The baker's map with a convex hole

TL;DR

This paper investigates the properties of convex holes in the baker's map, identifying conditions under which the survivor set has zero or positive Hausdorff dimension, and extending previous one-dimensional results to higher dimensions.

Contribution

It characterizes dimension and cycle traps for convex holes in the baker's map and extends prior work from one-dimensional cases to higher dimensions.

Findings

Interior convex holes are not dimension traps.

Positive Hausdorff dimension can occur for large measure holes.

Critical convex holes mark the boundary of trap properties.

Abstract

We consider the baker's map $B$ on the unit square $X$ and an open convex set $H\subset X$ which we regard as a hole. The survivor set $\mathcal J(H)$ is defined as the set of all points in $X$ whose $B$-trajectories are disjoint from $H$. The main purpose of this paper is to study holes $H$ for which $\dim_H \mathcal J(H)=0$ (dimension traps) as well as those for which any periodic trajectory of $B$ intersects $\overline H$ (cycle traps). We show that any $H$ which lies in the interior of $X$ is not a dimension trap. This means that, unlike the doubling map and other one-dimensional examples, we can have $\dim_H \mathcal J(H)>0$ for $H$ whose Lebesgue measure is arbitrarily close to one. Also, we describe holes which are dimension or cycle traps, critical in the sense that if we consider a strictly convex subset, then the corresponding property in question no longer holds. We also determine $\delta>0$ such that $\dim_H \mathcal J(H)>0$ for all convex $H$ whose Lebesgue measure is less than $\delta$. This paper may be seen as a first extension of our work begun in [3, 4, 6, 7, 13] to higher dimensions.

Tables

Table 1. Table 6.1. Upper and lower bounds on restricted subsets of 𝒯 ′ superscript 𝒯 ′ \mathcal{T}^{\prime}

Contains points	Symmetry	length	# of	Lower	Upper	Bound	$\epsilon$
		of words	polygon	bound	Bound	to search
			found
$(0,1/2),(1/2,1)$	mirror	8	2	$0.13793$	$0.13803$	$0.1381$	${10}^{-10}$
$(0,1/2),(1/2,1)$	none	8	6	$0.13532$	$0.13542$	$0.1355$	${10}^{-10}$
$(1/2,0),(1/2,1)$	rotational	8	2	$0.11891$	$0.12911$	$0.119$	${10}^{-10}$
$(1/2,0),(1/2,1)$	none	8	1253	$0.11802$	$0.12911$	$0.1182$	${10}^{-10}$

Equations

$$\mathcal{J}(H):=\{x:T^n(x)\notin H\text{for all}n\in \mathbb{Z}\}$$

$$\mathcal{J}(H):=\{x:T^n(x)\notin H\text{for all}n\ge 0\}$$

$$s_n$$

$$t_n$$

$$\mathfrak{s}(\mathbf{r})$$

$$\mathfrak{t}(\mathbf{r})$$

$$b-a<\frac{1}{4}\cdot \prod _{n=1}^{\infty }\left(1-2^{-2^n}\right)\approx 0.175092,$$

$$B(x,y)=\{\begin{array}{ll}(2x,\frac{y}{2})& \text{ if }0\le x<\frac{1}{2},\\ (2x-1,\frac{y+1}{2})& \text{ if }\frac{1}{2}\le x<1.\end{array}$$

$$R_0$$

$$R_1$$

$$x_n=k⟺B^{-n}(x,y)\in R_k.$$

$$\pi ((x_n{)}_{-\infty }^{\infty })=\left(\sum _{n=1}^{\infty }{x}_n{2}^{-n},\sum _{n=0}^{\infty }{x}_{-n}{2}^{-n-1}\right).$$

$$B(x,y)$$

$$B^{-1}(x,y)$$

$$\sigma (\dots y_3{y}_2{y}_1\cdot x_1{x}_2{x}_3\dots )$$

$${\sigma }^{-1}(\dots y_3{y}_2{y}_1\cdot x_1{x}_2{x}_3\dots )$$

$$\mathcal{J}(H):=\{(x,y):B^n(x,y)\notin H\text{for all}n\in \mathbb{Z}\}.$$

$$\{\dots 0101\cdot 0101\dots ,\dots 1010\cdot 1010\dots \}.$$

$${\dim }_{\mathtt{H}}(\pi (Y))=2h(Y).$$

$$2^{-2k(h-\epsilon )}\le \mu [x_{-k+1}\dots x_0\cdot x_1\dots x_k]\le 2^{-2k(h+\epsilon )},k\ge n(x).$$

$$2^{2k(h-\epsilon )}\cdot 2^{-ks}\le {\mathcal{H}}^s(\pi (S))\le 2^{2k(h+\epsilon )}\cdot 2^{-ks}.$$

$$A_n:=\left\{(u_k{)}_{k=-\infty }^{\infty }:(u_k)\text{ consists of concatenations of}10^n\text{and}10^{n+1}\right\}.$$

$$\tilde{\Omega }=$$

$$\{\dots 1^{b_0}01^{b_0}0\cdot 1^{b_0}01^{b_1}01^{b_2}\dots \mid 1\le b_i\le b_{i+1}\}\cup$$

$$\{\dots 0^{a_2}10^{a_1}10^{a_0}\cdot 10^{a_0}10^{a_0}\dots \mid 1\le a_i\le a_{i+1}\}.$$

$$\Omega ={\cup }_{k\in \mathbb{Z}}{\sigma }^k(\tilde{\Omega }).$$

$$\dots 0^{a_2}10^{a_1}10^{a_0}{1}^{b_0}01^{b_1}01^{b_2}\dots ,$$

$$\dots 1^{b_0}01^{b_0}01^{b_0}01^{b_0}01^{b_1}01^{b_2}\dots ,\mathrm{or}$$

$$\dots 0^{a_2}10^{a_1}10^{a_0}10^{a_0}10^{a_0}10^{a_0}\dots ,$$

$$y=0.10^{a_k}1\dots \le 0.10^{a_{k-1}}1\dots =\frac{1+x}{2}$$

$$y=0.1^{b_k}0\dots \le 1^{b_{k+1}}0\dots =2x.$$

$$\dots 10^{a_n}1$$

$$\dots 01^{b_{m-1}}$$

$$0^{s}10^{a_{\ell }}10^{a_{\ell -1}}1\dots 10^{a_0}{1}^{b_0}01^{b_1}0\dots 01^{b_m}01^t,$$

$$k$$

Full text

We show that any $H$ which lies in the interior of $X$ is not a dimension trap. This means that, unlike the doubling map and other one-dimensional examples, we can have $\mathrm{dim}_{\tt H}\mathcal{J}(H)>0$ for $H$ whose Lebesgue measure is arbitrarily close to one. Also, we describe holes which are dimension or cycle traps, critical in the sense that if we consider a strictly convex subset, then the corresponding property in question no longer holds.

We also determine $\delta>0$ such that $\mathrm{dim}_{\tt H}\mathcal{J}(H)>0$ for all convex $H$ whose Lebesgue measure is less than $\delta$.

This paper may be seen as a first extension of our work begun in [5, 6, 10, 11, 20] to higher dimensions.

Research of K. G. Hare was supported by NSERC Grant RGPIN-2014-03154. Computational support was provided by the Canada Foundation for Innovation.

1. The doubling map with a hole: recap

Open dynamical systems, i.e. the induced maps on the set of points which do not fall into a certain predetermined set (a hole) under iteration by a map, have been studied in numerous papers. The first discussion was in [16] where they considered an idealized game of billiards without friction, where there is a hole somewhere in the table. Depending upon the location or size of the hole, the rate of which random billiard balls will “escape” was considered. In [4] it was shown that not only does the size of the hole matter, but that the location of the hole can have huge impact upon the escape rate. The escape rate in this case can be well approximated by the set of periodic points of the map that lie within the hole, [3]. For a good history of these problems, see [8, 7]. These ideas have also be considered in theoretical computer science, in the context of combinatorics on words, in the study of unavoidable words, as first discussed in [18].

The set-up is as follows: let $T:X\to X$ be a map and $H$ be an open set. We define the survivor set for $H$ as follows:

[TABLE]

if $T$ is invertible and

[TABLE]

otherwise. If $T$ is invertible, or we wish to emphasis that we are only considering the forward orbits, we will use the notation $\mathcal{J}^{+}(H)$. In [10] P. Glendinning and the third author investigated the case $X=[0,1]$ and the doubling map $Tx=2x\bmod 1$, with $H=(a,b)$ with $0<a<b<1$. Here is a brief summary of the main results of that paper.

We need some definitions and basic results from combinatorics on words, which we borrow from [10]. See also see [13, Chapter 2] for a detailed exposition. For any two finite 0-1 words $u=u_{1}\dots u_{k}$ and $v=v_{1}\dots v_{n}$ we write $uv$ for their concatenation $u_{1}\dots u_{k}v_{1}\dots v_{n}$. We write $u^{n}=\underbrace{uu\dots u}_{n}$ as the concatenation of $u$ with itself $n$ times, and $u^{\infty}$ as the infinite concatenation of $u$ with itself. Let $w$ be a finite or infinite word. We say that a finite word $u$ is a factor of $w$ if there exists $k$ such that $u=w_{k}\dots w_{k+n}$ for some $n\geq 0$. For a finite word $w$ let $|w|$ stand for its length and $|w|_{1}$ stand for the number of 1s in $w$. The 1-ratio of $w$ is defined as $|w|_{1}/|w|$. For an infinite word $w_{1}w_{2}\dots$ the 1-ratio is defined as $\lim_{n\to\infty}|w_{1}\dots w_{n}|_{1}/n$, if it exists.

We say that a finite or infinite word $w$ is balanced if for any $n\geq 1$ and any two factors $u,v$ of $w$ of length $n$ we have $||u|_{1}-|v|_{1}|\leq 1$. An infinite word is called Sturmian if it is balanced and not eventually periodic. A finite word $w$ is cyclically balanced if $w^{2}$ is balanced. (And therefore, $w^{\infty}$ is balanced.) It is well known that if $u$ and $v$ are two cyclically balanced words with $|u|=|v|=q$ and $|u|_{1}=|v|_{1}=p$ and $\gcd(p,q)=1$, then $u$ is a cyclic permutation of $v$. Thus, there are only $q$ distinct cyclically balanced words of length $q$ with $p$ 1s.

We say that a finite or infinite word $u$ is lexicographically smaller than a word $v$ (notation: $u\prec v$) if either $u_{1}<v_{1}$ or there exists an $n\geq 1$ such that $u_{i}=v_{i}$ for $i=1,\dots,n$ and $u_{n+1}<v_{n+1}$.

For any $r=p/q\in\mathbb{Q}\cap(0,1)$ we define the substitution $\rho_{r}$ on $\{0,1\}$ as follows: $\rho_{r}(0)=\text{0-max}(r)$, the lexicographically largest cyclically balanced word of length $q$ with 1-ratio $r$ beginning with 0, and $\rho_{r}(1)=\text{1-min}(r)$, the lexicographically smallest cyclically balanced word of length $q$ with 1-ratio $r$ beginning with 1. In particular, $\rho_{1/2}(0)=01,\ \rho_{1/2}(1)=10$.

Consider some $(a,b)\subset(0,1)$ such that $\mathrm{dim}_{\tt H}\mathcal{J}^{+}(a,b)=0$. We then begin to move $a$ and $b$ towards each other until we get a survivor set of positive dimension. In doing so, we pass through a number of milestones, where we gain extra periodic orbits for $T$ which lie outside the interval. Namely, let $\bm{r}=(r_{1},r_{2},\dots)$ be a finite or infinite vector with each component $r_{i}\in\mathbb{Q}\cap(0,1)$. We define the sequences of 0-1 words parameterized by $\bm{r}$ as follows:

[TABLE]

In particular, if $r_{i}=1/2$ for all $i\leq n$, we have that $s_{n}$ is a truncated Thue-Morse word and $t_{n}$ is its mirror image (see Section 4 for a precise definition of the Thue-Morse word). Put

[TABLE]

Theorem 1.1**.**

[10, Theorem 2.13]** We have $\mathrm{dim}_{\tt H}\mathcal{J}^{+}(\mathfrak{s}(\bm{r}),\mathfrak{t}(\bm{r}))=0$ and moreover, $\mathrm{dim}_{\tt H}\mathcal{J}^{+}(a,b)>0$ for any $a>\mathfrak{s}(\bm{r}),b<\mathfrak{t}(\bm{r})$.

In other words, any infinite sequence of rationals from $(0,1)$ induces a route to chaos; moreover, it was shown in [10, Lemma 2.12] that $\mathcal{J}^{+}(s_{n}^{\infty},t_{n}^{\infty})\setminus\mathcal{J}^{+}(s_{n-1}^{\infty},t_{n-1}^{\infty})$ contains just one cycle, namely, $s_{n}^{\infty}$ (which is the same as $t_{n}^{\infty}$).

The reason behind this approach is that we need $\mathcal{J}^{+}(H)$ to be uncountable (and possibly, of positive Hausdorff dimension) for the induced map $T|_{\mathcal{J}^{+}(H)}$ to be of interest.

As a corollary of several results from [10], we have the following result, which, roughly speaking, implies that if a connected hole $H$ for the doubling map is too small, then its survivor set has positive Hausdorff dimension, whereas if it is too large, then it intersects all orbits.

Theorem 1.2** ([10]).**

Let $(a,b)\subset[0,1]$.

1. (1)

We always have $\mathrm{dim}_{\tt H}\mathcal{J}^{+}(a,b)>0$ if

[TABLE]

and this bound is sharp. 2. (2)

If $b-a>\frac{1}{2}$, then $\mathcal{J}^{+}(a,b)=\{0,1\}$.

The purpose of the present paper is to extend this approach to the baker’s map. As we will see, it works perfectly in the symmetric case $r_{i}\equiv 1/2$ (Section 4) and fails for an asymmetric case (Section 5). Having said that, the baker’s map generates a great deal of completely new effects in comparison with the doubling map (see, e.g., Section 3).

2. Baker’s map: preliminaries

Put $X=[0,1]^{2}$. The baker’s map $B:X\to X$ is the natural extension of the doubling map, conjugate to the shift map on the set of bi-infinite sequences. Namely,

[TABLE]

Put

[TABLE]

We associate each point $(x,y)\in[0,1)\times[0,1)$ with a bi-infinite sequence $(x_{n})_{-\infty}^{\infty}\in\{0,1\}^{\mathbb{Z}}$ as follows:

[TABLE]

(We cannot say what happens when $x=\frac{1}{2}$.) The map $\pi$ conjugating the left shift $\sigma:\{0,1\}^{\mathbb{Z}}\to\{0,1\}^{\mathbb{Z}}$ and $B$ is given by the formula

[TABLE]

In other words, if we write $x$ and $y$ in base 2 as $x=0.x_{1}x_{2}x_{3}\ldots$ and $y=0.y_{1}y_{2}y_{3}\ldots$, then

[TABLE]

We will denote the associated bi-infinite sequence as $\ldots y_{3}y_{2}y_{1}\cdot x_{1}x_{2}x_{3}\ldots$ We thus have

[TABLE]

We will use these two notations interchangeably. Let $H\subset X$; following Section 1, we define the survivor set for $H$ as follows:

[TABLE]

Let now $[u_{-k}\ldots u_{0}\cdot u_{1}\ldots u_{m}]$ denote the set of all $(x_{n})_{-\infty}^{\infty}\in\{0,1\}^{\mathbb{Z}}$ such that $x_{i}=u_{i}$ for $-k\leq i\leq m$ (a cylinder). See Figure 1 for the case $m=k=1$.

Definition 2.1**.**

• •

We will say that an open set $H$ is a complete trap if $\mathcal{J}(H)$ does not contain any points except, possibly, of the form $\pi(\dots 11110000\dots)$ or $\pi(\dots 00001111\dots)$, both orbits lying on the boundary of $X$.

• •

We will say that an open set $H$ is a cycle trap if $\mathcal{J}(\overline{H})$ does not contain any cycles of $B$.

• •

We will say that an open set $H$ is a dimension trap if $\mathrm{dim}_{\tt H}\mathcal{J}(H)=0$.

It is clear that any complete trap is both a cycle trap and a dimension trap. In this paper we wish to study those $H$ such that $H$ is a cycle trap (resp. dimension trap), and $H$ is convex. The only exception will be considered in Appendix. As we see below, if we do not restrict our holes to convex sets, then it is possible to have arbitrarily small complete traps.

Theorem 2.2**.**

For all $\varepsilon>0$ exists a connected non-convex complete trap $H$ such that its Lebesgue measure $\mathcal{L}$ is less than $\varepsilon$.

Proof.

Let $A_{1}$ be the set $(1/2,1)\times(0,1/2)$. Clearly, the only sequences that avoid $A_{1}$ are of the form $\dots 11110000\dots$, which implies that $A_{1}$ is a complete trap.

Now we choose $n$ such that $A_{1}$ can be completely tiled by $1/2^{n}\times 1/2^{n}$ cylinders $C_{i}=[b^{i}_{n}b^{i}_{n-1}\dots b^{i}_{1}\cdot a^{i}_{1}a^{i}_{2}\dots a^{i}_{n}]$. We see that there are $2^{2n}\mathcal{L}(A_{1})$ cylinders $C_{i}$. By choosing $n$ large enough, we make sure that the number of such cylinders is even.

Now we divide this set of cylinders into two sets of equal size, say $D_{1}$ and $D_{2}$. Let $A_{2}:=B^{n}(D_{1})\cup B^{-n}(D_{2})$. We claim that $A_{2}$ is a connected complete trap, and that $\mathcal{L}(A_{2})=\mathcal{L}(A_{1})-\mathcal{L}(A_{1})^{2}/4$.

To see that this is a complete trap, we observe if some element in the orbit of $(x,y)$, say $(x^{\prime},y^{\prime})$ is in $A_{1}$, then we have that either $B^{-n}(x^{\prime},y^{\prime})\in D_{1}$ or $B^{n}(x^{\prime},y^{\prime})\in D_{2}$. Therefore, the orbit of $(x,y)$ falls into $A_{2}$.

We observe that all images in $B^{n}(D_{1})$ are rectangles with width $1$ and height $1/2^{2n}$. That is, they are horizontal strips going from $x=0$ to $x=1$. We further observe that all images in $B^{-n}(D_{2})$ are rectangles with height $1$ and width $1/2^{2n}$. That is, they are vertical strips going from $y=0$ to $y=1$. Clearly each horizontal strip overlaps each vertical strip, hence this is a connected set.

Moreover, each horizontal strip overlaps each vertical strip in a block of size $1/2^{2n}\times 1/2^{2n}$. There are $\#(D_{1})\cdot\#(D_{2})=\left(2^{2n-1}\mathcal{L}(A_{1})\right)\cdot\left(2^{2n-1}\mathcal{L}(A_{1})\right)=2^{4n-2}\mathcal{L}(A_{1})^{2}$ such overlaps. Thus, the size of $A_{2}$ is the size of the vertical strips (which is $\mathcal{L}(A_{1})/2$), plus the size of the horizontal strips (which is $\mathcal{L}(A_{1})/2$), minus the size of the overlap (which is $2^{4n-2}\mathcal{L}(A_{1})^{2}/2^{4n}=\mathcal{L}(A_{1})^{2}/4$). Thus, $\mathcal{L}(A_{2})=\mathcal{L}(A_{1})-\mathcal{L}(A_{1})^{2}/4$.

We can repeat this process to get $A_{k}$, with $\mathcal{L}(A_{k})=\mathcal{L}(A_{k-1})-\mathcal{L}(A_{k-1})^{2}/4$. The fact that $\mathcal{L}(A_{k})\to 0$ as $k\to\infty$ completes the proof. ∎

Remark 2.3*.*

The existence of arbitrarily small complete traps for the shift has been known since the seminal paper [18]. The novelty of our Theorem 2.2 is the fact that our hole is connected.

See Figure 2 for $A_{1}$ and a possible choice for $A_{2}$ and $A_{3}$ from Theorem 2.2.

Henceforward we assume $H$ to be convex. Our first easy observation is that $\left\{\left(\frac{1}{3},\frac{2}{3}),(\frac{2}{3},\frac{1}{3}\right)\right\}$ is the 2-cycle for the baker’s map. As bi-infinite sequences, these correspond to the cycle of $\sigma$ given by

[TABLE]

Hence if $H$ is a cycle trap, then $\overline{H}$ must contain at least one of these two points. The next result shows that if $H$ is a dimension trap, then it must contain a part of the boundary of $X$. This helps to motivate the shapes of $H$ that we investigate. We recall that $A\subset\{0,1\}^{\mathbb{Z}}$ is called a subshift if it is closed under the product topology, and invariant under the shift operator. We first prove the claim which relates the topological entropy of a subshift to the Hausdorff dimension of its image.

Lemma 2.4**.**

Let $Y$ be a subshift of $\{0,1\}^{\mathbb{Z}}$ and let $h(Y)$ be its topological entropy ($\log$ base $2$). Then

[TABLE]

In particular, $h(Y)=0\iff\mathrm{dim}_{\tt H}(\pi(Y))=0$.

Proof.

111This proof was given by Anthony Quas on MathOverFlow [17], for which the authors are grateful to him.

Let $\mu$ stand for a measure of maximal entropy for $Y$. Then by the Shannon-McMillan-Breiman theorem, for any $\varepsilon>0$, we have a set $Y^{\prime}\subset Y$ of full $\mu$-measure and all $x=(x_{n})_{n=-\infty}^{\infty}\in Y^{\prime}$,

[TABLE]

Let $N_{0}$ be large enough that $\mu(S)>1/2$, where $S=\{x\in Y^{\prime}:n(x)\leq N_{0}\}$.

The set $\pi([x_{-k+1}\ldots x_{0}\cdot x_{1}\ldots x_{k}])$ is a $2^{-k}\times 2^{-k}$ square, so its $s$-Hausdorff measure $\mathcal{H}^{s}$ is $2^{-sk}$. Hence

[TABLE]

Now the claim follows from the definition of the Hausdorff dimension. ∎

Theorem 2.5**.**

Suppose $H$ is an interior hole; that is to say $\overline{H}\subset(0,1)\times(0,1)$. Then $\mathrm{dim}_{\tt H}\mathcal{J}(H)>0$.

Proof.

Define $A_{n}$ as the Cantor set $\{10^{n},10^{n+1}\}^{*}$. That is

[TABLE]

We claim that for every interior hole $H$, there exists an $n$ such that $A_{n}$ avoids $H$. Since the topological entropy of $A_{n}$ is clearly positive for each $n$, Lemma 2.4 will then yield the desired result.

Since $H$ is an interior hole, there exists $\varepsilon>0$ such that if $(x,y)\in H$ then $x,y>\varepsilon$. Pick $n$ such that $0.0^{\lfloor n/2\rfloor}1(0^{n}1)^{\infty}<\varepsilon$. This implies that if $(x,y)\in A_{n}$ then for all $i\in\mathbb{Z}$ we have that $(x_{i},y_{i})=B^{i}(x,y)$ has either $x_{i}<\varepsilon$ or $y_{i}<\varepsilon$. Hence the orbit of $B^{i}(x,y)$ avoids $H$ as required, which proves the result. ∎

Remark 2.6*.*

This means that, in a startling contrast to the one-dimensional case (see Theorem 1.2), there are holes of size arbitrarily close to full measure which are avoided by a set of positive Hausdorff dimension and also arbitrarily small connected holes whose survivor set contains only the two orbits on the boundary of $X$.

Remark 2.7*.*

By considering the Cantor set $\{10^{n},10^{n+1}\}^{*}$ we see from the above proof that we must contain a point of the form $\{(0,1/2^{n}),(1/2^{n},0)\}$ for some $n\geq 1$. Similarly by considering the Cantor set $\{01^{n},01^{n+1}\}^{*}$ we see that we must contain a point of the form $\{(1,1-1/2^{n}),(1-1/2^{n},1)\}$ for some $n\geq 1$. We show in Section 6 that if a convex set $H$ is a dimension trap such that $H$ is closed under the symmetry $(x,y)\leftrightarrow(1-y,1-x)$ then either $\mathcal{L}(H)>0.1381$ or $H$ contains the points $(0,1/2)$ and $(1/2,1)$ or it contains the points $(1/2,0)$ and $(1,1/2)$. This motivates the restriction considered in Section 3.

Similarly if a convex set $H$ is a dimension trap such that $H$ is closed under the symmetry $(x,y)\leftrightarrow(1-x,1-y)$ then either $\mathcal{L}(H)>0.13$ or $H$ contains the points $(0,1/2)$ and $(1,1/2)$ or it contains the points $(1/2,0)$ and $(1/2,1)$. This motivates the restriction considered in Section 4.

In view of the symmetry with respect to $y=x$, we divide this paper into two main sections. In Section 3, we consider dimension traps that contain $(0,1/2)$ and $(1/2,1)$. We show that the open convex polygon $\Delta$ with corners $(0,1),(1/2,1),(1/3,2/3)$ and $(0,1/2)$ is both a cycle trap and a dimension trap. We also give two polygons contained in $\Delta$ such that they are dimension traps that are in some sense optimal.

In Section 4, we consider traps and dimension traps that contain $(1/2,1/2)$. We show that the open parallelogram $P$ with vertices $(1/3,2/3),(1/2,1),(2/3,1/3)$ and $(1/2,0)$ is a cycle trap and describe all cycles for which there is an element lying on $\partial P$.

We then proceed to construct a sequence of nested polygons, the first of which is $P$ where each of these polygons is a dimension trap. The limiting polygon is the hexagon with vertices $(1/2,0),({\bm{t}},2{\bm{t}}),({\bm{t}},2-4{\bm{t}}),(1/2,1),(1-{\bm{t}},2-4{\bm{t}}),(1-4{\bm{t}},2{\bm{t}})$ with ${\bm{t}}$ being the Thue-Morse constant.

We also consider an asymmetrical family of parallelograms with vertices $(1/2,0),(a,2a),\linebreak(1/2,1),(b,2b-1)$, where the $a$ and $b$ are parameterized by the rationals $r\in(0,1)$. For all of these we prove that they are neither cycle nor dimension traps – with the exception of $r=\frac{1}{2}$, which corresponds to $P$. This shows that, unlike the doubling map (see Section 1), there appears to be no natural asymmetric “route to chaos” for the baker’s map.

Finally, in Section 6 we find $\delta>0$ such that any hole whose measure is less than $\delta$ is not a dimension trap (Theorem 6.2).

3. Traps which contain $(0,1/2)$ and $(1/2,1)$.

3.1. Symmetric cycle trap that contains $(0,1)$.

Denote by $\Delta$ the open convex polygon with corners $(0,1),(1/2,1),(1/3,2/3)$ and $(0,1/2)$. The results about this trap also hold for its reflection in the line $y=x$ by swapping maxima and minima as needed. One of these two traps is depicted in Figure 3.

Let

[TABLE]

Here we allow $a_{i}=\infty$ or $b_{j}=\infty$ or both. Let

[TABLE]

Lemma 3.1**.**

The orbit of a point does not fall into $\Delta$ if and only if the point is in $\Omega$.

Proof.

We first observe that $[11\cdot 00]\subseteq\Delta$. This implies that if a point in $\mathcal{J}(\Delta)$ has a subsequence $1^{n}0$ with $n\geq 2$ then this subsequence must be of the form $1^{n}01^{m}$. Further more as $\ldots 1^{m+1}\cdot 01^{m}0\ldots$ falls into $\Delta$ we see that if $\ldots 01^{n}\cdot 01^{m}0\ldots\not\in\Delta$ then $n\leq m$. Similarly if we have a subsequence $10^{n}$ we must have a subsequence $0^{m}10^{n}$ with $n\leq m$. Thus the only points that avoid $\Delta$ are points of the form

[TABLE]

where $1\leq a_{1}\leq a_{2}\leq a_{3}\leq\dots$ and $1\leq b_{1}\leq b_{2}\leq b_{3}\dots$.

Conversely, let a point be in $\Omega$. To prove that its orbit is disjoint from $\Delta$, it suffices to check the intersection of the orbit with the cylinder $[1\cdot 0]$ – see Figure 1. If we have the bi-infinite sequence is of the form $\ldots 0^{a_{k}}1\cdot 0^{a_{k-1}}1\ldots$ with $a_{k}\geq a_{k-1}$, then

[TABLE]

from which it follows that $(x,y)\not\in\Delta$. If instead the bi-infinite sequence is of the form $\ldots 1^{b_{k}}\cdot 01^{b_{k+1}}0\ldots$ then

[TABLE]

Again, it follows then that $(x,y)\not\in\Delta$. ∎

Corollary 3.2**.**

If a point does not fall into $\Delta$ then the corresponding bi-infinite 0-1 sequence is either eventually periodic both to the left and to the right (of which there are only countably many) or comes arbitrarily close to $(0,1/2)$ or $(1/2,1)$.

Proof.

If our sequence is not eventually periodic to the left or the right, then we have shifts of one of the two forms

[TABLE]

with $a_{n},b_{m}$ arbitrarily large. ∎

Corollary 3.3**.**

The convex polygon $\Delta$ is a cycle trap.

Proof.

The only cycles in $\Omega$ are of the form $(0^{n}1)^{\infty}$ or $(1^{n}0)^{\infty}$ for some $n$. All of these cycles intersect $\overline{\Delta}$. ∎

The next result shows that $\Delta$ is a dimension trap. This theorem is improved later by Theorem 3.5. It is included here because it introduces a key idea that is used in later proofs.

Theorem 3.4**.**

The convex polygon $\Delta$ is a dimension trap.

Proof.

By Lemmas 3.1 and 2.4, it suffices to show that $\mathrm{dim}_{\tt H}(\Omega)=0$. Let $\Omega_{n}=\{(x_{k})_{k=-n}^{n}:(x_{k})_{k=-\infty}^{\infty}\in\Omega\}$. We will show that $\lim\frac{1}{n}\log\#\Omega_{n}=0$, which will prove the result.

Consider a point in $\Omega_{n}$. We see that we can write this as

[TABLE]

where $1\leq a_{0}\leq a_{1}\leq\dots\leq a_{\ell}$, $1\leq b_{0}\leq b_{1}\leq\dots\leq b_{m}$, $0\leq s,t\leq n$. Here there may be no $a_{i}$ or no $b_{i}$, giving an empty sum. Let

[TABLE]

and similarly

[TABLE]

For fixed $k$, we see that the number of such points is bounded above by the number of partitions of $k-s$ and $2n-k-t$ respectively. Denote the number of partitions of $i$ by $p(i)$. This gives that (with $p(i)\equiv 0$ for $i\leq 0$)

[TABLE]

By the Hardy-Ramanujan formula, we have that $p(n)\sim\frac{1}{n}c_{1}\mathrm{e}^{c_{2}\sqrt{n}}$ as $n\to\infty$ for some some positive constants $c_{1}$ and $c_{2}$. Thus, $\frac{1}{n}\log\#\Omega_{n}\leq c_{3}\frac{\log n}{\sqrt{n}}$ for some constant $c_{3}$, from which the result follows. ∎

3.2. Symmetric and asymmetric dimension traps containing $(0,1)$.

Let $\Delta^{\prime}\subset\Delta$ be the quadrilateral with vertices

[TABLE]

and $\Delta^{\prime\prime}\subset\Delta$ the symmetric image of $\Delta^{\prime}$ under $(x,y)\to(1-y,1-x)$ with vertices

[TABLE]

– see Figure 4 for $\Delta^{\prime}$.

We claim that these two quadrilaterals are minimal convex dimension traps in the sense that if a set $\Delta_{0}\subset\Delta$ is a dimension trap, then $\Delta_{0}$ must contain the vertices $(0,1/2),(1/2,1),(0,1)$ and at least one of the vertices $(5/24,2/3)$ or $(1/3,19/24)$.

Theorem 3.5**.**

Let $\Delta_{0}\subset\Delta$ be a dimension trap, then either $\Delta^{\prime}\subset\Delta_{0}$ or $\Delta^{\prime\prime}\subset\Delta_{0}$.

Let $\mathrm{hull}(A)$ denote the convex hull of $A\subset\mathbb{R}^{2}$.

Corollary 3.6**.**

Let $\Delta_{0}\subset\Delta$ be a dimension trap symmetric with respect to $x+y=1$. Then $\mathrm{hull}(\Delta^{\prime}\cup\Delta^{\prime\prime})\subset\Delta_{0}$.

Proof of Theorem 3.5.

The fact that $\{(0,1/2),(1/2,1)\}\subset\Delta_{0}$ is contained in the proof of Theorem 2.5 and Remark 2.7. Similarly, we can show that $(0,1)\in\Delta_{0}$ by considering the Cantor set $\{0^{n}1^{n},0^{n+1}1^{n+1}\}^{*}$.

To see that one of $(5/24,2/3)$ or $(1/3,19/24)$ are in $\Delta_{0}$, consider the Cantor set given by

[TABLE]

Let $\varepsilon>0$; we see for $n$ sufficiently large that the cylinder $[(01)^{n}\cdot 0011(01)^{n}]\subseteq(5/24,2/3)+B_{\varepsilon}$, where $B_{\varepsilon}$ is the disc centred at the origin of radius $\varepsilon$. Furthermore, for $n$ sufficiently large we have $[(01)^{n}0011\cdot(01)^{n}]\subseteq(1/3,19/24)+B_{\varepsilon}$. Hence the Cantor set in question intersects a neighbourhood of these vertices. Moreover, for $n$ sufficiently large, we see that either $x\approx\frac{5}{24}$ for $x>\frac{1}{3}-\varepsilon$. Similarly either $y\approx 19/24$ or $y<2/3+\varepsilon$. This shows that a convex subset of $\Delta$ that is a dimension trap must contain either $(1/3,2/3)$ or $(5/24,2/3)$ or $(1/3,19/24)$. If this set contains $(1/3,2/3)$, then it has to be $\Delta$, and by convexity it would include the other two points. Otherwise, it has to contain one of the points $(5/24,2/3)$ or $(1/3,19/24)$, and we are done. ∎

Theorem 3.7**.**

The convex quadrilaterals $\Delta^{\prime}$ and $\Delta^{\prime\prime}$ are dimension traps.

Proof.

We will show that the projection of $\mathcal{J}(\Delta^{\prime})$ and $\mathcal{J}(\Delta^{\prime\prime})$ onto the $x$-axis both have dimension [math]. We then notice that under the symmetric $(x,y)\to(1-y,1-x)$ that $\Delta^{\prime}\leftrightarrow\Delta^{\prime\prime}$. From this we conclude that the projection of $\mathcal{J}(\Delta^{\prime})$ and $\mathcal{J}(\Delta^{\prime\prime})$ onto the $y$-axis both have dimension [math]. This will prove that both $\Delta^{\prime}$ and $\Delta^{\prime\prime}$ are dimension traps as required.

Projection of $\Delta^{\prime\prime}$ on the $x$-axis.

As with $\Delta$, we see that $[11\cdot 00]\subseteq\Delta^{\prime\prime}$. This follows from the fact that the $\pi$-image of this cylinder lies above $y=x+\frac{1}{2}$, which in turn lies inside $\Delta^{\prime\prime}$. This implies that $x$ cannot contain $1100$ as a substring.

Our first goal will be to show that either a point avoiding $\Delta^{\prime\prime}$ to the right is eventually of the form $01^{a_{1}}01^{a_{2}}0\ldots$ with $3\leq a_{1}\leq a_{2}\leq\dots$ or the point to the right is eventually of the form $\{10,110\}^{*}$. We will then show that this second case cannot happen.

We first claim that if $k^{\prime}<k$ and $k\geq 3$ then $(x,y)\in[1101^{k}\cdot 01^{k^{\prime}}0]\subseteq\Delta^{\prime\prime}$. It suffices to shows this for $k^{\prime}=k-1$ by monotonicity. We see that in this case that $y>\frac{19}{24}$, and hence we wish to show that the points $(x,y)$ lies above the line from $(1/2,1)$ to $(1/3,19/24)$. That is, we must show it lies above the line $y=\frac{5}{4}x+\frac{3}{8}$. We see that

[TABLE]

and similarly that

[TABLE]

This gives us that

[TABLE]

which proves the result.

We next claim that if $k^{\prime}<k-1$ and $k\geq 3$ then $(x,y)\in[101^{k}\cdot 01^{k^{\prime}}0]\subseteq\Delta^{\prime\prime}$. It suffices to show this for $k^{\prime}=k-2$ by monotonicity. We see that in this case that $y>\frac{19}{24}$, and hence we wish to show that the points $(x,y)$ lies above the line from $(1/2,1)$ to $(1/3,19/24)$. That is, we must show it lies above the line $y=\frac{5}{4}x+\frac{3}{8}$. We see that

[TABLE]

and similarly that

[TABLE]

This gives us that

[TABLE]

which proves the result.

This shows us that if we ever have $01^{k}01^{k^{\prime}}$ with both $k,k^{\prime}\geq 3$ then the sequence to the right is eventually of the form $01^{a_{1}}01^{a_{2}}0\ldots$ with $3\leq a_{i}\leq a_{i+1}$. The dimension of the projection of the set of such points is 0.

We next claim that $(x,y)\in[11011\cdot 010]\subseteq\Delta^{\prime\prime}$. We see that in this case that $y>\frac{19}{24}$, and hence we wish to show that the points $(x,y)$ lies above the line from $(1/2,1)$ to $(1/3,19/24)$. That is, we must show it lies above the line $y=\frac{5}{4}x+\frac{3}{8}$. We see that

[TABLE]

and similarly that

[TABLE]

This gives us that

[TABLE]

which proves the result.

This shows us that if we have $01^{k}0$ with $k\geq 3$ then either the sequence is eventually of the form $01^{a_{1}}01^{a_{2}}0\ldots$ with $3\leq a_{i}\leq a_{i+1}$, or the sequence to the right is eventually of the form $1110(110)^{\infty}$. Thus the projection of the set of points that contain $01^{k}0$ for some $k\geq 3$ has dimension [math].

Lastly we claim that $(x,y)\in[101011\cdot 010]\subseteq\Delta^{\prime\prime}$ or has a tail of the above mentioned form. We see that in this case that $y>\frac{19}{24}$, and hence we wish to show that the points $(x,y)$ lies above the line from $(1/2,1)$ to $(1/3,19/24)$. That is, we must show it lies above the line $y=\frac{5}{4}x+\frac{3}{8}$. We may assume at this point that there are no occurrences of $111$ after this subsequence, otherwise we are in the above mentioned form. That is, we must show it lies above the line $y=x+\frac{11}{24}$. We see that

[TABLE]

and similarly that

[TABLE]

This gives us that

[TABLE]

which proves the result.

Projection of $\Delta^{\prime}$ on the $x$-axis.

To prove that this projection has zero dimension, we first show that the set of $x$ in the projection that contain a substring $111$ has zero dimension. We then show that the set of $x$ in the projection that contain $000$ has zero dimension. Lastly, we show that the set of $x$ in the projection that do not contain the substrings $000$ or $111$ has zero dimension.

Let $(x,y)\in\mathcal{J}(\Delta^{\prime})$. The first observation that we make is that, again, $[11\cdot 00]\subset\Delta^{\prime}$. The next thing we observe is that we cannot have the $1^{k+2}01^{k}0$ or $11101^{k+1}01^{k}0$ as substrings, for all $k\geq 1$. Namely, $[1^{k+2}\cdot 01^{k}0]\subset\Delta^{\prime}$ and $[11101^{k+1}\cdot 01^{k}0]\subset\Delta^{\prime}$. The first claim follows from the fact this cylinder lies above $y=x+\frac{1}{2}$; let us prove the second one. Note that the line connecting $\left(\frac{5}{24},\frac{1}{3}\right)$ and $\left(\frac{1}{2},1\right)$ is $y=\frac{8x+3}{7}$. Then for the rectangle $\pi([11101^{k+1}\cdot 01^{k}0])$ we have

[TABLE]

whence $y>(8x+3)/7$, which is a direct check.

This implies that if we have a $1^{k}0$ as a substring, for $k\geq 3$, then it must be followed by a $1$ (as $1100$ is a forbidden substring). Thus, the substring will look like $1^{k}01$. Next, by the first observation, this must be followed by additional 1s, as $1^{k}01^{k-2}0$ is forbidden. Thus, the substring will look like $1^{k}01^{k-1}$. After this the sequence will look like either $1^{k}01^{k-1}0$ or $1^{k}01^{k}$. If we are in the second case, then our sequence is of the form $1^{k}01^{k_{2}}01^{k_{3}}0\dots$ where $3\leq k\leq k_{2}\leq k_{3}\leq\dots$ with the possibility that one of these $k_{i}$ is infinite. The projection of the points from this second case has zero dimension.

Thus, we can assume our word is of the form $1^{k}01^{k-1}0$. If $k-1\geq 3$ we can again repeat these argument, hence without loss of generality assume that $k=3$. Hence the substring which is not forbidden has to be of the form $1110110$.

We next claim that $111(011)^{n}010$ and $111(011)^{n}01110110$ are forbidden substrings for all $n\geq 1$. Namely, $[111(011)^{n}\cdot 010]\subset\Delta^{\prime}$ and $[111(011)^{n}0111\cdot 0110]\subset\Delta^{\prime}$. Let us prove the first one; we have $x\leq\frac{3}{8}$ and $y\geq\frac{6}{7}+\frac{1}{49}\cdot 8^{-n}>(8x+3)/7$. The second claim is proved in a similar way as are the rest of such claims. We leave the details to the interested reader.

This proves that the substrings will be of the form $111(011)^{\infty}$ or $111(011)^{n}01^{a_{1}}01^{a_{2}}01^{a_{3}}\dots$ with $3\leq a_{1}\leq a_{2}\leq a_{3}\leq\dots$. The projection of this set, again, is of zero dimension.

This shows that the set of $x$ in the projection containing a $111$ as a substring has zero dimension.

We now consider those $x$ with a substring $10^{k}1$ for $k\geq 1$. We now claim that both $10^{k}10^{k+2}$ and $10^{k}10^{k+1}100$ are forbidden for $k\geq 1$. This follows as $[10^{k}1\cdot 0^{k+2}]\subset\Delta^{\prime}$ and $[10^{k}1\cdot 0^{k+1}100]\subset\Delta^{\prime}$. Thus we see that we are either of the form $10^{k}10^{a}1$ with $a\leq k$, or $10^{k}10^{k+1}101$. If we are of the first form, and $a\geq 1$ we can repeat this argument. If instead we are of the form $10^{k}10^{k+1}101$ then we can again repeat this argument, with $k=1$. Thus we can assume that this is eventually contained within $\{100,110,10\}^{*}$.

Consider a word in $\{100,110,10\}^{*}$. We have that the words of the form $110(10)^{n}110(10)^{m}100$ and $110(10)^{n}100(10)^{m}100$ are forbidden for all $n,m\geq 0$, since $[110(10)^{n}11\cdot 0(10)^{m}100]\subset\Delta^{\prime}$ and $[110(10)^{n}1\cdot 00(10)^{m}100]\subset\Delta^{\prime}$. This tells us that our sequence is eventually of the form $110(10)^{n_{1}}110(10)^{n_{2}}110(10)^{n_{3}}110(10)^{n_{4}}\dots$ or $110(10)^{n_{1}}100(10)^{n_{2}}110(10)^{n_{3}}100(10)^{n_{4}}\dots$.

Next, $110(10)^{n}110(10)^{m}110$ is forbidden for $n\geq 0$ and $m\geq 2$, as well as for $n=0,m=1$. This follows as $[11011\cdot 010110]\subset\Delta^{\prime}$ and in general $[110(10)^{n}11\cdot 0(10)^{m}110]\subset\Delta^{\prime}$ for $n\geq 0$ and $m\geq 2$. This tells us that if our sequence of the form: $110(10)^{n_{1}}110(10)^{n_{2}}110(10)^{n_{3}}\dots$, then eventually either $n_{i}=0$ or $n_{i}=1$ for all $i$. The projection of this set will have zero dimension as well.

We further observe that $110(10)^{n}100(10)^{m}110$ is forbidden for $n\geq 0$, and $m\geq 1$ as well as for $n=m=0$. This follows as $[110(10)^{n}1\cdot 00(10)^{m}110]\subset\Delta^{\prime}$ for $n\geq 0$, and $m\geq 1$ as well as for $n=m=0$.

This tells us that if we are of the form $110(10)^{n_{1}}100(10)^{n_{2}}110(10)^{n_{3}}100(10)^{n_{4}}\dots$, then $n_{2}=n_{4}=n_{6}=\dots=0$. Hence we are of the form $100110(10)^{n_{1}}100110(10)^{n_{3}}100110\dots$. We observe that this is a forbidden word for all $n_{1},n_{3}\geq 0$.

This shows that the projection of set of $x$ in the alphabet $\{10,110,100\}^{*}$ that contains a $110$ is of zero dimension.

We observe that $100(10)^{n}100(10)^{m}100$ is forbidden for $n\geq 1$ and $m\geq 0$. This follows as $[100(10)^{n}1\cdot 00(10)^{m}100]\subset\Delta^{\prime}$ for for $n\geq 1$ and $m\geq 0$. Hence if a word contains no $110$ then it will be of the form $100(10)^{n_{1}}100(10)^{n_{2}}100(10)^{n_{3}}\dots$, and hence from the comment above, will eventually be of the form $(100)^{\infty}$. The projection of this set is clearly of zero dimension.

To summarize, if our sequence contains $01^{k}0$ with $k\geq 3$, then it is to the right eventually of the form $01^{a_{1}}01^{a_{2}}0\ldots$ with $3\leq a_{i}\leq a_{i+1}$, or to the right eventually of the form $1110(110)^{\infty}$. If instead it contains no occurrences of $01^{k}0$ with $k\geq 3$, then we see that the tail to the right has to be of the form $(011)^{\infty}$ or $(01)^{\infty}$. ∎

Thus, Theorems 3.5 and 3.7 yield that if we have a convex hole in the upper-left quadrant of the square and it is a dimension trap, then it has to contain $\Delta^{\prime}$ or $\Delta^{\prime\prime}$, which are the smallest dimension traps of this kind, with area $\frac{13}{96}\approx 0.135$.

Furthermore, as a corollary to this, we see that $\mathrm{hull}(\Delta^{\prime}\cup\Delta^{\prime\prime})$ is the smallest symmetric (with respect to $x+y=1$) dimension trap of this kind, with area $\frac{53}{384}\approx 0.1380208333$.

4. Cycle traps and dimension traps which contain $(1/2,1/2)$.

Let $P_{1}$ denote the open parallelogram with vertices $(1/3,2/3),(1/2,1),(2/3,1/3)$ and $(1/2,0)$ — see Figure 5.

This parallelogram may also be defined by the inequalities

[TABLE]

These inequalities have a clear symbolic meaning, so they are easy to check. For instance, let $x_{1}=1$; then the inequality $y<2x-1$ means $x_{0}x_{-1}x_{-2}\ldots\prec x_{2}x_{3}x_{4}\ldots$, and $y>3-4x$ means $x_{0}x_{-1}x_{-2}\ldots\succ\overline{x_{3}x_{4}x_{5}\ldots}$

Consider the Thue-Morse word ${\bm{t}}=t_{0}t_{1}t_{2}\ldots=0110\ 1001\ 1001\ 0110\ \ldots$ given by the recurrence $t_{2m}=t_{m},t_{2m+1}=1-t_{m}$ with $t_{0}=0$. Define ${\bm{t}}_{k}=(t_{0}t_{1}t_{2}\ldots t_{2^{k}-1})^{\infty}$. Let $P_{k}$ be the hexagon with vertices $(1/2,0),({\bm{t}}_{k},2{\bm{t}}_{k}),({\bm{t}}_{k},2-4{\bm{t}}_{k}),(1/2,1),(1-{\bm{t}}_{k},2-4{\bm{t}}_{k}),(1-4{\bm{t}}_{k},2{\bm{t}}_{k})$. This hexagon may also be defined by the inequalities

[TABLE]

We define $P_{\infty}=\lim_{k\to\infty}P_{k}$. See Figure 6.

There are a number of goals to this section.

• •

Show that for each $k\geq 1$ that there are only a finite number of cycles that avoid $\overline{P_{k}}$, and moreover that $P_{1}$ is a cycle trap.

• •

Show that for each $k\geq 1$ that $P_{k}$ is a dimension trap.

• •

Show that $P_{\infty}$ is a dimension trap.

Lemma 4.1**.**

No element of $\mathcal{J}(P_{k})$ can contain the factors

[TABLE]

Proof.

We observe that the second word for each case follows by the symmetric $(x,y)\to(1-y,1-x)$, so we will prove the first restriction only.

Case 1: Proof of (a):

We observe that if $n>\ell\geq 2$ then $[10^{\ell-1}\cdot 01^{n}0]\subseteq P_{k}$ for all $k$, from which the desired result follows.

Case 2: Proof of (b) and (c):

We observe that if $n>\ell\geq 1$ then $[10^{\ell-1}\cdot 01^{n}0]\subseteq P_{1}$. Further if $n\geq 3$ then $[10^{\ell-1}\cdot 01^{n}0]\subseteq P_{k}$ for all $k$.

Case 3: Proof of (d) and (e):

We observe that if $n\geq 2$ then $[1^{n}0^{n-1}\cdot 01^{n}01]\subseteq P_{2}\subseteq P_{1}$.

If in addition, $n\geq 3$ we have that $[1^{n}0^{n-1}\cdot 01^{n}01]\subseteq P_{k}$ for all $k\geq 3$.

Case 4: Proof of (f) and (g):

We observe that if $n\geq 2$ then $[1^{n}0^{n}\cdot 10^{n}10]\subseteq P_{2}\subseteq P_{1}$.

If in addition, $n\geq 3$ we have that $[1^{n}0^{n}\cdot 10^{n}10]\subseteq P_{k}$ for all $k\geq 3$. ∎

Corollary 4.2**.**

1. (1)

The only cycles of the baker’s map which do not intersect $P_{1}$ are of the form $(10^{n})^{\infty},(01^{n})^{\infty}$, $(1^{n}0^{n})^{\infty}$ or $(0^{n}10^{n}1^{n}01^{n})^{\infty}$. All of these intersect $\overline{P_{1}}$. Hence $P_{1}$ is a cycle trap. 2. (2)

The only cycles of the baker’s map which do not intersect $P_{2},P_{3},\dots$ are of the form $(10^{n})^{\infty},(01^{n})^{\infty}$, $(1^{n}0^{n})^{\infty}$ or $(0^{n}10^{n}1^{n}01^{n})^{\infty}$ or are contained in the Cantor set $\{01,001,011,0011\}^{*}$. Cycles of the form $(10^{n})^{\infty},(01^{n})^{\infty}$, $(1^{n}0^{n})^{\infty}$ or $(0^{n}10^{n}1^{n}01^{n})^{\infty}$, with $n\geq 2$ intersect $\overline{P_{k}}$ for $k\geq 2$.

Proof.

Consider a cycle of the form $(1^{a_{1}}0^{a_{2}}1^{a_{3}}\ldots 0^{a_{k}})^{\infty}$ with $k$ even. We will label this as $a_{1}a_{2}\ldots a_{k}$ for convenience. If all $a_{i}=1$, then we are done. Assume without loss of generality that $a_{1}=\min_{a_{i}\neq 1}a_{i}$.

By Lemma 4.1 (a), $a_{2}=1$ or $1<a_{2}\leq a_{1}$. Since $a_{1}$ is chosen minimally, we have $a_{2}=1$ or $a_{2}=a_{1}$. In the case $a_{2}=a_{1}$ we proceed to show that $a_{3}=1$ or $a_{3}=a_{1}$. If instead we have $a_{2}=1$ then by Lemma 4.1 (b) and (c) we have that $a_{3}\leq a_{1}$. Again, by minimality we have $a_{3}=a_{1}$ or $a_{3}=1$. If both $a_{2}=a_{3}=1$ and the cycle does not intersect $P_{1}$, then by Lemma 4.1 (b) we have $a_{2}=a_{3}=\dots=a_{k}=1$. By rotating this cycle, we get a contradiction with Lemma 4.1 (b), whence $a_{3}=a_{1}$.

If instead both $a_{2}=a_{3}=1$ and the cycle does not intersect $P_{k}$ for some $k\geq 2$ then by Lemma 4.1 (c) we have either that have $a_{2}=a_{3}=\dots=a_{k}=1$, or that there exists some $i$ such that $a_{i}=2$. By the minimality of $a_{1}$ we see then that this cycle is contained in $\{01,001,011,0011\}^{*}$.

By continuing in this fashion, we see that for all $i$ that $a_{i}=1$ or $a_{i}=a_{1}$. In the case of $P_{2},P_{3},\dots$ we will assume that $a_{1}\geq 3$, else all $a_{i}\in\{1,2\}$ as required.

It follows from Lemma 4.1 (b) and (c) that if for some $i$ we have $a_{i}=a_{i+1}=1$, then $a_{i}=1$ for all $i$. In other words, we cannot have two consecutive $1$s.

Lemma 4.1 (d) and (e) implies that if we have three consecutive $a_{1}$ in the sequence, then $a_{i}=a_{1}$ for all $i$.

Finally, if $a_{i}=a_{i+1}=a_{1}$ and $a_{i+2}=1$ then $a_{i+3}=a_{1}$ (as we do not have two consecutive $1$s). Furthermore, we then get that $a_{i+4}=a_{1}$ by Lemma 4.1 (f) and (g). Repeating this argument gives us the cycle $(0^{a_{1}}1^{a_{1}}01^{a_{1}}0^{a_{1}}1)^{\infty}$.

We see for $n\geq 2$ that all of these cycles intersect $\overline{P_{k}}$. Finally, we observe that the cycle $(01)^{\infty}$ intersects $\overline{P_{1}}$. ∎

Proposition 4.3**.**

The set $P_{1}$ is a dimension trap.

Proof.

Any sequence in $\mathcal{J}(P_{1})$ is of one of the forms

[TABLE]

where $a_{i}\leq a_{i+1}$ and $b_{i}\leq b_{i+1}$ for all $i$ and $a_{n}\leq b_{1}$.

This shows that these are eventually periodic in the left direction. Further, using a similar argument to that in Theorem 3.4 and Lemma 2.4 we conclude that the dimension of this set is zero. ∎

Proposition 4.4**.**

*The only cycle in $\mathcal{J}(\overline{P_{2}})$ is $\left\{\left(\frac{1}{3},\frac{2}{3}\right),\left(\frac{2}{3},\frac{1}{3}\right)\right\}$. *

Lemma 4.5**.**

Let $\mathsf{B}$ be the set of sequences that are eventually periodic to the left and are eventually of the form $01^{a_{1}}01^{a_{2}}0\ldots$, $10^{a_{1}}10^{a_{2}}10^{a_{3}}\ldots$ or $0^{a_{1}}1^{a_{2}}0^{a_{3}}1^{a_{3}}\ldots$ with $a_{1}\leq a_{2}\leq a_{3}\leq\dots$ in the right direction. Consider a sequence $a_{i}\in\{1,2\}$ corresponding to $0^{a_{1}}1^{a_{2}}0^{a_{3}}1^{a_{4}}\ldots$. No element of $\mathcal{J}(P_{2})\setminus\mathsf{B}$ written as $\ldots a_{-1}a_{0}a_{1}a_{2}\ldots$ with $a_{i}\in\{1,2\}$ may contain a factor

[TABLE]

Proof.

We note that the set $\mathsf{B}$ is of zero dimension, whence we can ignore this set without loss of generality. As we are assuming the sequence is not in $\mathsf{B}$ and in $\mathcal{J}(P_{2})$, we may assume that the sequence has left or right tail (or possibly both) of the form $\{011,01,001,0011\}^{*}$.

Case 1: Proof of (a):

This follows by noticing that that the rectangles $[.10010(010)^{\infty},.10010(110)^{\infty}]\times[.01(0)^{\infty},.01(1)^{\infty}]$ and $[.10010(0)^{\infty},.10010(1)^{\infty}]\times[.01(001)^{\infty},.01(101)^{\infty}]$ are contained within $P_{2}$. The symmetry $(x,y)\leftrightarrow(1-x,1-y)$ proves the equivalent statement swapping the roles of [math] and $1$ in the above statement.

Case 2: Proof of (b):

This is similar to case (a) by looking at the shift $\ldots 1001\cdot 10010\ldots$ with the restrictions that we are not in $\mathsf{B}$.

Case 3: Proof of (c):

This is similar to case (a) by looking at the shift $\ldots 10011\cdot 01101\ldots$ with the restrictions that we are not in $\mathsf{B}$.

Case 4: Proof of (d):

This is similar to case (a) by looking at the shift $\ldots(10)^{(m+3)/2}1\cdot 100(10)^{(m-1)/2}1001\ldots$ with the restrictions that we are not in $\mathsf{B}$.

Case 5: Proof of (e):

This is similar to case (a) by looking at the shift $\ldots 100(10)^{n/2}1\cdot 100(10)^{n+1}\ldots$ with the restrictions that we are not in $\mathsf{B}$. ∎

Proof of Proposition 4.4.

Consider a cycle $(0^{a_{1}}1^{a_{2}}\dots 1^{a_{k}})^{\infty}$. We can assume without loss of generality that all of the $a_{i}\leq 2$, for otherwise they are of one of the forms specified by Corollary 4.2. By Lemma 4.5 (b), we cannot have more than two 2s in a row, otherwise we have the cycle $(0011)^{\infty}$ which is of a form specified in Corollary 4.2. By this and Lemma 4.5 (a), if we have more than one 1, then it must be followed by exactly two 2s.

If all $2$s are isolated (i.e., if we do not have an occurrence of two $2$s in a row), then by Lemma 4.5 (a), we can have at most one $1$, and hence we have a word of the form $(011)^{\infty}$ or $(100)^{\infty}$, which are cycles of an admissible form. Hence we may assume that we have one occurrence of two $2$s in a row. By Lemma 4.5 (a) and (c) we see that all occurrences of $2$ must come in pairs.

Hence $a_{1}a_{2}a_{3}\dots$ looks like $221^{k_{1}}221^{k_{2}}\dots 221^{k_{n}}$ for some choice of $k_{i}$. If $k_{1}=2m>0$ is even, then by Lemma 4.5 (e) we have $k_{1}\geq k_{2}$. Furthermore, by Lemma 4.5 (d), $k_{2}$ cannot be odd. Continuing in this manner, we have that $k_{1}\geq k_{2}\geq\dots\geq k_{n}\geq k_{1}$, where all $k_{i}$ are even. This is of the form $(0011(01)^{m})^{\infty}$ or $(1100(10)^{m})^{\infty}$. A similar argument can be used if $k_{1}$ is odd to show that this must be of the form $(0011(01)^{m}01100(10)^{m}1)^{\infty}$. Both of these points lie on the boundary of $P_{2}$. ∎

Lemma 4.6**.**

Let $k\geq 2$ and $(x,y)\in P_{k-1}\setminus P_{k}$. Assume that the binary expansion of $x$ does not contain $000$ or $111$. If $(x,y)\in\mathcal{J}(P_{k})$ then $B^{2^{k-1}}(x,y)\in P_{k-1}\setminus P_{k}$. Furthermore $x=.{\bm{t}}_{k}^{\infty}$ or $x=.(\overline{{\bm{t}}_{k}})^{\infty}$.

Proof.

Let $(x,y)\in P_{k-1}\setminus P_{k}$ and $(x,y)\in\mathcal{J}(P_{k})$. By symmetry we may assume that $x<1/2$. Thus we have

[TABLE]

By the second inequality, $x=0.{\bm{t}}_{k-1}\ldots$; in particular, $(x,y)$ is contained in the polygon $\Pi_{k}$ with vertices $(x_{k-1},2x_{k-1}),(x_{k},2x_{k}),(x_{k},2-4x_{k}),(x_{k-1},2-4x_{k-1})$ where $x_{k-1}=0.{\bm{t}}_{k-1}^{\infty}$, and similarly for $x_{k}$. (We have that $\Pi_{k}$ is a trapezium for $k\geq 3$ and a triangle for $k=2$.) Since $x=0.{\bm{t}}_{k-1}\ldots$, we see that the first $2^{k-1}$ iterates of $B$ on $\Pi_{k}$ are continuous. Furthermore,

[TABLE]

where $y$ can take any value in $\Pi_{k}$. Hence, regardless if $k$ is even or odd, we see that the images of all four corners of the polygon are contained within $P_{k-1}$. Hence, as we picked $(x,y)\in\mathcal{J}(P_{k})$ this proves that $(x,y)\in P_{k-1}\setminus P_{k}$. This in turn proves that $x=0.{\bm{t}}_{k}^{\infty}$ as required. ∎

Corollary 4.7**.**

There are a finite number of cycles in $\mathcal{J}(\overline{P_{k}})$.

Proof.

Recall that $P_{1}$ is a cycle trap, whence $\mathcal{J}(\overline{P_{1}})$ does not contain any cycles. If a cycle intersects some $\overline{P_{\ell-1}}$ but not $\overline{P_{\ell}}$ for $\ell\leq k$ we see that it is of the form $\bm{t}_{\ell-1}^{\infty}$. Hence cycles in $\mathcal{J}(P_{k})$ are of the form ${\bm{t}}_{\ell}^{\infty}$ for $2\leq\ell\leq k-1$. ∎

Remark 4.8*.*

Thus, the sequence $\{P_{k}\}_{1}^{\infty}$ is a route to chaos for $B$ which is completely analogous to $\{(\bm{t}_{k}^{\infty},(\overline{\bm{t}_{k}})^{\infty})\}_{1}^{\infty}$ for the doubling map – see Section 1.

Proposition 4.9**.**

The set $P_{k}$ is a dimension trap for all $k$.

Proof.

Let $(x,y)\in\mathcal{J}(P_{k})$. We can assume without loss of generality that $(x,y)\not\in\mathcal{J}(P_{1})$, as the set of all such points is of zero dimension. This implies that there exists some shift of $(x,y)$ such that $(x^{\prime},y^{\prime})\in P_{1}$. Assume without loss of generality that $x^{\prime}<1/2$. Assume that $(x^{\prime},y^{\prime})\in P_{\ell-1}\setminus P_{\ell}$. Then we see that $x^{\prime}=0.{\bm{t}}_{a_{0}}^{b_{0}}{\bm{t}}_{a_{1}}^{b_{1}}\dots\bm{t}_{a_{n-1}}^{b_{n-1}}{\bm{t}}_{a_{n}}^{\infty}$ where $\ell\leq a_{0}<a_{1}<a_{2}\dots<a_{n}\leq k$. The set of such $x^{\prime}$ is clearly countable.

Either there is no first such occurrence of $(x^{\prime},y^{\prime})$ in which case the word is of the form

[TABLE]

or there is a first such occurrence. If there is a first such occurrence, we see that the word is as described by Proposition 4.3, which has a projection on to the $y$-axis of dimension [math].

Combining these gives that $P_{k}$ is a dimension trap. ∎

Theorem 4.10**.**

The set $P_{\infty}$ is a dimension trap.

Proof.

If a sequence is in $\mathcal{J}(P_{k})$, then the claim follows from Proposition 4.9. If a sequence is in $\mathcal{J}(P_{\infty})\setminus\mathcal{J}(P_{k})$ for all $k\geq 1$, then it has to be a concatenation of two types of blocks: of the form $1^{k_{i}}0^{k_{i+1}}$, where $k_{i}\geq k_{i+1}$, or of the form ${\bm{t}}_{k_{i}}$ for some $k_{i}\geq 2$ with $k_{i}\geq k_{i-1}$. This corresponds to a point on the boundary of $P_{\infty}$.

We see that a block of the first kind cannot appear after a block of the second kind. If we have blocks of the first kind, some of them end with $1^{j_{1}}01^{j_{2}}0\ldots$ with $j_{i}\geq j_{i+1}$. So, we have $2n+1=k_{i}+k_{i+1}+\dots+$ (possibly) $(j_{1}+1)+(j_{2}+1)+\dots$. In other words, we again have an ordered partition of $2n+1$, and the result follows from the Hardy-Ramanujan formula and Lemma 2.4, similarly to the proof of Theorem 3.4. If we have blocks of the second kind, then we have $n=2^{k_{i}}+2^{k_{i+1}}+\dots+$. Consequently, we again have a subset of ordered partition of $n$, and the result follows again from the Hardy-Ramanujan formula and Lemma 2.4. ∎

Proposition 4.11**.**

Any dimension trap contained in $P_{\infty}$ must contain the vertices $(\bm{t},\bm{t}),(1-\bm{t},\bm{t}),(1-\bm{t},1-\bm{t})$ and $(\bm{t},1-\bm{t})$.

Proof.

Consider the Cantor set $\{{\bm{t}}_{k},\overline{{\bm{t}}_{k}}\}^{*}$. We see that the point $(x,y)$, $x>1/2$ in the orbit of this Cantor set with minimal $x$ value has $x=\overline{{\bm{t}}_{k}}{\bm{t}}_{k}^{\infty}$. Similarly, the point $(x,y)$, $x<1/2$ in the orbit of this Cantor set with maximal $x$ value has $x={\bm{t}}_{k}\overline{{\bm{t}}_{k}}^{\infty}$. The possible $y$ values associated to these $x$ values are ${\bm{t}}_{k}\overline{{\bm{t}}_{k}}^{\infty}$ or $\overline{{\bm{t}}_{k}}{\bm{t}}_{k}^{\infty}$. The result follows by taking $k\to\infty$. ∎

Using the techniques in Section 6 we can show that if $P\subset P_{\infty}$ is a convex dimension trap, then $\mathcal{L}(P)\geq 0.11924$. Here, for comparison, $\mathcal{L}(P_{\infty})\approx 0.12911$.

5. Searching for asymmetric dimension traps containing $(1/2,1/2)$.

Let $r\in(0,1/2]\cap\mathbb{Q}$ and define $a=.\text{0-max}(r)^{\infty}$ and $b=.\text{1-min}(r)^{\infty}$. From [11] and the discussion in Section 1 we know that $\mathrm{dim}_{\tt H}(\mathcal{J}^{+}(a,b))=0$, and moreover that $\mathcal{J}^{+}(a,b)$ contains only a finite number of cycles. The simplest example of this is for $r=1/2$. In this case $a=.\text{0-max}(1/2)^{\infty}=1/3$ and $b=.\text{1-min}(1/2)^{\infty}=2/3$, and $\mathcal{J}(1/3,2/3)$ contains only the fix points $\{0,1\}$ and the 2-cycle $\{1/3,2/3\}$. In Section 4 we studied the analogous construction, there labelled $P_{1}$, of the more general open polygons $P_{r}$ with vertices given by

[TABLE]

Here again, $a=.\text{0-max}(r)^{\infty}$ and $b=.\text{1-min}(r)^{\infty}$. That is, $P_{1}$ of Section 4 is $P_{1/2}$ under this new notation. In this section we consider these more general $P_{r}$ for other rational numbers $r$, and show that the results analogous to those of the doubling map unfortunately do not hold.

Notation 5.1*.*

Let $r=\frac{c}{d}<\frac{1}{2}$. We define $r_{0}=\frac{c_{0}}{d_{0}}$ as the unique Farey neighbour222That is, a fraction with $|cd_{0}-c_{0}d|=1$. of $r$ such that $r<r_{0}$ and $d_{0}<d$.

We define $r_{k}=\frac{c_{k}}{d_{k}}=\frac{c_{k-1}+c}{d_{k-1}+d}$ as the $k^{th}$ Farey neighbour of $r$.

Lemma 5.2**.**

Let $r_{k}$ be defined as above.

1. (1)

$\lim_{k\to\infty}r_{k}=r$. 2. (2)

$\text{0-max}(r_{k})=(\text{0-max}(r))^{k-i}\text{1-min}(r_{i})=\text{0-max}(r_{i})(\text{0-max}(r))^{k-i}$* for $i=0,1,\dots k$.* 3. (3)

$\text{0-max}(r)^{\infty}\prec\dots\prec\text{0-max}(r_{2})^{\infty}\prec\text{0-max}(r_{1})^{\infty}\prec\text{0-max}(r_{0})^{\infty}$. 4. (4)

$\text{1-min}(r)^{\infty}\prec\dots\prec\text{1-min}(r_{2})^{\infty}\prec\text{1-min}(r_{1})^{\infty}\prec\text{1-min}(r_{0})^{\infty}$.

Example 5.3**.**

Consider as an example $r=\frac{3}{10}$. One can easily check that $r_{0}=\frac{1}{3}$. This gives $r_{1}=\frac{1+3}{3+10}=\frac{4}{13}$ and more generally, $r_{k}=\frac{1+3k}{3+10k}$.

In this case we see that $\text{0-max}(r)=0100100100$ and $\text{0-max}(r_{0})=010$, $\text{0-max}(r_{1})=0100100100100$, and in general, $\text{0-max}(r_{k})=(0100100100)^{k-1}100$. The other results can be easily verified for this example.

The proof follows from [10, Lemma 5.1] and is left as an exercise to the reader.

Theorem 5.4**.**

Let $r<\frac{1}{2}$. Then there exists a $k$ such that the Cantor set $\{\text{0-max}(r),\text{0-max}(r_{k})\}^{*}$ is disjoint from $P_{r}$.

Proof.

Let $x=.x_{1}x_{2}x_{3}\ldots$ and $y=.y_{1}y_{2}y_{3}\ldots$ be a shift of a point in this Cantor set. We see that if $x_{1}=1$, then by Lemma 5.2 part (4) we have $b\leq x$, with equality only if $x=(\text{1-min}(r))^{\infty}$. Hence the point avoids $P_{r}$ as required.

If $x_{1}=0$ and $y_{1}=0$, we have $x,y\leq(\text{0-max}(r_{k}))^{\infty}$. As the point $(a,a)$ lies outside of $P_{r}$, it is a bounded distance away from $P_{r}$. This, combined with the fact that $\text{0-max}(r_{k})\to a$ as $k\to\infty$, allows us to choose $k$ sufficiently large so that $(x,y)$ is bounded away from $P_{r}$.

Finally, assume that $x_{1}=0$ and $y_{1}=1$. Let $m<n$ such that $\text{0-max}(r)$ has length $n$ and $\text{0-max}(r_{1})$ has length $m$. Write

[TABLE]

This follows from Lemma 5.2. (It is worth noting that $m-n<n$.) Note that $\text{0-max}(r_{k})=\underbrace{\text{0-max}(r)\ldots\text{0-max}(r)}_{k-1}\text{0-max}(r_{1})$ from Lemma 5.2. By noticing that $\text{0-max}(r)$ and $\text{0-max}(r_{1})$ starts and ends with [math], we see that we are in one of three cases.

1. (1)

$x=.w_{s}w_{s+1}\ldots w_{n}w_{1}w_{2}\ldots w_{n}\ldots$ with $s\neq 1$, 2. (2)

$x=.w_{s}w_{s+1}\ldots w_{n}v_{1}v_{2}\ldots v_{m}\ldots$ with $s\neq 1$, 3. (3)

$x=.v_{s}v_{s+1}\ldots v_{m}w_{1}w_{2}\ldots w_{n}\ldots$ with $s\neq 1$.

Case (1) and (2). By construction, we have that

[TABLE]

and hence $x<a$.

Case (3). First assume that $s>n$. Then

[TABLE]

We clearly cannot have $w_{s-n}w_{s-n+1}\ldots w_{m-n+1}\succ w_{1}w_{2}\ldots w_{m-s+2}$. If

[TABLE]

then we are done. If instead $w_{s-n}w_{s-n+1}\ldots w_{m-n+1}=w_{1}w_{2}\ldots w_{m-s+2}$, then $w_{m-s+3}=1$ by [10, Lemma 5.1] and $w_{1}=0$, whence

[TABLE]

Next assume that $s\leq n$. We have

[TABLE]

Similar to the above, we clearly cannot have $w_{s}w_{s+1}\ldots w_{n}\succ w_{1}w_{2}\ldots w_{n-s+1}$. If $w_{s}w_{s+1}\ldots w_{n}\prec w_{1}w_{2}\ldots w_{n-s+1}$, then we are done. If instead we have $w_{s}w_{s+1}\ldots w_{n}=w_{1}w_{2}\ldots w_{n-s+1}$, then $w_{n-s+2}=1$ and $w_{1}=0$, whence $w_{s}w_{s+1}\ldots w_{n}w_{1}\prec w_{1}w_{2}\ldots w_{n-s+1}w_{n-s+2}$, as required. ∎

6. Lower bound for the size of a dimension trap

Let $\mathcal{T}$ denote the set of all convex open dimension traps. We define $\delta=\inf_{T\in\mathcal{T}}\mathcal{L}(T)$. Note that $\Delta,\Delta^{\prime}$ and $P_{\infty}$ are all in $\mathcal{T}$, and have area $\frac{13}{96},\frac{13}{96}$ and $0.129106155\dots$ respectively. Therefore, $\delta\leq 0.129106155\dots$. For the precise definition of $\Delta$ and $\Delta^{\prime}$ see Section 3 and for $P_{\infty}$ see Section 4. The goal of this section is to find a good lower bound for $\delta$.

By Remark 2.7, a dimension trap must contain one of $\{(0,1/2^{n}),(1/2^{n},0)\}$ for some $n\geq 1$, and must also contain one of $\{(1,1-1/2^{n}),(1-1/2^{n},1)\}$ for some $n\geq 1$. These are our starting “polygons” (although they are one-dimensional).

Let $a$ and $b$ be two finite words. We will assume throughout this section that $.a^{\infty}\neq.b^{\infty}$. Define the Cantor sets

[TABLE]

We further define the limit of this sequence of Cantor sets as follows:

[TABLE]

where $\mathcal{O}(\ldots y_{3}y_{2}y_{1}\cdot x_{1}x_{2}x_{3}\ldots)$ is the orbit of this point under $B$.

We see that if $T\in\mathcal{T}$, then for all $a,b,n$ we have $\mathcal{C}(a,b,n)\cap T\neq\emptyset$. Furthermore, for all $\varepsilon>0$ and all $a$ and $b$ we have that $\mathcal{C}(a,b)\cap(T+B_{\varepsilon})\neq\emptyset$, where, as above, $B_{\varepsilon}$ is the disc centred at the origin of radius $\varepsilon$. This allows us a way to estimate $\delta$ from below.

For some finite collection of $(a,b)$, consider the family $\mathcal{T}_{0}$ of convex polygons such that they contain at least one point from each $\mathcal{C}(a,b)$. We see that if $T\in\mathcal{T}$ then there exists a $T_{0}\in\mathcal{T}_{0}$ such that $T_{0}\subset T$. This gives us that $\delta\geq\inf_{T_{0}\in\mathcal{T}_{0}}\mathcal{L}(T_{0})$.

Here we take advantage of the fact that the forward and backward orbits of $\ldots bbb\cdot abbb\ldots$ tend to the orbits of $\ldots bbb\cdot bbb\ldots$. Thus, for any $\varepsilon>0$ there are only a finite number of points in orbit $\mathcal{O}(\ldots bbb\cdot abbb\ldots)$ that are not within $\varepsilon$ of the orbit of $\ldots bbb\cdot bbb\ldots$.

We will first look at a small example. Instead of looking at the full set $\mathcal{T}$, we will consider a restricted subset $\mathcal{T}^{\prime}\subset\mathcal{T}$. In particular, we will look at the subset of $\mathcal{T}^{\prime}$ where for each $T^{\prime}\in\mathcal{T}^{\prime}$ we have $(1/2,1),(1/2,0)\in T^{\prime}$ and $T^{\prime}$ is closed under the symmetry $(x,y)\leftrightarrow(1-x,1-y)$. We will wish to find a set $\mathcal{T}_{0}^{\prime}$ such for all $T^{\prime}\in\mathcal{T}^{\prime}$ with the above property either $\mathcal{L}(T^{\prime})\geq 0.13$ or there exists a $T_{0}^{\prime}\in\mathcal{T}_{0}^{\prime}$ with $T_{0}^{\prime}\subset T^{\prime}$. Clearly, we may assume that all $T_{0}^{\prime}\in\mathcal{T}_{0}^{\prime}$ satisfy the restriction that $(1/2,1),(1/2,0)\in T_{0}^{\prime}$ and $T_{0}^{\prime}$ is closed under the symmetry $(x,y)\leftrightarrow(1-x,1-y)$. By considering the infimum of $T_{0}^{\prime}$ in $\mathcal{T}_{0}^{\prime}$ we will show that if $T^{\prime}$ is a dimension trap with the above restrictions, then $\mathcal{L}(T^{\prime})\geq 293/2688-4\varepsilon\approx 0.1090029761$, where $\varepsilon=10^{-10}$. We see that $P_{\infty}$ is an example of such a $T^{\prime}$. Thus, the infimum of this subset is bounded above by $0.129106155\dots$. This is the reason why our search for all polygons with $\mathcal{L}(T^{\prime})\leq 0.13$ above is not unreasonable.

Next, one can check that

[TABLE]

When taking into account the symmetry, and the fact that we are searching for $T^{\prime}$ with $\mathcal{L}(T^{\prime})<0.13$, we see that only the point $(7/12,1/3)$ is utilized. This gives the polygon with vertices $(5/12,2/3)$, $(1/2,0)$, $(7/12,1/3)$ and $(1/2,1)$ with area $1/12<0.13$.

We next consider the limit Cantor set

[TABLE]

When taking into account the symmetry and the fact that we are searching for $T^{\prime}$ with $\mathcal{L}(T^{\prime})<0.13$, we see that the points $(29/48,1/3)$ and $(5/12,1/3)$. In the first case, this results in the polygon with vertices $(19/48,2/3)$, $(1/2,0)$, $(29/48,1/3)$ and $(1/2,1)$ with area $5/48<0.13$. and the second this results in the polygon with vertices $(5/12,2/3)$, $(5/12,1/3)$, $(1/2,0)$, $(7/12,1/3)$, $(7/12,2/3)$ and $(1/2,1)$ with area $1/9<0.13$.

We repeat this process again with these two polygons and with $\mathcal{C}(0,001)$, $\mathcal{C}(0,011)$, $\mathcal{C}(00,011)$, $\mathcal{C}(001,010)$, $\mathcal{C}(001,101)$, $\mathcal{C}(001,110)$ and $\mathcal{C}(011,100)$. (In fact we do this with all $a$ and $b$ where $|a|,|b|\leq 3$, but these were the only Cantor sets that contributed points outside of the polygons at this point of the algorithm.) We take advantage here of the observation that if we have $T_{1}^{\prime},T_{2}^{\prime}\in\mathcal{T}_{0}^{\prime}$ such that $T_{1}^{\prime}\subset T_{2}^{\prime}$, then we can remove $T_{2}^{\prime}$ from $\mathcal{T}_{0}^{\prime}$ and the results still hold. Even with this, and only looking at those $T_{0}^{\prime}$ with $\mathcal{L}(T_{0}^{\prime})<0.13$, this results in $13$ polygons, with areas ranging between $293/2688\approx.1090029762$ to $49/384\approx.1276041667$.

From this we can conclude that if $T$ is a dimension trap, $(1/2,0),(1/2,1)\in T$ and $T$ is closed under the symmetry $(x,y)\leftrightarrow(1-x,1-y)$, then $\mathcal{L}(T)\geq 293/2688-4\varepsilon\approx.1090029758$. This result can be strengthened by taking longer $a$ and $b$, as given in Table 6.1.

Here the $4\varepsilon$ arises because we are not using all points in $\mathcal{C}(a,b)$, only all of those points that are more than $\varepsilon$ away from one of the limit points, and one point within $\varepsilon$ of one of the limit points. If we consider all polygons in this restricted set, then it is possible that all of the vertices should be replaced by vertices that are $\varepsilon$ closer to the center of the polygon. This would result in a polygon that is at most $4\varepsilon$ smaller (as the polygons are convex, and contained within $[0,1]^{2}$).

We summarize the lower bounds given by such computations in Table 6.1.

Remark 6.1*.*

These proofs are computational in nature, and we summarize some details of these computations in Table 6.1. An interesting observation to be made is that if we assume $T$ is a convex dimension trap satisfying the symmetry $(x,y)\leftrightarrow(1-x,1-y)$ and $\mathcal{L}(T)\leq 0.13$, then by considering only words up to length 4, we have that $T$ must contain the points $(0,1/2)$ and $(1,1/2)$ or it contains the points $(1/2,0)$ and $(1/2,1)$. We note that $P_{\infty}$ is a convex dimension trap satisfying this property, hence the bound of $0.13$ is reasonable. All other $T$ are eliminated by this process by having too large a measure. See Figure 8 for the set of polygons in $\mathcal{T}_{0}$ with the restriction that they are closed under the symmetry $(x,y)\leftrightarrow(1-x,1-y)$ and $\mathcal{L}(T)\leq 0.13$, which avoid $\mathcal{C}(a,b)$ for $|a|,|b|$ less than $1$, $2$, $3$ or $4$.

Similarly, if we assume $T$ is a convex dimension trap satisfying the symmetry $(x,y)\leftrightarrow(1-y,1-x)$ and $\mathcal{L}(T)\leq 0.1381$, then by considering only words up to length 4 we have that $T$ must contain the points $(0,1/2)$ and $(1/2,1)$ or it contains the points $(1/2,0)$ and $(1,1/2)$. We again see that $\mathrm{hull}(\Delta^{\prime}\cup\Delta^{\prime\prime})$ satisfies this restriction, hence it is a reasonable region to search. More over, by length $7$ the resulting $T_{0}$ in $\mathcal{T}_{0}$ essentially look like $\mathrm{hull}(\Delta^{\prime}\cup\Delta^{\prime\prime})$. See Figure 9.

Lastly, if we assume $T$ is a convex dimension trap with $\mathcal{L}(T)\leq 0.1292$, with no restrictions on the symmetry, nor any restrictions on containing particular points, then by considering only words up to length $7$, we have that $T$ must one of $(0,1/2)$ or $(1/2,0)$ and one of $(1,1/2)$ or $(1/2,1)$. We see that $P_{\infty}$ is an example of such a dimension trap, hence this is a reasonable region to search. That is, such a search is equivalent to that done in row 2 or 4 of Table 6.1.

These observations help justify the restricted cases that are looked at in Sections 3 and 4.

In Table 6.1, we will call the symmetry $(x,y)\leftrightarrow(1-y,1-x)$ a mirror symmetry, and $(x,y)\leftrightarrow(1-x,1-y)$ a rotational symmetry.

Summing up, we obtain the following unconditional result.

Theorem 6.2**.**

If a convex hole $H$ has area less than $0.11802$, then $\mathrm{dim}_{\tt H}\mathcal{J}(H)>0$.

We note that this is not optimal. The smallest convex holes $H$ that we know with this property are $\Delta,\Delta^{\prime}$ and $P_{\infty}$ which have area larger than $0.12916155$. Hence it may be possible to improve this number.

7. Final remarks and open problems

It is worth noting that Theorem 2.5 and the first three claims in Section 3 have been proved in the first author’s thesis [6].

Theorems 2.2 and 2.5 have been generalized in the second and third authors’ recent paper [12]. The results are stated for subshifts; there are also applications to algebraic toral automorphisms of a certain type.

Some questions that arise out of this research include

1. (1)

Is it true that each cycle trap for $B$ is a dimension trap as well? It is known that there exist minimal subshifts (so, in particular, they do not have any periodic orbits) with positive topological entropy, which suggests that the answer is probably no. See also the discussion on MathOverFlow [15]. 2. (2)

Our approach to finding optimal asymmetric dimension traps from Section 5 has been a direct attempt to fuse [10] and Section 4 – and it did not work. Is there a natural family of asymmetric dimension traps leading to a generalization of $P_{\infty}$? 3. (3)

Put

[TABLE]

We have shown that $\delta>0.11802$. What is the actual value of $\delta$? 4. (4)

When considering $\delta(B)$ we restrict our search of a single convex hole $H$. We know that if we have no restrictions at all that the infimum will be [math], as shown in Theorem 2.2. What other restrictions on the search space would result in interesting bounds? For example, as suggested by the referee, what if $H$ was a union of $m$ disjoint convex sets for some fixed $m$?

Some possible directions for future related research include:

1. (1)

A hole $H$ is called supercritical if $\mathcal{J}(H)$ is countable, and $\mathcal{J}(H^{\prime})$ has positive dimension for any $H^{\prime}$ whose closure is in $H$. For the doubling map $T$ there exist non-trivial supercritical holes – see [20]. For instance, let $\mathsf{f}=010010100100101001010\dots$ denote the Fibonacci word defined as $\lim_{n\to\infty}\mathsf{f}_{n}$ given by $\mathsf{f}_{0}=0,\mathsf{f}_{1}=1$ and $\mathsf{f}_{n+1}=\mathsf{f}_{n}\mathsf{f}_{n-1}$ for $n\geq 0$. Then $H=(.01\mathsf{f},.10\mathsf{f})$ is supercritical for $T$. A more general construction involves Sturmian words. Describing supercritical holes for the baker’s map looks like an interesting problem. 2. (2)

One possible generalization of this paper would be the Fibonacci automorphism $F:\mathbb{T}^{2}\to\mathbb{T}^{2}$ given by the formula $F(x,y)=(x+y,y)\bmod\mathbb{Z}^{2}$. Put $\mathfrak{F}=\{(u_{n})\in\{0,1\}^{\mathbb{Z}}:(u_{n},u_{n+1})\neq(1,1)\ \text{for all}\ n\in\mathbb{Z}\}$. Then $(\mathfrak{F},\sigma)$ (equipped with the unique – Markov – measure of maximal entropy) is known to be metrically isomorphic to $(\mathbb{T}^{2},F)$ with the Haar/Lebesgue measure.

It was shown in [21] that there exists an arithmetic map $\varphi:\mathfrak{F}\to\mathbb{T}^{2}$ which conjugates the shift $\sigma$ and $F$ and has an expression which is very similar to $\pi$ given by (2.1). Namely,

[TABLE]

where $\tau=(1+\sqrt{5})/2$, i.e., the golden ratio. Given that the theory of critical holes for the $\beta$-transformation $[0,1)\to[0,1)$ given by $T_{\beta}x=\beta x\bmod 1$ is similar to that for the doubling case if $\beta$ is the golden ratio (see [5, 6]), this suggests that most results of Sections 3 and 4 could be probably transferred to the Fibonacci automorphism. (The holes in question should be probably chosen to be geodesically convex.) In particular, determining a good lower bound for $\delta(F)$ looks like an interesting question. 3. (3)

Other kinds of Pisot toral automorphisms ($=$ algebraic automorphisms of $\mathbb{T}^{m}$ given by matrices whose characteristic polynomial is irreducible and has a Pisot root) have been studied in [22] and [19] (including higher dimensions). Various results concerning $\beta$-transformations with holes – with $\beta>2$ [2] and $\beta<2$ [5, 6] – could be possibly used to build a consistent theory.

8. Appendix

In this section we consider a disconnected hole, for which the survivor set has a really nice structure.

Put

[TABLE]

(see Figure 10). Let $\mathcal{B}$ denote the set of bi-infinite balanced words. (That is, the words in $\{0,1\}^{\mathbb{Z}}$ whose every factor is balanced.) Let $\pi$ be as defined in equation (2.1).

Theorem 8.1**.**

With $\mathcal{H}$ and $\mathcal{B}$ as defined above:

1. (1)

We have

[TABLE] 2. (2)

Furthermore, if $u\in\mathcal{B}$ is aperiodic, then for any $\varepsilon>0$ there exists $n\in\mathbb{Z}$ such that $B^{n}(\pi(u))$ is at a distance less than $\varepsilon$ from the boundary of $\mathcal{H}$, so $\mathcal{H}$ is in some sense optimal with this property.

Proof.

333The proof is a result of a collective effort on MathOverFlow [14], and we are grateful to Ale De Luca and a number of anonymous colleagues who have contributed to it.

(1) Assume first that $u\in\mathcal{B}$. Then for any factor $w$ of $u$ we have that $\widetilde{w}:=REV(w)$ is a factor of $u$ as well (see [13]). Assume without loss of generality that $u_{0}=1,u_{-1}=0$; then we cannot have $u=\dots 0\widetilde{w}0\cdot 1w1\dots$, otherwise $u$ would not be balanced. Hence

[TABLE]

for all $N\in\mathbb{Z}$, which means that the orbit of $u$ stays outside $\mathcal{H}$.

Now assume $u\notin\mathcal{B}$; then there exists a $w$ such that $w=\widetilde{w}$ and that $0w0$ and $1w1$ are factors of $u$ (see [13, Proposition 2.1.3]). If $w$ is of minimal length $n$, say, then $u$ has at most $k+1$ factors of each length $k\leq n+1$ ([13, Proposition 2.1.2]). Moreover, $\{0w0,1w1\}$ are the only unbalanced factors of $u$ of length $\leq n+2$.

Consequently, the set of factors of length $n+2$ is contained in $\{0w0,1w1\}\ \cup\ \mathrm{CYCLE}(w01)\ \cup\ \mathrm{CYCLE}(w10)$, where $\mathrm{CYCLE}(x)$ stands for $x$ and all of its cyclic shifts. Also, every factor of $u$ of length $n+1$ except $0w$ and $1w$ always occurs followed by the same letter.

From this we can derive that there exists an $m\geq 0$ such that every occurrence of $0w0$ is followed by $(1w0)^{m}$ and every occurrence of $1w1$ is followed by $(0w1)^{m}$. If such $m$ is chosen to be maximal, then either

[TABLE]

or

[TABLE]

is a factor of $u$. Hence the orbit of $u$ falls into $\mathcal{H}$, since $[0(w01)^{m}w0\cdot 1w(10w)^{m}1]\subset\mathcal{H}$ as well as its mirror image.

(2) Since $u$ is aperiodic and balanced, it is Sturmian (see [13, Theorem 2.1.5]). Therefore, by the same theorem, it is a cutting sequence with an irrational slope.

Note that if we have a line $y=ax$, then clearly its cutting sequence in both directions is the same, i.e., they are each other’s reverses. Hence for any $n\geq 1$ there exists $\delta>0$ such that if we have $y=ax+\delta$, then $u_{i}=u_{1-i},\ 2\leq i\leq n$. Note also that if our straight line comes close to any grid point in $\mathbb{Z}^{2}$, we must have $u_{1}=1,u_{0}=0$ or vice versa. It follows from Diophantine approximations that any straight line with an irrational slope comes arbitrarily close to $\mathbb{Z}^{2}$, which implies that for any $\delta$ there exists $N\in\mathbb{Z}$ such that

[TABLE]

∎