TITLE: Why is $\partial\partial M=\varnothing$?
QUESTION [6 upvotes]: Why is the border of the border of an oriented differentiable $n$-dimensional Manifold $M$ empty, that is $$\partial\partial M = \emptyset?$$

REPLY [4 votes]: This simply says that the boundary of the manifold $M$ is a manifold without boundary. Indeed, by definition, it is locally homeomorphic to $\mathbb{R}^{n-1}$ by restriction of the atlas of $M$ to $\partial M$. Note this has nothing to do with the differential structure.
Note: I think it is worth noting that the boundary $\partial^\mathrm{man}$ of a manifold is a notion which is different from that of boundary in a general topological space $\partial^\mathrm{top}$. For instance, $\partial^\mathrm{man}(0,1)=\emptyset$ while $\partial^\mathrm{top}(0,1)=\{0,1\}$. To avoid confusion, some people call $\partial^\mathrm{top}$ the fronteer.
It is not true that $\partial^\mathrm{top}\partial^\mathrm{top} S=\emptyset$ in general. Precisely, in the case of a closed half-space, $\partial^\mathrm{top}\partial^\mathrm{top} S=\partial ^\mathrm{top}S\neq \emptyset$. For instance, 
$$
\partial^\mathrm{top} \{(x,y)\in\mathbb{R}^2\;;\;y\geq 0\}=\partial^\mathrm{man} \{(x,y)\in\mathbb{R}^2\;;\;y\geq 0\}=\{(x,y)\in\mathbb{R}^2\;;\;y=0\}=B
$$
and
$$
\partial^\mathrm{top}B=B\qquad\text{while}\qquad \partial^\mathrm{man}B=\emptyset
$$
as $B$ is homeomorphic to $\mathbb{R}$.