A compact set in $\Bbb R^n$ with smooth boundary is a manifold?

Question

Can a compact set $\Omega \subset \Bbb R^n$ with smooth boundary be considered as a smooth manifold with boundary?(Smooth boundary probably means $\partial\Omega$ is a $n-1$-smooth manifold?)

I think it's right. How can I prove it?

Answer 1

You'll need some extra assumptions to rule out some silly counterexamples. For example, let $D$ be a closed $n$-ball in $\mathbb R^n$, and let $S$ be an $(n-1)$-sphere that is disjoint from $D$. The $\Omega = D \cup S$ is compact, and its boundary is the union of two spheres, which is a smooth $(n-1)$-manifold. But $\Omega$ is not a manifold with boundary (unless you want to allow your manifolds to have different dimensions at different points, which I don't advise).

Another silly example is obtained by just taking $\Omega$ to be the sphere $S$ by itself. Then its boundary is the whole sphere $S$, which is a smooth $(n-1)$-manifold. In this case, in fact, $\Omega$ is a smooth manifold with boundary, whose boundary happens to be empty; but I suspect this is not what you had in mind. You're probably hoping for an $n$-manifold whose boundary in the manifold sense is the same as the boundary in the topological sense.

In order to get that kind of result, as I said, you're going to have to add a hypothesis. You might hope to get by with a weak hypothesis such as assuming $\Omega$ has nonempty interior; but that doesn't rule out my first example above. A common additional hypothesis in this setting is to assume that $\Omega$ is the closure of its interior.

Also, when you say $\partial \Omega$ is a smooth $(n-1)$-manifold, I assume you mean that it's a smooth, embedded $(n-1)$-dimensional submanifold of $\mathbb R^n$.

So with these assumptions, the result you want is true. Here's a theorem.

Theorem. Suppose $\Omega\subset \mathbb R^n$ is a compact subset that is the closure of its interior, and the topological boundary $\partial \Omega$ is a smooth embedded $(n-1)$-dimensional submanifold of $\mathbb R^n$. Then $\Omega$ is a smooth $n$-dimensional submanifold with boundary in $\mathbb R^n$, and its manifold boundary is the same as its topological boundary.

Sketch of Proof. If $x$ is an interior point of $\Omega$, then there's an open ball $B_\varepsilon(x)\subset \Omega$, and the identity map of $B_\varepsilon(x)$ is a smooth interior chart for $\Omega$ whose domain contains $x$.

If $x\in \partial \Omega$, then the fact that $\partial \Omega$ is a smooth $(n-1)$-dimensional submanifold means that there is an open set $U\subset\mathbb R^n$ containing $x$ and a diffeomorphism $\phi\colon U \to \widehat U$ (where $\widehat U$ is an open subset of $\mathbb R^n$) such that $\phi(U\cap \partial \Omega)$ is the slice $\widehat U \cap (\mathbb R^{n-1}\times \{0\})$. By shrinking $U$, we may also assume that $\widehat U$ is a ball centered at $\phi(x)$.

Now $\widehat U \smallsetminus \phi(U\cap \partial\Omega)$ has two connected components (both open half-balls), and therefore $U\smallsetminus \partial \Omega$ has two components diffeomorphic to half-balls; call them $H_1$ and $H_2$. The assumption that $\Omega$ is the closure of its interior guarantees that $\Omega$ has to contain a point in at least one of those open half-balls, say $H_1$. If $\Omega$ doesn't contain all of $H_1$, then there will be a point in $\partial \Omega\cap H_1$ (exercise), contradicting the fact that $H_1$ lies in the complement of $U\cap \partial\Omega$. Thus $\Omega$ contains all of $H_1$. If it contains a point in the other half-ball $H_2$, the same argument shows that it must contain all of $H_2$, which would imply that $U\subset \Omega$. But that would mean $x$ is an interior point of $\Omega$, contradicting our assumption that it's a boundary point. Thus $\phi(U\cap \Omega)$ is exactly the intersection of $\widehat U$ with the closed upper half-space $\mathbb R^{n-1}\times [0,\infty)$, which implies that $\phi$ restricts to a boundary chart for $\Omega$.