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In this answer it is remarked that the real projective plane minus one point is homeomorphic to the Möbius strip without boundary.

A normal Möbius strip is topologically equivalent to a real projective plane with the whole inside of a conic section left out. The conic section has become the single border of the strip. The plane is its single surface. Veblen & Young also describe this equivalence in their books on projective geometry (Vol.2, par. 33).

When we look at a Möbius strip of paper in Euclidean space, we observe two different "mirror versions" of the strip (depending on a right-handed or left-handed twist in the paper).

Question 1:
Is it correct to assume that the projective plane that it models also may be attributed with a certain "mirror version" with respect to its ambient space? As the ambient space, please consider Euclidean space extended with the plane at infinity.

When the conic section shrinks to one point the rest of the plane becomes a Möbius strip without boundary (or: a 1-point boundary).
When I model a Möbius strip by means of paper, then "without boundary" would mean an infinitely large piece of paper but still with the twist.

Question 2:
When question 1 is answered with yes, is it then also allowed to speak of the "mirror version" of a Möbius strip without boundary?

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Gerard
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  • Two things are unclear here: What do you mean "infinitely large"? (Do you mean metrically complete?) What do you mean by "chirality"? The definitions that I know are topological and have noting to do with metric properties. – Moishe Kohan Jun 11 '14 at 21:01
  • See my edit, I hope that may help. I meant topological behaviour, not metrical, but the mix comes from looking at a projective plane inside an Euclidean space. – Gerard Jun 12 '14 at 09:32
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    No, what is enantionirphic property? State precisely what it means without relying upon articles written by chemists. – Moishe Kohan Jun 14 '14 at 19:17
  • Edited again ... – Gerard Jun 15 '14 at 10:43
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    This still makes no sense. Consider reading a textbook on differential geometry and/or topology. Voting to close. – Moishe Kohan Jun 15 '14 at 15:32
  • Which part or sentence in the question is unclear? Just asking. – Gerard Jun 15 '14 at 17:26
  • Gerard: Both Questions 1 and 2 are meaningless. As I said, consider reading a textbook. What you are completely missing is a proper language that topology developed in the last 150 years. (Language which uses words like "homeomorphism", "isotopy", etc.) – Moishe Kohan Jun 16 '14 at 21:37
  • Let's see if I have this right. The OP is assuming some sort of correspondence between the Mobius strip and the real projective plane. Since the Mobius strip can be created either with a "left-handed twist" or "right-handed twist," the OP reasons that there are two different "versions" of the Mobius strip. Question 1 asks how these two different versions manifest themselves in the real projective plane -- and in particular whether there are in fact two "versions" of the real projective plane. @Gerard, is this what you're after? – Jesse Madnick Jun 17 '14 at 08:14
  • @Jesse Madnick: there is only one Moebius band, just it has infinitely many embeddings in 3space up to isotopy and only one embedding in real projective plane, up to isotopy. – Moishe Kohan Jun 17 '14 at 19:07
  • @studiosus: I'm aware. I was just trying to help clarify the OP's question, both so he could better understand it, and so that answerers would have a better idea what exactly he's asking (and where he's coming from / what mis-conceptions he may have). – Jesse Madnick Jun 17 '14 at 20:47
  • @Jesse: I tried myself but failed. My feeling is that this is hopeless. – Moishe Kohan Jun 18 '14 at 00:46
  • @JesseMadnick Your interpretation of question nr. 1 is correct apart from the fact that I am aware there is only one "version" of the projective plane, therefore I added "w.r.t. ambient space" - this is "embedding in 3space" of studiosus. – Gerard Jun 18 '14 at 12:00

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When we say 'without boundary', in this context we mean that the space is not closed (as a subspace of $\mathbb{R}^3$, say), in much the same way that the disc $\{z\mid \|z\|<1\}$ does not include its boundary, but $\{z\mid \|z\|\leq 1\}$ does include its boundary.

Depending on the context, if $X$ is a subset of a metric space then the boundary of $X$ is the closure of $X$ with the interior of $X$ removed. If $X$ is a manifold 'with boundary' then the boundary of $X$ is the set of all points which have a neighbourhood which is homemorphic to the half plane $\{(x,y)\in\mathbb{R}^2\mid y\geq 0\}$ and has no neighbourhood homeomorphic to $\mathbb{R}^2$.

Dan Rust
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  • Sorry but this doesn't help regarding my question about chirality. The context is the real projective plane minus one point, so the boundary has shrinked to this one point. – Gerard Jun 11 '14 at 20:46
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    What's chirality? – Dan Rust Jun 11 '14 at 23:54
  • Does the edit of my question help? – Gerard Jun 12 '14 at 09:33
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    @Gerard not really. You're mentioning conic sections for some reason, and then a piece of paper modeling the projective plane comes out of nowhere. You mention the word chirality like it has some kind of well-founded definition which it does not for topological spaces. I've attempted to explain to you where your flawed reasoning is in describing a mobius strip without boundary. You need to either spend more time learning standard definitions of these objects, or try to give a well-founded definition for these new terms you are using so you can communicate them to us. – Dan Rust Jun 12 '14 at 10:09