Closed ball not a manifold. My book on differential geometry claims that a closed ball in $\Bbb R^n$ can never be a differentiable manifold because of the boundary points. The book doesn't really give an explanation for why this is true. Could someone provide more details please? thanks
 A: The points on the boundary only have neighborhoods homeomorphic to an open ball in
$$\Bbb R^n_+ = \{(x_1, \dots, x_n) \in \Bbb R^n : x_1 \geq 0\}$$
centered on the boundary. Such a set is not open in all of $\Bbb R^n$.
For example, consider the closed unit ball $[-1, 1]$ in $\Bbb R$. Then any open neighborhood of $1$ has a connected component of the form $(1 - \varepsilon, 1]$, which is not open in $\Bbb R$. But the definition of an $n$-manifold requires that each point has an open neighborhood homeomorphic to an open set in $\Bbb R^n$; therefore $[-1,1]$ cannot be a manifold in the usual sense.
The closed ball in $\Bbb R^n$ is an example of a manifold with boundary.
A: a manifold is differentiable, if there exist an atlas for it. the atlas contains compatible charts (We call the pair $(U,f : U →R^n)$ a chart, $U$ a coordinate
neighborhood or a coordinate open set, and $f$ a coordinate map or a coordinate
system on U. f is a homeomorphism onto an open
subset of $R^n$) 
for boundary points we cannot define a chart (like above definition) because neighborhoods of these points are not homeomorphic to an open
subset of $R^n$.
