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

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 $$\Omega$$.

• +1. Good question. I thought before your question that a smooth manifold with boundary is a smooth Hausdorff, locally Euclidean topological space with boundary but it is not true!! Aug 9 '20 at 3:56
• Do you know what is the problem with the above definition? Aug 9 '20 at 3:58

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).
• In a manifold with boundary, each point is locally diffeomorphic to a half space $H$. Interior points are locally diffeomorphic to $H^\circ$ and boundary points are mapped to $\partial H$. However, I'm not sure how this is related to my question. Maybe you got it wrong? I want to prove that the topological boundary/interior and manifold boundary/interior coincide when $M$, as in the question, is properly embedded. Aug 9 '20 at 5:06