Connecting a compact subset by a simple curve

Question

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.

Answer 1

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.]

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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 $K=D\cup M$.

Answer 2

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.