Topological entropy

Remarks

The limit in $\varepsilon$ exists because the $\limsup$ grows as $\varepsilon$ decreases. The value depends only on the topology, not on the metric. Adler, Konheim and McAndrew defined it on any compact space through the growth of joins of open covers, and this gives the same number. Topologically conjugate maps have equal entropy, and $h_{\mathrm{top}}(f^k) = k \, h_{\mathrm{top}}(f)$ for $k \ge 1$. Example: the full one-sided shift $\sigma$ on $\set{0, \ldots, k-1}^{\N}$ with $d(x, y) = 2^{-\min\set{i : x_i \neq y_i}}$. Then $d_n(x, y) > 2^{-m}$ exactly when $x$ and $y$ differ at some coordinate $i < n + m - 1$, so $s(n, 2^{-m}) = k^{n + m - 1}$ and $h_{\mathrm{top}}(\sigma) = \log k$. A map with zero entropy: an isometry has $d_n = d$, so $s(n, \varepsilon)$ does not depend on $n$ and $h_{\mathrm{top}} = 0$.