TITLE: How does one prove that the Klein bottle cannot be embedded in $R^3$?
QUESTION [6 upvotes]: How does one prove that the Klein bottle cannot be embedded in $R^3$?
I'm talking about smooth embeddings.

REPLY [4 votes]: Any compact, smooth hypersurface in $\Bbb R^n$ is necessarily orientable. (Without fancy algebraic topology, this is a consequence of the Jordan-Brouwer Separation Theorem.) Indeed, more is true: A compact hypersurface in any simply connected manifold is orientable.