# Pair wise disjoint open balls

Let $(X,ρ)$ be a metric space and $A$ is a subset of $X$. How do I prove that there exists a family $(B(t,r_{t}))_{t\in\operatorname{Iso}A}$ of pair wise disjoint open balls?

• What is $IsoA$? – Chris Eagle Jul 1 '13 at 16:39
• Set of isolated points of A – Heisenberg Jul 1 '13 at 16:40
• Why do you think that such a family exists? – Carl Mummert Jul 2 '13 at 12:31
• came across it in a examination paper – Heisenberg Jul 2 '13 at 12:33
• You should include that context in the question - what exam was it? What level is it at? That will help others make their answers more focused, and help others who look at the question in the future. – Carl Mummert Jul 2 '13 at 12:34

Recall that a point $t \in A$ is isolated in $A$ if there is some ball $B(t, r_t)$ centered at $t$ of radius $r_t > 0$ which does not meet any other point of $A$. Now consider the family $\{B(t, r_t/10) : t \in \text{Iso}(A)\}$.
Proof: Suppose $x \in B(t_1, r_1/10) \cap B(t_2, r_2/10)$. WLOG, $r_1 \leq r_2$. But then $t_1 \in B(t_2, r_2)$!.
• It's very non-obvious why $r_t/10$ works for any choice of $r_t$. – Asaf Karagila Jul 1 '13 at 16:53
• @AsafKaragila what if we take $r_{t}=dist(t,A)/2$ – Heisenberg Jul 1 '13 at 16:57
• @Rajinda: $t\in A$ so $d(t,A)=0$. But even if it weren't (e.g. take $d(t,A')$ instead), there's no guarantee that the balls are pairwise disjoint. You need a better argument. E.g. one appealing to paracompactness for example, or regularity/normality of metric spaces. – Asaf Karagila Jul 1 '13 at 16:59