Topological entropy is finite? Let $f:X\rightarrow X$ be a continuous function on a compact metric space $X$ with distance $d$. For each $n\in \mathbb{N}$ we can introduce the distance on $X$ ,$d_n(x,y)=\max_{k=0,...,n-1}d(f^k(x),f^k(y))$, and then we define the topological entropy of $f$ to be $h(f)=\lim_{\epsilon\rightarrow 0}\lim \sup_{n\rightarrow \infty}\frac{1}{n}\log N(n,\epsilon)$, where $N(n,\epsilon)$ is the smallest number of balls of radius $\epsilon$ needed to cover $X$. Now it's clear that $N(n,\epsilon)$ is always finite since $X$ is compact. It also makes sense to me that the function $\epsilon\rightarrow \lim \sup_{n\rightarrow \infty}\frac{1}{n}\log N(n,\epsilon)$ is non-increasing , what I don't get is why this gives us that the topological entropy acually exists , couldn't I have something like $\lim \sup_{n\rightarrow \infty}\frac{1}{n}\log N(n,\epsilon)=\frac{1}{\epsilon}$?  Or can we have the topological entropy to be $\infty$?
Any help is appreciated.
 A: Let me give a partial answer for now. Maybe I can think of a complete answer soonish. (see edit below)
Actually, I do not recall anyone claiming that the topological entropy is always finite. It is certainly well-defined as a number in $\mathbb{R} \cup \{\infty\}$ since the limit in $\epsilon$ is monotone as you pointed out, but it may very well be infinite.
Unfortunately, I cannot think of an example with infinite entropy right now, but I am quite sure there must be one. (see edit below)
Instead, let me mention that there are some hypothesis that can guarantee finiteness of the entropy.
For instance, it suffices if your system is Lipschitz continuous and your space $X$ is somewhat nice.
Here, "nice" in the sense that the "ball dimension"
$$ BD(X) = \limsup_{\epsilon \rightarrow} \frac{\log(N(1,\epsilon))}{|\log(\epsilon)|} $$
is finite. If your space is, e.g., a manifold, then the ball dimension is exactly the dimension of the manifold.
Now, for a Lipschitz dynamical system with Lipschitz constant $L \geq 1$, we have $N(n,\epsilon) \leq N(1,\epsilon L^{-n})$ and, hence,
$$ \frac{1}{n} \log(N(n,\epsilon)) \leq \underbrace{\frac{|\log(\epsilon L^{-n})|}{n}}_{\rightarrow \log(L)} \underbrace{\frac{\log(N(1,\epsilon L^{-n}))}{|\log(\epsilon L^{-n})|}}_{\rightarrow BD(X)} \xrightarrow{n \rightarrow \infty}{ \log(L) BD(X)}. $$
Thus, the entropy is bounded by $\log(L)BD(X)$.
Another possible hypothesis is positive expansivitiy: if there exists a constant $\delta > 0$ such that $d_n(x,y) < \delta$ for all $n \geq 0$ implies that $x=y$, then the entropy is finite.
I will note give the details, but, given $0 < \gamma < \epsilon < \delta/2$, positive expansivity yields the existence of a constant $C(\gamma,\epsilon)$ such that for all $n \geq 0$
$$ N(n,2\gamma) \leq C(\gamma,\epsilon) N(n,\epsilon).$$
From this it follows that $\limsup_{n \rightarrow \infty} \frac{1}{n}\log(N(n,\epsilon))$ is independent of $\epsilon$ for $\epsilon < \delta/2$ and, hence, $h(f)$ is finite.
$\bf{EDIT:}$ The dynamical system on the circle $S^1 \rightarrow S^1$, $x \mapsto qx ~\mathrm{ mod }~ 1$ has entropy $\log(q)$.
Now consider the dynamical system $f \colon \mathbb{D} \rightarrow \mathbb{D}$ on the unit disc in the complex plane given by $f(0) = 0$ and
$$f(x) = |x| e^{\mathrm{arg}(x) * i/|x|} \qquad \text{for }x \ne 0.$$
Then restricted to each circle $\{|x| = r\}$, $r \in (0,1]$, the restricted system is conjugate to $S^1 \rightarrow S^1$, $x \mapsto x/r ~\mathrm{ mod }~ 1$.
Since the entropy of a system is greater than or equal to the entropy of the system restricted to any closed invariant subset, we conclude that $h(f) \geq \log(1/r)$ for any $r < 1$, i.e. $h(f) = \infty$.
