Closed ball not a manifold.

Solution 1:

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.

Solution 2:

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 $R^n$.