Show that for a properly embedded submanifold the manifold and topogoical boundary coincide

Question

Let $d\in\mathbb N$ and $M\subseteq\mathbb R^d$ be a $d$-dimensional properly embedded $C^1$-submanifold of $\mathbb R^d$. Let $\partial M$ and $M^\circ$ denote the manifold boundary and interior and $\operatorname{Bd}M$ and $\operatorname{Int}M$ denote the topological boundary and interior of $M$, respectively.

How can we show that $\partial M=\operatorname{Bd}M$ and $M^\circ=\operatorname{Int}M$?

Note that $M$ being properly empedded into $\mathbb R^d$ is equivalent to $M$ being $\mathbb R^d$-closed. So, $\operatorname{Bd}M=M\setminus\operatorname{Int}M$.

Let $x\in\partial M$. In order to prove $x\in\operatorname{Bd}M$, all we need to show is that every neighborhood of $x$ has a nonempty intersection with $M^c$.

There is a $C^1$-diffeomorphism from an $M$-open neighborhood $\Omega$ of $x$ onto an open subset $U$ of $\mathbb H^d:=\mathbb R^{d-1}\times[0,\infty)$ and $$u:=\phi(x)\in\partial\mathbb H^d=\mathbb R^{d-1}\times\{0\}\tag1.$$ Since $U$ is $\mathbb H^d$-open, $$U=V\cap\mathbb H^d\tag2$$ for some open subset $V$ of $\mathbb R^d$ and since $V$ is $\mathbb R^d$-open, $$B_\varepsilon(u)\subseteq V\tag3$$ for some $\varepsilon>0$. Now, clearly, $$B_\varepsilon(u)\cap\left(\mathbb R^d\setminus\mathbb H^d\right)\ne\emptyset\tag4.$$

But how can we conclude?

Note that $$\phi=\left.\tilde\phi\right|_\Omega\tag5$$ for some $\tilde\phi\in C^1(O,\mathbb R^d)$ for some $\mathbb R^d$-open neighborhood $O$ of $\Omega$.

Answer 1

As Jack Lee noted in his

Lee, John M., Introduction to smooth manifolds, Graduate Texts in Mathematics 218. New York, NY: Springer (ISBN 978-1-4419-9981-8/hbk; 978-1-4419-9982-5/ebook). xvi, 708 p. (2013). ZBL1258.53002.

on page 26:

Be careful to observe the distinction between these new definitions of the terms boundary and interior and their usage to refer to the boundary and interior of a subset of a topological space. A manifold with boundary may have nonempty boundary in this new sense, irrespective of whether it has a boundary as a subset of some other topological space. If we need to emphasize the difference between the two notions of boundary, we will use the terms topological boundary and manifold boundary as appropriate. For example, the closed unit ball $\overline{\Bbb B}^n$ is a manifold with boundary, whose manifold boundary is $\Bbb S^{n-1}$. Its topological boundary as a subset of $\Bbb R^n$ happens to be the sphere as well. However, if we think of $\overline{\Bbb B}^n$ as a topological space in its own right, then as a subset of itself, it has empty topological boundary. And if we think of it as a subset of $\Bbb R^{n+1}$ (considering $\Bbb R^n$ as a subset of $\Bbb R^{n+1}$ in the obvious way), its topological boundary is all of $\overline{\Bbb B}^n$. Note that $\Bbb H^n$ is itself a manifold with boundary, and its manifold boundary is the same as its topological boundary as a subset of $\Bbb R^n$. Every interval in $\Bbb R$ is a 1-manifold with boundary, whose manifold boundary consists of its endpoints (if any).

The nomenclature for manifolds with boundary is traditional and well established, but it must be used with care. Despite their name, manifolds with boundary are not in general manifolds, because boundary points do not have locally Euclidean neighborhoods.