Why the value of topological entropy of the identity map on a compact topological space is zero?

Question

I have tried to prove it using Adlar's definition of topological entropy. But I am little confused. Please help.

Answer 1

In the notation explained by this exposition of Adler's work, when $T$ is the identity map, $\mathcal{U}^n=\mathcal{U}$ and so $h(\mathcal U,T) = \lim\frac{N(\mathcal {U})}n=0$. Or, by this, $h(T^n)=nh(T)$ for all $n\ge 1$; since $T^2=T$ we have $h(T)=2h(T)$ so $h(T)=0$.

Answer 2

The identity is an isometry. So it has topological entropy $0$.

Answer 3

If $\mathcal{U}$ is an open cover, what can you say about $T^{-1} \mathcal{U}\vee \mathcal{U}$? What if you consider more iterates, how does the number of elements of that join change as you look at more iterates?