$\newcommand{\diam}{\operatorname{Diam}}$
$\newcommand{\R}{\mathbb{R}}$

**Definition:** Let $X$ be a metric space, $M \subseteq X$ is bounded if $\diam(M) < \infty$

Let $(X,d)$, $X \not= \emptyset$, be a metric space and $M \subseteq X$. Show that the following are equivalent:

(i) $M$ is bounded.

(ii) $\exists\_{x \in X}\left\{\exists\_{r > 0}\left[ M \subseteq B\_{r}(x)\right]
\right\}$

(iii) $\forall\_{x \in X}\left\{\exists\_{r > 0}\left[ M \subseteq B\_{r}(x)\right]\right\}$

**Attempt (following suggestions from you):**

Suppose $M$ is bounded. We want to see that $\exists\_{x \in X}\left\{\exists\_{r > 0}\left[ M \subseteq B\_{r}(x)\right]\right\}$.

(iii) $\rightarrow$ (ii)

Suppose that $\forall\_{x \in X}\left\{\exists\_{r > 0}\left[ M \subseteq B\_{r}(x)\right]\right\}$. We want to see that $\exists\_{x \in X}\left\{\exists\_{r > 0}\left[ M \subseteq B\_{r}(x)\right]\right\}$.

By hypothesis it is true for any $x \in X$. Therefore, in particular, it is true for some $x \in X$. Hence, $\exists\_{x \in X}\left\{\exists\_{r > 0}\left[ M \subseteq B\_{r}(x)\right]\right\}$.

(ii) $\rightarrow$ (i)

Suppose that $\exists\_{x \in X}\left\{\exists\_{r > 0}\left[ M \subseteq B\_{r}(x)\right]\right\}$. We want to see that $M$ is bounded. Suppose by contradiction that $M$ is not bounded.

By hypothesis above, we have that $\diam(M) = \infty$. So there exists $m\_{1},m\_{2} \in M$ such that they are arbitrary far. It is a contradiction! Because the ball $B\_{r}(x)$, which has a finite radius $r$, can not contain its points $m\_{1}$ and $m\_{2}$ but they are inside $M$.

How can I deduce (i) $\rightarrow$ (iii)?