The Möbius strip (without boundary) $S$ can be realized as a regular surface of $R^3$ (regular surface is meant in the sense of do Carmo's book). Using a 'manifold' language it can be proved that $S$ is an embedded submanifold of $R^3$. Therefore it is for itself a smooth manifold (that we realize as embedded into $R^3$).

My trouble is that we cannot obtain the Möbius strip with boundary as an embedded subanifold of $R^3$. If we want to realize the Möbius band with boundary it seems natural to consider the topological closure of $S$ as a subset of $R^3$ and then proving that this closed subset is an embedded submanifold with boundary into $R^3$. However this argument should fail since we know that no closed subset of $R^3$ cannot be homeomorphic to a non orientable $2$-manifold.

My question is : why does this argument should fail? I haven't checked the detail but i'm looking for an intuitive answer for it.

N.B: an embedded submanifold $M \subset R^3$ is defined as a subset such that for every $p \in M$ there exists a diffeomorphism $\phi:U \rightarrow V$, where $p \in U$, $\phi(p) =0$ and $U,V$ are open subsets of $R^3$ such that $\phi(U \cap M)=V \cap R^2$

• You could probably find this answer interesting. – A.P. May 6 '13 at 16:46

Let $M\subset \Bbb R^3$ be a $2$ dimensional submanifold with boundary (say the Möbius strip) and let $p\in M$ be a boundary point. Let $U$ be an open ball of $\Bbb R^3$ centered on $p$, like in your definition of embedded submanifold. Then $U\cap M$ will be an open neighbourhood of $p$ in $M$, diffeomorphic to an open half-disk $D$ in $\Bbb R^2$ ("diameter" included). Finally, observe that there is no open subset of $\Bbb R^2$ diffeomorphic to $D$, so there is no $V\subseteq \Bbb R^3$ open such that $\phi(U\cap M)=V\cap \Bbb R^2$.
Note that the difference between a manifold with an one without boundary lies exactly in the fact that every point of the former has a neighbourhood diffeomorphic to $\Bbb R^n$, while this fails for some points of the latter, which admit open neighbourhoods diffeomorphic to an open half-ball in $\Bbb R^n$.
• Thanks for your answer. I'm not sure about what you want to say... Have you proved that there are no embedded submanifolds in $R^n$? – user55449 May 6 '13 at 19:09
• No, I showed you why you cannot embed any manifold with boundary (orientable or not) in $\Bbb R^3$, at least with your definition of embedded submanifold. – A.P. May 6 '13 at 19:12
• Yes, but that's the scope of the result you cite, i.e. that every compact 2-dimensional submanifold of $\Bbb R^3$ is orientable. That is exactly the point made in the answer I linked in my comment to your question. Sorry if I wasn't clear about this. – A.P. May 6 '13 at 19:18