Diameter metric space 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.
 A: Here is a direct proof of (a).
First, since $E \subseteq \overline{E}$, we have $\{d(a,b) : a,b \in E\} \subseteq \{d(a,b) : a,b \in \overline{E}\}$, so
$$\phi(E) = \sup \{d(a,b) : a,b \in E\} \leq \sup \{d(a,b) : a,b \in \overline{E}\} = \phi(\overline{E}).$$
Next, let $a,b \in \overline{E}$ be arbitrary. There exist sequences $(x_n), (y_n)$ in $E$ such that $x_n \to a$ and $y_n \to b$ as $n \to \infty$. Now by the triangle inequality,
$$d(a,b) \leq d(a,x_n) + d(x_n,y_n) + d(y_n,b) \leq d(a,x_n) + \phi(E) + d(y_n,b)$$
for all $n$. Thus,
$$d(a,b) \leq \inf \{d(a,x_n) + \phi(E) + d(y_n,b) : n \in \mathbb{N}\} = \phi(E).$$
Since $a$ and $b$ were arbitrary, $\phi(E)$ is an upper bound for $\{d(a,b) : a,b \in \overline{E}\}$. Therefore,
$$\phi(E) \geq \sup \{d(a,b) : a,b \in \overline{E}\} = \phi(\overline{E}).$$
We have now shown $\phi(E) \leq \phi(\overline{E})$ and $\phi(E) \geq \phi(\overline{E})$, so we conclude that $\phi(E) = \phi(\overline{E})$. $\square$
A: 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 $d(z, x) \leq d(z, y) + d(y, x) \leq 2\varepsilon$.
A: First we show that if $E \subset B$ then $\phi(E) \leq \phi(B)$. This follows directly:
$$\phi(E) = \sup_{x,y \in E} d(x,y) \leq \sup_{x,y \in B}d(x,y) = \phi(B).$$
Hence, $\phi(E) \leq \phi(\overline{E})$ follows immediately.
Going the other way (i.e., showing $\phi(E) \geq \phi(\overline{E})$): Cover every point in $E$ by balls of some fixed radius $\epsilon > 0$:
$$E = \bigcup_{x \in E} \{x\} \subset \bigcup_{x\in E} B_\epsilon(x)$$
Next note that $\overline{E} \subset \bigcup_{x\in E} \overline{ B_{\epsilon/2}(x) }\subset \bigcup_{x\in E} B_\epsilon(x) $ and by the triangle inequality, $\phi\left( \bigcup_{x\in E} B_\epsilon(x) \right) \leq \phi(E) + 2\epsilon$.
So by applying the first result (with $B \equiv \bigcup_{x\in E} B_\epsilon(x)$),
$$ \phi(\overline{E}) \leq \phi\left( \bigcup_{x\in E} B_\epsilon(x) \right) \leq \phi(E) + 2\epsilon $$
as this holds for arbitrary $\epsilon > 0$, let $\epsilon \to 0^+$ and hence $\phi(\overline{E}) \leq \phi(E)$.
