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Given a compact set $K$ in $\mathbb{R}^2$ assume that $0$ lies in the unbounded component of $K^c$ (the complement of $K$).

I am struggling to show the following claim and I would appreciate any help/advices on how to establish this:

(Claim:) There is simply connected open set $\Omega$ containing $K$ and with $0 \notin \Omega$.

EDIT: I wonder if the following works: If I take the union of balls centered at points of $K$ (and with radius so small that no ball contains 0) and denote this by $U$. Then can't I take the interior of a contour surrounding $K$ in $U$ as the definition of my $\Omega$?

Digitallis
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P.Jo
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    Glue together small open balls not containing $0$ centered at every point in the compact set. Glue in neighborhoods of all the bounded components of complement (it suffices to consider a large enough ball at each point in bounded part of complement and intersect with the open cover from before). I don't see how compactness is at all relevant here, merely closedness – Brevan Ellefsen Jun 10 '23 at 09:42
  • thanks for the hint. But which part of the "gluing together" ensures that the resulting set is simply! connected - not merely connected? – P.Jo Jun 10 '23 at 10:44
  • And one further question: How do I ensure that a intersection of connected sets remains connected? (this seems to be needed in the proof you indicated) – P.Jo Jun 10 '23 at 11:10
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    Intuitively it's null homologous, because we can deformation retract along the balls. More formally, it's a standard lemma in complex analysis that a subset of the plane is simply connected iff both it and its complement are connected in the Riemann Sphere, which is clearly the case here. (the cover for $K$ is obviously connected, and outside of a large enough ball containing $K$ the complement is just the plane so is connected). An intersection of path connected sets is path connected by definition: every point reaches every other point via a path. – Brevan Ellefsen Jun 11 '23 at 09:55
  • Here's a standard proof of the complex analysis result. I'd have to think about the truth of the generalization of your question to $\mathbb R^3$, but I'd think the Alexander Horned Sphere would yield a counterexample. – Brevan Ellefsen Jun 11 '23 at 09:58
  • Thank you very much for these explanations. I am making some progress in understanding this - however: maybe you could elaborate further on your proposes construction: E.g. If I take the union of small balls at every point of the compact set, then I don't see what guarantees connectedness of the union. Suppose I have a compact set K' and a single point x outside and define K as the union of K' with the single point x. Then K is compact but certainly, not every union of balls around the points of K is connected. – P.Jo Jun 18 '23 at 12:16
  • You're right that I was assuming the compact set is connected to use that complex analysis lemma (which is usually done via the Riemann Mapping Theorem).. If it's not connected already, then we can connect the components one-by-one using arcs avoiding $0$ and leaving the complement connected. Explicitly, you should just be able to take straight line paths for every point in the plane except the line through $0$ to the compact set, on which we use a.... – Brevan Ellefsen Jun 19 '23 at 01:04
  • .... slightly curved path (say, to the left; by connecting all components along this line first then bridging this component to the main compact set, we ensure $0$ remains in the unbounded component). Now just cover with small enough balls to avoid $0$ and pull back via compactness to finite cover and glue it all together to get a connected cover of your compact set which excludes $0$ and has connected complement so we're done. Note also the following questions which yours is duplicate to and have answers similar to mine: – Brevan Ellefsen Jun 19 '23 at 01:06
  • https://math.stackexchange.com/q/4469338/269764 $$$$ https://math.stackexchange.com/q/1141922/269764 $$$$ https://math.stackexchange.com/q/4621685/269764 – Brevan Ellefsen Jun 19 '23 at 01:07

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