A Self-Similar Dendrite with One-Point Intersection and Infinite Post-Critical Set

TL;DR

This paper constructs a specific self-similar dendrite in the plane with a one-point intersection property and a dense, countable post-critical set, revealing complex topological and dynamical features.

Contribution

It introduces a novel example of a self-similar system with unique intersection and post-critical set properties not previously documented.

Findings

The attractor is a plane dendrite containing [0,1].

The post-critical set is countable and dense in a Cantor set.

The system exhibits one point intersection property.

Abstract

We build an example of a system $\mathcal{S}$ of similarities in $\mathbb{R}^2$ whose attractor is a plane dendrite $K\supset [0,1]$ which satisfies one point intersection property, while the post-critical set of the system $\mathcal{S}$ is a countable set whose natural projection to $K$ is dense in the middle-third Cantor set.

Equations

$$S_j(x,y)=((x+j)/3,y/3),j=0,1,2\text{mbox}and{S}_h(x,y)=(-hy+c,hx)$$

$$\frac{3^{-n}h}{(c-c_{\mathbf{j}})}<\frac{ch}{h^2}\text{mbox}or{h}^2<3^n(c-c_j)c$$

$$\frac{h^2}{3^n}<(c_{\mathbf{j}}^{\prime }-c)\text{mbox}or{h}^2<3^n(c_{\mathbf{j}}^{\prime }-c)$$

Full text

MSC classification: 28A80

1 Introduction

Let ${\mathcal{S}}=\{S_{1}\ldots S_{m}\}$ be a system of contraction maps in $\mathbb{R}^{n}$. A non-empty compact set $K$ $\in$ $\mathbb{R}^{n}$ satisfying $K=S_{1}(K)\cup\cdots\cup S_{m}(K)$ is called invariant set, or attractor, of the system ${\mathcal{S}}$. The uniqueness and existence of the attractor $K$ is provided by Hutchinson’s Theorem [3].

Let $I=\{1,2,\ldots,m\}$, $I^{*}=\bigcup_{n=1}^{\infty}I^{n}$ be the set of all finite $I$-tuples and $I^{{\infty}}=\{{\bf\alpha}=\alpha_{1}\alpha_{2}\ldots,\ \ \alpha_{i}\in I\}$ be the index space and $\pi:I^{{\infty}}\rightarrow K$ be the address map.

A system ${\mathcal{S}}$ satisfies open set condition(OSC) if there is nonempty open $O$ such that $S_{i}(O){\subset}O$ and $S_{i}(O)\cap S_{j}(O)$ = ${\varnothing}$ for any $i,j\in I,i\neq j$ [3, 5]. We say that the system ${\mathcal{S}}$ satisfies one point intersection property [1] if for any $i,j\in I,i\neq j$, $\#(S_{i}(K)\cap S_{j}(K))\leq 1$.

Let ${\mathcal{C}}$ be the union of all $S_{i}(K)\cap S_{j}(K)$, $i,j\in I,i\neq j$. The post-critical set ${\mathcal{P}}$ of the system ${\mathcal{S}}$ is the set of all $\alpha\in I^{{\infty}}$ such that for some ${\bf j}\in I^{*}$, $S_{\bf j}(\alpha)\in{\mathcal{C}}$ . In other words, ${\mathcal{P}}=\{\sigma^{k}(\alpha)|\alpha\in{\mathcal{C}},k\in\mathbb{N}\}$, where the map $\sigma^{k}:I^{{\infty}}\to I^{{\infty}}$ is defined by $\sigma^{k}(\alpha_{1}\alpha_{2}\ldots)=\alpha_{k+1}\alpha_{k+2}\ldots$.

A system ${\mathcal{S}}$ is called post-critically finite(pcf) [4] if its post-critical set is finite. This obviously implies finite intersection property.

Our aim is to show that the converse need not be true even in the case of plane dendrites. We construct an example of pcf system ${\mathcal{S}}$, whose attractor is a dendrite $K{\subset}\mathbb{R}^{2}$, satisfying one point intersection property.

So we prove the following

Theorem 1**.**

There is a system ${\mathcal{S}}=\{S_{1},S_{2},S_{3},S_{h}\}$ in $\mathbb{R}^{2}$, whose attractor $K$ is a dendrite, which satisfies OSC and 1-point intersection property and has infinite post-critical set whose projection to $K$ is dense in the middle-third Cantor set.

2

Construction

Take a system ${\mathcal{S}}=\{S_{0},S_{1},S_{2},S_{h}\}$ of contraction similarities of $\mathbb{R}^{2}$, defined by

[TABLE]

and let $K$ be the attractor of ${\mathcal{S}}$. Here $c$ is infinite base-3 fraction beginning with 0.11 and containing all finite tuples, consisting of 0 and 2: $c=0.110200220020222\ldots$

We will show that if $h$ is sufficiently small, then all the images  $S_{i_{1}i_{2}\ldots i_{n}h}(K)$, $i_{k}=0,1,2$, are disjoint.

Put $I=\{0,1,2\}$ and denote by $I^{*}=\bigcup_{n=1}^{\infty}I^{n}$ the set of all tuples formed by $\{0,1,2\}$. Consider the images of $c$ under the maps $S_{{{\bf j}}},{{\bf j}}\in I^{*}$. Using base-3 fractions, we can write them as  $S_{{{\bf j}}}(c)=3^{-n}c+{0.j_{1}\ldots j_{n}},\ \ {{\bf j}}\in I^{*}$, so ${(c_{{\bf j}})}_{k}=j_{k}$ for $k\leq n$ and ${(c_{{\bf j}})}_{k}=c_{k-n}$ for $k>n$.

\Delta$$B$$A_{{{\bf j}}}$$h^{2}$$ch$$c$$c_{{{\bf j}}}$$c^{\prime}_{{{\bf j}}}$$\Delta_{{{\bf j}}}$$(0,0)$$(1,0)$$\ (c,h)$$\Delta^{\prime}_{{{\bf j}}}Figure 1.

Let $D$ be a triangle with vertices $\{(0,0),(1,0),(c,h)\}$ and $\Delta=S_{h}(D)$. Since $c$ is not rational, the point $c_{{{\bf j}}}=S_{{{\bf j}}}(c)$ is not equal to $c$ for any ${{\bf j}}\in I^{*}$. So either $c_{{{\bf j}}}<c$ or $c_{{{\bf j}}}>c$.

For $c_{{{\bf j}}}<c$, $\Delta_{{{\bf j}}}\cap\Delta\neq{\varnothing}$ if $A_{{{\bf j}}}(c_{j},h3^{-n})$ lies in $\Delta$. To avoid this, the slope of the line $Bc$ has to be steeper than that of $A_{{{\bf j}}}c$ (See Fig 1.):

[TABLE]

similarly, for $c<c^{\prime}_{{{\bf j}}}$, we have to require that $S_{{{\bf j}}}(B)$ does not lie in $\Delta$

[TABLE]

So we need to estimate $3^{n}(c-c_{{{\bf j}}})$ and $3^{n}(c^{\prime}_{{{\bf j}}}-c)$.

Case 1. If $c_{{{\bf j}}}<c$, there are the following possibilities:

(a) $j_{1}\ldots j_{n}$ = $i_{1}\ldots i_{n}$. Then $(n+1)$-th entry $(c_{{{\bf j}}})_{n+1}=i_{1}=1$ . Since $i_{n+1}>1$, then $i_{n+1}=2$. So $c_{{{\bf j}}}<0.i_{1}\ldots i_{n}12$, $c>0.i_{1}\ldots i_{n}20$, then $c-c_{{{\bf j}}}>{3^{-n-2}}$.

(b) $j_{1}\ldots j_{k}$ = $i_{1}\ldots i_{k}$ for some $k<n$ and $j_{k+1}<i_{k+1}$. Since the only entries allowed here are 0 and 2, so $j_{k+1}=0$ and $i_{k+1}=2$. So $c>0.i_{1}\ldots i_{k}2$ and $c_{{{\bf j}}}<0.i_{1}\ldots i_{k}1$, therefore $c-c_{{{\bf j}}}<3^{-k-1}.$

Case 2. $c<c^{\prime}_{{{\bf j}}}$, then

(a) $j_{1}\ldots j_{n}=i_{1}\ldots i_{n}$. Since $(c_{{{\bf j}}})_{n+1}=i_{1}=1$ , $c<c^{\prime}_{{{\bf j}}}$ implies $i_{n+1}=0$, so $c^{\prime}_{{{\bf j}}}>0.i_{1}\ldots i_{n}1102$, $c<0.i_{1}\ldots i_{n}0(2)$, then $c-c^{\prime}_{{{\bf j}}}>3^{-n-2}$.

(b) $j_{1}\ldots j_{k}=i_{1}\ldots i_{k}$ for some $k<n$ and $j_{k+1}>i_{k+1}$. Then $j_{k+1}=2$, $i_{k+1}=0$, so $c^{\prime}_{{{\bf j}}}>0.i_{1}\ldots i_{k}2$, $c<0.i_{1}\ldots i_{k}1$, so $c^{\prime}_{j}-c>3^{-k-1}$.

(c) if $n=1$ and $j_{1}=1$ i.e. $c_{{{\bf j}}}=0.11102002\ldots$, then $c_{{{\bf j}}}-c>0.0001121$.

Therefore $(c_{{{\bf j}}}-c)3^{n}>4/81.$

(d) if $n=2$ and $j_{1}j_{2}=11$, i.e. $c_{{{\bf j}}}=0.11110200$, we similarly get $c_{{{\bf j}}}-c>0.000201$. Thus, $(c_{{{\bf j}}}-c)3^{n}>2/9.$

So if $h\leq 2/9$ the inequalities (1) and (2) are satisfied. Further we show that if $h\leq 2/9$, the system ${\mathcal{S}}$ satisfies open set condition(OSC) and one point intersection property and the attractor $K$ is a dendrite.

3 Proof of the Theorem.

Lemma 2**.**

If $h\leq 2/9$, for any ${{\bf i}},{{\bf j}}\in I^{*}$, $\Delta_{{{\bf i}}}\bigcap\Delta_{{{\bf j}}}={\varnothing}$.

Proof. Let ${{\bf i}}=i_{1}\ldots i_{n}$, ${{\bf j}}=j_{1}\ldots j_{m}$. If $j_{1}\ldots j_{k}=i_{1}\ldots i_{k}$ for some $k<n$ and $j_{k+1}\neq i_{k+1}$, then $\Delta_{{{\bf i}}}\bigcap\Delta_{{{\bf j}}}=S_{i_{1}\ldots i_{k}}(\Delta_{i_{k+1}\ldots i_{n}}\bigcap\Delta_{j_{k+1}\ldots j_{m}})$. It follows from $\Delta_{i_{k+1}\ldots i_{n}}{\subset}S_{i_{k+1}}(D)$ and $\Delta_{j_{k+1}\ldots j_{m}}{\subset}S_{j_{k+1}}(D)$ that $\Delta_{i_{k+1}\ldots i_{n}}$ and $\Delta_{j_{k+1}\ldots j_{m}}$ are disjoint. Thus $\Delta_{{{\bf i}}}\cap\Delta_{{{\bf j}}}={\varnothing}$.

If $m>n$ and $i_{s}=j_{s}$ for $s=1,\ldots,m$, then $\Delta_{{{\bf i}}}\bigcap\Delta_{{{\bf j}}}=S_{{{\bf i}}}(\Delta\bigcap\Delta_{j_{n+1}\ldots j_{m}})$. By the construction, if $h\leq 2/9$, $\Delta\bigcap\Delta_{j_{n+1}\ldots j_{m}}={\varnothing}$. $\blacksquare$

1/3$$2/3$$D$$\Delta$$D_{0}$$D_{1}$$D_{2}Fig. 2.

Lemma 3**.**

The system ${\mathcal{S}}$ satisfies one point intersection property and open set condition(OSC).

Proof. Let $\dot{D},\dot{\Delta}$ be the interiors of $D$ and $\Delta$. Define $O=\dot{\Delta}\cup\bigcup_{{{\bf i}}\in I^{*}}S_{{{\bf i}}}(\dot{\Delta})$. Obviously, for $i\in I$, $O_{i}=S_{i}(O){\subset}O$. Moreover, $O_{h}=S_{h}(O){\subset}\dot{\Delta}{\subset}O$.

Observe that with the only exception $S_{1}(\dot{D})\cap S_{h}(\dot{D})\neq{\varnothing}$, the sets $S_{0}(\dot{D}),S_{1}(\dot{D})$, $S_{2}(\dot{D})$ and $S_{h}(\dot{D})$ are disjoint. Since $O{\subset}\dot{D}$, the same is true for the sets $O_{0},O_{1},O_{2}$ and $O_{h}$. But $O_{h}{\subset}\dot{\Delta}$, so $O_{h}\bigcap O_{1}={\varnothing}$ too, therefore $O_{0},O_{1},O_{2},O_{h}$ are disjoint and (OSC) is fulfilled.

It follows from Lemma 2 that $\Delta\bigcap S_{1}(\overline{O})=\{c\}$ and therefore $S_{1}(K)\bigcap S_{h}(K)=\{c\}$, which implies one point intersection property.$\blacksquare$

Lemma 4**.**

The system ${\mathcal{S}}$ is post critically infinite and its post critical set is dense in the middle-third Cantor set $\mathcal{C}$.

Proof. Let $y=.y_{1}y_{2}...,y_{i}\in\{0,2\}$ be base 3 representation for some point from the middle-third Cantor set $\mathcal{C}$. Since the representation of $c$ contains all possible tuples of symbols 0 and 2, then for any $n\in\mathbb{N}$ there is $k=k(n)$ such that $c_{k+i}=y_{i}$ for $i=1,\ldots,n$. Therefore $|\sigma_{k}(c)-y|<3^{-n}$. So, the sequence $\sigma_{k(n)}(c)$, converges to the point $y\in\mathcal{C}$. $\blacksquare$

To finish the proof of the Theorem 1, we need only to check that the set $K$ is a dendrite. Let $\Delta_{0}=\bigcup\limits_{{{\bf i}}\in I^{*}}S_{{\bf j}}(\Delta)\cup\Delta\cup[0,1]$. This set is compact and it is simply-connected, because the sets $S_{{\bf j}}(\Delta)$ are disjoint. It is a strong deformation retract of the set $D$. Define $\Delta_{k+1}=\bigcup\limits_{{{\bf i}}\in I^{*}}S_{{\bf j}}*S_{h}(\Delta_{k})\cup S_{h}(\Delta_{k})\cup[0,1]$. The sets $\Delta_{k}$ form a nested sequence of compact simply-connected sets, each being a strong deformation retract of the previous one. Then the intersection $\bigcap\limits_{k=1}^{{\infty}}\Delta_{k}=K$ is a strong deformation retract of the set $D$. By Kigami’s theorem [4] it is locally connected and arcwise connected. Since the interior of $K$ is empty, it contains no simple closed curve, therefore it is a dendrite [2, Theorem 1.1]. $\blacksquare$