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.



Connecting a compact subset by a simple curve
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 $K=D\cup M$.


