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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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    [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

2 Answers 2

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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...)