Why is $0$ excluded in the definition of the projective space for a vector space? 
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!
 A: 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. Is there a reason to care about it?
One reason we care about the space of lines through the origin is that it is a rich arena for discovering interesting theorems and examples.
In general, projective space is a more natural setting for algebraic geometry than affine space. For instance, theorems have fewer special cases - the most natural one being that two lines always intersect in the projective plane. Others include: Bezout's theorem, the classification of plane conics, 27 lines on a cubic, etc.
There are other reasons it is nice, which have to do with it being compact. Along those lines, 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. This is related to how we can use projective space to resolve singularities via blow-ups, by remembering the tangent line along which a point enters the singularity. All of these are natural situations where we care about the space of lines as a geometric object.
Maybe there are natural situations where it also makes sense to include a separate $0$ point, which is the limit of the other points. That doesn't sound too far fetched to me, especially thinking about the blow-up example.
A: 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.
