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An $n$-dimensional topological manifold is a second countable Hausdorff space such that every point has a neighborhood which is homeomorphic to $\mathbb{R}^n$. Let $n \ge 2$. Let $M$ denote an $n$-dimensional topological manifold, and let $\Omega \subset M$ denote a pre-compact open subset which is homeomorphic to $\mathbb{R}^n$. Is the topological boundary of $\Omega$ in $M$ connected? Is this fact stated in any published reference?

This answer has a nice example and picture showing that the boundary need not be path connected.

Geoffrey Sangston
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    I think, though I haven't checked the details, that in $\mathbb R^n$ you can use Alexander duality to show that the zeroth cech cohomology of the boundary is rank 1, which means it has one quasicomponent, which in the compact hausdorff case is the same as a connected component. – Cheerful Parsnip Sep 30 '23 at 18:35
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    Just follow my proof here: https://math.stackexchange.com/questions/3056322/do-contractible-homology-manifolds-have-one-end/3057042#3057042. – Moishe Kohan Sep 30 '23 at 18:40
  • Right.......... – Moishe Kohan Sep 30 '23 at 19:40

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See comments.

The boundary components $\{K_c\}_{c = 1}^C$ of the closure of a pre-compact open set $\Omega$ in a topological manifold $M$ are separated by disjoint connected open sets in $M$, and for each $c = 1, \ldots, C$ a sequence $\{U_{c, k}\}_{k = 1}^\infty$ of such open sets can be chosen with limits $\bigcap_{k = 1}^\infty U_{c, k}$ equal to each given boundary component. The union $\bigcup_{c = 1}^C U_{c,k}$ of the sets at each index covers $\partial \Omega$, and the complements of these unions in $\Omega$ forms a particular sequence of compact sets $\{\Omega \backslash \bigcup_{c = 1}^C U_{c,k}\}_{k = 1}^\infty$ exhausting $\Omega$. Each $\Omega \backslash \bigcup_{c = 1}^C U_{c,k}$ is compact because $\bigcup_{c = 1}^C U_{c,k}$ is an open set covering $\partial \Omega$, so $\Omega \backslash \bigcup_{c = 1}^C U_{c,k} = \overline{\Omega} \backslash \bigcup_{c = 1}^C U_{c,k}$ is a closed subset of the compact set $\overline{\Omega}$

It follows that the rank plus one of the limit defined in Moishe Kohan's answer is greater than or equal to the number of connected components of $\partial \Omega$. Moishe Kohan shows this rank is $0$ for the set $\Omega$ in question.

Geoffrey Sangston
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