Let $X$ be a metric space and let $r>0$. Then either there is a countable collection of balls of radius $r$ that covers $X$. The full theorem is:
Let $X$ be a metric space and let $r>0$. Then either there is a countable collection of balls of radius $r$ that covers $X$, or there is an uncountable collection of disjoint balls in $X$ of radius $r/2$.
The professor suggested that we might need Zorn's Lemma to prove this theorem, but I had no idea, and I want to know what this theorem wants to show in geometry.
Thanks!
 A: I think the idea behind this problem is the following: Let $d$ denote the metric on $X$ and let $B(x,c):=\{y \in X: d(y,x) < c\}$ (the open ball with center $x \in X$ and radius $c>0$). Fix $r >0$ and let $Z \subseteq X$ be such that
${\cal C}:=\{B(z,r/2): z \in Z\}$ is a collection of balls such that

*

*$z_1,z_2 \in Z$, $z_1 \not= z_2$ $\Rightarrow$ $B(z_1,r/2) \cap B(z_2,r/2)= \emptyset$,

*${\cal C}$ is maximal in the sense that adding any ball $B(x,r/2)$ for some $x \notin Z$ leads to $B(z_0,r/2) \cap B(x,r/2) \not= \emptyset$ for some $z_0 \in Z$.

The existence of such a maximal collection can be proved by Zorn's Lemma by ordering the set of all collections with property 1. by inclusion. Now given such a maximal collection you can show that
$$
X=U:=\bigcup_{z \in Z} B(z,r),
$$
since if $x \in X$ there are two cases: If $x \in Z$ then $x \in U$, trivially. If $x \in X \setminus Z$ then $B(z_0,r/2) \cap B(x,r/2) \not= \emptyset$ for some $z_0 \in Z$. Let $y \in B(z_0,r/2) \cap B(x,r/2)$. Then
$$
d(x,z_0) \le d(x,y) + d(y,z_0)< r/2 + r/2 = r,
$$
that is $x \in B(z_0,r)$. Hence $x \in U$. Thus $\{B(z,r): z \in Z\}$ covers $X$. If there is no countable collection of balls of radius $r$ that covers $X$ then $Z$ and with that too ${\cal C}$ is uncountable.
A: If you assume Ball of radius $N < \infty$ is compact in $X$ and distance between any pair of points is finite in $X$ then $X$ can be covered by a countable collection of balls of radius $r$. Here is the proof: Take $N$ an integer, take Ball of radius $N$, $B_N$ in X centered around some fixed point $x$. Then by compactness this $B_N$ can be covered by a finite collection $B_N(x) = \cup_{i = 1}^{r(N)} B_r(x_i)$. Now form a set $S_N = \{x_1,...,x_{r(N)}\}$. then as $N \rightarrow \infty$, the total collection of such sets $X = \cup_{N=1}^{\infty} B_N(x) = \cup_{N=1}^{\infty} \cup_{i = 1}^{r(N)} B_r(x_i)$ is a cover for $X$ with a countable collection of sets since $\cup_{N=1}^{\infty} S_N$ is countable as it is a countable union of countable sets. So now the problem is now reduced to the case, when $B_N(x)$ is not compact or distance between some pair of points is $\infty$. I dont think latter is an interesting case. So consider when $B_N$ is not compact and complete the proof.
A case where $B_N$ with $N>(r+1)$ is not compact: Take some uncountable collection of points $X$. Define $d(x,y) = r+1$ if $x \neq y$ and $d(x,y) = 0$ iff $x=y$. This is a metric and with this metric $B_N(x)$ is not compact as $B_N(x) = X$ (full space) and $B_r(x)$ contains only a singleton i.e., $B_r(x) = \{x\}$
