Suppose $d\in \{3,4,\dotsc\}$ and $A\subseteq \mathbb{R}^d$ is non-empty, open and connected with its complement $A^c$ connected too and $\text{int}(A^c)\neq \emptyset$. Its boundary $S:=\partial A$ is then [connected too](https://math.stackexchange.com/questions/170337/connectedness-of-the-boundary?noredirect=1&lq=1). Let $A^{(2)}$ be the set of 2-dimensional affine planes within the affine space $\mathbb{R}^d$. For any $x\in S$, define
\begin{gather\*}
S\_x:=\bigcup\_{n=1}^\infty S\_x^{(n)} \\
S\_x^{(1)}:=\{y\in S\mid\exists H\in A^{(2)}\mathrel:\text{ $x$ and $y$ are in the same connected component of }H\cap S\},\\
S\_x^{(n+1)}:=\{y\in S|\,\exists z\in S\_x^{(n)}:\,y\in S\_z^{(1)}\}.
\end{gather\*}
Is $S\_x=S$ for all $x \in S$? If not, is $\{S\_x\mid x\in S\}$ a countable partition of $S$?



Can one explore a surface along ‘piecewise planar’ curves?
Consider the set
$S=\{\,(t,t^2,t^3)\in\mathbb R^3\mid\,t\in\mathbb R\}$.
Clearly $S$ is connected, and so is its complement $A=\mathbb R^3\setminus S$.


Note that each plane has at most 3 points of intersection with $S$.
It follows that $S\_x=\{x\}$ for any $x\in S$.


The only problem is the set $A^c=S=\partial A$ has empty interior, but it is easy to fix by removing from $A$ a closed ball centered at a point in $S$.


