[Math] simple bijection between $\mathbb{R}^2$ and lines on a plane

Question

The two sets have the same cardinality as $\mathbb{R}$, so there is a bijection, but I can't come up with a simple function or prove that one does not exist.

Answer 1

I will do this by using first a bijection from lines in the plane to lines in the plane that do not go through the origin (this is very similar to the standard bijection $[0,1]\to (0,1]$, if you're familiar with that). Then a bijection from lines that do not contain the origin to the plane with the origin taken away. Finally, a bijection from $\Bbb R^2\setminus\{0\}$ to the plane.

For each line through the origin, find its normal vector in the direction most upwards (the direction with positive $y$-coordinate) (for the $y$-axis, use the vector $(1,0)$). Then take each of those lines and move them one unit in that normal direction. Then take any line it landed on, and move that one unit in the same direction, and so on.

In other words, if you prescribe to each line in the plane a normal vector in this manner, then each line whose distance to the origin is an integer, and where the origin lies on the opposite side from the normal vector, is to be moved one unit in the direction of its normal vector. This gives a bijection between the lines in the plane and the lines in the plane that do not go through the origin.

Now, for each line in the plane that doesn't go through the origin, it has a unique representation on the form $ax+by = 1$ where $a$ and $b$ are not both zero, and for each such pair $a, b$, we get a unique line. Let each line correspond to the point $(a, b)\in \Bbb R^2$, and we have a bijection between lines that do not go through the origin to the plane without the origin.

Finally, a bijection between $\Bbb R^2\setminus\{0\}$ and $\Bbb R^2$, I think you can find yourself.