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In the application of Runge type theorems on the approximation of functions with some regularity on a neighborhood of a compact, it is interesting to know whether the complement of a compact has bounded components or even is connected.

Suppose $F$ is a closed and unbounded set in $\mathbb{R}^m$ with $m>1$. For $r>0$ let $F_r$ be the set of all points in $\mathbb{R}^m$ having a distance $\:=r\:$ to $\,F$. We set $A=(F\cup F_r\cup F_R)\cap B$, where $0<r<R$ and $B$ is a closed ball. Suppose that no set here is empty.

My question: Are there some known conditions on $F$ so that the complement of $A$ in $\mathbb{R}^m$ is connected? Any reference or suggestion is welcome.

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    $\begingroup$ What is the purpose of introducing the sets A, B, and $F_r, F_R$, when your question doesn't seem to invoke them at all? $\endgroup$
    – Willie Wong
    Sep 29, 2021 at 18:39
  • $\begingroup$ The set B is there to make all other sets compact. But I do need $F_r$, $F_R$ and $F$. My function is regular on a neighborhood of $F$ as it is. $\endgroup$
    – M. Rahmat
    Sep 29, 2021 at 21:53
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    $\begingroup$ This sounds related to Karol Borsuk's results covered in the Appendix of the Eilenberg-Steenrod monograph. Also, the still more powerful Alexander-Pontryagin duality is here of interest. $\endgroup$
    – Wlod AA
    Sep 30, 2021 at 9:50
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    $\begingroup$ I still don't understand. Your question is "Are there some known conditions on $F$ so that the complement of $F$ in $\mathbb{R}^m$ is connected?" The other sets introduced play absolutely no role in that question. Did you mean to ask about whether the complement of $A$ is connected? $\endgroup$
    – Willie Wong
    Sep 30, 2021 at 19:07
  • $\begingroup$ Yes, you are right! I corrected. Thanks. $\endgroup$
    – M. Rahmat
    Oct 1, 2021 at 10:17

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