TITLE: Manifold embedded in $R^{n+1}$ with a submanifold that doesn't embed in $R^n$
QUESTION [32 upvotes]: Suppose you have a closed $m$-dimensional manifold $M$, which embeds in $\mathbb{R}^{n+1}$ for some $n$. Can it have a closed submanifold $N$ (of dimension strictly smaller than $m$) which does not embed in $\mathbb{R}^n$? (By which I mean there is no embedding, not just that the restriction/projection of the first one doesn't work.)
I'm not sure how important the dimension of $N$ is; it seems like codimension 1 would be the easiest place to find an example, but I'm also interested in higher codimension examples. In fact, is there an upper limit to the codimension in which such examples can exist?
I would particularly like a codimension 1 example where both $M$ and $N$ are orientable. Or an example (orientable or not) for $n=2$ or $3$, but maybe there are obstructions in low dimensions...
I am primarily thinking about smooth manifolds, but examples in the topological category would also be interesting.

REPLY [17 votes]: Here is a simple example, using $2$-manifolds and $3$-manifolds. 
A Klein bottle does not embed in $\mathbb R^3$.  But it is a submanifold of a 3-manifold that embeds in $\mathbb R^4$.  For the $3$-manifold take the unit normal bundle of the standard embedding of the Klein bottle in $\mathbb R^4$.  This bundle has a section, so it contains the Klein bottle.  This bundle's Euler class is zero.