Visibility of Cartesian products of Cantor sets

TL;DR

This paper characterizes the set of slopes for which lines are completely outside the Cartesian product of Cantor sets generated by a specific IFS, analyzing its Hausdorff dimension and topological properties.

Contribution

It provides a complete description of the visibility set for Cartesian products of Cantor sets, including its Hausdorff dimension and topological features.

Findings

Determined the Hausdorff dimension of the visibility set V.

Established topological properties of V.

Analyzed a related slicing problem.

Abstract

Let $K_{\lambda}$ be the attractor of the following IFS \begin{equation*} \{f_1(x)=\lambda x, f_2(x)=\lambda x+1-\lambda\}, \;\;0<\lambda<1/2. \end{equation*} Given $\alpha \geq 0$, we say the line $y=\alpha x$ is visible through $K_{\lambda}\times K_{\lambda}$ if $$ \{(x, \alpha x): x\in \mathbb R\setminus \{0\}\}\cap ((K_{\lambda}\times K_{\lambda}))=\emptyset. $$ Let $V=\left \{\alpha \geq 0: y=\alpha x \mbox{ is visible through } K_{\lambda}\times K_{\lambda} \right \}$. In this paper, we give a completed description of $V$, e.g., its Hausdoff dimension and its topological property. Moreover, we also discuss another type of visible problem which is related to the slicing problem.

Equations

$$\{f_1(x)=\lambda x,f_2(x)=\lambda x+1-\lambda \},0<\lambda <1/2.$$

$$\{(x,\alpha x):x\in \mathbb{R}\setminus \{0\}\}\cap ((K_{\lambda }\times K_{\lambda }))=\varnothing .$$

$$\{(x,\alpha x):x\in \mathbb{R}\setminus \{0\}\}\cap F=\varnothing .$$

$$\{f_1(x)=\lambda x,f_2(x)=\lambda x+1-\lambda \},0<\lambda <1/2,$$

$$K_{\lambda }=f_1\left(K_{\lambda }\right)\cup f_2\left(K_{\lambda }\right).$$

$$V=\{\alpha \ge 0:y=\alpha x\text{mbox}isvisiblethrough{K}_{\lambda }\times K_{\lambda }\}.$$

$$\alpha \notin \frac{K_{\lambda }}{K_{\lambda }\setminus \{0\}}:=\left\{\frac{x}{y}:x,y\in K_{\lambda },y\ne 0\right\}.$$

$$V=[0,+\infty )\setminus \frac{K_{\lambda }}{K_{\lambda }\setminus \{0\}}.$$

$$V=(0,+\infty )\setminus \bigcup _{k=-\infty }^{\infty }{\lambda }^k\left[1-\lambda ,\frac{1}{1-\lambda }\right].$$

$$l_{\theta }=\{(x,(\tan \theta )x):x\in \mathbb{R}\}.$$

$$V_{\theta }(K_{\lambda }\times K_{\lambda })=\{(x,y)\in K_{\lambda }\times K_{\lambda }:((x,y)+l_{\theta })\cap (K_{\lambda }\times K_{\lambda })=\{(x,y)\}\}.$$

$$Pro{j}_{\theta }(x,y)=y-x\tan \theta .$$

$$Pro{j}_{\theta }(K_{\lambda }\times K_{\lambda })=\{y-x\tan \theta :(x,y)\in K_{\lambda }\times K_{\lambda }\},$$

$$Pro{j}_{\theta }(V_{\theta }(K_{\lambda }\times K_{\lambda }))=\{y-x\tan \theta :(x,y)\in V_{\theta }(K_{\lambda }\times K_{\lambda })\}.$$

$$g_1(x)$$

$$g_2(x)$$

$$g_3(x)$$

$$g_4(x)$$

$$H_i=g_i(E)\cap g_{i+1}(E),1\le i\le 3,$$

$$T_{i_1\cdots i_v}(a_i)\in H,T_{j_1{j}_2\cdots j_w}(b_i)\in H,$$

$$Pro{j}_{\theta }(V_{\theta }(K_{\lambda }\times K_{\lambda }))=\{a\in [-\tan \theta ,1]:♯\{(x,(\tan \theta )x+a)\cap (K_{\lambda }\times K_{\lambda })\}=1\}$$

$$F(I,J)=F(\tilde{I},\tilde{J}),$$

$$K\cap [A,B]={\cap }_{n=k_0}^{\infty }{G}_n.$$

$$F(K\cap [A,B],K\cap [A,B])={\cap }_{n=k_0}^{\infty }F(G_n,G_n).$$

$$G_n={\cup }_{1\le i\le t_n}{I}_{n,i}\text{mbox}forsome{t}_n\ge 1,$$

$$F(I,J)=F(\tilde{I},\tilde{J}),$$

$$F(G_n,G_n)$$

$$\tilde{I}=[a,a+\lambda t]\cup [a+t-\lambda t,a+t],\tilde{J}=[b,b+\lambda t]\cup [b+t-\lambda t,b+t].$$

$$f(\tilde{I},\tilde{J})=J_1\cup J_2\cup J_3\cup J_4,$$

$$\begin{array}{rl}& J_1=\left[\frac{a}{b+t},\frac{a+\lambda t}{b+t-\lambda t}\right]=:[r_1,s_1]\\ & J_2=\left[\frac{a}{b+\lambda t},\frac{a+\lambda t}{b}\right]=:[r_2,s_2]\\ & J_3=\left[\frac{a+t-\lambda t}{b+t},\frac{a+t}{b+t-\lambda t}\right]=:[r_3,s_3]\\ & J_4=\left[\frac{a+t-\lambda t}{b+\lambda t},\frac{a+t}{b}\right]=:[r_4,s_4].\end{array}$$

$$r_3-r_2=\frac{a+t-\lambda t}{b+t}-\frac{a}{b+\lambda t}=\frac{t(1-\lambda )(b-a+t\lambda )}{\left(b+t\lambda \right)\left(b+t\right)}\ge 0.$$

Full text

[TABLE]

Given $\alpha\geq 0$, we say the line $y=\alpha x$ is visible through $K_{\lambda}\times K_{\lambda}$ if

[TABLE]

Let $V=\left\{\alpha\geq 0:y=\alpha x\mbox{ is visible through }K_{\lambda}\times K_{\lambda}\right\}$. In this paper, we give a completed description of $V$, e.g., its Hausdoff dimension and its topological property. Moreover, we also discuss another type of visible problem which is related to the slicing problems.

1 Introduction

Projections, sections, geodesic curves and visiblity are the main problems in geometry measure theory. It is related to many aspects of fractal geometry, for instance, the arithmetic sum of two self-similar sets is indeed the projectional problem [9, 14]; sections of some fractal sets are connected to the multiple representations of real numbers [12]; geodesic curves on fractal sets are distinct from the classical differential manifolds [6]. For more results on these problems see [19, 20, 3, 22, 21, 23, 18] and references therein. In this paper, we shall consider the visiblity of the Cartesian products of some Cantor sets.

Given $\alpha\geq 0$ and some subset $F\subset\mathbb{R}^{2}$, we say the line $y=\alpha x$ is visible through $F$ if

[TABLE]

The concept of “visiblity” was investigated by many scholars. Nikodym [13] constructed a subset $F$ of $\mathbb{R}^{2}$ such that every point of $F$ is visible from two diametrically opposite directions. In convex geometry, Krasnosel [7] offered a beautiful criterion which enables us to check whether the entire boundary of a compact set of $\mathbb{R}^{2}$ is visible from an interior point. Falconer and Fraser [8] proved that for a class of plane self-similar sets when the attractor $F$ has Hausdorff dimension greater than $1$ then the Hausdorff dimension of the visible subset is $1$. The readers can find more related results in [15, 10, 1, 5].

In this paper, we shall analyze the following self-similar set. Let $K_{\lambda}$ be the attractor with the IFS

[TABLE]

i.e.,

[TABLE]

Let

[TABLE]

It is easy to verify that the line $y=\alpha x$ is visible through $K_{\lambda}\times K_{\lambda}$ if and only if

[TABLE]

Thus,

[TABLE]

By $A^{o}$ we denote the set of interior points of $A$, by $m(A)$ we denote the Lebsgue measure of $A$. In this paper, we obtain the following results.

Theorem 1.1**.**

Let $K_{\lambda}$ be given by (1). Then

• (1)

When $\dfrac{3-\sqrt{5}}{2}\leq\lambda<1/2$, $V=\emptyset$;

• (2)

When $\dfrac{1}{3}\leq\lambda<\dfrac{3-\sqrt{5}}{2}$,

[TABLE]

• (3)

When $0<\lambda<\dfrac{1}{3}$, $V^{o}\neq\emptyset$. In particular, when $\dfrac{1}{4}<\lambda<\dfrac{1}{3}$, $([0,+\infty)\setminus V)^{o}\neq\emptyset$; when $0<\lambda\leq\dfrac{1}{4}$, $m([0,+\infty)\setminus V)=0$ and $\dim_{H}([0,+\infty)\setminus V)=\dfrac{\log 4}{-\log\lambda}$.

There are mainly two types of visible problem [8]. Now, we shall consider another one. First, we introduce some definitions. Let $l_{\theta}$ denote the line going through the origin in direction $\theta\in(0,\pi/2),$ that is,

[TABLE]

Given $\theta\in(0,\pi/2)$. The visible part of $K_{\lambda}\times K_{\lambda}$ is defined as follows:

[TABLE]

Let $(x,y)\in K_{\lambda}\times K_{\lambda}$. Define

[TABLE]

In other words, we project a point $(x,y)$ to the $y$-axis in direction $\theta.$ Moreover, we also define the following sets.

[TABLE]

[TABLE]

Generally, $Proj_{\theta}(K_{\lambda}\times K_{\lambda})$ is not an interval. In what follows, we always assume that $E=Proj_{\theta}(K_{\lambda}\times K_{\lambda})=[-\tan\theta,1]$, which is a natural assumption [3]. Clearly, $E$ is the attractor of the following IFS,

[TABLE]

For the IFS of $E$, i.e. $\{g_{i}\}_{i=1}^{4}$, define $T_{j}(x):=g_{j}^{-1}(x)$ for $x\in g_{j}(E)$ and $1\leq j\leq 4$. We denote the concatenation $T_{i_{n}}\circ\ldots\circ T_{i_{1}}(x)$ by $T_{i_{1}\ldots i_{n}}(x)$. Let

[TABLE]

i.e. we define $H=H_{1}\cup H_{2}\cup H_{3}=[a_{1},b_{1}]\cup[a_{2},b_{2}]\cup[a_{3},b_{3}]$. The following two propositions are motivated by the results in open dynamical systems.

Proposition 1.1**.**

Suppose that $Proj_{\theta}(K_{\lambda}\times K_{\lambda})=[-\tan\theta,1]$. For any $[a_{i},b_{i}],1\leq i\leq 3$, if there are some $i_{1}\cdots i_{v},j_{1}j_{2}\cdots j_{w}$ such that

[TABLE]

then

[TABLE]

is a graph-directed self-similar sets with the strong separation condition, where $\sharp(\cdot)$ denotes the cardinality.

Analogously, we have the following result.

Proposition 1.2**.**

$Proj_{\theta}(K_{\lambda}\times K_{\lambda})=[-\tan\theta,1]$. For any $[a_{i},b_{i}],1\leq i\leq 3$, if all the possible orbits of $a_{i}$ and $b_{i}$ hit finitely many points, then apart from a countable set

[TABLE]

is a graph-directed self-similar sets with the open set condition.

Propositions 1.1 and 1.2 give a sufficient conditition which allows us to calculate the dimension of the slicing set $\{a\in[-\tan\theta,1]:\sharp\{(x,(\tan\theta)x+a)\cap(K_{\lambda}\times K_{\lambda})\}=1\}$.

The paper is arranged as follows. In section 2, we give the proof of Theorem 1.1. In section 3, we give the proofs of Propositions 1.1 and 1.2. Finally, we give some remarks.

2 Proofs of Main results

Before we prove Theorem 1.1, we give some definitions, and prove a useful lemma. Let $E=[0,1]$. For any $(i_{1},\cdots,i_{n})\in\{1,2\}^{n}$, we call $f_{i_{1},\cdots,i_{n}}([0,1])=(f_{i_{1}}\circ\cdots\circ f_{i_{n}})([0,1])$ a basic interval of rank $n,$ which has length $\lambda^{n}$. Denote by $E_{n}$ the collection of all these basic intervals of rank $n$. Suppose $A$ and $B$ are the left and right endpoints of some basic intervals in $E_{k}$ for some $k\geq 1$, respectively. Denote by $G_{n}(\subset E_{n})$ the union of all the basic intervals of rank $n$ which are contained in $[A,B].$ Let $I$ be a basic interval with rank $n$. Define $\widetilde{I}=f_{1}(I)\cup f_{2}(I)$.

Lemma 2.1**.**

Let $F:U\rightarrow\mathbb{R}$ be a continuous function, where $U\subset\mathbb{R}^{2}$ is a non-empty open set. Suppose $A$ and $B$ are the left and right endpoints of some basic intervals in $G_{k_{0}}$ for some $k_{0}\geq 1$ respectively such that $[A,B]\times[A,B]\subset U.$ Then $K\cap[A,B]=\cap_{n={k_{0}}}^{\infty}G_{n}$. Moreover, if for any $n\geq k_{0}$ and any two basic intervals $I,J\subset G_{n}$, such that

[TABLE]

then $F(K\cap[A,B],K\cap[A,B])=F(G_{k_{0}},G_{k_{0}}).$

Proof.

By the construction of $G_{n}$, i.e. $G_{n+1}\subset G_{n}$ for any $n\geq k_{0}$, it follows that

[TABLE]

The continuity of $F$ yields that

[TABLE]

Without loss of generality, we may assume that

[TABLE]

where $I_{n,i}$ is a basic interval in $G_{n}$. By the condition in lemma, i.e. for any $n\geq k_{0}$ and any two basic intervals $I,J\subset G_{n}$, such that

[TABLE]

it follows that

[TABLE]

Therefore, $F(K\cap[A,B],K\cap[A,B])=F(G_{k_{0}},G_{k_{0}}).$ ∎

Lemma 2.2**.**

Let $f(x,y)=\dfrac{x}{y}$, and $I=[a,a+t],J=[b,b+t]$ be two basic intervals. If $1/3\leq\lambda<1/2$, and $b\geq a\geq 1-\lambda,$ then $f(\tilde{I},\tilde{J})=f(I,J).$

Proof.

Note that

[TABLE]

Therefore,

[TABLE]

where

[TABLE]

Note that $f(I,J)=[r_{1},s_{4}]$. In the following, we verify that $f(I,J)=J_{1}\cup J_{2}\cup J_{3}\cup J_{4}$

Since $b\geq a\geq 1-\lambda$ and $\lambda\geq\frac{1}{3}$, we have

[TABLE]

Now it suffices to check that

[TABLE]

We have

[TABLE]

and

[TABLE]

Finally,

[TABLE]

If $b\neq a$, then $b>a+t$. Therefore, we have

[TABLE]

which leads to $s_{3}-r_{4}\geq 0$. However, if $a=b$, then

[TABLE]

Thus, we finish checking that $f(I,J)=J_{1}\cup J_{2}\cup J_{3}\cup J_{4}=[r_{1},s_{4}]=\left[\dfrac{a}{b+t},\dfrac{a+t}{b}\right]$. ∎

Lemma 2.3**.**

We have

[TABLE]

Proof.

From Lemmas 2.1 and 2.2 it follows that if $\lambda\geq\frac{1}{3}$

[TABLE]

Each $x\in K_{\lambda}$ can be uniquely represented as

[TABLE]

Note that $x\in f_{2}(K_{\lambda})$ if and only if $x_{1}=1-\lambda$. Thus each $x\in K_{\lambda}\setminus\{0\}$ is of form

[TABLE]

Thus for any two $x=\lambda^{m}x^{*},y=\lambda^{n}y^{*}\in K_{\lambda}\setminus\{0\}$ with $x^{*},y^{*}\in f_{2}(K_{\lambda})$ one has

[TABLE]

Thus

[TABLE]

It is easy to check that $\{0\}\cup\bigcup_{k=-\infty}^{\infty}\lambda^{k}\left[1-\lambda,\dfrac{1}{1-\lambda}\right]=[0,+\infty)$ when $\dfrac{3-\sqrt{5}}{2}\leq\lambda<1/2$, and intervals $\lambda^{k}\left[1-\lambda,\dfrac{1}{1-\lambda}\right]$ are pairwise disjoint when $1/3\leq\lambda<\dfrac{3-\sqrt{5}}{2}$. ∎

Pourbarat [16], making use of the thickness of the Cantor sets, proved the following result.

Theorem 2.5**.**

If $\dfrac{\lambda^{2}}{(1-2\lambda)^{2}}>\lambda$, then $\dfrac{K_{\lambda}}{K_{\lambda}\setminus\{0\}}$ contains an interior point.

Lemma 2.6**.**

If $\dfrac{1}{4}<\lambda<1/3$, then $\dfrac{K_{\lambda}}{K_{\lambda}\setminus\{0\}}$ contains an interior point.

Proof.

If $\dfrac{1}{4}<\lambda<1/3$, then $\dfrac{\lambda^{2}}{(1-2\lambda)^{2}}>\lambda$. Therefore, $\dfrac{K_{\lambda}}{K_{\lambda}\setminus\{0\}}$ contains an interior point by Theorem 2.5. ∎

Lemma 2.7**.**

If $0<\lambda<\dfrac{3-\sqrt{5}}{2}$, then $V$ has an interior point.

Proof.

Note that $f_{2}(K_{\lambda})\subset[1-\lambda,1]$. Thus, by the argument in Lemma 2.3, we have

[TABLE]

Note that the intervals $\left[\lambda^{k}(1-\lambda),\dfrac{\lambda^{k}}{1-\lambda}\right]$ for $k\in\mathbb{Z}$ are pairwise disjoint when $0<\lambda<\dfrac{3-\sqrt{5}}{2}$. Therefore, $V$ has an interior point by (2). ∎

Lemma 2.8**.**

If $0<\lambda<\dfrac{1}{4}$, then $\dfrac{K_{\lambda}}{K_{\lambda}\setminus\{0\}}$ has Lebesgue measure zero.

Proof.

If $0<\lambda<\dfrac{1}{4}$, then

[TABLE]

We note that for any $X,\,Y\subseteq\mathbb{R}$, we have $Y-X=\Pi_{\frac{\pi}{4}}(X\times Y)$, where $\Pi_{\frac{\pi}{4}}(X\times Y)$ denotes the projection of $X\times Y$ on the $y$ axis along lines having $\frac{\pi}{4}$ angle with the $x$ axis. Therefore,

[TABLE]

where we use the fact that $\dim_{H}E=\dim_{H}\ln E$ and $\dim_{P}E=\dim_{P}\ln E$. for a bounded set $E\subset(0,+\infty)$. Thus, $\dfrac{K_{\lambda}}{K_{\lambda}\setminus\{0\}}$ has Lebesgue measure zero. ∎

We will use a result given by Simon and Solomyak [17].

Theorem 2.10**.**

Let $\Lambda$ be a self-similar $1$-set in $\mathbb{R}^{2}$ with the open set condition, which is not on a line. Then

[TABLE]

where

[TABLE]

Lemma 2.9**.**

$\dfrac{K_{1/4}}{K_{1/4}\setminus\{0\}}$* has Lebesgue measure zero.*

Proof.

Note that when $\lambda=1/4$, $\Lambda=K_{\lambda}\times K_{\lambda}$ is a self-similar set with the following IFS

[TABLE]

[TABLE]

Clearly, the above IFS satisfies the open set condition. Therefore, the Hausdorff dimension of $\Lambda$ is 1, and $0<\mathcal{H}^{1}(\Lambda)<\infty$. Let

[TABLE]

The Lebesgue measure of $\Gamma$ is [math] due to Theorem 2.10. Let

[TABLE]

Clearly, $m(\Gamma_{1})=m(\Gamma)=0$. The metric on $\Gamma_{1}$, denoted by $d_{1}$, is the arc metric. It is well known that on $S^{1}$, the arc metric is equivalent to the Euclidean metric. Let

[TABLE]

The metric on $\Gamma_{2}$ the Euclidean metric (we denote it by $d_{2}$). We define the the map

[TABLE]

by

[TABLE]

The map $\phi$ is indeed mapping a point on $S^{1}$ into its associated polar angle in the polar coordinate system. Therefore, we may define $\phi$ in another way as follows: define

[TABLE]

Clearly, $\phi$ is well-defined, and it is a bijection. Moreover, we shall prove that $\phi$ is a Lipschitz map, i.e. there exists some constant $L>0$ such that

[TABLE]

Note that $d_{2}(\phi(\vec{a}),\phi(\vec{b}))=d_{2}(\theta_{\vec{a}},\theta_{\vec{b}})$, and that

[TABLE]

Now, $m(\Gamma_{2})=0$ follows from $\phi(\Gamma_{1})=\Gamma_{2}$, $m(\Gamma_{1})=0$, and $\phi$ is Lipschitz. Theorefore, $m\left(\dfrac{K_{1/4}}{K_{1/4}\setminus\{0\}}\right)=0.$ ∎

The following is from Bárány [4].

Theorem 2.11**.**

Let $\Lambda$ be an arbitrary self-similar set in $\mathbb{R}^{2}$ not contain in any line. Suppose that $g:\mathbb{R}^{2}\to\mathbb{R}$ is a $C^{2}$ map such that

[TABLE]

for any $(x,y)\in\Lambda$. Then

[TABLE]

Lemma 2.10**.**

When $0<\lambda\leq\dfrac{1}{4}$, $\dim_{H}([0,+\infty)\setminus V)=\dfrac{\log 4}{-\log\lambda}$.

Proof.

By the argument in Lemma 2.3, we have

[TABLE]

Thus

[TABLE]

Clearly, $\Lambda=f_{2}(K_{\lambda})\times f_{2}(K_{\lambda})$ is a two-dimensional self-similar set which is not contained in any line. Let $g(x,y)=\dfrac{x}{y}$, then

[TABLE]

for any $(x,y)\in\Lambda$. Therefore, in terms of Theorem 2.11,

[TABLE]

Hence, if $0<\lambda\leq 1/4$, then

[TABLE]

∎

Proof of Theorem 1.1.

Theorem 1.1 (1) and (2) follows from Lemmas 2.3. Theorem 1.1 (3) follows from 2.6, 2.7, 2.8, 2.9 and 2.10. ∎

3 Visible sets, slicing sets and open dynamical systems

In this section, we give the proofs of Propositions 1.1 and 1.2. Define

[TABLE]

It is easy to see that the length of $P_{ij}$ is $\lambda(1+\tan\theta).$ In what follows, we always assume that $E=Proj_{\theta}(K_{\lambda}\times K_{\lambda})=[-\tan\theta,1]$. Clearly, in terms of $P_{ij}$, $E$ is the attractor of the following IFS,

[TABLE]

In other words,

[TABLE]

For any $x\in E$, there exists a sequence $(i_{n})_{n=1}^{\infty}\in\{1,\ldots,4\}^{\mathbb{N}}$ such that

[TABLE]

We call such a sequence a coding of $x$. Usually, the coding of $x$ is not unique. If $x$ has a unique coding, then we call $x$ a univoque point. Write $U_{1}$ for all the univoque points of $E$ with respect to the IFS $\{g_{i}\}_{i=1}^{4}.$

The following result is proved in [2, Lemma 2.1] which states that any self-similar set can be regarded as a topological dynamical system. For the IFS of $E$, i.e. $\{g_{i}\}_{i=1}^{4}$, define $T_{j}(x):=g_{j}^{-1}(x)$ for $x\in g_{j}(E)$ and $1\leq j\leq 4$. We denote the concatenation $T_{i_{n}}\circ\ldots\circ T_{i_{1}}(x)$ by $T_{i_{1}\ldots i_{n}}(x)$.

Lemma 3.1**.**

Let $x\in K.$ Then $(i_{n})_{n=1}^{\infty}\in\{1,\ldots,m\}^{\mathbb{N}}$ is a coding for $x$ if and only if $T_{i_{1}\ldots i_{n}}(x)\in K$ for all $n\in\mathbb{N}.$

Motivated by Lemma 3.1, we may define the orbits of the points of $K$.

Definition 3.2**.**

Let $x\in K$ with a coding $(i_{n})_{n=1}^{\infty}$, we call the set

[TABLE]

an orbit set of $x$, where $T_{i_{0}}(x)=x.$

It is easy to see that for different codings, the orbits of $x$ may be distinct.

The set $Proj_{\theta}(V_{\theta}(K_{\lambda}\times K_{\lambda}))$ can be viewed as a slicing set, i.e.

[TABLE]

where $\sharp(\cdot)$ denotes the cardinality. The following lemma is trivial.

Lemma 3.2**.**

Suppose that $Proj_{\theta}(K_{\lambda}\times K_{\lambda})=[-\tan\theta,1]$. Then

[TABLE]

is exactly the univoque set of $E$ under the IFS $\{g_{i}\}_{i=1}^{4},$ i.e.

[TABLE]

Proof.

The proof is trivial. We leave it to readers. ∎

If $\theta\in[\pi/4,\pi/2)$, let $H_{i}=g_{i}(E)\cap g_{i+1}(E),1\leq i\leq 3$, i.e.

[TABLE]

Let $H=H_{1}\cup H_{2}\cup H_{3}=[a_{1},b_{1}]\cup[a_{2},b_{2}]\cup[a_{3},b_{3}]$. Similarly, if $\theta\in(0,\pi/4)$, then we can also define the $H_{i}=[a_{i},b_{i}],1\leq i\leq 3.$ Now the following result is a corollary of the main result of [2].

Proposition 3.1**.**

Suppose that $Proj_{\theta}(K_{\lambda}\times K_{\lambda})=[-\tan\theta,1]$. For any $[a_{i},b_{i}],1\leq i\leq 3$, if there are some $i_{1}\cdots i_{v},j_{1}j_{2}\cdots j_{w}$ such that

[TABLE]

then

[TABLE]

is a graph-directed self-similar sets with the strong separation condition.

The following is a corollary of the main result of [11].

Proposition 3.2**.**

$Proj_{\theta}(K_{\lambda}\times K_{\lambda})=[-\tan\theta,1]$. For any $[a_{i},b_{i}],1\leq i\leq 3$, if all the possible orbits of $a_{i}$ and $b_{i}$ hit finitely many points, then apart from a countable set

[TABLE]

is a graph-directed self-similar sets with the open set condition.

4 Some remarks

The main idea of this paper is to establish a connection between the visible problem and arithmetic on the fractal sets. Our idea can be implemented for other overlapping self-similar sets. Similar results can be obtained if we replace the line $y=\alpha x$ in the definition of $V$, by some parabolic curves or hyperbolic curves. Nevertheless, for these cases, the analysis can be difficult. We shall discuss these problems in another paper.

Acknowledgements

The work is supported by National Natural Science Foundation of China (Nos.11701302,

11671147). The work is also supported by K.C. Wong Magna Fund in Ningbo University.