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Suppose $K$ is a compact subset of $\mathbb{R}^m$ ($m>1$), and $0<r<R$ are fixed numbers. Let $A$ be the set of points having a distance $<R$ and $>r$ from $K$. My questions are

  1. If $K$ and its complement are connected, is it possible that $A$ have an infinite number of components?

  2. If yes, which are some extra conditions that imposed to $K$ will exclude that $A$ can have an infinite number of components?

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  • $\begingroup$ 1) Yes, think about embedding the universal cover of $S^1 \vee S^1$ in the plane, i.e. $m=2$. Choose any embedding where the ends get close to touching. 2) Sure. Isn't your item (2) one such condition? $\endgroup$
    – Ryan Budney
    Commented Feb 18, 2022 at 5:37
  • $\begingroup$ @RyanBudney : Where can one learn about necessary conditions and sufficient conditions for the existence of a universal cover, and about how to construct a universal cover when it is known to exist? $\endgroup$
    – Iosif Pinelis
    Commented Feb 18, 2022 at 14:35
  • $\begingroup$ @IosifPinelis: Most introductory algebraic topology textbooks have the construction. The method I am thinking of is as a quotient space of a path-space construction. Once you see it, it has a very intuitive nature to it. I came across a homeless guy in New York (no formal mathematical training) who had near a hundred hand sketches of embeddings of finite-sheeted and universal covers of graphs into the plane. I imagine this could be turned into a fun semi-complicated game, like latin squares / sudoku. $\endgroup$
    – Ryan Budney
    Commented Feb 18, 2022 at 17:22
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    $\begingroup$ I don't know anything cute and minimal. But my first inclination would be to have a "nice" geometric regular neighbourhood. By "nice" I mean something like smooth or PL. And by "geometric" I'm talking about using the distance function you use in your question to define the regular neighbourhood. So maybe you could get away with an NDR pair $(\Bbb R^n, K)$ or something like that? $\endgroup$
    – Ryan Budney
    Commented Feb 18, 2022 at 19:03
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    $\begingroup$ This is not a topological problem because $A$ can have infinitely many components with $K$ being the image of an injective curve $\gamma:[0,3]\to\mathbb{R}$: you can take $\gamma(\frac{1}{n})=(\frac{1}{n},0)$, $\gamma(3-\frac{1}{n})=(\frac{1}{n},1)$ for natural $n\geq1$, $\gamma(0)=(0,0)$, $\gamma(3)=(0,1)$ and define $\gamma$ in the rest of the domain so that it stays outside of the rectangle $[0,1]\times[-2,2]$ (and $r=1/2,R=\infty$). $\gamma$ can probably be chosen to be $C^\infty$ as well. $\endgroup$
    – Saúl RM
    Commented Feb 19, 2022 at 18:15

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