Suppose $(A, d)$ is a metric space with a non-empty subset $E \subseteq A$. Let $S = \{d(a, b) : a, b \in E\}$. Define $\phi (E)$ be the diameter of $E$ so that $\phi (E) = \sup S$, and $\phi(E) = \infty$ if $S$ is unbounded. Prove the following:
(a) If $E \subseteq A$ is non-empty then $\phi(E) = \phi(\overline{E})$ (the diameter of $E$ is equal to that of the closure of $E$).
(b) If $\phi (E) < \epsilon$ for some $\epsilon > 0$ and $E \cap B_{\epsilon}(x) \neq \emptyset$ for some $x$, then $E \subseteq B_{2\epsilon}(x)$.
For (a) I came up with a proof by contradiction but I am curious as to whether there are other proofs. After much time spent, I still cannot figure out a proof for (b). Any assistance or alternatively proofs are appreciated.
Best Answer
For part (b): Following the comments of Martin R and myself, we pick a point $y \in E \cap B_\varepsilon(x)$, since the latter is non-empty. Then for any point $z \in E$, we have $d(z, y) \leq \varepsilon$ (because $\phi(E) = \varepsilon$), and we also have $d(y, x) \leq \varepsilon$ (since $y \in B_\varepsilon(x)$). Therefore, by the triangle inequality, we must have $d(z, x) \leq d(z, y) + d(y, x) \leq 2\varepsilon$.