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In a compact metric space $(X,d)$, for a given $\epsilon>0$, if $(x_j)_{j \in J}$ is a family of points of $X$ such that the balls $B(x_j, \epsilon)$ are pairwise disjoint, does it automatically follow that $J$ is finite?

Motivation: I was working through an exercise stating that if $(O_i)_{i \in I}$ is a family of disjoint open sets then $I$ is at most countable, so the above problem was the first idea that I had. Eventually I gave up on it and solved my exercise differently but I would like to know if that property even holds.

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Assume the contrary that $J$ is not finite.

Let $M$ be the closure of $(x_j)_{j\in J}$. As $M$ is a closed subset of the compactum $X$, itself is compact too. Now, the system $(B(x_j, \epsilon))_{j\in J}$ is an open cover of $M$. Hence, there are $j_1,\dotsc,j_n\in J$ such $$ M \subseteq \bigcup_{i=1}^n B(x_{j_i}, \epsilon). $$ Let $j\in J\setminus\{ j_1, \dotsc, j_n \}$. Then, it follows $$ x_j \in M \subseteq \bigcup_{i=1}^n B(x_{j_i}, \epsilon). $$ Thus, $B(x_j, \epsilon)$ is not disjoint to $B(x_{j_i}, \epsilon)$ for a suitable $1\le i \le n$.

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This property is called "totally bounded". You can use the compactness of X to prove it as follows: Let $ S= \{ B(x_j, \epsilon ) \} _{j \in J}$ where $B(x_j, \epsilon) \ne B(X_k, \epsilon)$ when $j \ne k$ .Let $T= \cup \{ B(y, \epsilon /2) | y \in X - \cup S \}$. Now $C=(S \cup \{ T \} ) - \{ \phi \}$ is an open cover of $X$ and it is irreducible: No proper subset of $C$ is a cover of $X$. So $C$ is finite ,because $X$ is compact. So $S- \{ \phi \}$ is finite. So $S$ is finite. So $J$ is finite. If you do not assume $B(x_j, \epsilon) \ne B(x_k, \epsilon)$ for distinct $j,k$ then $J$ can be infinite but I'm sure it's what you meant.

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