Minimal Cantor Type Sets on Discrete Dynamical Systems

Abstract

Cantor sets on topological spaces can be obtained as minimal sets with respect to discrete dynamical systems (d.d.s.), $(X, f)$ where X is a compact metric space and f could be an homeomorphism or a non-invertible continuous map. In general, given a Cantor type set or simply a Cantor set in a compact metric space, it is difficult to obtain a d.d.s. having it as minimal. Examples of Cantor sets that are minimal or non-minimal can be obtained using the ternary Cantor set on the real interval and constructing appropriate continuous maps. In the case of the minimality of such Cantor sets, every point of them are uniformly recurrent points for f. Other examples of minimal Cantor sets like the shift space $\Sigma^{2}$ or subsets of it are given for homeomorphisms. The whole Smale horseshoe is non-minimal for an adequate homeomorphism or one of its subsets is minimal for it.

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10.1 Introduction and Some Definitions

In what follows in Sects. 10.1, 10.2, and 10.3, we will concentrate mainly in discrete dynamical systems (d.d.s.) of the form $(I, f),$ where $I = [0, 1]$ and $f: I \rightarrow I$ is continuous. The main aim is to study the behavior of the orbits of all points in I by f. The orbit generate by the point $x \in I$, denoted by $Orb_{f}(x) = (f^{n})_{n=0}^{\infty}$, where $f^{n} = f(f^{n-1})$ and f ⁰ is the identity in I.

A point $x \in I$ is periodic of minimal period p if p is the minimum positive integer for which is $f^{p}(x) = x$, when $p = 1$ we say the point is fixed. The point x is eventually periodic, if there exists a positive integer m for which $f^{m}(x)$ is periodic. The knowledge of the periodic points allow us to state a criterium for chaotic behavior. We will say that the d.d.s. $(I, f)$ is chaotic if it has a periodic point of a period not a power of two which is equivalent to have positive topological entropy (see [2]). We will say that a point $x \in I$ is approximately periodic if for every $\epsilon> 0$ there exists a periodic point y and a positive integer N such that

$$\mid f^{n}(x) - f^{n}(y)\mid < \epsilon for all \hspace{0.1cm}n> N$$

We say a point $x \in I$ is asymptotically periodic for $(I, f)$ if $lim_{n\rightarrow \infty}f^{nm}(x)$ exists for some $m \in N$. The system $(I, f)$ is uniformly non-chaotic if every point $x \in I$ is approximate periodic. The same system is strongly non-chaotic if every $x \in I$ is asymptotically periodic. Thus, if a system is strongly non-chaotic, then it is uniformly non-chaotic. It can be proved also that if a d.d.s. is uniformly non-chaotic, then it is non-chaotic (see [3]).

A point $x \in I$ is uniformly recurrent if for every open set U containing x, there exists a positive integer $N = N(U)$ such that if $f^{m}(x) \in U$ with $m \geq 0$, then $f^{m+k}(x) \in U$ for some k holding $0 < k \leq N$. It is clear that a strongly recurrent point is one recurrent with bounded return times. A simple sufficient condition for a point x to be uniformly recurrent is to be regularly recurrent, that is, for each open set U containing x there exists a positive integer $N = N(U)$ such that $f^{kN}(x) \in U$ for all $k> 0$.

10.2 Construction and Properties of the Ternary Cantor Set

We start recalling (see [12]) the well-known construction of the ternary Cantor set on the unit interval $I = [0, 1]$ of the real line. First, we remove the open interval of length $1/3$ from the center of I and we denote the remaining open set by I ₁, $I_{1} = [0, \frac{1}{3}] \bigcup [\frac{2}{3}, 1]$. We continue with the process of removing from the center of each new created subinterval the open interval whose length is one third of the length subinterval to define inductively the kth set I _k ; I _k is a union of $2^{k}$ subintervals of length $3^{-k}$ and $(I_{k})_{k}$ is a monotone decreasing sequence of compact sets. The limit of such sequence is

$$C = \bigcap_{k=1}^{\infty}I_{k}$$

is a compact set called the ternary Cantor or Cantor set, which will be denoted by C. We think of this set as a porous set. More precisely, the connected component at each point of C is the singleton set of the point itself. We say that C is a totally disconnected set. C is also perfect which means that it is closed and contains no isolated point. As a consequence, there exists a ball center at each point of C with arbitrary small radius such that its circumference does not intersect C. Therefore, it follows that the topological dimension of C is zero (see [12]).

The ternary Cantor set is of measure zero since it is covered by interval of length $\frac{2}{3^{n}}$ which is arbitrarily small for large enough n. Although C has no length, it contains many points of I. Obviously 0 and 1 belongs to it. Similarly, both endpoints of any deleted middle third also belong to it. After all, they will never end up in the middle third of a remaining subinterval at any stage of the construction.

Surprisingly, the end points compose only a nonsignificative portion of the points of C. For example the number $1/4$, although never is an endpoint of any subinterval in the construction, belongs to C. To see it, it is useful to prove the following result,

Theorem 10.1

The ternary Cantor set consists of all points in I that can be represented in base 3 using only the digits 0 and 2.

Proof

(see [1]) Express the numbers between 0 and 1 in base-3 representation. For any point from I, this representation is unique except for points with a finite base-3 representation. By a finite base-3 representation we mean that the ternary digit a _n is non-zero and $0 = a_{n+1} = a_{n+2} = \ldots$. In this case a point $r \in [0, 1]$ is represented by exactly two base-3 expansions:

$$r =.a_{1}a_{2}...a_{n} =.a_{1}a_{2}...(a_{n} - 1)222...$$

The subinterval $(\frac{1}{3}, \frac{2}{3})$ consists of the points whose base-3 representation satisfy $a_{1} = 1$. The number $1/3$ can be expressed in two ways, as $.1 = 0\overline{2}$ in base 3. As a consequence, the set $I_{1} = [0, \frac{1}{3}]\bigcup[\frac{2}{3}, 1]$ consists of all numbers in I that can be represented in base 3 with $a_{1} = 0$ or 2. Similarly, the set

$$I_{2} = [0, \frac{1}{9}] \bigcup [\frac{2}{9}, \frac{1}{3}] \bigcup [\frac{2}{3}, \frac{7}{9}] \bigcup [\frac{8}{9}, 1]$$

from the second step of the Cantor set construction is the set that consists of all points having representations with a ₁ and a ₂ being either 0 or 2. We can ask what the analogous property is for I _n and then ask what property a number must have if it is simultaneously in all the I _n , that is, if it belongs to C.

For example, the base-3 point $r =. \overline{02}$ belongs to C. To see it, note that

$$r = 0 \times 3^{-1} + 2 \times 3^{-2} + 0 \times 3^{-3} + 2 \times 3^{-4} + \cdots = \frac{2}{9}\frac{1}{1 - \frac{1}{9}} = \frac{1}{4}$$

As mentioned above, some points have two base-3 representations, for example, $\frac{1}{3}$ can be expressed as either $.1\bar{0}$ or .1 in ternary expansion. However, each point in C has exactly one representation including no $1' s$. Using now the binary representation of points in C, and the Cantor selection method, it is possible to prove that C is an uncountable set (see for example [1]).

C has the following curious property (see [12]): for an arbitrary real number $x \in [-1, 1]$ , there exist numbers $y,z \in C$ such that $x = y - z$. We indicate this property by $[-1, 1] \subset C - C$. Since C has zero Lebesgue measure, we compare it with the known property due to Steinhaus: If the Lebesgue measure of A is positive, then A - A contains some neighborhood of the origin.

The set L of Liouville numbers is even thinner than C but satisfies the above property. In [7], Erdös proved that $L - L = R$. Before proving such property, it was known that the Hausdorff dimension of L was zero (see [8]). To this respect, assume that while L is a set of second category in the sense of Baire, C is of thefirst category.

10.3 An Interval Map for Which the Ternary Cantor Set is Minimal

In [3], Block and Coppel give an example of an interval map having interesting dynamical properties concerning chaoticity and which is an elaborate variant of another previously constructed by Delahaye in [4].

Example

(see [3])

Let $f \in C(I)$ defined by

$$f(0) = \frac{2}{3}, f(1) = 0,$$

$$f(1 - \frac{2}{3^{k}}) = \frac{1}{3^{k - 1}}, f(1 - \frac{1}{3^{k}}) = \frac{2}{3^{k + 1}} (k \geq 1)$$

and in the rest of the interval we connect the points. The result is a piecewise linear map.

More precisely, the map is

$$f(x) = x + \frac{2}{3} for \hspace{0.1cm}0 \leq x \leq \frac{1}{3}$$

$$f(x) = \frac{16}{9} - \frac{7x}{3} for\hspace{0.1cm} \frac{1}{3} \leq x \leq \frac{2}{3}$$

$$f(x) = \frac{f(3x - 2)}{3} for \hspace{0.1cm}\frac{2}{3} \leq x \leq 1.$$

From definition, we have that for any $0 \leq x \leq 1$

Fig. 10.1

Minimal piecewise linear map

$$f^{2}(\frac{x}{3}) = \frac{f(x)}{3} (1)$$

and for iteration is

$$f^{2n}(\frac{x}{3}) = \frac{f^{n}(x)}{3} for\, any\, n \geq 1.$$

It is immediate that the map has a unique fixed point $x_{0} = 8/15$ and also that for any $x \neq x_{0}$ is $f^{p}(x) \in (0, \frac{1}{3})$ for some $p = p(x) \geq 0$. If $x \neq x_{0}$, it is immediate that $f^{m}(x) \in (0, \frac{1}{3})$ for some positive integer $m = m(x) \geq 0$. It means that any periodic orbit with period greater than one, has a point in $(0, \frac{1}{3})$.

Since f maps the interval $[0, \frac{1}{3}] = I$ onto $[\frac{2}{3}, 1] = J$ and vice versa, it follows that the orbit of x must be of even period. To see it, since $f(I) = J$ and $f(J) = I$, then $f^{2}(I) = I$, that is f ² has in I a fixed point which is a periodic point of f of even period (see for example, [3]). In addition, using $(1)$, $\frac{x}{3}$ is periodic if and only if x is periodic. Then, if x has period n, $\frac{x}{3}$ has period 2n.

Let y be a point of period $n = 2^{d}q$, where $q> 1$ is odd and let $d \geq 1$. By what was said previously, assume $y \in (0, \frac{1}{3})$. Then $x = 3y$ has period $n/2$. Repeating the argument, we obtain a point of odd period q, which is a contradiction. Then the topological entropy of f is zero and then the map is non-chaotic. Similarly, we find that if there exists a unique periodic orbit of period n, then there exists a unique orbit of period 2n. Therefore, f has a unique orbit of period $2^{d}$ for every $d \geq 0$.

Let $I^{1}_{0} = [0, \frac{1}{3}]$, $I^{1}_{1} = [\frac{2}{3}, 1]$ and suppose we have defined $2^{j}$ pairwise disjoint closed intervals $I^{j} = [\alpha_{i}^{j}, \alpha_{i}^{j} + \frac{1}{3^{j}}]$ (with $0 \leq i \leq 2^{j}$) of length $\frac{1}{3^{j}}$. Then we introduce $2^{j+1}$ pairwise disjoint closed intervals $I_{i}^{j+1}$ by setting

$$I_{i}^{j+1} = [\alpha_{i}^{j}, \alpha_{i}^{j} + \frac{1}{3^{j+1}}], I_{i+2^{j}}^{j+1} = [\alpha_{i}^{j} + \frac{2}{3^{j+1}}, \alpha_{i}^{j} + \frac{1}{3^{j}}]$$

for $0 \leq i \leq 2^{j}$. In this way it is possible to define by induction, the intervals $I_{i}^{j} (0 \leq i \leq 2^{j})$ for every positive integer j. The left points $\alpha_{i}^{j}$ of $I_{i}^{j}$ are rational numbers of the form

$$\frac{b_{1}}{3} + \frac{b_{2}}{3^{2}} +...+ \frac{b_{j}}{3^{j}}$$

with $b_{j} = 0 \hspace{0.1cm}or \hspace{0.1cm}2$. From the definition of f it is easy to see that the $2^{j}$ intervals $I_{i}^{j}$ are permuted cyclically by f. Therefore, the orbit of period $2^{j} (j> 0)$ is contained in $\bigcup_{i}I_{i}^{j}$.

For any $x_{0} \in I$, with $x_{0} \neq \overline{x}$, it is possible to choose $m_{0} \geq 0$ such that $x_{1} = f^{m_{0}}(x_{0}) \in I_{0}^{1}$. If x ₁ is not periodic, then neither is $3x_{1}$ and as a consequence we can choose $m_{1} \geq 0$ such that $f^{m_{1}}(3x_{1})/3 \in I_{0}^{1}$. Then we construct $x_{2} = f^{m_{0} + 2m_{1}}(x_{0}) = f^{2m_{1}}(x_{1}) = f^{m_{1}}\frac{(3x_{1})}{3} \in I_{0}^{2}$. Following with this procedure, we see that, for any $x \in I$, is either x eventually periodic or, for every positive integer j, $f^{n}(x) \in \bigcup_{i}I_{i}^{j}$, for all large n. It follows that x is approximately periodic, and therefore, f is a uniformly non-chaotic map.

Besides the previous geometric introduction of the ternary Cantor set C ₀, it can also be described [3] as the set of all points in I whose ternary expansion contain only 0’s and 2’s (including points like 1 whose expansion can be expressed by $0.2222\ldots$). From the above constructions, it is immediate to see that $I_{a_{1}}^{1} \supset I_{a_{2}}^{2}\supset\ldots$ is a nested sequence of intervals of type $I_{i}^{j}$, then $\bigcap I_{a_{j}}^{j} = y$ where y belongs to C ₀. Moreover, each point of C ₀ can be obtained in this way. This means that the closure of the orbit of 0 by f is C ₀ and in fact this is the case for all $x \in I$ not being eventually periodic. Thus, C ₀ is a minimal set of f and

$$R(f) = P(f) \bigcup C_{0}$$

In addition, since each point of C ₀ is regularly recurrent, it follows (see [3]) that $f|C_{0}$ is topologically conjugate to theadding machine transformation ([3]) τ. Such transformation acts in the following way. If $x \in C_{0}$ has the ternary expansion $x = \sum_{i=0}^{\infty}\frac{2b_{i}}{3^{i+1}}$, then f(x) has the ternary expansion $f(x) = \sum_{i=0}^{\infty}\frac{2c_{i}}{3^{i+1}}$ determined by the rule $\gamma = \beta + 1$, where $\beta = (b_{0}, b_{1},\ldots)$ and $\gamma = (c_{0}, c_{1},\ldots)$.

In the next result we state that any Cantor set on I can be obtained as minimal of a map g.

Theorem 10.2

Let $C \subset I$ be a Cantor set. Then there exists a continuous interval map g such that C is minimal with respect to it.

Proof

Let consider the interval map f constructed in the previous paragraphs and C ₀ the ternary Cantor set. Using the result in topology saying that all Cantor sets are homeomorphic (see for example [10]), there exists an homeomorphism $h: C \rightarrow C_{0}$. Since the complement in I of C is a union of countably many open sets, C results minimal with respect to the map g defined when $g(x) = h^{-1}\circ f\circ h(x)$ when $x \in C_{0}$ and linearly on any component of the complement of C.

The result cannot be extended to general metric spaces, since in such setting, a Cantor set is a subset without isolated points and totally disconnected but we do not know the structure of the complement of the Cantor set and as consequence it is not possible to define a map in such general setting. But recently in maps from the unit square $Q = [0, 1]^{2}$ into itself called triangular, that is, continuous maps $F(x,y) = (f(x), g(x,y))$ with f and g continuous, the above map has been used to close some open problems in a so called Sharkovsky’s program for triangular maps (see [5] and [6]).

10.4 A Minimal Cantor Set in the Full Shift $\Sigma^{2}$ and in the Smale Horseshoe

The full shift on two symbols $\Sigma^{2}$ is the Cantor space composed of all bi-sequences of the form $(...s_{-n}...s_{-1}\cdot s_{0} s_{1}s_{2}...s_{n}...),$ where $s_{i} \in{0,1}$ for $i \in Z$. It is well-known that in the thirties, Marston Morse showed that there exists an element in such space that is uniformly recurrent under the shift map σ, where

$$\sigma (...s_{-n}...s_{-1}s_{0}\cdot s_{1}s_{2}...s_{n}...) = (...s_{-n}...s_{-1}\cdot s_{0} s_{1}s_{2}...s_{n}...)$$

Such bi-sequence is called the $\mathit{Morse \hspace{0.1cm} sequence}$. If $(X, h)$ is a d.d.s. where X is a metric space and $h: X \rightarrow X$ a homeomorphism, then it is well-known (see for example [3]) that if $M \subseteq X$ is a minimal set, then every $x \in X$ is uniformly recurrent, and conversely, if $x \in X$ is uniformly recurrent, then the closure of its orbit, $Orb_{h}(x)$ is a minimal set of X. Such results are also valid when $f: X \rightarrow X$ is a continuous map and we use forward orbits. Since the Morse sequence is not periodic we have that its orbit contains infinite distinct elements.

Since the closure of the Morse sequence, M is closed in $\Sigma^{2}$, then M is a Cantor set and additionally it is minimal.

In [11] is given a detailed account of the construction of a two-dimensional d.d.s. which contains an invariant set called the Smale horseshoe denoted by Λ. In this previous sections it is proved that there exists a homeomorphism $\Phi: \Lambda \rightarrow \Sigma^{2}$. Since Cantor sets are kept by homeomorphisms, then $\Phi^{-1}(M)$ the inverse image by such homeomorphism of the closure of Morse sequence, is also an example of a minimal Cantor set.

In fact the d.d.s. $\left(\Sigma^{2}, \sigma \right)$ contains copies of all possible orbits of all d.d.s.. This fact is proved in [9] by the result

Theorem 10.3

Let $Orb_{f}(x)$ be the orbit by a homeomorphism h on a metric space X . Then there is a conjugation $j: Orb_{f}(x) \rightarrow{0, 1}^{Z}$ such that $\sigma j = j h.$

10.5 Non-Minimal Cantor Sets

To have non-minimal Cantor sets in d.d.s. is a usual situation since it is sufficient to have a periodic orbit included in the Cantor set to be non-minimal. In such cases, it is unusual to have a Cantor set as a unique invariant set. This is the situation that arises in the following example. Let the d.d.s. $(R, f)$ and,

$$f(x) = 3x\quad {\it if}\quad 0 \leq x \leq \frac{1}{2}$$

and

$$f(x) = 3(1 - x)\quad {\it if}\quad \frac{1}{2} \leq x \leq 1$$

the orbit of any point x ₀ from the open interval $(\frac{1}{3}, \frac{2}{3})$ holds $lim _{n \rightarrow \infty} f^{n}(x_{0})_{n=0}^{\infty} = -\infty$. But all open subintervals of I that are pre-images of it hold the same property. The rest of I is composed of points belonging to the ternary Cantor set included 0 which is a fixed point of f. Outside I all points hold the former asymptotic property. The unique invariant set is precisely the ternary Cantor set C.

Fig. 10.2

Non-minimal piecewise linear map and the orbit of a point starting close to zero

Similarly, for the Smale horseshoe, the set $\Phi(\Lambda) = \Sigma^{2}$ is a Cantor set but non-minimal, since it contains infiniteperiodic orbits (for a detailed account of such details, see [11]).