma+(\ell/k)|<k^{-2}$. Taking $k>\beta/\epsilon$, we get $|\gamma+(\ell/k)|<\epsilon/(k\beta)$. Continued... – Gerry Myerson Oct 18 '18 at 05:46
  • But this is $|(\alpha/\beta)+(\ell/k)-(x/(k\beta))|<\epsilon/(k\beta)$, or $|k\alpha+\ell\beta-x|<\epsilon$. – Gerry Myerson Oct 18 '18 at 05:49
  • It seems that you are using the letter $k$ in two different capacities. First you choose a suitable $k$ to define $\gamma$. Then you find large $\ell$ and $k$ by applying Dirichlet. This is causing confusion. – caffeinemachine Oct 18 '18 at 15:32
  • Dirichlet says there are arbitrarily large $k$, and any $k$ large enough will do to define $\gamma$. To put it another way, first work out how large $k$ has to be for $\gamma$ to work, then use Dirichlet to get $k$ that large. – Gerry Myerson Oct 18 '18 at 22:33

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