Removing a line from the projective plane

Question

Consider the real projective space $\mathbb{R}\mathbb{P}^{2}$ and suppose we remove a line from it. Why are we left with the Euclidean plane?

Answer 1

Your "lines through the origin" can be thought of as triples $(a,b,c)$ with $a$, $b$, $c$ not all zero, and with the equivalence relation $(a,b,c)\sim k(a,b,c)$ for $k\ne 0$. Lines in the projective plane correspond to planes through the origin. Try removing the line corresponding to the $x$-$y$ plane. Can you use the equivalence relation to map the remaining triples to points in the Euclidean plane?

Answer 2

We can visualize $\Bbb P^2$ minus a line (plane in $\Bbb R^3$) to be the top half of a sphere (the bottom half is just an identical copy of it). Put a plane on top of and tangent to our sphere, parallel to the plane we cut out of it. To map our space to this hovering plane we simply project rays out from the origin to the points on the open upper hemisphere, and wherever it these points end up hitting on the hovering plane is where we say they are sent.