
The Moebius Band
I've got a conceptual question (that depends on definitions involved being unambiguous).
Is the Moebius strip as we tend to imagine it in a surface? (If so I would like to see the FORM of the intervals the parameter set)
Is it a compact surface if we imagine it with the edge curve? We imagine a compact set of …but which definition of a compact surface, if any, is appropriate here?
Note that the Jordan Curve theorem allows to conclude that a compact surface in is orientable.
And remember that the Moebius band is nonorientable, although compact (!), as a 2dim (quotient) manifold of a closed rectangle.
The question is essentially this: How does one not become a victim of a misconception?

Re: The Moebius Band

Re: The Moebius Band
You say the Moebius strip is compact. If you mean by that the actual bounded strip that one cuts and pastes from paper and holds in ones hand, then, yes, of course it exists in 3 dimensions. You are holding it in your hand and you are in 3 dimensions, aren't you?
But if you want to make a full surface of it by extending the edges infinitely then it cannot be embedded in 3 dimensions but then it is no longer compact.

Re: The Moebius Band
Hello, I somehow didn't receive a notification about the reply which is pretty weird.
The question was essentially this:
Consider the parametrization
$$F: D \rightarrow \bb{R}^3,\,F(\varphi,t)=c(\varphi)+tv(\varphi)$$ with
$$c(\varphi)=(\cos\varphi,\sin\varphi,0), \, v(\varphi)=\left(\cos\frac{\varphi}{2}\cos \varphi,\cos\frac{\varphi}{2}\sin \varphi,\sin\frac{\varphi}{2}\right)$$
If we take $$D=(0,2\pi)\times(\frac{1}{2},\frac{1}{2})$$ then we conform with the definition of surface theory. But are we allowed to take $$D=[0,2\pi]\times(\frac{1}{2},\frac{1}{2})$$ and speak of the image of this set under F as of a surface? Are we allowed to take even $$D=[0,2\pi]\times[\frac{1}{2},\frac{1}{2}]$$ and speak of a compact surface?