Hi everyone: Suppose $A\subset \mathbb{R}^{m}$ ($m>1$) is a closed set with empty interior. Which are the necessary and sufficient conditions that $A$ have a neighborhood $V$ such that every bounded component of $ \mathbb{R}^{m}\setminus A$ has a point of $\mathbb{R}^{m}\setminus V$?


closed as off-topic by Dima Pasechnik, Loïc Teyssier, user1688, RP_, Stefan Kohl Mar 12 '17 at 10:11

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    $\begingroup$ Necessary and sufficient for the existence of such a $V$ is the existence of such a $V$. You need to be more specific. $\endgroup$ – user1688 Mar 12 '17 at 7:49
  • $\begingroup$ What do you mean more specific? The set $A$ is closed and has no interior points. $\endgroup$ – M. Rahmat Mar 12 '17 at 20:00
  • $\begingroup$ I added more conditions. $\endgroup$ – M. Rahmat Mar 12 '17 at 20:12

Suppose $A$ is compact. Then necessary and sufficient is that there is some $\varepsilon > 0$ such that every bounded component of $\mathbb R^n \backslash A$ contains a ball of radius $\varepsilon$.

  • $\begingroup$ Correct, but this is not necessary: the union of the spheres $S(n,1/n)$ may have a neighborhood $V$, as explained but the your $\varepsilon$ does not exist. What is the necessary condition? $\endgroup$ – M. Rahmat Mar 12 '17 at 18:31
  • $\begingroup$ @M.Rahmat Unless I misunderstand your notation, that example is not compact. $\endgroup$ – Robert Israel Mar 12 '17 at 19:11
  • $\begingroup$ You are right, this is why I accepted you partial answer. How about the general case, where $A$ is just closed? $\endgroup$ – M. Rahmat Mar 12 '17 at 19:57

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