Arithmetic on Moran sets

TL;DR

This paper investigates the arithmetic properties of Moran sets, establishing conditions under which the continuous image of two such sets contains an interior, based on derivatives of a function and the structure of the sets.

Contribution

It provides new conditions involving derivatives and Moran set parameters that guarantee the image of two Moran sets has an interior, extending understanding of their arithmetic properties.

Findings

The image of Moran sets under certain functions contains an interior.

Conditions relate derivatives of the function to Moran set parameters.

Results apply to a broad class of Moran sets with convex hull [0,1].

Abstract

Let $(\mathcal{M}, c_k,n_k)$ be a class of Moran sets. We assume that the convex hull of any $E\in (\mathcal{M}, c_k,n_k)$ is $[0,1]$. Let $A,B$ be two non-empty sets in $\mathbb{R}$. Suppose that $f$ is a continuous function defined on an open set $U\subset \mathbb{R}^{2}$. Denote the continuous image of $f$ by \begin{equation*} f_{U}(A,B)=\{f(x,y):(x,y)\in (A\times B)\cap U\}. \end{equation*} In this paper, we prove the following result. Let $E_1,E_2\in(\mathcal{M}, c_k, n_k)$. If there exists some $(x_0,y_0)\in (E_1\times E_2)\cap U$ such that $$\sup_{k\geq 1}\left\{1-c_kn_k\right\}<\left\vert \frac{\partial _{y}f|_{(x_{0},y_{0})}}{\partial _{x}f|_{(x_{0},y_{0})}}\right\vert <\inf_{k\geq 1}\left\{\dfrac{c_k}{1-n_kc_k}\right\},$$ then $f_U(E_1, E_2)$ contains an interior.

Equations

$$f_U(A,B)=\{f(x,y):(x,y)\in (A\times B)\cap U\}.$$

$$\underset{k\ge 1}{\sup }\left\{1-c_k{n}_k\right\}<\left|\frac{{\partial }_{y}f{|}_{(x_0,y_0)}}{{\partial }_{x}f{|}_{(x_0,y_0)}}\right|<\underset{k\ge 1}{\inf }\left\{\frac{c_k}{1-n_k{c}_k}\right\}.$$

$$f_U(A,B)=\{f(x,y):(x,y)\in (A\times B)\cap U\}.$$

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

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

$$S_x=\left\{(y_1,y_2):y_1-y_2=x,(y_1,y_2)\in C\times C\right\}.$$

$$U_r=\{x:♯(S_x)=r\},r\in {\mathbb{N}}^{+}.$$

$$0<{\mathcal{H}}^s(U_1)<\infty ,{\mathcal{H}}^s(U_{2^k})=\infty ,k\in {\mathbb{N}}^{+},$$

$$\{f_1(x)=\lambda x,f_2(x)=\lambda x+c-\lambda ,f_3(x)=\lambda x+1-\lambda \},$$

$$\sqrt{K}+\sqrt{K}=[0,2]$$

$$\sqrt{c}+1\ge 2\sqrt{1-\lambda },$$

$$\frac{K}{K}=\left\{\frac{x}{y}:x,y\in K,y\ne 0\right\}=\left[0,\infty \right).$$

$$u=x_1+x_2=x_3-x_4=x_5\cdot x_6=x_7÷x_8=\sqrt{x_9}+\sqrt{x_{10}};$$

$$D_k=\{({\sigma }_1,\cdots ,{\sigma }_k):{\sigma }_j\in {\mathbb{N}}^{+},1\le {\sigma }_j\le n_j,1\le j\le k\}.$$

$$D={\cup }_{k\ge 0}{D}_k.$$

$$\mathcal{F}=\{T_{\sigma }\subset T:\sigma \in D\}$$

$$int(T_{\sigma \ast i})\cap int(T_{\sigma \ast j})=\varnothing ,i\ne j,$$

$$E={\cap }_{k\ge 1}^{\infty }{\cup }_{\sigma \in D_k}{T}_{\sigma }$$

$$\underset{k}{\sup }\left\{1-c_k{n}_k\right\}<\left|\frac{{\partial }_{y}f{|}_{(x_0,y_0)}}{{\partial }_{x}f{|}_{(x_0,y_0)}}\right|<\underset{k}{\inf }\left\{\frac{c_k}{1-n_k{c}_k}\right\}.$$

$$E_k={\cup }_{\sigma \in D_k}{T}_{\sigma },E={\cap }_{k=1}^{\infty }{E}_k,$$

$$E_1={\cap }_{k=1}^{\infty }{E}_k^{(1)},E_2={\cap }_{k=1}^{\infty }{E}_k^{(2)}.$$

$$F(I_1,I_2)=F(\widetilde{I_1},\widetilde{I_2}),$$

$$E_1\cap [A,B]={\cap }_{n=k_0}^{\infty }{G}_n^{(1)},E_2\cap [C,D]={\cap }_{n=k_0}^{\infty }{G}_n^{(2)}.$$

$$F(E_1\cap [A,B],E_2\cap [C,D])={\cap }_{n=k_0}^{\infty }F(G_n^{(1)},G_n^{(2)}).$$

$$F(G_n^{(1)},G_n^{(2)})$$

$$\underset{k}{\sup }\left\{1-c_k{n}_k\right\}<\left|\frac{{\partial }_{y}f{|}_{(x_0,y_0)}}{{\partial }_{x}f{|}_{(x_0,y_0)}}\right|<\underset{k}{\inf }\left\{\frac{c_k}{1-n_k{c}_k}\right\}.$$

$$\tilde{I}={\cup }_{i=1}^{n_k}{I}_i,\tilde{J}={\cup }_{j=1}^{n_k}{J}_j.$$

$$f(\tilde{I},\tilde{J})={\cup }_{i=1}^{n_k}{\cup }_{j=1}^{n_k}f(I_i,J_j).$$

$$(c_1\cdots c_k){\partial }_{x}f(\xi ,\eta )\ge (c_1{c}_2\cdots c_{k-1}-n_k{c}_1\cdots c_k){\partial }_{y}f(\xi ,\eta ).$$

$$\frac{{\partial }_{y}f{|}_{(x_0,y_0)}}{{\partial }_{x}f{|}_{(x_0,y_0)}}<\underset{k}{\inf }\left\{\frac{c_k}{1-n_k{c}_k}\right\},$$

Full text

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In this paper, we prove the following result. Let $E_{1},E_{2}\in(\mathcal{M},c_{k},n_{k})$. If there exists some $(x_{0},y_{0})\in(E_{1}\times E_{2})\cap U$ such that

[TABLE]

Then $f_{U}(E_{1},E_{2})$ contains an interior.

1 Introduction

Given two non-empty sets $A,B\subset\mathbb{R}$. Define $A*B=\{x*y:x\in A,y\in B\}$, where $*$ is $+,-,\times$ or $\div$ (when $*=\div$, $y\neq 0$). We call $A*B$ the arithmetic on $A$ and $B$. Generally, we may define the arithmetic on $A$ and $B$ in terms of some functions. Suppose that $f$ is a continuous function defined on an open set $U\subset\mathbb{R}^{2}$. Denote the continuous image of $f$ by

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For simplicity, we still call $f_{U}(A,B)$ the arithmetic on $A$ and $B$. Arithmetic on the fractal sets has strong connections with many different problems in geometry measure theory and dynamical systems [30, 26]. For instance, in geometry measure theory, the visible problem is related to the division on the fractals [6, 11, 18]. The main reason is due to the following observation. Let $K\subset[0,1]$ be a fractal set. Given $\alpha\geq 0$, we say the line $y=\alpha x$ is visible through $K\times K$ if

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

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The arithmetic sum of two Cantor sets was studied by many scholars. There are many results concerning with this topic, see [2, 3, 4, 8, 12, 15, 20] and references therein. It is an important problem in homoclinic bifurcations [19]. Palis [19] posed the following problem: whether it is true (at least generically) that the arithmetic sum of dynamically defined Cantor sets either has measure zero or contains an interval. This conjecture was solved in [2]. Motivated by Palis’ conjecture, it is natural to investigate when the sum of two Cantor sets contains some interiors. Newhouse [27] proved the following thickness theorem. Given any two Cantor sets $C_{1}$ and $C_{2}$, if $\tau(C_{1})\tau(C_{2})>1$, where $\tau(C_{i}),i=1,2$ denotes the thickness of $C_{i},i=1,2,$ then $C_{1}+C_{2}$ contains some interiors. However, Newhouse thickness theorem cannot handle a general function $f$, i.e. whether $f(C_{1},C_{2})$ contains an interior or not.

To date, there are not so many results concerning with the arithmetic on the fractal sets [1, 23, 24]. The first result of this direction, to the best of our knowledge, is due to Steinhaus [23] who proved the following interesting result: $C-C=[-1,1]$, where $C$ is the middle-third Cantor set. Equivalently, Steinhaus proved that for any $x\in[-1,1]$, there are some $x_{1},x_{2}\in C$ such that $x=x_{1}-x_{2}.$ Recently, Athreya, Reznick and Tyson [1] considered the multiplication on the middle-third Cantor set. They proved that $17/21\leq\mathcal{L}(C\cdot C)\leq 8/9$, where $\mathcal{L}$ denotes the Lebesgue measure. Jiang and Xi [13] proved that $C\cdot C$ indeed contains infinitely many intervals. In [14], Jiang and Xi considered the representations of real numbers in $C-C=[-1,1]$, i.e. let $x\in[-1,1]$, define

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and

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They proved that $\dim_{H}(U_{r})=\dfrac{\log 2}{\log 3}$ if $r=2^{k}$ for some $k\in\mathbb{N}$. Moreover,

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where $s=\dfrac{\log 2}{\log 3}$. $U_{3\cdot 2^{k}}$ is an infinitely countable set for any $k\geq 1$, where $\dim_{H}$ and $\mathcal{H}^{s}$ denote the Hausdorff dimension and Hausdorff measure, respectively. For more results, see [14]. In [25], Tian et al. defined a class of overlapping self-similar sets as follows: let $K$ be the attractor of the IFS

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where $f_{1}(I)\cap f_{2}(I)\neq\emptyset,(f_{1}(I)\cup f_{2}(I))\cap f_{3}(I)=\emptyset,$ and $I=[0,1]$ is the convex hull of $K$. This class of self-similar set was investigated by many scholars, see [7, 9, 16, 17, 28, 29, 30]. Tian et al. $K\cdot K=[0,1]$ if and only if $(1-\lambda)^{2}\leq c$. Equivalently, they gave a necessary and sufficient condition such that for any $x\in[0,1]$ there exist some $y,z\in K$ such that $x=yz$. Moreover, Ren, Zhu, Tian and Jiang [21] proved that

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if and only if

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where $\sqrt{K}+\sqrt{K}=\{\sqrt{x}+\sqrt{y}:x,y\in K\}$. If $c\geq(1-\lambda)^{2}$, then

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As a consequence, they proved that the following conditions are equivalent:

• (1)

For any $u\in[0,1]$, there are some $x,y\in K$ such that $u=x\cdot y;$

• (2)

For any $u\in[0,1]$, there are some $x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},x_{7},x_{8},x_{9},x_{10}\in K$ such that

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• (3)

$c\geq(1-\lambda)^{2}$.

In this paper, we shall consider similar problems on the Moran sets. The Moran sets are, in certain sense, random. Nevertheless, any self-similar set with the open set condition is a Moran set [10]. Now we give the definition of a class of Moran set. Let $\{n_{k}\}\subset\mathbb{N}^{+}$ be a sequence(we assmue that $n_{k}\geq 2$). For any $k\in\mathbb{N}^{+}$, write

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Define

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We call $\sigma\in D$ a word. For simplicity, we let $D_{0}=\emptyset.$ If $\sigma=(\sigma_{1},\cdots,\sigma_{k})\in D_{k}$, $\tau=(\tau_{1},\cdots,\tau_{m})\in D_{m}$, then we define the concatenation $\sigma*\tau=(\sigma_{1},\cdots,\sigma_{k},\tau_{1},\cdots,\tau_{m})\in D_{k+m}$. Let $T=[0,1]$ and $\{c_{k}\}$ be a positive real sequence with $c_{k}n_{k}<1,k\in\mathbb{N}^{+}$, we say the class

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has the Moran structure if the following conditions are satisfied:

• (1)

for any $\sigma\in D$, $T_{\sigma}$ is similar to $T$, i.e. there exists a similitude $S_{\sigma}:\mathbb{R}\to\mathbb{R}$ such that $S_{\sigma}(T)=T_{\sigma};$

• (2)

for any $k\geq 0$ and $\sigma\in D_{k}$, $T_{\sigma*1},T_{\sigma*2},\cdots,T_{\sigma*n_{k+1}}$ is a subset of $T_{\sigma}$ and

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where $int(A)$ denotes the interior of $A$, for simplicity, we denote by $\widetilde{T_{\sigma}}=\cup_{i=1}^{n_{k+1}}T_{\sigma*i}$;

• (3)

for any $k\geq 1$ and $\sigma\in D_{k-1}$, $\dfrac{|T_{\sigma*i}|}{|T_{\sigma}|}=c_{k}$, and the convex hull of $T_{\sigma*i}$ and $T_{\sigma}$ coincide for any $1\leq i\leq n_{k}$, where $|A|$ denotes the diameter of $A$.

Suppose $\mathcal{F}=\{T_{\sigma}\subset T:\sigma\in D\}$ has the Moran structure, then we call

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a Moran set. We denote by $(\mathcal{M},c_{k},n_{k})$ all the Moran sets generated by the Moran structure $\mathcal{F}.$ By the third condition, it is easy to see that the convex hull of any $E$ from $(\mathcal{M},c_{k},n_{k})$ is $[0,1].$

Now we are ready to state the main result of this paper.

Theorem 1.1**.**

Let $E_{1},E_{2}\in(\mathcal{M},c_{k},n_{k})$. If there exists some $(x_{0},y_{0})\in(E_{1}\times E_{2})\cap U$ such that

[TABLE]

Then $f_{U}(E_{1},E_{2})$ contains an interior.

The paper is arranged as follows. In section 2, we prove two basic lemmas and give a proof of Theorem 1.1. In section 3, we give some remarks.

2 Proof of Theorem 1.1

In this section, we shall prove Theorem 1.1. First, we give some definitions and prove two useful lemmas.

For any $k\geq 1$, denote by $E_{k}$ the union of basic intervals when we construct a Moran set $E$, i.e.

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where $T_{\sigma}$ is called a basic interval with rank $k$. It is easy to check that the length of any basic interval with rank $k$ is $c_{1}c_{2}\cdots c_{k}.$ Let $[A,B]\subset[0,1]$, where $A$ and $B$ are the left and right endpoints of some basic intervals in $E_{k}$ for some $k\geq 1$, respectively. $A$ and $B$ may not in the same basic interval. In the following lemma, we choose $A$ and $B$ in this way. Let $F_{k}$ be the collection of all the basic intervals in $[A,B]$ with length $c_{1}c_{2}\cdots c_{k},k\geq k_{0}$ for some $k_{0}\in\mathbb{N}^{+}$, i.e. the union of all the elements of $F_{k}$ is denoted by $G_{k}=\cup_{i=1}^{t_{k}}I_{k,i}$, where $t_{k}\in\mathbb{N}^{+}$, $I_{k,i}\subset E_{k}\cap[A,B]$. Clearly, by the definition of $G_{n}$, it follows that $G_{n+1}\subset G_{n}$ for any $n\geq k_{0}.$

Lemma 2.1**.**

Let $E_{1},E_{2}\in(\mathcal{M},c_{k},n_{k})$, i.e.

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Assume $F:\mathbb{R}^{2}\to\mathbb{R}$ is a continuous function. Suppose $A$ and $B$ ($C$ and $D$) are the left and right endpoints of some basic intervals in $E_{k_{0}}^{(1)}$($E_{k_{0}}^{(2)}$) for some $k_{0}\geq 1$, respectively. Then $E_{1}\cap[A,B]=\cap_{n={k_{0}}}^{\infty}G_{n}^{(1)},E_{2}\cap[C,D]=\cap_{n={k_{0}}}^{\infty}G_{n}^{(2)}$. Moreover, if for any $n\geq k_{0}$ and any basic intervals $I_{1}\subset G_{n}^{(1)},I_{2}\subset G_{n}^{(2)}$, we have

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then $F(E_{1}\cap[A,B],E_{2}\cap[C,D])=F(G_{k_{0}}^{(1)},G_{k_{0}}^{(2)}).$

Proof.

We assume that $G_{n}^{(i)}=\cup_{1\leq j\leq t_{n}^{(i)}}I_{n,j},i=1,2.$ By the construction of $G_{n}^{(i)},i=1,2$, it is clear that $G_{n+1}^{(i)}\subset G_{n}^{(i)}$ for any $n\geq 1.$ Therefore,

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In terms of the continuity of $F$, we conclude that

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Therefore,

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Consequently, $F(E_{1}\cap[A,B],E_{2}\cap[C,D])=F(G_{k_{0}}^{(1)},G_{k_{0}}^{(2)})$ follows immediately from the identity (1) and $F(G_{n}^{(1)},G_{n}^{(2)})=F(G_{n+1}^{(1)},G_{n+1}^{(2)})$ for any $n\geq k_{0}.$ ∎

Lemma 2.2**.**

Let $I=[a,a+t],J=[b,b+t]$ be two basic intervals in $G_{k-1}^{(1)}$ and $G_{k-1}^{(2)}$, respectively. If there exists some $(x_{0},y_{0})\in(E_{1}\times E_{2})\cap(I\times J)\cap U$ such that

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Then $f(I,J)=f(\widetilde{I},\widetilde{J})$.

Proof.

Without loss of generality, we assume that $\partial_{x}f|_{(x_{0},y_{0})}>0,\partial_{y}f|_{(x_{0},y_{0})}>0$. For other cases, we may consider the new function $F(x,y)=f(x,1-y)$ or $-f(x,y)$. By the definition of $\widetilde{I}$ and $\widetilde{J}$, we have

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Moreover, $t=|I|=|J|=c_{1}\cdots c_{k-1}$, where $|A|$ denotes the length of $A$. Therefore, we have

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We first prove that for any $1\leq i\leq n_{k}$, $\cup_{j=1}^{n_{k}}f(I_{i},J_{j})$ is an interval. By the construction of Moran set, it suffices to prove that $f(P_{1})\geq f(P_{2})$, see the second picture of Figure 1, that is, it remains to prove that there exists some $(\xi,\eta)\in E_{1}\times E_{2}$ contained in the neighbour of $(x_{0},y_{0})$ such that

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However, this is clear due to the condition

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and the assumption $\partial_{x}f,\partial_{y}f$ are continuous. Next, we prove that

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is an interval. Analogously, we need to show that $f(P_{3})\geq f(P_{4})$, see the third picture of Figure 1. Indeed, it only remains to prove that there is some $(\xi_{1},\eta_{1})\in E_{1}\times E_{2}$ which lies in the neighbour of $(x_{0},y_{0})$ such that

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However, the above inequality follows from the condition

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and $\partial_{x}f,\partial_{y}f$ are continuous. Therefore, we have proved that $f(I,J)=f(\widetilde{I},\widetilde{J})$. ∎

Proof of Theorem 1.1.

Theorem 1.1 follows immediately from Lemmas 2.1 and 2.2. ∎

3 Final remark

In Lemma 2.1, we note that if some basic intervals of $E_{k}$ intersects, then similar result as Theorem 1.1 can be obtained. We leave it to the readers.

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.