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 $
Best Answer
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 $\Bbb R^n$.