Direct proof for any homeomorphism on $[0,1]$ having zero entropy

Question

Let $(X, d)$ be a compact metric space and $f: X → X$ be a continuous map. For each natural number $n$, a new metric $d_n$ is defined on X by the formula

$$ d_{n}(x,y)=\max\{d(f^{i}(x),f^{i}(y)):0\leq i<n\}. $$

Given any $ε > 0$ and $n ≥ 1$, two points of $X$ are $ε$-close with respect to this metric if their first n iterates are $ε$-close. This metric allows one to distinguish in a neighborhood of an orbit the points that move away from each other during the iteration from the points that travel together. A subset $E$ of $X$ is said to be $(n, ε)$-separated if each pair of distinct points of $E$ is at least $ε$ apart in the metric $d_n$. Denote by $\text{Sep}(n, ε)$ the maximum cardinality of an $(n, ε)$-separated set. The topological entropy of the map $f$ is defined by

$$ h(f)=\lim _{\epsilon \to 0}\left(\limsup _{n\to \infty }{\frac {1}{n}}\log \text{Sep}(n,\epsilon )\right). $$

Note: In this definition, $\text{Sep}(n,\epsilon)$ can be replaced by $\text{Cov}(n,\epsilon)$ (the minimal cardinality of $\epsilon$-covering w.r.t. $d_n$: each member in the covering has diameter no more than $\epsilon$) or by $\text{Span}(n,\epsilon)$ (the minimal cardinality of $\epsilon$-Spanning w.r.t. $d_n$: each point in $X$ is within the $\epsilon$ distance of the spanning set)

The definitions above are due to Bowen and Dinaburg in 1970s. Credit: Wikipedia.

My questions is: Simply using this definition (as well as basic properties from it), without introducing other definitions of topological entropy in more general settings, can we show that Any homeomorphism of the interval $[0,1]$ is of zero topological entropy?

This fact is taken from Walters' Introduction to Ergodic Theory, page 180, Corollary 7.14.1, where he uses that any homeomorphism of unit circle is of zero entropy and more general definitions of topological entropy. But I only want to use this metric definition.

Since Walters proves that any homeomorphism of unit circle is of zero entropy with only elementary methods and Sahiba Arora kindly typed it here. If you can use this result together with just the definitions above to show any homeomorphism of $[0,1]$ is of zero entropy, I am also happy with it!

Any references will also be greatly appreciated!

Answer 1

Below I present two proofs; the first one uses some technology (although classical by now), so may fail to answer the question, whereas the second one only uses the definition of topological entropy via spanning sets and a counting argument.

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Proof with Technology: Consider the topological identification map $\pi: [0,1]\to\mathbb{T}$ associated to the equivalence relation $x\sim y\iff x-y\in\mathbb{Z}$. $\pi$ is one-to-one everywhere except at the boundary, where it's two-to-one. If $f\in\operatorname{Homeo}([0,1])$, then either $f(0)=0 \,\&\, f(1)=1$ xor $f(0)=1 \,\&\, f(1)=0$. In any event, if $x\sim y$, then $f(x)\sim f(y)$, thus $f$ descends to a continuous map $\underline{f}:\mathbb{T}\to\mathbb{T}$ which is a homeomorphism (that fixes $\pi(0)$) since $\left(\underline{f}\right)^{-1}=\underline{f^{-1}}$. Then by a formula by Bowen (which I mentioned in this previous answer: https://math.stackexchange.com/a/4205931/169085) and the fact that homeomorphisms of the circle have zero topological entropy, we have

\begin{align*} 0\leq \operatorname{topent}(f) \leq \sup_{y\in \mathbb{T}} \operatorname{topent}(f;\pi^{-1}(y)). \end{align*}

Each fiber $\pi^{-1}(y)$ is finite, so the RHS of this inequality vanishes, giving that $\operatorname{topent}(f)=0$.

Note that this is an argument analogous to the argument Walters presents in his book.

I should mention that Bowen's formula used here only uses some counting arguments.

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Proof by Counting: Let's use the definition of entropy using spanning sets. Let $f:[0,1]\to[0,1]$ be a homeomorphism. $f$ has to preserve the boundary; let's assume $f(0)=0 \,\&\, f(1)=1$ and so that $f$ is monotonically increasing. Fix $n\in\mathbb{Z}_{\geq1}$ and $\epsilon\in \mathbb{R}_{>0}$. Consider the set $S'=\left\{\dfrac{1}{p},\dfrac{2}{p},\cdots,\dfrac{p-1}{p}\right\}$ with $p\in\mathbb{Z}_{>1}$ be the smallest such that $\dfrac{1}{p}<\epsilon$, and put $S=\bigcup_{k\in0}^{n-1} f^{-k}(S')$.

We claim that $S$ is an $(n,\epsilon)$-spanning set for $f$. Indeed, let $x\in[0,1]$. Then there are at most two elements $s_1,s_2\in S$ that are the closest in $S$ to $x$, i.e. $d(x,S)=d(x,s_1)=d(x,s_2)$. Let us choose one of them and write instead $s_x\in S$ (one may choose the smaller one if an algorithm is needed). $S$ being an $(n,\epsilon)$-spanning set for $f$ is equivalent to $d_n^f(x,s_x)<\epsilon$. Suppose otherwise. Then there is some $k\in\{0,1,...,n-1\}$ such that $d(f^k(x),f^k(s_x))\geq \epsilon$ (in particular $x\neq s_x$; wlog let's assume $x\lneq s_x$). Then for some $s'_x\in S'$, we have $f^k(x)\lneq s'_x \lneq f^k(s_x)$. Since $f^k$ is a homeomorphism, its restriction is also a homeomorphism $[x,s_x]\to [f^k(x),f^k(s_x)]$; consequently we have $x\lneq f^{-k}(s'_x) \lneq s_x$, a contradiction since $f^{-k}(s'_x)\in S$ but $S$ has no points closer to $x$ than $s_x$. Thus $S$ is an $(n,\epsilon)$-spanning set for $f$.

Therefore we have

$$\operatorname{minspan}(f;n,\epsilon)\leq \#(S)=\#\left(\bigcup_{k\in0}^{n-1} f^{-k}(S')\right)\leq \sum_{k=0}^{n-1}\#(S')= n(p-1)\leq\dfrac{n}{\epsilon},$$

(in the last inequality we used the fact that since $p$ is the smallest integer greater than $1$ whose reciprocal is less than $\epsilon$, $p-1\leq\frac{1}{\epsilon}<p$)

whence $\limsup_{n\to\infty}\dfrac{1}{n}\log(\operatorname{minspan}(f;n,\epsilon))=0$ and consequently $\operatorname{topent}(f)=0$.

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I should mention that the second argument I presented above can be generalized to even unbounded intervals. For instance de Vries' book Topological Dynamical Systems contains such a generalization. I report from there (p.391, Lem.8.3.1):

Theorem: Let $I\subseteq \mathbb{R}$ be a path connected subset, and $f:I\to I$ be continuous. If $f$ is injective and if for any compact interval $J\subseteq I$ and positive integer $n$ there exists a number $\Lambda(J,n)\in\mathbb{R}_{>0}$ such that

1. For any $k\in\{0,1,...,n-1\}$, the length of the interval $f^k(J)$ is not more than $\Lambda(J,n)$, and
2. $\limsup_{n\to\infty} \dfrac{1}{n}\log(\Lambda(J,n))=0$,

then $\operatorname{topent}(f)=0$.

In this case instead of using one set $S'$ uniformly for all iterates $k\in\{0,1,...,n-1\}$, one should choose variable subsets $S'_k\subseteq f^k(J)$ and then replace $S$ with $\bigcup_{k=0}^n f^{-k}(S'_k)$. Of course, the combinatorics get slightly gnarlier.