I'm reading "Topological entropy bounds measure-theorettic entropy", by L.W. Goodwyn.


[enter link description here](https://www.ams.org/journals/proc/1969-023-03/S0002-9939-1969-0247030-3/S0002-9939-1969-0247030-3.pdf)


After Proposition 2, he mentions that "finite closed cover can yield entropy strictly greater than topological entropy". I was looking for the reference to this (Topological entropy and expansive cascades - Dissertation) but i couldn't find it.


I would appreciate if can help me with an example or the reference.



Example of finite closed cover with entropy strictly greater than topological entropy
You can find an example of a cascade which has topological entropy equal to zero, and which has a finite closed cover having entropy equal to $\log 2$ in the [article](https://www.jstor.org/stable/1995916) "The Product Theorem for Topological Entropy" by L.W. Goodwyn.


Let $\hslash (T) = \sup h(\alpha, T)$, where the supremum is taken over all finite closed covers. We have that $\hslash (T)$ may not be equal to $h(T)$. In fact, that's what usually happens.


I don't know if the example presented in the dissertation is the same as in the article. But the idea of the example should be the same.


For a positive integer $m$, let $C\_m$ denote the set of all bisequences $(\ldots, x(- 1), x(0), x(1),\ldots)$ of points in $I^m$, i.e., $C\_m = (I^m)^{\mathbb{Z}}$. We let $C\_m$ have the product topology and we define $\sigma\_m: C\_m \to C\_m$ by


$$\sigma(x)(n)=x(n+1),$$


and let $\pi\_m :C\_m \to I^m$ be the projection $\pi\_m(x) = x(0)$ for $x \in C\_m$.


The example you are looking for will be a subcascade of the sequence cascade $(C\_1, \sigma\_1)$. For each positive integer $m$, we define $X\_m$ to be the set of all bisequences $x= (\ldots, -x(-1), x(0), x(1),\ldots) \in C\_1$ satisfying the requirement that there are no more than $m$ integers $i$ such that $|x(i) -\frac{1}{2}|>1/m$. Every $X\_m$ is closed and $\sigma\_1$-invariant. Define


$$X= \bigcap\_{m=1}^{\infty} X\_m,\qquad T=\sigma\_1|\_{X}\quad\text{and}\quad \pi=\pi\_1|\_X.$$


So you have $h(\alpha,T)=0$ (*check*), the topological entropy of $T$ is zero. To see that $\hslash (T) > h(T)$, define $F\_0 = \{x \in X: x(0) \in [0,\frac{1}{2}]\}$ and $F\_1 = \{x\in X : x(0) \in [\frac{1}{2},1]\}$, and let $\beta=\{F\_0,F\_1\}$. We have $h(\beta,T)=\log 2$.


