TITLE: Can a set of Hausdorff codimension 2 disconnect a connected open set?
QUESTION [15 upvotes]: Consider a connected open set $U\subset \Bbb R^n$ (or a Riemannian manifold if you're ambitious), and $S\subset U$ closed and with Hausdorff dimension $\le n-2$. Is $U\setminus S$ connected? If not, does $\dim S\le n-3$ work? What is the optimal dimension?
It seems surprisingly hard even in the case of $U=\Bbb R^2$. There exist uncountable zero-dimensional subsets of $\Bbb R^2$, so one cannot use this classical result. I think that if one proves it in the $n=2$ case one can do induction to higher $n$ via some sort of slicing argument (maybe using Fubini). 
Even more generally, consider $R\subset \Bbb R^n$ connected and $\dim R=k$. Can $S\subset R$, $\dim S=k-2$ (or maybe $k-3$) disconnect $R$?

REPLY [6 votes]: $U\setminus S$ is path connected as soon as $\mathrm{dim} \,S<n-1$.
Let us prove the local version, that is for $U$ equal to a ball. Suppose you want to connect $x,y\in U\setminus S$. Since $S$ is closed there are two open balls $B_x$ and $B_y$ around $x$ and $y$ respectively that are entirely contained in $U\setminus S$, and in particular are path connected. Let $\ell$ be the line through $x$ and $y$. Let $\pi$ be the projection of $S$ into a $(n-1)$-dimensional subspace $V$ orthogonal to $\ell$. Since $\pi$ is $1$-Lipschitz it does not increase the Hausdorff dimension, therefore $\pi(S)$ has dense complement inside $V$ (otherwise it would have at least dimension $n-1$). This means that you can find $z$ such that the line $\pi^{-1}(\{z\})$ intersects both $U_x$ and $U_y$ but does not intersect $S$, and you are done.
For $U$ arbitrary open set, fix any $x\in U\setminus S$ and let $U_x$ be the set of points $y$ in $U$ such that you can path-connect  $x$  to points arbitrarily close to $y$ (with a path contained in $U\setminus S$). Note that $y$ could also be a point of $S$. Then, by the local path-connectedness on balls proved above, $U_x$ is both open and closed in $U$ (check it making two cases, depending on whether $y\in S$ or not) and therefore is the whole $U$. This allows to conclude.
The same argument works inside a manifold, because locally it is like $\mathbb{R}^n$.
As for the last question, there should be at least an assumption on the dimension of $R$ around any point, or something like that, otherwise just take as $R$ two disjoint balls joined by a segment, and $S$ a point inside the segment.