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 $R$?

  • $\begingroup$ When $S$ is a submanifold you can show $U\setminus S$ is path-connected using a tubular neighbourhood (and the fact that the complement of the zero section in the normal bundle is connected). I have no idea how to answer the general case. $\endgroup$ Apr 11 '18 at 2:39
  • $\begingroup$ @AnthonyCarapetis I'm interested in the case when $S$ is the singular set of a current, so the behavior could be something much worse than a submanifold. $\endgroup$
    – Ryan Unger
    Apr 11 '18 at 3:07
  • $\begingroup$ Yeah, I figured. Perhaps tag this [geometric-measure-theory]. $\endgroup$ Apr 11 '18 at 3:20
  • $\begingroup$ @AnthonyCarapetis Done. $\endgroup$
    – Ryan Unger
    Apr 11 '18 at 3:23

$U\setminus S$ 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.

  • $\begingroup$ Hey, do you happen to also have a reference for this statement? $\endgroup$
    – Dennis
    Mar 18 '20 at 20:54

Partial answer: If $S$ is assumed compact, the result is true.

The Cech cohomological dimension is lower than (or equal to) the covering dimension, which in turn is lower than (or equal to) the Hausdorff dimension.

By Poincaré-Alexander duality, $$H_1(U,U-S) \simeq \check{H}^{n-1}(S)=0 .$$

From the following fragment of the reduced long exact sequence of $(U,U-S)$: $$H_1(U,U-S) \to \widetilde{H}_0(U-S) \ \to \widetilde{H}_0(U),$$ we get that $\widetilde{H}_0(U-S)$ is zero (since the left and right are $0$), and thus $U-S$ is connected.

  • $\begingroup$ Do you use the fact that $U$ is open here? I'm not very familiar with this stuff. $\endgroup$
    – Ryan Unger
    Apr 11 '18 at 18:02
  • $\begingroup$ @0celo7 Yes, I do, for the sake of $U$ being a manifold. Poincaré-Alexander duality holds for manifolds (depending on the version you want, you may need to require the manifold to be orientable. But this is not needed in this case, since taking $\mathbb{Z}/2\mathbb{Z}$ coefficients is enough for the result. Furthermore, open sets of Euclidean space are trivially orientable, so this issue is not relevant for your question anyway). $\endgroup$
    – Aloizio Macedo
    Apr 12 '18 at 0:54

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