Topological Entropy and Chaos

Topological Entropy and Li-Yorke Chaos

"Topological entropy of maps on the real line"

Let \(X\) be a Hausdorff topological space and let \(f: X\to X\) be a continuous self-map
on \(X\). The pair \((X,f)\) is called a dynamical system.
A subset \(K\subset X\) is said to be invariant by \(f\) if \(f(K)\subset K\) and it is strictly invariant by \(f\) if \(f(K)=K\).

We summarize some properties of the topological entropy below.

Theorem 1. Let \(X\) and \(Y\) be two (metric) compact topological sets and let \(f: X \to X\) and \(g: Y \to Y\) be two continuous maps. Then the following properties are held:

(a) Let \(\phi:X\to Y\) be continuous such that \(g\circ \phi=\phi\circ f\). Then:
(a1) If the map \(\phi\) is injective, then \(h(f)\le h(g)\).
(a2) If the map \(\phi\) is surjective, then \(h(f)\ge h(g)\).
(a3) If the map \(\phi\) is bijective, then \(h(f)=h(g)\).

(b) Suppose that \(X=\cup_{i=1}^nX_i\), where \(X_i\) are compact and invariants by \(f\). Then \(h(f)=\max\{h(f|_{x_i})\}\)

(c) For any integer \(n \geqslant 0\) it is hold \(h\left(f^n\right)=n h(f)\).

(d) Let \(f \times g: X \times Y \rightarrow X \times Y\) be defined by \((f \times g)(x, y)=(f(x), g(y))\) for all \((x, y) \in\) \(X \times Y\). Then \(h(f \times g)=h(f)+h(g)\).

(e) If \(f\) is a homeomorphism, then \(h(f)=h\left(f^{-1}\right)\).

(f) Let \(\varphi: X \rightarrow Y\) be a continuous surjective map such that \(\varphi \circ f=g \circ \varphi\). Then \(\max \left\{h(g), \sup \left\{h\left(f, \varphi^{-1}(y)\right): y \in Y\right\}\right\} \leqslant h(f) \leqslant h(g)+\sup \left\{h\left(f, \varphi^{-1}(y)\right): y \in Y\right\}\).

(g) If \(f: X \rightarrow Y\) and \(g: Y \rightarrow X\) are continuous, then \(h(f \circ g)=h(g \circ f)\).

(h) Let \(f: X \rightarrow Y, g: Y \rightarrow X\) be continuous and let \(F: X \times Y \rightarrow X \times Y\) be defined by \(F(x, y)=(g(y), f(x))\) for all \((x, y) \in X \times Y\). Then

\[\begin{equation} h(F)=h(f \circ g)=h(g \circ f). \end{equation} \]

(i) If \(X_\infty=\cap_{n\ge 0}f^n(X)\) then \(h(f)=h(f|_{X_\infty})\).

(j) \(h(f)=h(f_{\Omega(f)})\) where \(x\in \Omega(f)\) if for all neighborhood \(U\) of \(x\) there is \(n\ge 0\) such that \(f^n(U)\cap U\ne \emptyset\)(\(\Omega(f)\) is called non-wandering set of \(f\)).

A dynamical system (X,f) is called minimal if X does not contain any non-empty, proper,  
closed sf-invariant subset. 

In such a case we also say that the map f itself is minimal.


The following conditions are equivalent: 
-$(X,f)$ is minimal,
-every orbit is dense in $X$, 
-$\omega_f(x)=X$ for every $x\in X$. 

Definition of topological entropy on matric space

For continuous maps on a metric space \((X,f)\) the topological entropy of \(f\) is defined by

\[\begin{equation} {\rm ent}(f):=\sup\{h(f|_K): K\subset X, \text{compact and invariant by}\, f\}. \end{equation} \]

By Theorem 1(i) we have

\[\begin{equation} {\rm ent}(f)=\sup\{h(f|_K):K\in \mathcal{K}(X,f)\} \end{equation} \]

where \(\mathcal{K}(X,f)\) is the family of all the compact subsets of \(X\) which are strictly invariant by \(f\). Notice that this definition makes sense when \(X\) is matric or simply a topological space.

Explanation/interpretation (3):

Any compact strictly\(f\)-invariant set \(K\) determines uniquely a strictly \(f\)-invariant closed set \(K_\infty=\cap_{n\ge 0}f^n(K)\in \mathcal{K}(X,f)\) such that \(h(f|_K)=h(f|_{K_\infty})\).
Conversely, any \(K\in \mathcal{K}(X,f)\) we have \(K=K_\infty\) .So

\[\begin{equation}\begin{aligned} &\{K_\infty=\cap_{n\ge 0}f^n(K):K \text{ compact and}\, f-\text{invariant}\}\\ & = \{K: K \in \mathcal{K}(X,f)\}. \end{aligned} \end{equation} \]

Therefore,

\[\begin{equation} \begin{aligned} {\rm ent}(f)&=\sup \{h(f|_K):K\subset X, K \text{compact and} \, f-\text{invariant}\}(Th1(i))\\ &=\sup \{h(f|_{K_\infty}):K\subset X, K\text{compact and} \, f-\text{invariant}\}\\ &= \sup \{h(f|_K):K\in \mathcal{K}(X,f)\} \end{aligned} \end{equation} \]

i.e., (3) holds.

Remark.

由上述定义, 为了计算\(f\)的拓扑熵，我们需要

1. 找集合\(\{K:K \text{compact and} \,f-\text{invariant}\}\),
2. 再对每一个\(K\) 计算\(h(f|_K)\),
3. 最后取它们的上确界得到ent(\(f\)).

值得注意的是：如果我们认为\(f|_K\)是当\(x\in K\), \(f|_K(x)=f(x)\), 其它地
方, \(f|_K\)没有定义. 那么有可能\(f\)没有紧不变集，如\(f(x)=\mu x(x-1)\), \(\mu>4\), 定义在全实轴上, 如果\(K\)是紧且\(f\)-不变的, 那么\(K=\{0\}\). 按照上述定义计算ent(\(f\))=0, 但是这个映射有Li-Yorke混沌. 所以这并不是人们期望的结果. 但在后面我们会看到有\(f|_K\)的其它定义被给出，克服了这个困难.

We start by studying some properties of the new definition of topological entropy, which is summed up in the next result.

Theorem 2 Let \(X\) and \(Y\) be two matric spaces and let \(f: X\to X\) and \(g: Y\to Y\) be continuous.

(a) Let \(\phi : X\to Y\) be continuous map such that \(\phi\circ f=g\circ \phi\). Then
(a1) If the map \(\phi\) is injective then ent(\(f\))\(\le\) ent(\(g\)).
(a2) If the map \(\phi: X\to Y\) is surjective and such that the set \(\phi^{-1}(K)\) is compact for all compact \(K\subset X\), then ent(\(f\))\(\ge\) ent(\(g\)).
(a3) If the map \(\phi\) is bijective then ent(\(f\))=ent(\(g\)).

(b) ent(\(f^n\))\(= n\) ent(\(f\)).

(c) Let \(f:X\to Y\) and \(g:Y\to X\) then ent(\(f\circ g\))\(=\) ent (\(g\circ f\)).

(d) Let \(F:X\times Y\to X\times Y\) be defined by \(F(x,y)=(f(x),g(y))\) then ent(\(F\))\(=\) ent(\(f\))+ent(\(g\)).

(e) Let \(f:X\to Y\) and \(g:Y\to X\) then \(F(x,y)=(g(y),f(x)):X\times Y\to X\times Y\) satisfies ent(\(F\))=ent(\(f\circ g\))=ent (\(g\circ f\))

(f) The conclusion that if \(X=\cup_{i=1}^nX_i\) with \(X_i\) compact \(f\)-invariant then ent(\(f\))\(=\max\{h|_{X_i}:i=1\cdots n\}\) no holds in general.

(g) Let \(\phi:X\to Y\) be continuous, surjective map and \(\phi^{-1}(K)\) is compact for any \(K\) compact in \(Y\), then \(\max\{{\rm ent}(g),\sup\{h(f,\phi^{-1}(y)):y\in Y\}\}\le h(f)\le {\rm ent}(g)+\sup\{h(f,\phi^{-1}(y)):y\in Y\}.\)

(h) Let \(f:X\to X\) be a homeomorphism then ent(\(f\))=ent(\(f^{-1}\)).

(i) Let \(X_\infty=\cap_{n\ge 0}f^n(X)\). Then ent(\(f\))=ent(\(f|_{X_\infty}\)).

(j) ent(\(f\))=ent(\(f|_{\Omega (f)}\)), where \(x\in \Omega(f)\) if for any neighborhood \(U\) of \(x\) there exist positive integer \(n\) such that \(U\cap f^n(U)\ne \emptyset\).

Proof.

(a)

Since \(\phi \circ f=g\circ \phi\), \(f(K)=K\) implies \(g (\phi(K))=\phi(K)\), i.e., \(\{\phi(K):K\in \mathcal{K}(X,f)\}\subset \mathcal{K}(Y,g)\).

(a1) If \(\phi\) is injective then

\[\begin{equation} \begin{aligned} \text{ent}(f)&=\sup\{h(f|_K): K \in \mathcal{K}(X,f)\},(\text{by Th}1(a1))\\ &\le\sup\{h(g|_{\phi(K)}): K \in \mathcal{K}(X,f)\}\\ &\le \{h(g|_{L}): L \in \mathcal{K}(Y,g)\}\\ &=\text{ent}(g). \end{aligned} \end{equation} \]

(a2) \(\phi \circ f=g\circ \phi\) implies \(\phi^{-1}(K)\in \mathcal{K}(X,f)\) for \(K\in \mathcal{K}(Y,g)\).
If \(\phi\) is surjective then

\[\begin{equation} \begin{aligned} \text{ent}(g)&=\sup\{h(g|_K): K \in \mathcal{K}(X,g)\},(\text{by Th}1(a2))\\ &\le \sup\{h(f|_{\phi^{-1}(K)}): K \in \mathcal{K}(X,g)\}\\ &\le \sup\{h(f|_{L}): L \in \mathcal{K}(X,f)\}\\ &=\text{ent}(f). \end{aligned} \end{equation} \]

(a3) It is immediate consequence of (a1) and (a2).

(b)

We prove ent(\(f^n\))=\(n\) ent (\(f\)). Since \(\mathcal{K}(X,f)\subset \mathcal{K}(X,f^n)\) and
\( h((f|_K)^n)=h(f^n|_K), \)
\(\left\{(f|_K)^n=f|_K\circ f|_K\cdots f|_K\circ f|_K=f|_{f^{n-1}(K)}\circ \cdots f|_{f(K)}\circ f|_K=f^n|_K\right\} \)

\[\begin{equation} \begin{aligned} n\,\text{ent}(f)&=\sup\{n h(f|_K):K\in \mathcal {K}(X,f)\}\\ &=\sup \{h((f|_K)^n):K\in \mathcal {K}(X,f)\} \\ &=\sup \{h(f^n|_K):K\in \mathcal {K}(X,f)\} \\ &\le \sup \{h((f|_K)^n):K\in \mathcal {K}(X,f^n)\} \\ &=\text{ent}(f^n). \end{aligned} \end{equation} \]

On the other hand, for any \(K\in \mathcal(X,f^n)\), we have \(K_n=\cup_{i=0}^{n-1}f^i(K)\in \mathcal{K}(X,f).\)
Hence,
\( \{K_n:K\in \mathcal{K}(X,f^n)\}\subset \mathcal{K}(X,f) \)
and

\[\begin{equation} \begin{aligned} n\,\text{ent}(f)&=n\,\sup\{h(f|_K):K\in \mathcal{K}(X,f)\}\\ &\ge n\, \sup\{h(f|_{K_n}):K\in \mathcal{K}(X,f^n)\}\\ &\ge \, \sup\{h(f^n|_{K_n=\cup_{i=0}^{n-1}f^i(K)}):K\in \mathcal{K}(X,f^n)\}(by Th1(b))\\ &\ge \sup\{h(f^n|_K):K\in \mathcal{K}(X,f^n)\}\\ &=\text{ent}(f^n). \end{aligned} \end{equation} \]

(c)

We prove that if \(f:X\to Y\) and \(g:Y\to X\) are continuous then ent (\(f\circ g\))\(=\) ent (\(g\circ f\)).

\(f\circ g(K)=K \Rightarrow g\circ f(g(K))=g(K)\Rightarrow g(K)\in \mathcal{K}(X,g\circ f)\)

\( \Rightarrow \{g(K):K\in \mathcal{K}(Y,f\circ g)\}\subset \mathcal{K}(X,g\circ f) \)

\[\begin{equation} \begin{aligned} \text{ent} (f\circ g)&=\sup\{h(f\circ g|_K):K\in \mathcal{K}(Y,f\circ g)\} (by Th1(g))\\ &=\sup\{h(g\circ f|_{g(K)}):K\in \mathcal{K}(Y,f\circ g)\} \\ &\le \sup\{h(g\circ f|_{L}):L\in \mathcal{K}(Y,f\circ g)\}\\ &=\text{ent}(g\circ f). \end{aligned} \end{equation} \]

By a symmetrical reasoning, we prove that ent(\(f\circ g\))\(\le\) ent(\(g\circ f\)).

(d)

We prove that if \(F:X\times Y\to X\times Y\) be defined by \(F(x,y)=(f(x),g(y))\) then ent(\(F\)) \(=\) ent(\(f\))+ent(\(g\)).

Let \(\Pi_1(x,y)=x,\Pi_2(x,y)=y\). For any \(K\), we have \(K\subset K_1\times K_2\), where \(K_1=\Pi_1(K),K_2=\Pi_2(K)\).

For \(K_1\in \mathcal{K}(X,f), K_2\in \mathcal{K}(Y,g)\), we have \(K_1\times K_2 \in \mathcal{K}(X\times Y,F)\).
Therefore,

\( \{ K_1\times K_2: K_1\in \mathcal{K}(X,f), K_2\in\mathcal{K}(Y,g)\subset \mathcal{K}(X\times Y, F) \} \)
and

\[\begin{equation} \begin{aligned} \text{ent}(F)&=\sup \{h(F|_{K}):K\in \mathcal{K}(X\times Y,F)\}\\ &\ge\sup\{h(F|_{K_1\times K_2}):K_1\in \mathcal{K}(X,f),K_2\in\mathcal{K}(Y,g)\} (by Th1(d))\\ &=\sup\{h(f|_{K_1})+h(g|_{K_2}):K_1\in \mathcal {K}(X,f),K_2\in\mathcal{K}(Y,g)\}\\ &=\text{ent}(f)+\text{ent}(g). \end{aligned} \end{equation} \]

On the other hand, let \(i: K\to K_1\times K_2\) is inclusion, i.e., \(i(x)=x\), an injective.

\[F|_{K_1\times K_2}:K_1\times K_2, F|_K:K\to K, i:K\to K_1\times K_2, injective, \]

and

\[ F|_{K_1\times K_2}\circ i=i\circ F|_K\circ . \]

By Th1(a2)

\[ h(F|_K)\le h(F|_{K_1\times K_2}). \]

Then

\[\begin{equation} \begin{aligned} \text{ent}(F)&=\sup\{h(F|_{K}):K\in \mathcal{K}(X\times Y,F)\}\\ &\le \sup\{h(F|_{K_1\times K_2}):K_1\in \mathcal{K}(X,f),K_2\in \mathcal{K}(X,g)\} (by Th1(d))\\ &=\sup\{h(f|_{K_1})+h(f|_{K_2}):K_1\in \mathcal{K}(X,f),K_2\in \mathcal{K}(X,g)\}\\ &\le \sup\{h(f|_{K_1}):K_1\in \mathcal{K}(X,f)\}+\sup\{h(f|_{K_2}):K_2\in \mathcal{K}(Y,g)\}\\ &=\text{ent}(f)+\text{ent}(g). \end{aligned} \end{equation} \]

(e)

We prove that if \(f:X\to Y\) and \(g:Y\to X\) are continuous then \(F(x,y)=(g(y),f(x)):X\times Y\to X\times Y\) satisfies

\[\text{ent}(F)=\text{ent}(f\circ g)=\text{ent}(g\circ f) \]

\[\begin{equation} \begin{aligned} 2\text{ent}(F)&=\text{ent}(F^2) =(g\circ f(x),f\circ g(y)) (\text{by what just proved (d)}) \\ &=\text{ent}(g\circ f)+\text{ent}(f\circ g)=2\text{ent}(g\circ f)=2\text{ent}(f\circ g) \end{aligned} \end{equation} \]

Thus
\( \text{ent}(F)=\text{ent}(g\circ f)=\text{ent}(f\circ g). \)

(f)

We prove that the conclusion that if \(X=\cup_{i=1}^nX_i\) with \(X_i\) compact \(f\)-invariant then ent
(\(f\))\(=\max\{h|_{X_i}:i=1\cdots n\}\) no holds in general.

Let \(f:X \to X\) be a minimal homeomorphism defined on a compact set with positive topological entropy.

For example:

Let \(x\in X\). \(X_1=\)FullOrb\(_f(x)=\{f^n(x):x\in \mathbb{Z}\}\) and \(X_2=X\backslash X_1\).
It is clear that both sets are invariant by \(f\) and that they do not have compact invariant subsets (in case they have compact invariant subsets, the map \(f\) is not minimal).
Therefore, ent(\(f|_{X_i}\))=0, for \(i=1,2\), while ent(\(f\))=\(h(f)>0\).

(g)

Let \(\phi:X\to Y\) be continuous, surjective map and \(\phi^{-1}(K)\) is compact for any \(K\)
compact in \(Y\), then

\[\max\{{\rm ent}(g),\sup\{h(f,\phi^{-1}(y)):y\in Y\}\}\le h(f)\le {\rm ent}(g) +\sup\{h(f,\phi^{-1}(y)):y\in Y\}. \]

(h)

We prove that if \(f:X\to X\) be a homeomorphism then

\[\text{ent}(f)=\text{ent}(f^{-1}). \]

Since \(f(K)=K\) implies \(f^{-1}(K)=K\),

\[\begin{equation} \begin{aligned} \text{ent}(f)&=\sup\{h(f|_K):K\in \mathcal{K}(X,f)\}(by Th1(e))\\ &=\sup\{h(f^{-1}|_K):K\in \mathcal{K}(X,f)\}(by Th1(e))\\ &=\sup\{h(f^{-1}|_K):K\in \mathcal{K}(X,f^{-1})\}(by Th1(e))\\ &=\text{ent}(f^{-1}). \end{aligned} \end{equation} \]

(i)

We prove that if let \(X_\infty=\cap_{n\ge 0}f^n(X)\). Then

\[\text{ent}(f)=\text{ent}(f|_{X_\infty}). \]

For any \(K\in \mathcal{K}(X,f)\), we have \(K\subset X_\infty\). Therefore,

\[f(K)=K\Leftrightarrow f|_{X_\infty}(K)=K \]

\[\text{ent}(f)=\sup\{h(f|_K):K\in \mathcal{K}(X,f)\}=\sup\{h(f_{X_\infty}|_K):K\in \mathcal{K}(X,f|_{{X_\infty}})\}=\text{ent}(f|_{X_\infty}) \]

(j)

We prove that \(\text{ent}(f)=\text{ent}(f|_{\Omega (f)})\),
where \(x\in \Omega(f)\) if for any neighborhood \(U\) of \(x\)
there exist positive integer \(n\) such that \(U\cap f^n(U)\ne \emptyset\).

\[f|_{\Omega(f)}:\Omega(f)\to \Omega(f), f:X\to X, i:\Omega(f)\to X(\text{inclusion map,injective}) \]

satisfy the exchange diagram

Since \(i\) is injective, by (a1) in the theorem,

\[\text{ent}(f|_{\Omega(f)})\le \text{ent}(f). \]

We prove the converse inequality. For any \(K\in \mathcal{K}(X,f)\), \(K\cap \Omega(f)\) is \(f_{\Omega(f)}\)-invariant.
So

\[\{K\cap \Omega(f):K\in \mathcal{K}(X,f)\}\subset \{K:K\,\text{is}\, f_{\Omega(f)}-\text{invariant}\}. \]

Moreover,

\[\begin{equation} \begin{aligned} \text{ent}(f)&=\sup\{h(f|_K):K\in \mathcal{K}(X,f)\}(by Th1 (j))\\ &=\sup\{h((f|_K)|_{\Omega(f)}):K\in \mathcal{K}(X,f)\}\\ &=\sup\{h(f|_I):I=K\cap\Omega(f)(f|_{\Omega(f)}-\text{invariant}),K\in \mathcal{K}(X,f)\} \end{aligned} \end{equation} \]

\[( \{I=K\cap\Omega(f)~\text{is}~f|_{\Omega(f)}-\text{invariant},~K\in \mathcal{K}(X,f)\}\subset \{J\subset \Omega(f): J~\text{is}~f|_{\Omega(f)}-\text{invariant}\}) \]

\[\begin{equation} \begin{aligned} (15)&\le \sup\{h(f|_J):J\subset \Omega(f)~\text{is}~f|_{\Omega(f)}-\text{invariant}\}\\ &= \sup\{h((f|_{\Omega(f)})|_J):J\subset \Omega(f)~\text{is}~f|_{\Omega(f)}-\text{invariant}\}\\ &=\text{ent}(f|_{\Omega(f)}). \end{aligned} \end{equation} \]

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