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For a vector space $V$, $P(V)$ is defined to be $(V \setminus \{0 \}) / \sim$, where two non-zero vectors $v_1, v_2$ in $V$ are equivalent if they differ by a non-zero scalar $λ$, i.e., $v_1 = \lambda v_2$.

I wonder why vector $0$ is excluded when considering the equivalent classes, since $\{0\}$ can be an equivalent class too? Thanks!

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I've been asking myself the same thing in my geometry class a few weeks ago! It might have to do with the fact that 0 is in the field already? I am curious to the answers. –  user12205 Dec 24 '11 at 23:54
    
I think that if you ask this question, you might as well ask why we're defining an equivalence relation in this way—what good is it? –  Dylan Moreland Dec 24 '11 at 23:57
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@DylanMoreland: That is exactly my point. I thought it might be because of some consequences following the definitions that I am not aware of, although both definitions seem valid to me for now. –  Tim Dec 25 '11 at 0:00
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This is not a deep explanation, but if you look at $\mathbf P (\mathbf R^n)$, then the equivalence classes can be identified with the points on the unit sphere with the antipodal points identified. If you include $\mathbf 0$ as a separate class by itself, then you get the sphere as before, plus the additional point $\mathbf 0$. The first one seems slightly simpler to me. –  Srivatsan Dec 25 '11 at 0:07
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This answer may provide some intuition on the projective plane and why we start with 3-space excluding zero. –  Arturo Magidin Dec 25 '11 at 2:25

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up vote 8 down vote accepted

You could do this, but the resulting space would not be as useful.

For example, suppose $V$ is $\mathbb{R}^n$ equipped with its usual topology. Then the projective space $P \mathbb{R}^n$ can be made into a topological space by giving it the quotient topology. If you include 0 as in your suggestion, the projective space would not be Hausdorff in this topology; in fact, the only open neighborhood of the equivalence class $\{0\}$ is the entire quotient space.

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