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?

  • $\begingroup$ What is $IsoA$? $\endgroup$ – Chris Eagle Jul 1 '13 at 16:39
  • $\begingroup$ Set of isolated points of A $\endgroup$ – Heisenberg Jul 1 '13 at 16:40
  • $\begingroup$ Why do you think that such a family exists? $\endgroup$ – Carl Mummert Jul 2 '13 at 12:31
  • $\begingroup$ came across it in a examination paper $\endgroup$ – Heisenberg Jul 2 '13 at 12:33
  • $\begingroup$ 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. $\endgroup$ – 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 $t_1 \in B(t_2, r_2)$!.

  • 1
    $\begingroup$ It's very non-obvious why $r_t/10$ works for any choice of $r_t$. $\endgroup$ – Asaf Karagila Jul 1 '13 at 16:53
  • $\begingroup$ Thanks. It doesn't. That was careless. $\endgroup$ – hot_queen Jul 1 '13 at 16:55
  • $\begingroup$ @AsafKaragila what if we take $r_{t}=dist(t,A)/2$ $\endgroup$ – Heisenberg Jul 1 '13 at 16:57
  • $\begingroup$ @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. $\endgroup$ – Asaf Karagila Jul 1 '13 at 16:59
  • 1
    $\begingroup$ I added a (very simple) proof. $\endgroup$ – hot_queen Jul 1 '13 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.