---

Source. Laczkovich (1984), Lemma 2, printed pp. 112–114 (PDF pp. 4–6).

Statement

Let $N\ge2$ be an integer, and let $\alpha$ be an irrational real number with bounded regular continued-fraction partial quotients. For every real $b$, there is a finite $H\subseteq G_\alpha=\mathbb Z\alpha+\mathbb Z$ such that

$$G_\alpha\cap(-\infty,b]\subseteq H^{(N)}. \tag{1}$$

Here $H^{(N)}$ is the backward closure.

Exact external input. Use the convergent errors, recurrence, and denominator bounds in Continued-fraction inputs.

Source precision. On p. 113 the source chooses a step involving either $q_j$ or $q_{j-1}$, but subsequently writes $|m|=|n|-iq_j$ in both cases and uses $q_j$ in both factors of the denominator comparison. That identity does not hold in the second case. Its printed auxiliary constant is $(K+1)^2N^2$. The proof below uses the chosen denominator $q$ explicitly and enlarges that constant to $(K+1)^3N^2$ throughout the seed and induction. It proves the same finite-existence statement (1). This is a compilation-supplied proof repair, not an author-issued erratum and not a disproof of Lemma 2.

Bears on. Problem 1125, through Theorem 2.

Proof

Write $\alpha=[a_0;a_1,a_2,\ldots]$, and choose $K\ge1$ such that $a_i\le K$ for every $i\ge1$. Put

$$C=(K+1)^3N^2$$

and define the finite seed

$$H=\{n\alpha+k:n,k\in\mathbb Z,\ |n|\le Na_1,\ b-N\le n\alpha+k\le b+C\}. \tag{2}$$

There are finitely many allowed $n$, and finitely many integers $k$ in the displayed interval for each $n$, so $H$ is finite.

For each fixed $n$ with $|n|\le Na_1$, set $k_0=\lfloor b+C-n\alpha\rfloor$. The $N$ consecutive points $n\alpha+k_0-r$, $0\le r\le N-1$, all lie in the seed interval: the smallest is strictly larger than $b+C-N\ge b-N$. By backward closure with step $1$, every point of $n\alpha+\mathbb Z$ at most $b+C$ belongs to $H^{(N)}$. This includes the needed integers, for $n=0$.

We prove, by induction on the positive integer $|n|$, the stronger assertion

$$n\alpha+k\le b+\frac C{|n|} \quad\Longrightarrow\quad n\alpha+k\in H^{(N)} \qquad(k\in\mathbb Z). \tag{3}$$

For $1\le|n|\le Na_1$, the preceding paragraph proves (3), since $C/|n|\le C$.

Now let $t=|n|>Na_1$ and assume (3) for smaller positive absolute coefficients. Choose the largest $j$ for which $q_j<t/N$. It exists, since $q_1=a_1<t/N$, and is finite because the denominators are unbounded. In particular $j\ge1$. Maximality and the denominator bound give

$$\frac{t}{(K+1)N}\le q_j<\frac tN. \tag{4}$$

Among $j$ and $j-1$, choose an index $r$ such that the error $q_r\alpha-p_r$ has the sign opposite to $n$. Consecutive errors have opposite signs. Write $q=q_r$, $p=p_r$, and set

$$h= \begin{cases} q\alpha-p,&n<0,\\ p-q\alpha,&n>0. \end{cases}$$

Then $h\in G_\alpha$ and

$$0<h<\frac1{q_j},\qquad \frac{t}{(K+1)^2N}\le q\le q_j<\frac tN. \tag{5}$$

For $r=j-1$, the error bound gives $h<1/q_j$ directly; for $r=j$, it gives $h<1/q_{j+1}\le1/q_j$. The lower bound for $q$ follows from (4) and $q_{j-1}\ge q_j/(K+1)$, also valid for $j=1$.

Suppose $x=n\alpha+k\le b+C/t$. For each $1\le i\le N$ write

$$x+ih=m\alpha+\ell,\qquad m=n-\operatorname{sgn}(n)\,iq,\quad \ell=k+\operatorname{sgn}(n)\,ip.$$

Since $iq\le Nq<t$,

$$0<|m|=t-iq<t. \tag{6}$$

Moreover, (4)–(5) yield

$$Cqq_j\ge t^2>t(t-iq).$$

All denominators are positive. Multiplying the last strict inequality by $i/[q_jt(t-iq)]$ gives

$$ih<\frac i{q_j} < C\left(\frac1{t-iq}-\frac1t\right). \tag{7}$$

Consequently

$$x+ih<b+\frac C{t-iq}=b+\frac C{|m|}.$$

The induction hypothesis puts every $x+ih$ in $H^{(N)}$. Its closure property then puts $x$ there as well, proving (3). Every noninteger point of $G_\alpha$ at most $b$ satisfies the hypothesis of (3); integers at most $b$ were already included. This proves (1).