Let $f:X\to X$ be uniformly continuous on the metric space (X,d). For $E,F\subset X$ we say that $E\, (n,\varepsilon)$-spans $F$ (with respect to $f$), if for each $x\in F$ there is an $y\in E$ so that $d(f^k(x),f^k(y))\leq \varepsilon$ for all $0\leq k<n$. We let $r_n(f,\varepsilon)$ denote the minimum cardinality of a set which $(n,\varepsilon)$-spans $F$. If $K$ is compact, then the continuity of $f$ guarantees $r_n(f,\varepsilon)<\infty$. For $K$ compact we define $$\overline{r}_f(K,\varepsilon)=\limsup_{n\to\infty}\frac{1}{n}\log r_n(K,\varepsilon)$$ and \begin{equation} h(f,K)=\lim_{\varepsilon\to 0}\overline{r}_f(K,\varepsilon)\hspace{1cm}(1) \end{equation} Finally let $$h(f)=\sup_{K}h(f,K)\hspace{1cm}(2)$$ where $K$ varies over all compact subsets of X. If $X$ is compact is clear that $h(f)=h(f,X)$. The definition (2) is given by R. Bowen to define the topological entropy of $f$.
Now, define for $x\in X$ $$\Phi_{\varepsilon}(x)=\bigcap_{n\geq 0}f^{-n}B_{\varepsilon}(f^n(x))=\{y\in X:\,d(f^n(x),f^n(y))\leq \varepsilon,\,\,\forall n\geq 0\}$$
I need to know if for $X$ not necessarily compact the set $\Phi_{\varepsilon}(x)$ is compact.
The reason: I'm reading Entropy-Expansive Maps by R. Bowen and he defines $$h^*(\varepsilon)=\sup_{x\in X}h(f,\Phi_{\varepsilon}(x))$$ I'm confusing about $h(f,\Phi_{\varepsilon}(x))$. I don't know if he is referring to definition given in (1) (in that case $\Phi_{\varepsilon}(x)$ must be compact) or simply the topological entropy of $f$ restricted to $\Phi_{\varepsilon}(x)$ ($f:\Phi_{\varepsilon}(x)\to \Phi_{\varepsilon}(x)$).
I would appreciate some comments