How Many Colours Will You Need If Colouring Polygons Enclosed By Straight Lines? There is an infinite plane.
I can draw straight infinitely long lines in the plane(mind you,only finitely many).
Now,I want to colour the polygons.
The colours of adjacent polygons must not be the same(if the polygons only share the same point,then they aren't adjacent).
How many colours do I need to colour the polygons enclosed by the straight line(including the infinite ones)?
Example
I think the answer is two, but I couldn't prove it.
Can anyone prove it for me(or prove that it needs 3 or more colours)?

If you only allow the lines to be actual, infinitely-long-in-both-directions lines instead of bounded line segments, then two colors is all you need.
To see that it works, let each line be given by an equation like $ax+by+c=0$. Then, for any finite collection of lines, take the expression
$$
(a_1x+b_1y+c_1)(a_2x+b_2y+c_2)\cdots
$$
where the left-hand side for each line is included, and included only once. Color red where this expression is positive, and blue where it's negative. Any two polygons that share a border must have opposite signs, because exactly one of the factors changes sign when you move from one polygon to the other: the factor that corresponds to the line that they have in common.
For an infinite collection of lines, the proof doesn't work directly. But if you add all the lines one by one, say in order of whichever is closest to the origin, we see that no region will ever conflict with its immediate neighbors by the above argument. So even for infinitely many lines (as long at
as we require that any bounded region of the plane only intersects finitely many of them), this works.
If you allow infinitely many lines "close to" one another, this problem becomes ill-posed. Consider, for instance, having infinitely many lines $\ell_n:y-\frac1n=0$, for $n\in \Bbb N$. Which color does the region below the $x$ axis get?
