# Questions in topology/geometry

Hurkyl said:
The mobius strip (without boundary) is a 2-dimensional connected nonorientable manifold without boundary, yet its usual embedding in R^3 has a connected complement.
But it does have a boundary as given by the topology of R^3, i.e. there are points not in the Moebius strip such that any ball centered at them will always intersect both the Moebius strip and its complement in R^3.

I'm taking here the definition of "a hypersurface without boundary" to be an embedding of an (n-1)-manifold in R^n such that its complement in R^n is open. Whether that is standard or not, that's what I mean for the claim that I make above.

Hurkyl
Staff Emeritus
Gold Member
Just FYI, a "manifold without boundary" (or just manifold) is a manifold for which there is a neighborhood that looks like R^n for every point.

A manifold with boundary is one for which every point has a neighborhood that looks like R^n or a half-subspace of R^n.

For example, (0, 1) is a 1-d manifold without boundary, and [0, 1) is a 1-d manifold with boundary. (Of course, so is (0, 1), I think)

The conclusion looks right, for embeddings of a (n-1)-manifold as a closed subset of R^n.

mathwonk