How do we construct an n-dimensional projective space $P^n$? The most immediate way is to take an $(n + 1)$-dimensional vector space $V^{n+1}$, and regard our space $P^n$ as the space of the 1-dimensional vector subspaces of $V^{n+1}$. (These 1-dimensional vector subspaces are the lines through the origin of $V^{n+1}$.)A straight line in $P^n$ (which is itself an example of a $P^1$) is given by a 2-dimensional subspace of $V^{n+1}$ (a plane through the origin), the collinear points of $P^n$ arising as lines lying in such a plane (Fig. 15.15).
Source: Road to Reality by Roger Penrose
I actually can not believe the bolded statement. If I try to draw it myself, I seem to get only a triangular region:
Top down view:
I suppose this maybe due to canvas only being finite size.. suppose it was very big, even then I think only half of a plane could be seen (artist would have to turn around to see part of plane behind him).
What am I missing here? I have seen the same statement written in other place as well, so I don't think it is actually wrong.
What you seem to be missing here is everything that came after Figure 15.13 and before the quoted paragraph, which refers to Figure 15.15.
Projective geometry is not a set of instructions for how to paint a painting. Painting involves the art of perspective, which has only a slight relationship to projective geometry.
(I could go into a rant here about the propriety of using perspective painting as an explanation of projective geometry, rather than just an inspiration for it, but I'll save that for another time.)
The subject of projective geometry is not the painting, not the painter, and not the scene with the mile marker alongside the road, the tree on the other side, and the hill in the distance. The subject of projective geometry is a projective space, which in its purest sense is a completely abstract thing, but which can be modeled (as explained in the quoted paragraph) by a bundle of straight lines through a single point. Each straight line in that bundle of lines is the model of one single point of the projective space, which is why it takes an $(n+1)$-dimensional vector space to model an $n$-dimensional projective space in this manner.
Referring to Prof. Hwang's answer to your earlier question, if you use Penrose's construction to make a model of a $2$-dimensional projective plane using a vector space $V^3$ with the origin at the point labeled "eye", the two gray planes in the second figure represent lines in the projective plane.
We can capture two-dimensional Euclidean "pictures" (of a sort) of these lines by intersecting a viewing plane with these gray planes. If the viewing plane is the one labeled $Z=-1$ then we get two parallel lines. The same happens for any other plane not parallel to (but not coinciding with) the line along which the gray planes intersect. If we choose any other plane, the lines where the gray planes intersect the other plane are non-parallel and intersect somewhere.
You always lose something from the projective plane when you make one of these "pictures", however. Namely, you have no place to plot the line in the gray plane that is parallel to your viewing plane. So you have a "picture" of all the points on the projective line except one.
Note that in the "pictures" of the gray planes where the lines in the picture intersect, the lines don't just meet at the "horizon" and then stop. No, the lines in the picture go straight through the intersection, making an X with infinitely long legs in all four directions. The parts of the X "above the horizon" are what the artist would paint if he looked behind himself and then painted what he saw in the exact opposite direction from where he saw it. This would make a highly unintuitive, fantastical painting, but it's correct projective geometry, which is one of the reasons I said projective geometry has only a slight relationship with the art of perspective.