Ex: Show that (d+1)(r-1)+1 is sharp, there are (d+1)(r-1) points that cannot be partitioned into r sets.
Geometric Selection Theorems
Before, we prove a Ham-Sandwich type-thm. (See also http://www.borisbukh.org/buckbuck.html)
Lemma[Buck-Buck]: Given a finite point set P in the plane, there are 3 concurrent lines
that cut P into 6 parts, each containing >=n/6-1 points in its interior.
Proof: The statement follows if we prove that any continuous probability measure for which any open set has pos measure, can be cut into 6 equal parts as above.
For every fi angle, there is a unique halving line, l(fi).
Using continuity, there is a unique point on each line, p(fi), thru which there is a second line such
that the two regions between this line and l(fi) contain 1/6 of the measure each.
(Here we consider the two regions that are in a pos oriented p,l,2.line triangle.)
Using continuity, uniqueness of p(fi) and antipodality, we get a third line.
Boros-Furedi thm: Given a point set P in the plane in general position with n points,
there is a point that is in >=2/9{n \choose 3}-O(n^2) of the triangles determined by P.
Proof[Bukh]: Use above lemma.
For any 6 points in different parts, at least 8 of their triangles contain the intersection of the 3 lines.
Rest is calculation; 8(n/6)^6/(n/6)^3=2/9n^3-O(n^2).
Note: 2/9 is the best possible. In higher dim we do not know exact value.
Hw: a) Exists c>0 such that given a point set with n points and a line family with n lines, we can select cn points and cn lines
such that all the selected points are in the same cell (i.e. no two are separated by a selected line).
b) Using the above, prove that exists c>0 given n segments and n lines we can select cn of each such that
selected lines are segments either all intersect, or none of them do.
c) Using the above, prove the same for n red and n blue segments.
First selection lemma: For all d, there is a c>0 such that given a point set P in d-dim in general position with n points,
there is a point that is in >=c{n \choose d+1} of the d-simplices determined by P.
(For another proof, see Matousek book chapter 9.)
Proof: We use fractional Helly, so it is enough to prove that there is an a such that
an a fraction of the d+1-tuples of d-simplices intersect.
Using Tverberg and Caratheodory, any (d+1)^2 points can be divided into d+1 intersecting simplices.
This implies that a>= 1/{(d+1)^2 \choose d+1\times d+1} (if n is at least (d+1)^2).
Second selection lemma: If we have a d+1-uniform hypergraph on n points in d-dim with >= a{n \choose d+1} edges, then
there is a point that is in >= ca^s {n \choose d+1} edges, where c>0 and s depend only on d.
Eg: 1-dim we have a{n \choose 2} intervals with n endpoints, then there is a point in ca^s{n \choose 2} intervals.
Hw: Prove that in 1-dim s=2 (upper and lower bound)!
Proof sketch [of 2. sel. lemma]: We again use the fractional Helly, so we want many d+1-tuples of d-simplices that intersect.
Problem is that from (d+1)^2 points the intersecting simplices might not be in our hypergraph (Turan-type extreme).
What we can guarantee is a complete d+1-partite, with t vertices in each class, T(d+1,t), using:
Thm[Erdos-Simonovits]: H is d+1-uniform with >= a{n \choose d+1} edges, then H contains ca^{t^{d+1}}n^{(d+1)t}
different copies of T(d+1,t) for some c(d,t)>0 constant (if a is constant). (For proof, see Matousek.)
For each T(d+1,t) we can use colorful Tverberg to get r=d+1 intersecting colorful d-simplices (enough if t>4d).
Rest is double counting, as before, to get ca^{t^{d+1}}n^{(d+1)^2} d+1-tuples of d-simplices that intersect, done.
Same type lemma: For all d, m, exists c>0 for any X_1,..,X_m in gen pos in d-dim we can select Y_i from X_i such that
its size is >=c|X_i| and all y_i from Y_i have the m-tuple (y_1,..,y_m) has the same order type (these are called transversal).
Proof: 1, It is enough to prove for m=d+1 (by repeatedly applying lemma for d+1).
2, If there is no hyperplane intersecting C_0,..,C_d convex sets, then all colorful d-simplices have same order type.
3, If for each I from {0,..,d} there is a hyperplane separating {C_i: i from I} from {C_i: i not from I},
then there is no hyperplane intersecting all C_i (using Radon on hyperplane that would intersect all).
(Ex: Moreover, converse also holds.)
Until 3 is true for all I for convex hulls of Y_i, bisect with Ham-Sandwich the Y_1,..,Y_d
and take the part that has at least half of the points of Y_0.
This gives c=2^{-2^d} for m=d+1 and this raised to the {m \choose d+1} power for bigger m's.
Ex: Prove pos-frac Erdos-Szekeres: For all k exists c>0 from any point set with n points we can select k disjoint sets,
each of size cn, such that any transversal is in convex position.