Dichotomy in the number of regions on a plane formed by an infinite number of lines

Question

I'm reading Knuth's Concrete Mathematics and we are dealing with recurrence relations. He proves that the number of regions $L_n$ formed by $n$ lines on a plane is $L_n=\frac{n(n+1)}{2}$.

I don't have the formal proof for it, but I have an intuition that an infinite number of lines in two dimensions forms a plane - because there are an infinite number of points that can live on these lines and therefore every single point in 2-D space lives on these lines and so we have a curve.

Under the tentative assumption that this postulate is true, I'm wondering what the behavior of $\lim\limits_{x \to \infty} L_n$ is, i.e. how many regions are formed by an infinite number of lines on a plane.

A region is defined as a set of points not on the line segments that surround them. If with an infinite number of lines you cover the entire plane, i.e. the number of regions approaches $0$, but by the equation given the number of regions approaches $\infty$, there is a dichotomy. I'm not sure how to understand this or if I am missing something.

Answer 1

I have an intuition that an infinite number of lines in two dimensions forms a plane - because there are an infinite number of points that can live on these lines and therefore every single point in 2-D space lives on these lines

It depends on how many lines you have. If you have a countably infinite collection of lines, then you will have covered $0$ area no matter how you do it, and so there will be a lot of points missed. This is vaguely related to your correct idea that $\displaystyle{\lim_{n\to\infty}}L_n=\infty$, though that limit only talks about the maximum numbers of regions for finite numbers of lines.

If you have uncountably many lines, then it depends how you place them. For example, if they're all vertical and pass through points of the Cantor set then you've missed infinitely many regions (in fact, countably infinitely many vertical strips). But if you place a vertical line at every horizontal coordinate, you'll cover your region completely.

This uncountable case is very far removed from $\displaystyle{\lim_{n\to\infty}}L_n$ since we're not talking about the limit of the number of regions of finitely many lines, or the countably many lines (which could be thought of as a "limit" of finitely many lines), but many more lines than that.

Answer 2

A concrete example of infinitely many lines which dramatically fail to cover the plane: let $L$ be the collection of all lines of the form $y=mx+b$ with $m, b$ rational. Then as long as $\alpha\over\beta$ is irrational (and $\beta\not=0$), no line in $L$ will go through the point $(\alpha, \beta)$.

Let $A$ be the set of points not on some line in $L$. The (countably many) lines in $L$ divide $A$ into regions, each consisting of one point (exercise: if $p\not=q$ are distinct points in $A$, then there is a line in $L$ going between $p$ and $q$). However, even though there are countably many lines, there are uncountably many of these regions (that is, $A$ is uncountable)! This is the same sort of thing that's going on with the fact that the reals are uncountable, but have a countable dense subset (the rationals).

Bottom line: be careful about limits of infinite processes!