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Let $M = \mathbb{R}^2$ (or more generally, let $M$ be a topological manifold) and let $\Omega$ be an open set in $M$.

I'm considering the following regularity conditions for the boundary of $\Omega$:

  1. $\overline{\Omega}$ is a topological manifold with boundary (in other words, around each boundary point, $\Omega$ looks like a Euclidean half-space --up to homeomorphism).
  2. $\Omega$ and $M \setminus \overline{\Omega}$ are locally connected near each boundary point of $\Omega$.

My question is: does 2. imply 1.?

NB: Maybe my definitions are not precise enough, let me rephrase more formally. Let $\delta \Omega = \overline{\Omega} \setminus \Omega$ denote the boundary of $\Omega$.

  1. For any $x \in \delta \Omega$, there exists a homeomorphism $\varphi : U \to V$ where $U$ is an open set in $M$ containing $x$ and $V$ is an open set in $\mathbb{R}^2$ containing the origin, such that $\varphi(U \cap \Omega) = \{(x,y) \in V, ~ y \geqslant 0\}$.
  2. For any $x \in \delta \Omega$ and for any open set $U \subset M$ containing $x$, there exists an open set $V \subset U$ containing $x$ such that $V \cap \Omega$ and $V \cap (M \setminus \overline{\Omega})$ are connected.

Also maybe I should ask that $\Omega$ is "regular" in the sense that it is equal to the interior of its closure.

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up vote 1 down vote accepted

2 does not imply 1 in a general topological manifold. Here's a counterexample for the case $M = \mathbb{R}^3$.

For $n \in \mathbb{N}$, let $A_n$ be the circle in $\mathbb{R}^2$ centered at $\left(\frac{1}{2^n}, 0\right)$ and of radius $\frac{1}{2^{n+1}}$. Considering $\mathbb{R}^2$ as a subset of $\mathbb{R}^3$ in the standard way, let $B_n$ be the set of points (in $\mathbb{R}^3$) of distance no more than $\frac{1}{10^n}$ from $A_n$. Essentially, each $B_n$ is an $\epsilon$-neighborhood of $A_n$ in $\mathbb{R}^3$.

Set $\Omega = \bigcup B_n$. $\Omega$ is essentially an infinite chain of circles (not actually linked circles, but rather circles that are welded together) that get smaller and smaller as you approach the origin. Note also that $\Omega$ is the interior of its closure.

Furthermore, $\Omega$ and $M - \overline{\Omega}$ are connected near the boundary; every boundary point besides the origin looks like a manifold boundary point, and near the origin both the connected circles and their complement are connected. But $\Omega$ is not simply connected in any neighborhood of the origin, which means $\overline{\Omega}$ is not a manifold with boundary.

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I see, nice counter-example. Seems clear that it extends to any dimension. Except dimension 2! Do you think 2. implies 1. is true in dimension 2? (NB: it reminds me of this fact: semilocally simply connected implies locally connected in dimension 2 (for locally path connected sets), but not higher dimensions) – Seub Jan 30 '14 at 1:06
Yes, I would suspect it is true in dimension 2, essentially because of Alexander duality. If you insist the complement of $\Omega$ is connected, it probably means that $\Omega$ is simply connected, and there's just not much wiggle room once you know you've got a connected, simply connected subset of $\mathbb{R}^2$. – MartianInvader Jan 30 '14 at 1:23

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