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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.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.

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.