We follow Terence Tao's blog and different chapters (4,6) from Matousek book.
Given n lines in the plane (not necessarily in general position), they divide it into at most n^2 parts.
Any line intersects (the interior) of at most O(n) regions. Also the total number of sides of these O(n) regions is also O(n). The key is to realize that each line contains at most 1 upper left edge that is not top left (a edge to the left of the topmost vertex that is not immediately adjacent to the top most vertex).
Ex1: Given n lines in the plane, show that you can properly color the cells formed by these lines black and white. (In a proper coloring adjacent cells don't have the same color. Cells that share an edge are adjacent).
Another way to obtain the Szemeredi Trotter is to apply Cutting Lemma as stated below. Cutting Lemma can be generalized to higher dimensions.
Cutting lemma: Given n lines, L, in the plane, we can take O(r) lines, R, plus some vertical segments, that subdivide the plane into O(r^2) cells such that the interior of any cell is incident to at most O(n/r) lines.
Proof: Using random sample from L - first we select each line with probability r/n.
Sidenote experiment: Suppose we select each integer from 1 to n with prob r/n.
Can we bound the probability of the event that more than 100r numbers are selected?
Yes, the Markov inequality gives us at most 1% chance for this.
Can we bound the probability that there is a run of m unselected numbers?
As in Ramsey theory, n bad events, Pr[each]<(1-r/n)^m, so Pr[all]< n(1-r/n)^m.
If m = 100n/r log n, then this is very small.
Back to proof. If each line from L is in R with prob r/n, then from sidenote we know that with high probability, each side of each cell determined by R intersects at most 100n/r log n line from L. (Since on n lines we have at most n^2 possible runs of a fixed length.) This is close. The problems - log n and some cells might have many sides.
First, take care of big cells. Wlog, suppose no line is vertical.
Divide cells with many sides by vertical segments starting from their vertices.
Each vertex gives at most two segments, so still O(r^2) cells.
These new segments also intersect at most 100n/r log n lines, as we can use sidenote for them too. (At most n^2 intersections give n^2 vertical lines, n^3 runs.)
Finally we have to get rid of the log n. Back to sidenote:
How many runs of unselected longer than kn/r do we expect?
Each contains one of 2r/k fixed kn/2r long intervals unselected.
For each the chance to be unselected is at most (1-r/n)^(kn/2r)<exp(-k).
So expected number of long runs is O(rexp(-k)).
We claim that expected number of segments with more than kn/r lines intersecting them is O(r^2exp(-k)). For non-vertical segments, on each of n line rexp(-k) and each line is selected with probability r/n. For vertical segments, chance is exp(-k) and each is selected with probability at most (r/n)^2. We are done with the claim.
To finish proof of cutting lemma, We use induction on n. The cells that are intersected by more than Cn/r lines, we recut them. Induction hypothesis: if n'<n, then the plane is cut into Br'^2 cells with at most Cn'/r' lines intersecting each.
We use this for n'=2kn/r lines and r'=2k for each cell with more than kn/r lines but less than 2kn/r, where k>C. This in total gives sum_{k>C} O(Bk^2) k^2exp(-k) more cells. So the expected number of all cells created is O(r^2)+B sum_{k>C} O(k^4exp(-k)). If we choose B big enough compared to constand hidden in the first O(.), and then C big enough compared to B, we get that this is less than Br^2.
Hw:Â Prove weak version of Szemeredi-Trotter: Incidencies < O(mn^{1/2}+n).
A very easy weaker cutting lemma. Let s = 6rlogn lines be randomly selected (with replacement, so each line may be selected multiple number of times). Triangulate the regions obtained randomly. If there are at least n/r lines in a particular triangle (we call such a triangle dangerous), then the probability that this triangle is not stabbed (one of these lines is not selected) = (1-1/r)^s = exp(-6log n) = n^(-6). Total number of possible triangles = n^6. So with a positive probability all dangerous triangles are stabbed.
Ex2: Prove the following 'circular' version of cutting lemma: Given n circles, C, in the plane, let S be a random subset (selection with replacement) of s=Crlogn circles. Draw a vertical line upwards and downwards from each vertex (point of intersection of circles). Show that there are O(s^2) regions/circular trapezoids. Show that if C is a sufficiently large constant, then with a positive probability S intersects all possible trapezoids with more than n/r circles in their interior.