In the book "A Course in Metric Geometry" by (2x Burago + Ivanov) in the proof of Hopf$-$Rinow$-$Cohn-Vossen theorem, page 53, The authors want to show that a compact ball $\overline{B}_R(p)$ of radius $R$ and center $p$ is properly contained in an open ball $\overline{B}_{R + \varepsilon}(p)$ for some $\varepsilon > 0$. They assume a complete locally compact length (metric) space.
By the local compactness they cover $\overline{B}_R(p)$ with precompact open balls $\{B_{r(x)}(x)\}_{x \in \overline{B}_R(p)}$. Then by compactness they choose any finite subcover $\{B_{r_i}(x_i)\}_{i=1}^k$ thereof.
From the book:
- Since the open ball $B_R(p)$ is pre-compact, the closed ball $\overline{B}_R(p)$ is compact. (Recall that a closed ball in a length space is the closure of the respective open ball.) Now we show that a ball $B_{R+\varepsilon}(p)$ is pre-compact for some $\varepsilon > 0$. Since $X$ is locally compact, for every $x \in B_R(p)$ there is an $r(x) > 0$ such that the ball $B_{r(x)}(x)$ is pre-compact. Choose a finite subcover $B_{r(x_i)}(x_i)$ out of the cover of $\overline{B}_R(x)$ by these balls. The union of these balls is pre-compact and contains the ball $B_{R+\varepsilon}(p)$ for $\varepsilon = \min r_i > 0$.
Somehow the authors conclude that $\varepsilon = \min r_i$, which doesn't seem to work even for $\mathbb{R}^2$. For that take in $\mathbb{R}^2$, with the Euclidean metric, a unit ball and cover it with four unit balls equally separated from each other and from the center of the first.
My Questions:
Is the authors' conclusion for the $\varepsilon$ correct and how is that if it is ?
Otherwise how to prove that there is such $\epsilon$ ?
I will define $r(x)$ so that each ball $B(x,2r(x))$ is compact (note the extra factor of two comparing to authors' choice). Then, I will follow the authors' definition of $\epsilon=\min_i r(x_i)$. Let's show that this choice works.
Take any $x\in B(p, R+\epsilon)$. Then, since $X$ is a length metric space, there exists $y\in \bar{B}(p, R)$ such that $d(x,y)<\epsilon$. Since the balls $B(x_i, r(x_i))$ cover $\bar{B}(p, R)$, there exists $i$ such that $d(x_i, y)\le r(x_i)$. By the choice of $\epsilon$ and the triangle inequality, it follows that $d(x_i, x)< r(x_i) + \epsilon\le 2r(x_i)$. Hence, the precompact balls $B(x_i, 2r(x_i))$ cover the ball $B(p, R+\epsilon)$. In particular, $B(p, R+\epsilon)$ is precompact.
Added by @Physor: For the existence of such $y$.
Since $X$ is a length space then for any $x,y \in X$ and $\varepsilon > 0$ there is a constant-speed rectifiable $\gamma:[0,1] \to X$ joining $x,y$ such that $L(\gamma)<d(x,y) + \varepsilon$. Assuming $R_1,R_2>0$ are such that $d(x,y) < R_1 + R_2$, then for all $t \in [0,1]$ we have $d(x,y) \le d(x,\gamma(t)) + d(\gamma(t),y) \le L(\gamma|_{[0,t]}) + L(\gamma|_{[t,1]}) = L(\gamma) \le d(x,y) + \varepsilon$. This implies with $\varepsilon = R_1 + R_2 - d(x,y)$ that $L(\gamma|_{[0,t]}) + L(\gamma|_{[t,1]}) \le R_1 + R_2$. By the continuity and monotonicity of the map $t \mapsto L(\gamma|_{[0,t]})$ and the intermediate-value theorem there is $a \in (0,1)$ such that $L(\gamma|_{[0,a]}) = R_1$ and thus $L(\gamma|_{[a,1]}) < R_2$. Similarly there is a $\delta > 0$ such that $L(\gamma|_{[a-\delta,1]}) < R_2$ and hence $L(\gamma|_{[0,a-\delta]}) < R_1$. Thus we have $d(x,\gamma(a-\delta)) < R_1$ and $d(\gamma(a-\delta),y) < R_2$, i.e. $\gamma(a-\delta) \in B_{R_1}(x) \cap B_{R_2}(y)$.