Let $K$ be a compact set in $\mathbb{R}^n$ and let $U$ be a bounded open set that contains $K$. You may assume both are connected.

Can we always find an open $V$ such that $K\subset V\subset\overline{V}\subset U$ such that $U\backslash K$ retracts on $U\backslash V$?

For example, if we somehow find $V$ such that $\overline{V}$ is homeomorphic to the closed ball, then choosing any point of $x\in K$ to be its center, we can radially repel all the points of $V\backslash\{x\}\supset V\backslash K$ onto $\partial V$ and leave every other point fixed.

I tried to cover $K$ with cubic neighborhoods and run an induction over the number of cubes, but it is getting too messy. Perhaps there is a more clever way.

  • $\begingroup$ I hope that simplicial partitions of $\mathbb R^n$ could be of help. In other words the methods of PL-topology should work. $\endgroup$ – Taras Banakh May 10 at 18:27
  • $\begingroup$ @TarasBanakh I am not familiar with the PL-topology. Perhaps you could illustrate your idea in dimension $2$, with a triangulation of the plane? $\endgroup$ – erz May 10 at 22:32
  • $\begingroup$ Divide the plane into squares, then each square into two triangles. Take the partition into squares so small that a single square does not intersect simultaneously $K$ and the complement to $U$. Divide each triangle by barycentric subdivision into 6 smaller triangles with a vertex in a center of the triangle. Take the union $T$ of all triangles of the barycentric subdivision that intersect $K$. Take the union of all triangles of the barycyntric subdivision that intersect $T$ and put $V$ be its interior in the plane. It seems that $V$ is the neighborhood of $K$ you are looking for. $\endgroup$ – Taras Banakh May 11 at 7:49
  • $\begingroup$ For basic definitions of PL-topology look the first pages of this paper maths.ed.ac.uk/~v1ranick/papers/pltop.pdf (this was the first what google suggested for the search "PL-topology"). $\endgroup$ – Taras Banakh May 11 at 7:53
  • $\begingroup$ @TarasBanakh I thought for a while about separating the plane into squares or triangles, but still not sure how they help. What is the significance of the specific triangulation that you suggested? $\endgroup$ – erz May 15 at 14:30

Let us show how to find such a retraction for $n=2$ (I do not know if this method generalizes to higher dimensions).

Given a compact set $C\subset\mathbb R^2$ and an open neighborhood $U\subseteq\mathbb R^2$ of $C$, choose a triangulation on $\mathbb R^2$ so fine that no triangle of the triangulation meets $C$ and $\mathbb R^2\setminus U$ simultaneously.

Replacing the triangulation by a finer triangulation, we can assume that for each triangle $T$ with $T\not\subseteq C$, one vertex of $T$ does not belong to $C$.

How to find such a triangulations? Assuming that $T\not\subseteq C$, we can find an interior point $v$ of $T$ that does not belong to $C$ and replace the triangle $T$ by 3 subtriangles having $v$ as a vertex.

Also we can assume that either $T\subseteq\mathbb R^2\setminus C$ or $T$ has a vertex in $C$. Assuming that $T$ has no vertices in $C$ but $T\cap C\ne\emptyset$, we can choose a point $c\in T\cap C$ and replace the triange $T$ by two or three triangles having $c$ as a vartex.

Therefore, we lose no generality assuming that each triangle $T$ of the triangulation has one of the following properties:

1) $T\subseteq C$;

2) $T\cap C=\emptyset$;

3) $T$ has one vertex in $C$ and one vertex outside of $C$;

4) If two vertices $u,v$ of $T$ do not belong to $C$, then the side $[u,v]$ does not intersect $C$.

A triangle $T$ will be called difficult if it has one vertex say $u$ outside $C$, two vertices $v,w$ in $C$ and the side $[v,w]$ is not a subset of $C$. In this case choose any point $c[v,w]\in [v,w]\setminus C$. The points $c[v,w]$ can be chosen so that for two difficult triangle sharing the common side $[v,w]$ the point $c[v,w]$ is the same.

Now for every triangle $T$ of the triangulation we define a function $r_T\setminus C:T\to T\setminus C$ such that $r_T\circ r_T=r_T$ as follows. In case (1), let $r_T$ be the empty map and in case (2) $r_T$ be the idenity map of $T$. In the remaining cases, the triangle $T$ has one vertex in $C$ and one vertex outside of $C$. If the triangle $T$ is not difficult, then it has two vertices $u,v$ such that the side $[u,v]$ either is contained in $C$ or is disjoint with $C$. If $[u,v]$ is contained in $C$, then let $r_T:T\setminus C\to\{w\}$ be the constant map into the unique vertex $w\notin C$ of $T$.

If $[u,v]\cap C=\emptyset$, then the third vertex $w$ of $T$ belongs to $C$ and we can apply the Urysohn lemma to find a function $r_T:T\setminus C\to[u,v]$ such that $r_T[[w,u]\setminus C]=\{u\}$, $r_T[[w,v]\setminus C]=\{v\}$, and $r_T(x)=x$ for every $x\in [u,v]$.

It remains to consider the case of a difficult triangle $T$. Such a triangle has one vertex $u$ outside of $C$, two vertices $v,w$ in $C$ and the point $c[v,w]\in [v,w]\setminus C$. Two cases are possible.

1) There exists a path $\gamma:[0,1]\to T\setminus C$ such that $\gamma(0)=u$ and $\gamma(1)=c[v,w]$. We can assume that $\gamma$ is injective and hence its image $A_T=\gamma[0,1]$ is an arc with endpoints $u$ and $c[v,w]$. Using the Urysohn Lemma, we can find a continuous function $r_T:T\setminus C\to A_T$ such that $r_T[([u,v]\cup[u,w])\setminus C]\subseteq \{u\}$, $r_T[[v,w]\setminus C]\subseteq\{c[v,w]\}$ and $r_T(a)=a$ for every $a\in A_T$.

2) No such a path $\gamma$ exists. Then the points $u$ and $c[v,w]$ belong to distinct connected components of $T\setminus C$. In this case we can choose a continuous map $r_T:T\setminus C\to\{u,c[v,w]\}$ such that $r_T[([u,v]\cup[u,w])\setminus C]\subseteq\{u\}$ and $r_T[[v,w]\setminus C]\subset\{c[v,w]\}$.

The definitions of the maps $r_T$ ensure that they agree on the intersections of their domains. Consequently, the union $r=\bigcup_T r_T$ of these maps is a continuous function $r:\mathbb R^2\setminus C\to\mathbb R^2\setminus C$ such that $r\circ r=r$. So, $r$ is a retraction onto the closed subset $F$ which can be written as the union of the triangles of the triangulation that do not intersect $C$, some vertices of the triangles that intersect $C$ and the arcs $A_T$ of difficult triangles (of the first type).

The choice of the triangulation $T$ (as sufficiently fine) implies that $V=\mathbb R^2\setminus F$ is a neighborhood of $C$ with $\bar V\subset U$. Then $r{\restriction}U\setminus C$ is the required retraction of $U\setminus C$ onto $U\setminus V$.

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  • $\begingroup$ this is not much different from what i came up on my own, and here is where i was stuck all along: how to retract $\bar{\sigma}\backslash C$ onto $\partial\sigma\backslash C$? (by the way, i assume you meant $\sigma\in K^{(3)}$, right?) First, the latter set may be empty, even if the former is not (this can be solved by further dividing the cell). But what if the latter set is disconnected, while the former is connected? For example, imagine a triangle for which $C$ is an arc between two points in the triangle, but ventures into a neighboring triange. $\endgroup$ – erz May 17 at 2:53
  • $\begingroup$ @erz I corrected the definition of $K^{(n)}$ so $K^{(2)}$ is now defined correctly as the set of 2-dimensional simplexes. Concerning the retraction, then indeed, there is a problem in my argument. I will think how to correct it. $\endgroup$ – Taras Banakh May 17 at 8:01
  • $\begingroup$ @erz I have totally rewritten the solution but only for dimension 2. I am not sure if this approach will work for higher dimensions. One can try to understand what is going on for wild compact sets: horned Alexander sphere, wild Cantor set, Antoine necklage, etc. $\endgroup$ – Taras Banakh May 17 at 10:59
  • $\begingroup$ Thank you for your answer! Yeah, I also don't see how to apply this method to $n=3$: when you try to purify an edge, you spoil a face. $\endgroup$ – erz May 19 at 1:12

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