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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.2011-12-24
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    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?2011-12-24
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    @Jeroem: Thanks for sharing the same feeling! "0 is in the field", do you mean that the vector 0 is in the vector space V, or the scalar 0 is in its base field K?2011-12-24
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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.2011-12-25
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    @JeroenVaelen: Do you use some books or notes or other references for your geometry class?2011-12-25
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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.2011-12-25
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    @Tim offline notes written by the professor. To get back to your question, we defined a projective space: $P(V) = \{kv \mid v \in V\setminus \{0\}\}$ for some field $k$. The minus zero might have to do with the fact that $0\in k$. This still implies $P(\{0\}) = \{0\}$. You can look at elements of a projective space as lines that go through $0$. This would not be the case with $k0$. I am guessing that is the reason.2011-12-25
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    I was having a discussion with a friend about the subject and he basically said the same thing as @Srivatsan.2011-12-25
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    [This answer](http://math.stackexchange.com/questions/13763/elliptic-curves-and-points-at-infinity/13767#13767) may provide some intuition on the projective plane and why we start with 3-space excluding zero.2011-12-25

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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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Projective space is supposed to parametrize lines through the origin. A line is determined by two points, so a line through the origin is determined by any nonzero vector.

As Nate's explains, you can certainly include 0, but you will get a different space.

The reasons we care to parametrize lines through the origin is deep (by hich I mean supported by many nontrivial theorems and examples). One answer is that projective space is a more natural setting for (algebraic) geometry than affine space, in the sense that theorems fewer less special cases (Bezout's theorem or the classification of plane conics, 27 lines on a cubic, etc.). We can think of projective space as a natural compactification of affine space, which is designed to catch points that wander off to infinity by assigning to their limit the direction they wandered off in.

(I know this is an old question, but maybe somebody has a similar question, and I think this answer is sufficiently different from the others...)