Topological entropy of a homeomorphism of $[0,1]$

Question

Let $f : [0, 1] \to [0, 1]$ be a homeomorphism. Show that the topological entropy of f is 0, i.e $$h_{\text{top}}(f) = 0.$$

Can someone give me a hint?

Thank you!

Answer 1

The entropy is not zero because $f$ is a homeomorphism (for example, all toral automorphisms are homeomorphisms, but many have nonzero entropy).

This has instead to do with the specific space. Notice that if you start with a cover $\mathcal U$ by intervals, the preimage $f^{-1}\mathcal U$ has exactly the same number of intervals (since $f$ is strictly increasing or strictly decreasing). Hence, $$ {\rm card} (\mathcal U\vee f^{-1}\mathcal U)\le 2{\rm\, card\,} \mathcal U. $$ By induction you have $$ {\rm card} (\mathcal U\vee \cdots\vee f^{-n+1}\mathcal U)\le n{\rm\, card\,} \mathcal U $$ and so $$ \lim_{n\to\infty}\frac1n H(\mathcal U\vee \cdots\vee f^{-n+1}\mathcal U)=0. $$ It follows that the topological entropy is zero by taking the supremum over all covers by intervals (indeed it suffices to take covers by intervals since they generate the topology).