Let $X$ be a finite topological space and $T\colon X\to X$ continuous. As the title already suggests, I am wondering if the topological entropy of $T$, denoted by $h(X,T)$, then is $0$.
As far as I see, this is indeed the case. Before giving my reason, let's give me the definition of topological entropy given by Adler, Konheim and McAndrew (see Peter Walters' An Introduction to Ergodic Theory).
Definition 1 If $\alpha$ is an open cover of X let $N(\alpha)$ denote the number of sets in a finite subcover of $\alpha$ with smallest cardinality. We define the entropy of $\alpha$ by $H(\alpha)=\log N(\alpha)$.
Definition 2 If $\alpha$ is an open cover of $X$ and $T\colon X\to X$ is a continuous map then the entropy of $T$ relative to $\alpha$ is given by $h(T,\alpha)=\lim_{n\to\infty}\frac{1}{n}H\left(\bigvee_{i=0}^{n-1}T^{-i}\alpha\right)$, where $\bigvee_{i=1}^{n}\alpha_i$ is the join of any finite collection of open covers of $X$.
Definition 3 If $T\colon X\to X$ is continuous, the topological entropy of $T$ is given by $h(T)=\sup_{\alpha}h(T,\alpha)$, where $\alpha$ ranges over all open covers of $X$.
Now, let $\alpha$ be any open cover of $X$. Since $X$ is finite, $N\left(\bigvee_{i=0}^{n-1}T^{-i}\alpha\right)\le\text{card}(X)<\infty$. Hence $$ 0\le h(T,\alpha)\le \lim_{n\to\infty}\frac{1}{n}\log\text{card}(X)=0, $$ implying that $h(T,\alpha)=0$, implying that $h(T)=\sup_{\alpha}h(T,\alpha)=0$.
Please let me know if I am right. Thank you.