Compact manifolds and orientability I've a doubt about compact manifolds and orientability.
I know that Compact Manifolds in $\mathbb{R^3}$ are orientable.
My questions is:
The statement above is valid only for compact manifolds without boundary (in this case, closed manifolds)?
I'm asking this because I'd read that the Möbius-Strip with its boundary is a compact manifold.
Can someone explain me this?
Thanks.
 A: There are two notions here: manifold, and manifold-with-boundary. If you add the boundary circle to a Möbius strip, you get a manifold-with-boundary, which is not a manifold. The statement that all compact manifolds embedded in $\mathbb R^3$ is true. For $0$ and $1$-manifolds, it follows because all $0$ and $1$-manifolds are orientable. For surfaces it follows because any connected compact surface in $\mathbb R^3$ divides $\mathbb R^3$ into two pieces, which takes a little work to prove. (I like Alexander duality as a proof.) Then you can orient the surface by taking an outward-pointing normal vector at each point and using the right-hand rule to orient the tangent-plane at each point.
The statement is false for $\mathbb R^4$. You can embed the projective plane in $\mathbb R^4$. 
A: It is pretty simple, actually. For some mathematicians the definition of manifold covers only those without boundary, while others distinguish between manifolds with or without boundary.
As you correctly observed, the Möbius strip is a non-orientable manifold with boundary which can be embedded in $\Bbb R^3$. Moreover it is a compact topological space, since for example it is a closed subspace of $\Bbb P^2_{\Bbb R}$.
