n-dimensional submanifold of $\mathbb{R}^n$ Can an n-dimensional submanifold of $\mathbb{R}^n$ be a closed set in/of $\mathbb{R}^n$? And how to show the opposite if that's not the case ?
 A: The unit ball $B^n:=\{x \in \mathbb{R}^n \mid \Vert x \Vert \leq 1\}$ is a $n$-dimensional submanifold of $\mathbb{R}^n$, although with boundary.
A: I believe that if we exclude $\mathbf{R}^n$ from consideration – viewed as an $n-$dimensional submanifold of itself – we should find that the only closed $n-$dimensional submanifolds of $\mathbf{R}^n$ are manifolds with (nonempty) boundary. That is, they are locally diffeomorphic to $\mathbf{H}^n$, the upper half space, such that there exists some $p\in \mathbf{R}^n$ with $\psi(p)\in \partial \mathbf{H}^n$, for all charts $(U,\psi)$ containing $p$.
For example, the closed solid ball, $\overline{\mathbf{B}^n}\subset \mathbf{R}^n$ is an $n-$dimensional manifold with boundary. It is worth noting, however, that a manifold with boundary is not a $C^\infty$ manifold in the usual sense. 
As for how to show this, I would recommend contemplating what it means for an $n-$manifold $M$ to be closed in $\mathbf{R}^n$. This implies that its complement is open. If we assume that $M^c$ is an open subset, then are neighborhoods of the topological boundary points of $M$ locally diffeomorphic to $\mathbf{R}^n$? I think you will find that they are not.
