# Geometry of complements to compacts of codimension 2

Let $$K\subset \mathbb{R}^n$$ be a (nonempty) compact of covering dimension $$\le n-2$$. In particular, $$K$$ does not separate $$\mathbb{R}^n$$ (even locally). I will equip $$M=\mathbb{R}^n-K$$ with the distance function $$d$$ associated with the flat Riemannian metric restricted from $$\mathbb{R}^n$$. The metric space $$(M,d)$$ is incomplete, let $$(\bar{M}, \bar{d})$$ denote its Cauchy-completion. Since the identity embedding $$(M,d)\to \mathbb{R}^n$$ is distance-decreasing, it extends to a continuous map $$f: \bar{M}\to \mathbb{R}^n$$.

Question. Is $$f$$ injective? Is it a homeomorphism?

Less formally: If points in $$M$$ are close in the Euclidean metric, does it follow that they can be connected by a short path in $$M$$?

Remark. The answer is positive if I assume that Hausdorff dimension of $$K$$ is less than $$n-1$$. Namely, in this case $$d$$ equals the Euclidean distance.

• What are the examples of the length metric space $X$ and $K\subset X$ - compact that does not locally separate $X$, such that the length metric on $X\backslash K$ is different from the subspace metric? – erz Feb 8 at 4:02
• What if $K$ is a Cantor space? even $n=2$ (where $K$ has to be topologically trivial) sounds not obvious to me. – YCor Feb 8 at 5:04
• @YCor: Yes, even this case is unclear although I may have an argument in the 2-dimensional case (it was studied by people in robotics and I think their robot-navigation algorithm yields a positive answer). – Misha Feb 8 at 12:25
• @erz: I do not have an example although I suspect that a totally disconnected subset of $R^2$ of positive area would give such an example. – Misha Feb 8 at 12:53

I think it is possible to construct a totally disconnected compact $$K\subset \mathbb R^2$$ such that $$f:\overline{M}\to\mathbb R^2$$ would send at least two points to the origin. I will make a certain assumption about a polygonal neighborhood around a polygonal path. I can try to fill this step in or find a reference for that step if you aren't convinced by it.

I will describe a procedure to construct a descending chain of compact sets $$K_n$$ whose intersection will be the required $$K.$$

Start with the square $$[-1,1]\times [-1,1]$$ and remove all points $$(x,y)$$ with $$|y|<|x|.$$ Call the result $$K_0.$$ Removing these points ensures that $$\overline{M}$$ will actually contain new points as limits of the sequence $$(t,0)$$ as $$t\to 0^+$$ or $$t\to 0^-.$$ Let $$U_n$$ be a sequence of open balls such that $$(0,0)\not\in \overline{U_n},$$ and such that $$\{U_n\}$$ generates the topology of $$\mathbb R^2\setminus\{(0,0)\},$$ and such that each ball in the sequence is repeated infinitely often. I want to ensure that:

1. $$U_n\not\subset K_n$$
2. $$U_{n}\cap (\mathbb R^2\setminus K_n)$$ is path-connected
3. $$\inf_{t>0}\{d_{\mathbb R^2\setminus K_n}((-t,0),(t,0))\}>1$$ where $$d_S$$ is the path metric for a subset $$S$$ of the Euclidean plane
4. $$K_n$$ is a filled polygon - a regular closed set whose boundary is a finite union of line segments

Suppose that $$K_{n-1}$$ has been constructed. To ensure condition 1, if $$U_n\subset K_{n-1},$$ remove an open square contained in the interior of $$U_n\cap K_{n-1}.$$

To ensure condition 2, it suffices to remove a component at a time. So, given a set $$K'$$ satisfying conditions 3 and 4, we need to produce $$K''\subset K'$$ satisfying conditions 3 and 4 and such that $$U_n\cap (\mathbb R^2\setminus K'')$$ has fewer connected components than $$U_n\cap (\mathbb R^2\setminus K').$$

Pick a simple polygonal path $$P$$ of length $$1000$$ joining two connected components of $$U_n\cap (\mathbb R^2\setminus K'),$$ and lying in the interior of $$U_n\cap K'$$ except at the endpoints of $$P,$$ which each lie in the interior of an edge of $$K.$$ Because $$P$$ is so long, removing $$P$$ from $$K'$$ would not decrease the distances considered for property 3. But $$P$$ is a closed set so this would ruin property 4. I want to say that for small enough $$\epsilon>0,$$ the set $$K'_\epsilon:=K'\setminus\{(x+a,y+b)\mid (x,y)\in P, |a|,|b|<\epsilon\}$$ satisfies condition 3. (With $$K'_\epsilon$$ instead of $$K_n.$$) The set $$K'_\epsilon$$ is a filled polygon whose vertices vary continuously in $$\epsilon$$ for small enough $$\epsilon.$$ The modifications occur at a bounded distance away from the origin, so the $$\inf$$ considered for property 3 won't care about the origin. It seems geometrically obvious to me that for small $$\epsilon,$$ a path of length less than 2 (say) with endpoints outside $$K'$$ cannot make use of the set removed in the definition of $$K'_\epsilon.$$ This is where I'm making an assumption. But given that assumption, we're done by setting $$K''=K'_\epsilon.$$

That completes the description of $$K_n$$ (apart from the assumption described above).

Any path in $$M=\mathbb R^2\setminus K$$ will lie in some finite stage $$M=\mathbb R^2\setminus K_n,$$ which shows that the points $$\lim_{t\to 0^+}(t,0)$$ and $$\lim_{t\to 0^-}(t,0)$$ in $$\overline{M}$$ lie at distance at least $$1.$$

$$\mathbb R^2\setminus K$$ is locally path connected and dense. The intersection $$K=\bigcap K_n$$ will be totally disconnected. Indeed any neighborhood of a point $$k\in\mathbb R^2$$ contains a loop winding around $$k$$ and lying in $$\mathbb R^2\setminus K.$$ The intersection of $$K$$ with the interior of this curve is clopen in $$K,$$ showing that $$K$$ has a base of clopen sets.

• Thank you! I will look at this example later today/tomorrow. – Misha Feb 8 at 18:08