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
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I also wanted to rigorously check out this detail but I am a little bit disappointed by the answers. We want to proove that there exists no homeomorphism between $U\cap \overline{B}_n$ ($U$ an open neighborhood of a boundary point of the closed n-dimensional ball $B_n$) and an open subset of $\mathbb{R}^n$. What I consider a real "proof" can be found in this post – Noix07 Nov 01 '16 at 10:36
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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.
Henry T. Horton
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Thank you for the nice response, the one dimensional example was very illuminating. – Micheal Mar 24 '13 at 17:29
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Just because the neighborhood is homeomorphic to something that is not open in the ambient space doesn't a priori mean it can't also be homeomorphic to something else that is. – hmakholm left over Monica Nov 17 '17 at 19:22
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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 $R^n$.
Sepideh Bakhoda
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