11
$\begingroup$

Let $K$ be a compact subset of $\mathbb R^n$ with $n\ge 2$ (say if you like $n=2$, which is possibly sufficiently representative).

Q: Does there exist a closed simple curve $u:\mathbb S^1\to\mathbb R^n $ such that $K\cup u(\mathbb S^1 )$ is connected?

The set $K$ may have uncountably many connected components, and $u$ has to meet them all. Yet this does not seem a serious obstruction. For instance, the cartesian square of the Cantor set can be connected by some simple self-similar curve (necessarily of infinite length; in fact I think of dimension at least $4/3$), e.g. just connecting suitably the four main square clusters between them by segments, and then iterating.

$\endgroup$
7
  • 1
    $\begingroup$ In the case of $n=2$ one possible approach is to show that there exists a totally disconnected compact subset $C$ of $K$ intersecting every component of $K$. Such $C$ is necessarily contained in a Jordan curve since there is a self-homeomorphism of $R^2$ sending $C$ to a subset of the x-axis. $\endgroup$ Jan 10, 2021 at 12:06
  • 1
    $\begingroup$ Do you know the answer when $K$ is totally disconnected? Then for $n=2$ the answer is yes (some self homeomorphism of the plane maps $K$ into a fixed line). However there are wild Cantor subsets in higher dimension. $\endgroup$
    – YCor
    Jan 10, 2021 at 14:45
  • $\begingroup$ I don't know the answer in the case of a totally disconnected $K$, but I would believe it is affirmative, by a construction that mimics the curve described above in the OP for the squared Cantor set. We may define a curve iteratively this way. First we join the finitely many connected components (if more than one) of the uniform neighborhood $N_\epsilon(K):=\{x: \rm{dist}(x,K)\le \epsilon\}$, where $\epsilon=1$, parametrizing the arcs on disjoint closed intervals of $[0,1]$. $\endgroup$ Jan 10, 2021 at 15:25
  • $\begingroup$ Then we proceed filling the gaps and connecting the components of $N_\epsilon(K)$ for $\epsilon=1/2,1/4,1/8\dots.$ Since $K$ is totally disconnect, the diameters of the components tend to $0$, which should prevent the lost of continuity and injectivity as would happen in your counterexample. $\endgroup$ Jan 10, 2021 at 15:25
  • 2
    $\begingroup$ Maybe the case of $K$ totally disconnected would make a good follow-up question. $\endgroup$
    – YCor
    Jan 10, 2021 at 22:49

2 Answers 2

11
$\begingroup$

Not always.

Let $K$ be a subset of an ambient space $V$ ($V=\mathbf{R}^2$ is fine, but doesn't matter) that is the closure of a discrete subset $D$, such that $K-D$ is homeomorphic to a segment. This exists in $\mathbf{R}^n$ for $n\ge 2$.

Then every closed subset of $V$ that meets every component of $K$ has to contain all $D$, and hence contains its closure, and hence contains $S$. But if $j:[0,1]\to C$ is an injection of a segment in a circle, the interior of $j([0,1])$ in $C$ is equal to exactly $j(\mathopen]0,1\mathclose[)$; in particular, $j([0,1])$ can't have empty interior in $C$.

But if $C$ were a circle within $V$ meeting every connected component of $K$, we would have $D\subset C$, hence $K\subset C$. Since $j(S)=S$ has empty interior in $K$ and $K\subset C$, it has empty interior in $C$. This is a contradiction with the above fact.

[Edit: I initially described $K$ as subset of the sine curve, but this doesn't matter and complicates the description.]


Minor variant: let $M$ be any compact subset with empty interior, which is not homeomorphic to any subset of a circle (e.g., the whole sine example in the plane, a sphere in a higher space). Let $D$ a discrete subset of $\mathbf{R}^n$ whose set of accumulation points is exactly $M$ (this exists). Then no subset of $\mathbf{R}^n$ homeomorphic to $C$ meets every component of the compact subset $K=D\cup M$.

$\endgroup$
1
2
$\begingroup$

Here is a proof that the answer is positive for a totally disconnected $K$. I will prove it in the case $n\ge 3$ since for $n=2$ there exists a homeomorphism $h: R^2\to R^2$ sending $K$ to a subset of the unit circle. I will also assume that the set $K$ is infinite (since the claim is clear otherwise).

First of all, since $K$ is compact and totally disconnected, there exists a compact subset $C\subset S^1$ and a homeomorphism $f: C\to K$. I will be using the following:

Lemma 1. Let $A\subset R^n, n\ge 3$, be a compact subset of covering dimension $\le 1$ (e.g. a subset homeomorphic to a subset of ${\mathbb R}$). Then $A$ does not locally separate $R^n$.

The proof is a direct application of the Alexander duality and I will omit it.

I will extend $f=h_0$ to topological embedding of $S^1$ as follows. I enumerate the components $I_j$ of $S^1-C$, $j\in {\mathbb N}$, set $C_0=C$ and $$ C_j:= C\cup I_1\cup ... \cup I_j. $$ I first extend $h_0$ inductively to topological embeddings $h_j: C_j\to R^n$ such that $h_{i}$ is the restriction of $h_{i+1}$.

Assume that $h_j$ is defined. Let $B_{j+1}$ denote the unique closed round ball in $R^n$ of the diameter equal to the diameter of $h(\partial I_{j+1})$ and containing the 2-point set $h(\partial I_{j+1})=\{x^\pm_{j+1}\}$.

Lemma 2. There exists a topological arc $\alpha_{j+1}\subset B_{j+1}$ connecting the points $x^\pm_{j+1}$ and otherwise disjoint from $h_j(C_j)$.

Proof. Take a biinfinite sequence $(y_i)_{i\in {\mathbb Z}}$ in $B_{j+1} \setminus h_j(C_j)$ such that $$ \lim_{i\to\pm \infty} y_i= x^\pm_{j+1}. $$ Then (since $h_j(C_j)$ is (at most) 1-dimensional) use inductively Lemma 1 to connect successive points in the sequence $(y_i)_{i\in {\mathbb Z}}$ by smooth simple arcs in $B_{j+1} \setminus h_j(C_j)$ whose diameters converge to zero as $i\to\pm \infty$. qed

Next, extend $h_j$ to $I_{j+1}$ by a homeomorphism parameterizing the arc $\alpha_{j+1}$. The result is $h_{j+1}$. By the construction, it is a topological embedding.

Together, the maps $h_j, j\in {\mathbb N}$, define an injective map $h: S^1\to R^n$ such that $h|_{C_j}=h_j$. Continuity of $h$ follows from the fact that diameters of the sets $h_j(I_j)$ converge to $0$ as $j\to\infty$. Hence, $h$ is the required topological embedding.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.