Understanding the definition of a closed manifold

Question

Let $D\subset R^n$ be a bounded domain with smooth boundary. Is $\partial D$ a closed manifold?

Answer 1

Actually we run into some technical difficulties here. There exists two different concepts of a boundary, namely

1. The boundary $\partial U$ of a subset $U\subset X$ of a topological space $X$, which is defined as $\partial U = \overline U \cap \overline{(X\setminus U)}$.
2. A manifold $M^n$ with boundary $\partial M$, which is defined to be locally homeomorphic to the Euclidean half-plane $\mathbb R^n_{\geq 0}$. Its boundary $\partial M$ is the set of all points which are mapped by a local homeomophism to the boundary of the Euclidean half-plane.

To make the difference clear: The topological boundary of an open disk is a circle whereas the boundary of the open disk, seen as a manifold, is empty. Additionally, for an open set $U$, the top. boundary of the top. boundary of $U$, i.e. $\partial\partial U = \overline{\partial U} \cap \overline{(X\setminus \partial U)}=\partial U$, because $\partial U$ is a closed set. For manifolds, $\partial^2 = 0$ (corner stone of singular homology).

So the difference is that 1. relies on an embedding space, whereas 2. is an intrinsic definition.

Back to the question: In my understanding, a domain $D$ is an open connected subset. That means that as a manifold, it does not have a boundary. Its topological boundary $\partial D$ is a submanifold of $\mathbb R^n$ and hence it can also be a smooth one. One could avoid mixing the two concepts, by requiring the equivalent assumption that $\overline D$ is a smooth manifold with boundary. Last, a closed manifold is a compact manifold with empty boundary.

So let's show that $\partial\partial \overline D=\emptyset$. This is a basic statement and for sure there is a proof for this somewhere around here. However, it is really easy.

First, $\overline D$ is compact (closed and bounded) and $\partial \overline D$ is hence also compact. By definition, $\partial \overline D$ is locally homeomorphic to $\partial\mathbb R^n_{\geq 0}$. But the boundary of the Euclidean half-plane $\partial\mathbb R^n_{\geq 0}$ is just a $(n-1)$-dimensional subspace of $\mathbb R^n$ and thus it is homeomorphic to $\mathbb R^{n-1}$. Therefore $\partial\partial \overline D$ is locally homeomorphic to $\partial\mathbb R^{n-1}=\emptyset$ and hence $\partial\partial \overline D=\emptyset$.