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In a paper I am reading, the following is considered obvious:

Let $K$ be a compact and connected subset of $\,\mathbb R^2$, with $\mathbb R^2\smallsetminus K$ also connected, and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected $($and $K$ not necessarily connected$)$ and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.

I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries.

Any ideas?

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    $\begingroup$ Have you tried considering $\varepsilon$-neighborhoods of $K$ with $\varepsilon$ small enough ? I bet such neighborhoods are simply connected if $K$ is (when $\varepsilon$ is small enough). $\endgroup$ – Loïc Teyssier Feb 11 '15 at 8:21
  • $\begingroup$ If $K$ is sufficiently smooth this is a consequence of the tubular neighborhood theorem (because $K$ is a retract of a sufficiently small $\epsilon$-neighborhood). $\endgroup$ – Francesco Polizzi Feb 11 '15 at 8:48
  • $\begingroup$ If the boundary of $K$ is sufficiently smooth, then the tubular neighbourhood works. But when $K$ has a non-smooth boundary, one can construct a counterexample, where for every $\varepsilon>0$, the corresponding tubular neighbourhood is not simply connected. $\endgroup$ – smyrlis Feb 11 '15 at 9:43
  • $\begingroup$ In fact, $K$ is retract of some (open) neighborhood if and only if it is weak locally contractible, see my answer below. $\endgroup$ – Francesco Polizzi Feb 11 '15 at 10:16
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    $\begingroup$ What is the paper’s definition of “simply connected”? As demonstrated in Gabriel Drumond-Cole’s answer, the “more generally” part is not actually more general if one defines simply connected by null-homotopy of loops, as this does not imply connectedness of the complement of $K$. Perhaps the authors intend the latter as their working definition? $\endgroup$ – Emil Jeřábek Feb 11 '15 at 13:48
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The Warsaw circle is compact and simply connected but there are obvious neighborhoods with no simply connected open refinement. This provides a counterexample for the question as originally worded but does not have connected complement.

It can be realized as the union of the following: $$\left\{(x,\sin\frac{1}{x}):0<x<1\right\}$$ $$0\times[-2,1]$$ $$[0,1]\times -2 $$ $$1\times[-2,\sin 1]$$

picture of the above union

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    $\begingroup$ Smyrlis, what is your definition of `simply connected'? $\endgroup$ – Paul Fabel Feb 11 '15 at 13:50
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    $\begingroup$ This is called Warsaw circle. It is a standard example of 1-dimensional continuum whose homotopy groups are all trivial, but which is neither contractible nor locally contractible. en.wikipedia.org/wiki/Continuum_%28topology%29 $\endgroup$ – Francesco Polizzi Feb 11 '15 at 14:10
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    $\begingroup$ Maybe you should add the connected complement requirement to the question. $\endgroup$ – Gabriel C. Drummond-Cole Feb 11 '15 at 14:14
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    $\begingroup$ @Francesco I don't see why the comb has an open neighborhood with no simply connected open subspace. $\endgroup$ – Gabriel C. Drummond-Cole Feb 11 '15 at 14:20
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    $\begingroup$ @Francesco, sure, of course it's not locally connected, but I think it's essentially obvious that it's not a counterexample to the question. $\endgroup$ – Gabriel C. Drummond-Cole Feb 11 '15 at 14:35
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This is only a partial answer, however it is too long for a comment.

What you want is actually true for compact subsets $K \subset \mathbb{R}^n$, i.e. in any dimension, under the additional assumption that $K$ is locally contractible, i.e. that any point $x \in K$ has a contractible neighborhood.

This is a consequence of the following result, that can be found in Hatcher's book Algebraic Topology, Theorem A.7. page 525 in the Appendix.

Theorem. A compact subspace $K \subset \mathbb{R}^n$ is a retract of some (open) neighborhood $V$ if and only if it is locally contractible in the weak sense that for each $x \in K$ and each neighborhood $A$ of $x$ in $K$ there exists a neighborhood $B \subset A$ suct that the inclusion map $B \hookrightarrow A$ is nullhomotopic.

Now, if $K$ is retract of some neighborhood, it is also a retract of any smaller neighborhood, just by restriction of the retraction. So, if the assumption of the theorem is satisfied, you can find a simply connected open subset $V$ such that $K \subset V \subset U$.

This does not completely answer the question because there are compact, simply connected subspaces of $\mathbb{R}^n$ that are not locally contractible. An example in $\mathbb{R}^2$ is the comb space, which is compact, contractible (hence simply connected) but not locally connected, hence not locally contractible. An example in $\mathbb{R}^3$ is given by the cone over the Hawaiian earring, which is contractible but not even locally simply connected.

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Perhaps the theorem at the common intersection of the various ambiguous interpretations of the inquiry at hand is the following:

If the planar open set U contains a continuum K and if the planar complement of K is connected, then there exists a connected and simply connected open set V such that K is a subset of V, and V is a subset of U.

This follows, for example, from the fact that K is cellular, the nested intersection of closed topological planar disks. Surround K by a slightly larger disk D, so that D is contained in U, and the interior of D is the coveted open set V.

For a barehanded argument of the cellularity claim, prove by induction that if C is a planar simple closed curve, and if C is the cancatanation of vertical and horizontal segments of length 1, then C bounds a topological disk.

To obtain the mentioned disk D, tile the plane by very small squares, and extract a curve C from the boundaries of the tiled squares.

Of course the argument sketched above does not use any heavy artillery such as the Jordan curve theorem or the Schoenflies theorem.

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  • $\begingroup$ Interesting use of language here (twice) that I don't recall seeing before: "If [X], if [Y], then [Z]." Is this equivalent to "If [X] and [Y], then [Z]"? Or maybe "If [X] such that [Y], then [Z]". $\endgroup$ – Allen Hatcher Feb 11 '15 at 15:07
  • $\begingroup$ I admit that I don't see how to easily choose the curves so that the intersection of their `interiors' is $K$ without some artillery. $\endgroup$ – Mathieu Baillif Feb 11 '15 at 19:26
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    $\begingroup$ If we allow the Riemann mapping theorem, take a conformal map from the (punctured) open unit disk minus (0,0) onto the complement of K. The images in the plane of the circles centered at (0,0) of radius 1/n will do. $\endgroup$ – Paul Fabel Feb 11 '15 at 21:07
  • $\begingroup$ Thanks, Paul. That's what I had in mind, but the Riemann mapping theorem is in my mind heavy artillery, that's why I thought there was an "elementary" argument that I had not seen. $\endgroup$ – Mathieu Baillif Feb 11 '15 at 21:41
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    $\begingroup$ If the complement of $K$ is connected, just consider the set of points of the complement to which you can drag an open disk of fixed very small radius from infinity without touching $K$. It is a closed set and its complement $V$ is open, contains $K$, and is simply connected just because to drag the center of the disk inside a loop, you need to drag it to the boundary of the loop first (though you will need some kind of induction argument to show it with full rigor and completely from scratch). $\endgroup$ – fedja Feb 13 '15 at 23:58
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Pick a point $x_i$ in every connected component $X_i$ of $\mathbb{R}^2 \backslash U$. Since $\mathbb{R}^2 \backslash K$ is connected and open, it is path connected. Fix some distinguished $x_0 \in \mathbb{R}^2 \backslash K$, and create a path $\gamma_i$ connecting $x_i$ to $x_0$ in $\mathbb{R}^2 \backslash K$. Let $V$ be $U$ minus the union of these paths. As long as there are a finite number of $X_i$, $V$ will be open and all its connected components will be simply connected. So you just need to show that you can replace $U$ with a small open subset containing $K$ whose complement has finitely many connected components.. A proof eludes me right now.

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