Lattice Paths and Submonoids of $\mathbb Z^2$

Abstract

We study a number of combinatorial and algebraic structures arising from walks on the two-dimensional integer lattice. To a given step set $X\subseteq \mathbb Z^2$, there are two naturally associated monoids: $\mathscr {F}_X$, the monoid of all X-walks/paths; and $\mathscr {A}_X$, the monoid of all endpoints of X-walks starting from the origin O. For each ${A\in \mathscr {A}_X}$, write $\pi _X(A)$ for the number of X-walks from O to A. Calculating the numbers $\pi _X(A)$ is a classical problem, leading to Fibonacci, Catalan, Motzkin, Delannoy and Schröder numbers, among many other well-studied sequences and arrays. Our main results give relationships between finiteness properties of the numbers $\pi _X(A)$, geometrical properties of the step set X, algebraic properties of the monoid $\mathscr {A}_X$, and combinatorial properties of a certain bi-labelled digraph naturally associated to X. There is an intriguing divergence between the cases of finite and infinite step sets, and some constructions rely on highly non-trivial properties of real numbers. We also consider the case of walks constrained to stay within a given region of the plane. Several examples are considered throughout to highlight the sometimes-subtle nature of the theoretical results.

Appendix (joint with Stewart Wilcox): Combination (V)

In Sect. 2.9 we used Theorem 2.24 to limit the ostensibly possible combinations of finiteness conditions and geometric properties of step sets to just ten. One combination was shown never to occur (cf. Proposition 2.36), and all but one of the other combinations have been exemplified by various step sets whose locations are listed in Table 1. The purpose of this appendix is to construct a step set with the final combination (see Example A.5). The construction relies crucially on the existence of very specific sequences of real numbers (see Proposition A.2), and we turn to these first.

In all that follows, we fix a positive irrational number $\xi$, and we denote by

$$\begin{aligned} M = \{ {a+b\xi } : {a,b\in \mathbb Z,\ a+b\xi \ge 0} \} \end{aligned}$$

the additive monoid consisting of all non-negative $\mathbb Z$-linear combinations of 1 and $\xi$. Note that a or b might be negative in $a+b\xi \in M$, but we require $a+b\xi$ itself to be non-negative. So M is a submonoid of $\mathbb R_{\ge 0}$, and is dense in $\mathbb R_{\ge 0}$; see the claim in the proof of Lemma 2.15. Since 1 and $\xi$ are linearly independent over $\mathbb Q$, there is a well defined (and surjective) monoid homomorphism

$$\begin{aligned} \phi :M\rightarrow \mathbb Z\qquad \text {given by}\qquad \phi (a+b\xi )=b. \end{aligned}$$

For $a,b\in \mathbb R$, we write [a, b] and (a, b) for the closed and open intervals of all $x\in \mathbb R$ satisfying $a\le x\le b$ or $a<x<b$, respectively; we also write [a, b) and (a, b] for the half-open intervals, with the obvious meanings. If $\Sigma \subseteq \mathbb R$, we will also write $[a,b]_\Sigma =[a,b]\cap \Sigma$, with similar notation for other kinds of intervals; for example, if $a,b\in \mathbb Z$ and $a\le b$, then $[a,b]_\mathbb Z=\{a,a+1,\ldots ,b\}$. If x is a real number, we will write $(\!(x)\!) = x-\lfloor x\rfloor$ for the fractional part of x.

Lemma A.1

There is a mapping $\mathbb P\rightarrow \mathbb P:k\mapsto p_k$ such that

$$\begin{aligned} \phi ^{-1}\big ( [p,p+p_k]_\mathbb Z\big ) \cap (\alpha ,\alpha +\tfrac{1}{k})\not =\emptyset \qquad \text {for all}~\, p\in \mathbb Z\quad \text {and}~\, \alpha \in \mathbb R_{\ge 0}. \end{aligned}$$

Proof

Fix some $k\in \mathbb P$. By the claim in the proof of Lemma 2.15, there exists $l\in \mathbb P$ and $a\in \mathbb Z$ such that $0<l\xi -a<\frac{1}{k}$. Let $p_k\in \mathbb P$ be arbitrary so that

$$\begin{aligned} p_k > l(1+\tfrac{1}{l\xi -a}). \end{aligned}$$

Now suppose we are given $p\in \mathbb Z$ and $\alpha \in \mathbb R_{\ge 0}$. Define

$$\begin{aligned} t = 1 + \left\lfloor \frac{ (\!(\alpha -p\xi )\!)}{l\xi -a}\right\rfloor . \end{aligned}$$

Then

$$\begin{aligned} 1 \le t \le 1 + \frac{ (\!(\alpha -p\xi )\!)}{l\xi -a}< 1 + \frac{1}{l\xi -a} < \frac{p_k}{l}. \end{aligned}$$

(A.1)

We also claim that

$$\begin{aligned} 0< t(l\xi -a) - (\!(\alpha -p\xi )\!) < \frac{1}{k}. \end{aligned}$$

(A.2)

Indeed, for the inequality $0 < t(l\xi -a) - (\!(\alpha -p\xi )\!)$, note that if we write $\beta =l\xi -a$ and $\gamma =(\!(\alpha -p\xi )\!)$, then we have

$$\begin{aligned} t\beta - \gamma = (1+\lfloor \tfrac{\gamma }{\beta }\rfloor )\beta - \gamma = \beta \big (1 - (\tfrac{\gamma }{\beta }-\lfloor \tfrac{\gamma }{\beta }\rfloor )\big ) = \beta \big (1-(\!(\tfrac{\gamma }{\beta })\!)\big )>\beta (1-1)=0, \end{aligned}$$

while for the inequality $t(l\xi -a) - (\!(\alpha -p\xi )\!) \le l\xi -a$, we continue from above to obtain

$$\begin{aligned} t\beta - \gamma = \beta \big (1-(\!(\tfrac{\gamma }{\beta })\!)\big ) \le \beta = l\xi -a < \frac{1}{k}. \end{aligned}$$

Now that we have established (A.2), adding $\alpha$ throughout gives

$$\begin{aligned} \alpha< t(l\xi -a) +\alpha - (\!(\alpha -p\xi )\!) < \alpha +\frac{1}{k}. \end{aligned}$$

Since $\alpha -(\!(\alpha -p\xi )\!)=\alpha -(\alpha -p\xi )+\lfloor \alpha -p\xi \rfloor =p\xi +\lfloor \alpha -p\xi \rfloor$, it follows that

$$\begin{aligned} \alpha< t(l\xi -a) + p\xi + \lfloor \alpha -p\xi \rfloor < \alpha +\tfrac{1}{k}. \end{aligned}$$

That is,

$$\begin{aligned} \alpha< b + c\xi < \alpha +\tfrac{1}{k} \qquad \text {where}\qquad b=\lfloor \alpha -p\xi \rfloor -ta \quad \text {and}\quad c=tl+p. \end{aligned}$$

So $b+c\xi \in (\alpha ,\alpha +\frac{1}{k})$, and also $b+c\xi \in \phi ^{-1}\big ([p,p+p_k]_\mathbb Z\big )$ since $\phi (b+c\xi )=c=tl+p$ clearly satisfies $p\le c$, while $c\le p+p_k$ follows from $t<\frac{p_k}{l}$ which is itself part of (A.1).$\square$

In what follows, we fix the mapping $\mathbb P\rightarrow \mathbb P:k\mapsto p_k$ from Lemma A.1. In fact, by suitably increasing each $p_k$ if necessary, we may assume that $p_1<p_2<\cdots$.

In what follows, for any subset $\Sigma$ of $\mathbb R$, we write $S_n(\Sigma )=\{ {\sigma _1+\cdots +\sigma _n} : {\sigma _1,\ldots ,\sigma _n\in \Sigma } \}$ for the set of all sums of n elements of $\Sigma$. Clearly if $\Sigma$ is finite, then $|S_n(\Sigma )|\le |\Sigma |^n$.

For each $l\in \mathbb P$, we define

$$\begin{aligned} B(l) = \phi ^{-1}\big ((-l,l)_\mathbb Z\big )\cap [0,l)\subseteq M \qquad \text {and}\qquad n_l = l+l^3\in \mathbb P. \end{aligned}$$

Note that the “[0, l)” in the definition of B(l) is not “$[0,l)_\mathbb Z$”; in particular, B(l) contains non-integers. We clearly have $B(1)\subseteq B(2)\subseteq \cdots$, and we also have $M=\bigcup _{l\in \mathbb P}B(l)$. Indeed, for the latter, if $\alpha \in M$, then $\alpha \in B(l)$ for any l greater than both $\alpha$ and $|\phi (\alpha )|$. We aim to prove the following:

Proposition A.2

There exist sequences $\alpha _i,\beta _i,\gamma _i$ $(i\in \mathbb P)$ of elements of M satisfying:

• (i) $\lim _{i\rightarrow \infty }\alpha _i=0$,

• (ii) $\lim _{i\rightarrow \infty }\gamma _i=1$,

• (iii) $\gamma _i>1 \text { for all}\,\, i\in \mathbb P$,

• (iv) $\beta _i+\gamma _i=4 \text { for all}\,\, i\in \mathbb P$, and

• (v) $S_n(\Sigma )\cap B(l)=\emptyset$ for all $l\in \mathbb P$ and $n>n_l$, where $\Sigma =\{ {\alpha _i,\beta _i,\gamma _i} : {i\in \mathbb P} \}$.

To prove the proposition, we will construct the $\alpha _i$ series shortly, and after that the $\beta _i,\gamma _i$ series inductively. We will write $\mathbf{A}=\{ {\alpha _i} : {i\in \mathbb P} \}$ and $\mathbf{A}_k=\{ {\alpha _i} : {i\in \{1,\ldots ,k\}} \}$ for each $k\in \mathbb P$, and similarly define the sets $\mathbf{B}$, $\mathbf{C}$, $\mathbf{B}_k$ and $\mathbf{C}_k$. (Of course these sets are only well-defined once their elements have been specified.)

For $k\in \mathbb P$, define

$$\begin{aligned} R_k = (2k+p_k+1) \left( 1+ \sum _{n=0}^{n_k} n_k(3k)^n\right) \in \mathbb P, \end{aligned}$$

noting that $R_1<R_2<\cdots$. For each $k\in \mathbb P$, let $\alpha _k\in M\cap (\frac{1}{k},\frac{2}{k})$ be such that $\phi (\alpha _k)>k(1+R_k)$; such an element $\alpha _k$ exists by Lemma A.1.

We will now inductively construct $\beta _k,\gamma _k$ ($k\in \mathbb P$) satisfying $\beta _k+\gamma _k=4$, $\gamma _k\in (1,1+\frac{1}{k}]$ and

$$\begin{aligned} S_n(\mathbf{A}\cup \mathbf{B}_k\cup \mathbf{C}_k)\cap B(l)=\emptyset \qquad \text {for all }l\in \mathbb P\text { and }n>n_l. \end{aligned}$$

For the base of the induction, we set $\beta _1=\gamma _1=2$. We must show the following:

Lemma A.3

With the above notation, we have $S_n(\mathbf{A}\cup \{2\})\cap B(l)=\emptyset$ for all $l\in \mathbb P$ and $n>n_l$.

Proof

Suppose to the contrary that there exists $\varepsilon \in S_n(\mathbf{A}\cup \{2\})\cap B(l)$ for some $l\in \mathbb P$ and $n>n_l$. Then there exist integers $c_i,d\in \mathbb N$ ($i\in \mathbb P$) such that

$$\begin{aligned} \varepsilon = \sum _{i\in \mathbb P}c_i\alpha _i+2d \qquad \text {and}\qquad \sum _{i\in \mathbb P} c_i + d = n . \end{aligned}$$

In particular, recalling the definition of B(l), we have $l>\varepsilon>c_i\alpha _i>\frac{c_i}{i}$ for each $i\in \mathbb P$, so that $c_i<il$ for each i. Similarly $l>2d\ge d$. Again recalling the definition of B(l), we also have

$$\begin{aligned} \sum _{i\in \mathbb P}c_i\phi (\alpha _i) = \phi (\varepsilon )<l. \end{aligned}$$

But $\phi (\alpha _i)>i(1+R_i)$ for all i, so it follows that $\phi (\alpha _i)\ge 0$ for all $i\in \mathbb P$, and that $\phi (\alpha _i)>l$ for $i>l$. This gives $c_i=0$ for all $i>l$. Putting all of the above together, we have

$$\begin{aligned} l+l^3&= n_l< n = \sum _{i\le l} c_i + d <(l+2l+\cdots +l^2)+l=\frac{l^2(l+1)}{2}+l \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \le \frac{l^2(l+l)}{2}+l = l^3+l, \end{aligned}$$

a contradiction.$\square$

Now suppose $k>1$, and that we have defined the sequences $\beta _i,\gamma _i$ as desired for all $i<k$. Let $K>k$ be such that

$$\begin{aligned} R_K > |\phi (\beta _i)|, |\phi (\gamma _i)| \qquad \text {for all }i<k. \end{aligned}$$

Define the sets

$$\begin{aligned} \Omega = \bigcup _{n=0}^{n_K}\bigcup _{t=1}^{n_K} \frac{\phi \big ( S_n(\mathbf{A}_K\cup \mathbf{B}_{k-1}\cup \mathbf{C}_{k-1})\big )}{t} \qquad \text {and}\qquad \Gamma = \Omega \cup (-\Omega ). \end{aligned}$$

Note that

$$\begin{aligned} |\Gamma | \le 2|\Omega | \le 2 \sum _{n=0}^{n_K}\sum _{t=1}^{n_K} \big |S_n(\mathbf{A}_K\cup \mathbf{B}_{k-1}\cup \mathbf{C}_{k-1})\big | \le 2\sum _{n=0}^{n_K}n_K(3K)^n. \end{aligned}$$

It quickly follows that $\big (|\Gamma |+1\big )(2K+p_k+1) < 2R_K$, and so there exists an integer $p\in \mathbb Z$ such that

$$\begin{aligned}{}[p,p+2K+p_k]_\mathbb Z\subseteq (-R_K,R_K)_\mathbb Z\setminus \Gamma . \end{aligned}$$

(A.3)

By Lemma A.1, we may fix some

$$\begin{aligned} \gamma _k \in \phi ^{-1}\big ( [p+K,p+K+p_k]_\mathbb Z\big ) \cap (1,1+\tfrac{1}{k}) \qquad \text {and we also put}\qquad \beta _k=4-\gamma _k. \end{aligned}$$

Since $\phi (\gamma _k)\in [p+K,p+K+p_k]_\mathbb Z\subseteq [p,p+2K+p_k]_\mathbb Z\subseteq (-R_K,R_K)_\mathbb Z$, we have $|\phi (\gamma _k)|<R_K;$ since $\phi (\beta _k)=-\phi (\gamma _k)$, it follows that $|\phi (\beta _k)|<R_K$ as well. We also claim that

$$\begin{aligned} |\phi (\gamma _k)\pm \omega | > K \qquad \text {for all } \omega \in \Omega . \end{aligned}$$

(A.4)

Indeed, we have $\phi (\gamma _k)\in [p+K,p+K+p_k]_\mathbb Z$, so the set of all integers of distance at most K from $\phi (\gamma _k)$ is contained in $[p,p+2K+p_k]_\mathbb Z$, and by (A.3) the latter interval is disjoint from $\Gamma$. Thus, for any $\omega \in \Omega$, since ${\mp }\omega \in \Gamma$, it follows that the distance from $\phi (\gamma _k)$ to ${\mp }\omega$ is greater than K: i.e., $|\phi (\gamma _k)-({\mp }\omega )|>K$, completing the proof of (A.4).

Lemma A.4

With the above notation, we have $S_n(\mathbf{A}\cup \mathbf{B}_k\cup \mathbf{C}_k)\cap B(l)=\emptyset$ for all $l\in \mathbb P$ and $n>n_l$.

Proof

Suppose to the contrary that there exists $\varepsilon \in S_n(\mathbf{A}\cup \mathbf{B}_k\cup \mathbf{C}_k)\cap B(l)$ for some $l\in \mathbb P$ and $n>n_l$. Then there exist integers $c_i,d_i,e_i\in \mathbb N$ such that

$$\begin{aligned} \varepsilon = \sum _{i\in \mathbb P}c_i\alpha _i + \sum _{i=1}^k(d_i\beta _i+e_i\gamma _i)\in B(l) \qquad \text {and}\qquad \sum _{i\in \mathbb P}c_i + \sum _{i=1}^k(d_i+e_i) = n. \end{aligned}$$

Since $\beta _k+\gamma _k=\beta _1+\gamma _1$, we may assume without loss of generality that $d_k=0$ or $e_k=0$. But we note that $d_k$ and $e_k$ cannot both be zero, or else then $\varepsilon \in S_n(\mathbf{A}\cup \mathbf{B}_{k-1}\cup \mathbf{C}_{k-1})\cap B(l)$, contradicting the assumption that $\beta _i,\gamma _i$ ($i=1,\ldots ,k-1$) have the desired properties. As in the proof of Lemma A.3, we have $c_i<il$ for all $i\in \mathbb P$.

Case 1. Suppose first that $d_k=0$, so that $e_k>0$ as just noted. Also, since each $\beta _i,\gamma _i>1$ and each $\alpha _i>0$, and since $\varepsilon \in B(l)$, we have $\sum _{i=1}^k(d_i+e_i)<\sum _{i=1}^k(d_i\beta _i+e_i\gamma _i)\le \varepsilon <l$. Next note that

$$\begin{aligned}&l > \phi (\varepsilon ) = \sum _{i\in \mathbb P}c_i\phi (\alpha _i) + \sum _{i=1}^k\big (d_i\phi (\beta _i)+e_i\phi (\gamma _i)\big ) \\&\qquad \qquad \ge \sum _{i\in \mathbb P}c_i\phi (\alpha _i) - \sum _{i=1}^k\big (d_i|\phi (\beta _i)|+e_i|\phi (\gamma _i)|\big ), \end{aligned}$$

from which it follows that

$$\begin{aligned}&\sum _{i\in \mathbb P}c_i\phi (\alpha _i)< l + \sum _{i=1}^k\big (d_i|\phi (\beta _i)|+e_i|\phi (\gamma _i)|\big ) \nonumber \\&\qquad \qquad \quad< l + \sum _{i=1}^k(d_iR_K+e_iR_K) = l + R_K\sum _{i=1}^k(d_i+e_i) < l(1+R_K). \end{aligned}$$

(A.5)

We now consider two subcases.

Case 1.1. Suppose $l\ge K$. Then (A.5) gives

$$\begin{aligned} l(1+R_K) > \sum _{i\ge K}c_i\phi (\alpha _i) \ge \sum _{i\ge K}c_ii(1+R_i) \ge \sum _{i\ge K}c_i(1+R_K) \ \ \Rightarrow \ \ \sum _{i\ge K}c_i<l. \end{aligned}$$

From this it follows that

$$\begin{aligned} l+l^3 = n_l&< n = \sum _{i<K}c_i + \sum _{i\ge K}c_i + \sum _{i=1}^k(d_i+e_i) \\&< (l+2l+\cdots +(K-1)l)+l+l = l \frac{K(K-1)}{2} + 2l \le \frac{l^3}{2}+2l. \end{aligned}$$

But $l+l^3<\frac{l^3}{2}+2l$ implies $l^2<2$, a contradiction since $l\ge K>1$.

Case 1.2. Now suppose $l<K$. For $i\ge K$ we have $\phi (\alpha _i)>i(1+R_i)\ge K(1+R_K)$. Together with (A.5), it follows that for any such i,

$$\begin{aligned}&c_iK(1+R_K) \le c_i\phi (\alpha _i)<l(1+R_K) <K(1+R_K) \end{aligned}$$

so that $c_i=0$ for all $i\ge K$.

Setting $t=e_k\ge 1$, we have

$$\begin{aligned} \varepsilon -t\gamma _k = \sum _{i<K}c_i\alpha _i + \sum _{i<k}(d_i\beta _i+e_i\gamma _i) \in S_{n-t}(\mathbf{A}_K\cup \mathbf{B}_{k-1}\cup \mathbf{C}_{k-1}). \end{aligned}$$

But also

$$\begin{aligned} n-t< n&= \sum _{i<K}c_i + \sum _{i=1}^k(d_i+e_i)< (l+2l+\cdots +(K-1)l)+l \\&\qquad \qquad \qquad \qquad \qquad = l\frac{K(K-1)}{2}+l<\frac{K^3}{2}+K<n_K, \end{aligned}$$

and $t=e_k\le \sum _{i\in \mathbb P}c_i + \sum _{i=1}^k(d_i+e_i) = n<n_K$. So it follows that $\phi (\varepsilon )-t\phi (\gamma _k)\in t\Omega$, say $\phi (\varepsilon )-t\phi (\gamma _k)=t\omega$. Then by (A.4),

$$\begin{aligned} |\phi (\varepsilon )| = t|\phi (\gamma _k)+\omega | >tK \ge K. \end{aligned}$$

But also from $\varepsilon \in B(l)$, we have $|\phi (\varepsilon )|<l<K$, so we have arrived at a contradiction again.

Case 2. The case in which $e_k=0$ and $d_k>0$ is almost identical, since $\phi (\beta _k)=-\phi (\gamma _k)$. $\square$

We are now ready to tie together the loose ends.

Proof of Proposition A.2

With respect to the sequences $\alpha _i,\beta _i,\gamma _i$ ($i\in \mathbb P$) constructed above, conditions (i)–(iv) are immediate, while (v) follows from the fact that

$$\begin{aligned}&S_n(\Sigma )\cap B(l) = S_n(\mathbf{A}\cup \mathbf{B}\cup \mathbf{C})\cap B(l) = \bigcup _{k\in \mathbb P}\big (S_n(\mathbf{A}\cup \mathbf{B}_k\cup \mathbf{C}_k)\cap B(l)\big )\\ \end{aligned}$$

for all $n,l\in \mathbb P$. $\square$

We now use Proposition A.2 to construct a step set $X\subseteq \mathbb Z_\times ^2$ with combination (V); cf. Table 1.

Example A.5

In all that follows, we keep the notation above: in particular, the irrational number $\xi >0$, the monoid $M=\{ {a+b\xi } : {a,b\in \mathbb Z,\ a+b\xi \ge 0} \}$ and the sequences $\alpha _i,\beta _i,\gamma _i$ $(i\in \mathbb P)$. Also let

$$\begin{aligned} N = \big \{ {(a,b)\in \mathbb Z^2}: {a+b\xi \ge 0} \big \} \end{aligned}$$

be the additive submonoid of $\mathbb Z^2$ consisting of all lattice points on or above the line $\mathscr {L}$ with equation ${x+\xi y=0}$. The map

$$\begin{aligned} \psi :N\rightarrow M:(a,b)\mapsto a+b\xi \end{aligned}$$

is clearly a surjective monoid homomorphism. In fact, $\psi$ is an isomorphism, as injectivity follows quickly from the irrationality of $\xi$. For each $i\in \mathbb P$, let

$$\begin{aligned} A_i = \psi ^{-1}(\alpha _i) ,\qquad B_i = \psi ^{-1}(\beta _i) ,\qquad C_i = \psi ^{-1}(\gamma _i) , \end{aligned}$$

and put $X=\{ {A_i,B_i,C_i} : {i\in \mathbb P} \}$. Also let $E=(1,0)=\psi ^{-1}(1)$. We claim that:

• (i) X does not satisfy the SLC,

• (ii) X satisfies the LC,

• (iii) X does not have the FPP,

• (iv) X has the BPP.

First note that (ii) is clear, as $\mathscr {L}$ itself witnesses the LC (as $\xi$ is irrational, the only lattice point on $\mathscr {L}$ is O). Item (iii) follows quickly from the fact that $\beta _i+\gamma _i=4$ for all $i\in \mathbb P$; indeed, since $\psi$ is an isomorphism, this implies that $B_i+C_i=\psi ^{-1}(4)=(4,0)=4E$ for all i, and hence $\pi _X(4E)=\infty$.

Fig. 18

A subset of the step set from Example A.5 (not to scale). Some X-walks of the form $B_jC_j$ from $\Pi _X(4E)$ are shown in red (color figure online)

To establish the remaining items, first define $\eta =\sqrt{1+\xi ^2}$. For $A=(u,v)\in N$ write $\delta (A)$ for the (perpendicular) distance from A to $\mathscr {L}$. Then by (2.7), and since $u+v\xi \ge 0$ as $A\in N$, we have

$$\begin{aligned} \delta (A) = \frac{u+v\xi }{\sqrt{1+\xi ^2}} = \frac{\psi (A)}{\eta }. \end{aligned}$$

Next, let $\mathscr {L}'$ and $\mathscr {L}''$ be the lines obtained, respectively, by sliding $\mathscr {L}$ a distance of $\frac{1}{\eta }$ or $\frac{3}{\eta }$ units into the half-plane on the side of $\mathscr {L}$ containing X (or, equivalently, by sliding $\mathscr {L}$ by 1 or 3 units to the right, since $\delta (E)=\frac{1}{\eta }$). This is all shown in Fig. 18. Now,

$$\begin{aligned} \lim _{i\rightarrow \infty }\delta (A_i) = \lim _{i\rightarrow \infty }\frac{\alpha _i}{\eta }=0. \end{aligned}$$

This shows that X contains points arbitrarily close to $\mathscr {L}$, and consequently that:

• (v) no line parallel to $\mathscr {L}$ witnesses the SLC.

We also have

$$\begin{aligned} \lim _{i\rightarrow \infty }\delta (C_i) = \lim _{i\rightarrow \infty }\frac{\gamma _i}{\eta }=\frac{1}{\eta }\qquad \text {and}\qquad \lim _{i\rightarrow \infty }\delta (B_i) = \lim _{i\rightarrow \infty }\frac{\beta _i}{\eta }= \lim _{i\rightarrow \infty }\frac{4-\gamma _i}{\eta }= \frac{3}{\eta }. \end{aligned}$$

This means that the points $C_1,C_2,\ldots$ approach $\mathscr {L}'$ from the right, while $B_1,B_2,\ldots$ approach $\mathscr {L}''$ from the left. Since the points $C_1,C_2,\ldots$ are between the lines $\mathscr {L}'$ and $\mathscr {L}''$, and since a bounded region of $\mathbb R^2$ contains only finitely many lattice points, the y-coordinates of $C_1,C_2,\ldots$ are unbounded, either above or below or both; it follows (since $B_i=4E-C_i$ for all i) that the y-coordinates of $B_1,B_2,\ldots$ are unbounded below or above or both, respectively. Thus, X contains points between $\mathscr {L}'$ and $\mathscr {L}''$ with arbitrarily large positive and negative y-coordinates, and it quickly follows that:

• (vi) $\mathscr {L}$ is the only line through O that witnesses the LC.

Items (v) and (vi), together with Lemma 2.9(ii), show that X does not satisfy the SLC. Figure 18 depicts all the above, but only showing subsequences $A_i$ ($i\in I$), and $B_j,C_j$ ($j\in J$) with monotone y-coordinates (with the y-coordinates of $A_i,C_j$ increasing, and those of $B_j$ decreasing).

Finally, the BPP follows quickly from the properties of the $\alpha _i,\beta _i,\gamma _i$ sequences. Indeed, let $D\in \mathscr {A}_X$ be arbitrary, and fix some $w\in \Pi _X(D)$. Write $w=F_1\cdots F_k$, where $F_1,\ldots ,F_k\in X$, so that ${D=F_1+\cdots +F_k}$. Now consider the real number ${\psi (D)\in M}$, and let $l\in \mathbb P$ be such that $\psi (D)\in B(l)$; the set $B(l)\subseteq M$ was defined just before Proposition A.2. Now, $\psi (D)=\psi (F_1)+\cdots +\psi (F_k)$, and $\psi (F_1),\ldots ,\psi (F_k)$ all belong to $\Sigma =\{ {\alpha _i,\beta _i,\gamma _i} : {i\in \mathbb P} \}$. This means that $\psi (D)\in S_k(\Sigma )\cap B(l)$, and so Proposition A.2 gives $\ell (w)=k\le n_l=l+l^3$. This shows that the set $\{ {\ell (w)} : {w\in \Pi _X(D)} \}$ is contained in $\{1,\ldots ,n_l\}$, and hence is finite. Since $D\in \mathscr {A}_X$ was arbitrary, the BPP has been established.