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For example Möbius strip can be embedded but it is not closed. Klein bottle is closed but cannot be embedded.

I've seen statements that such 2-manifold does not exists in various sources. However, I have seen no justification for such a result. Does it follow from some more general theorem in an obvious way?

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This is known not to be possible, and by an obvious argument: If a closed surface is embedded in $\mathbb{R}^3$, then the complement of the image has two connected components, one of which is bounded and can therefore be regarded as the 'inside'. This will define an orientation of the surface. – Robert Bryant Nov 12 at 16:01
Oops! I meant to assume that the surface is connected, which is where the number of components of the complement is determined. – Robert Bryant Nov 12 at 16:02
@Enok, see the answers of mathoverflow.net/questions/18987/… – jc Nov 12 at 16:02
Oh, and there is a stronger result that says that a non-orientable surface (compact or not) cannot be embedded as a closed subset of $\mathbb{R}^3$. This is not quite as obvious, but Milnor and Stasheff provide a proof in their classic book Characteristic Classes. – Robert Bryant Nov 12 at 16:04
@Enok, given that you assert that the Klein bottle cannot be embedded, your proof for that ought to extend easily to other non-orientable surfaces. – jc Nov 12 at 16:06

closed as too localized by jc , Igor Rivin, HW, Agol, Chris Gerig Nov 12 at 18:51

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