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I have come across this sentence when I was reading something about algebraic topology:

The space of all lines on a plane is an open Möbius Band.

I don't quite understand this, can anyone explain this to me? Thank You.

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Please do your research first: en.wikipedia.org/wiki/M%C3%B6bius_strip which contains a subsection precisely on this. –  Chris Gerig Jul 30 '12 at 8:31

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From Wikipedia:

The open Möbius band is formed by deleting the boundary of the standard Möbius band. It is constructed from the set $S = \{ (x,y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1 \text{ and } 0 < y < 1 \}$ by identifying (glueing) the points $(0,y)$ and $(1,1−y)$ for all $0 < y < 1$.

Occurrence and use in mathematics

The space of unoriented lines in the plane is diffeomorphic to the open Möbius band.[6]

To see why, notice that each line in the plane has an equation $ax + by + c = 0$ for fixed constants $a$, $b$ and $c$. We can identify the equation $ax + by + c = 0$ with the point $(a,b,c)$. However, the line given by $ax + by + c = 0$ is also given by $λ(ax + by + c) = (λa)x + (λb)y + (λc) = 0$ for all $λ ≠ 0$.

These equations, which give the same line, are identified with the points $(λa,λb,λc)$. Thus, the space of lines in the plane is a (proper) subset of the real projective plane; where the equation $ax + by + c = 0$ corresponds to the point $(a:b:c)$ in homogeneous coordinates. It is only a subset because some equations of the form $ax + by + c = 0$ do not give lines. We need to disallow $a = b = 0$ to be sure that the equation $ax + by + c = 0$ does indeed give a line. The space of unoriented lines in the plane is given by deleting the point $(0:0:c) = (0:0:1)$ from the real projective plane. This space is exactly the open Möbius band.

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Come on, at least write up the formulae in LaTeX even if you're going to copy-paste the rest... –  Zhen Lin Jul 30 '12 at 11:56
    
I almost got it, but I don't understand that why "The space of unoriented lines in the plane is given by deleting the point (0:0:c)=(0:0:1) from the real projective plane. This space is exactly the open Möbius band." - could you help? Thanks –  1LiterTears Jun 12 '13 at 14:36
    
I would really appreciate any intuition someone could offer about the particular question raised by 1LiterTears...why is the real projective plane with a single point removed topologically equivalent to the mobius band? –  8one6 May 19 at 11:46

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