4
$\begingroup$

Let $C$ be a compact and totally-disconnected subspace of the $n$-sphere $\mathbb{S}^n$, where $n\geq 2$.

Question: Must the Freudenthal compactification of $\mathbb{S}^n \setminus C$ be homeomorphic to $\mathbb{S}^n$?

The case where $C$ is homeomorphic to the Cantor set is interesting.

For $n=2$ the answer is yes, see e.g. Corollary 1.5. in Georgakopoulos - On planar Cayley graphs and Kleinian groups. For $n=3$ I suspect it is still easy using a triangulation of $\mathbb{S}^n \setminus C$.

Improve this question
$\endgroup$
6
  • $\begingroup$ The terminology is so evocative! $\endgroup$
    – LSpice
    Commented Jul 31, 2022 at 18:58
  • 2
    $\begingroup$ The space of ends will be homeomorphic to $C$, and the end compactification will be homeomorphic to $S^n$. Since it’s disconnected, we can find a finite open cover by at least two disjoint connected open sets. Then one checks that these give neighborhoods of the ends (puncturing by the Cantor set can’t disconnect them). Now repeat with the Cantor sets inside each of these open sets. $\endgroup$
    – Ian Agol
    Commented Aug 1, 2022 at 3:36
  • $\begingroup$ @IanAgol: I'm not convinced. To find a neighborhood of an end $\omega$, we need to display a compact subset $K$ of $\mathbb{S}^n \setminus C$ that separates $\omega$ from some other end, don't we? How do you find such $K$? It is important that $K$ be disjoint from $C$. $\endgroup$
    – Agelos
    Commented Aug 1, 2022 at 20:51
  • 1
    $\begingroup$ @IanAgol: ... I got convinced, ignore the previous comment. $\endgroup$
    – Agelos
    Commented Aug 2, 2022 at 21:21
  • 1
    $\begingroup$ The right question is whether the embedding $S^n-C\to C$ is the Freudenthal compactification (not just whether there is an arbitrary homeomorphism) and the answer is yes too. In addition it holds in an arbitrary compact connected topological manifold of dimension $\ge 2$, boundary allowed (and probably anything connected and locally homeomorphic to simplicial complexes with no isolated vertex/edge). $\endgroup$
    – YCor
    Commented Aug 15, 2022 at 15:59

1 Answer 1

Reset to default
1
$\begingroup$

The answer is yes. The key ingredient is that given any $x,y\in C$, we can find a compact connected $K\subset \mathbb{S}^n$ separating $x$ from $y$. To see this, notice that as $C$ is totally-disconnected, we can decompose $C$ into disjoint open sets $C_x \ni x, C_y \ni y$. Let $\epsilon:= d(C_x, C_y)$, and cover $C$ by (finitely many) balls of radius $\epsilon/3$. Then we can take the complement of this cover to be the desired set $K$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .