2. Problems in Ramsey theory typically ask a question of the form: "how many elements of some structure must there be to guarantee that a particular property will hold?“ Here we consider geometric Ramsey-type results about finite point sets in the plane.

3. For every choice of natural numbers p,r,n, there exists a natural number N such that whenever X is an N-element and is an arbitrary coloring of the system of all p-element subsets of X by r colors, then there is an n-element subset Y ⊆ X such that all p-tuples in have the same color.

4. We can now use Ramsey theorem, with p=r=2, where is interpreted as the edge set of the complete graph K N on N vertices. Ramsey’s theorem asserts that if each of the edges of K N is colored red or blue, we can always find a complete subgraph on n vertices with all edges red or all edges blue. N=5N=6 n=3 Definition: a complete graph K N is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. It is easy to see that K N has edges.

5. Example (finite convex independent set). Define S={A,B,C,D,E,F}. S is a finite convex independent set. WLOG, Let’s look at vertex A. A∊SA∊S A ∉ conv(S\{A})

6. Among any 5 points in the plane in general position (3 points not lying on a single straight line L), we can find 4 points forming a convex independent set. If the convex hull has 4 or 5 vertices, we are done.

7. Otherwise, we have a triangle with two points inside. The two interior points together with two of the points of the triangle define a convex quadrilateral. How many points do we need in order to create a convex independent set with 5 points? 9 points. This problem is a more complex to solve (there are more cases). We will now prove a general theorem.

8. For every natural number k, there exists a number N(k) such that any N(k)- point set X ⊂ R 2 in general position contains a k-point convex independent subset. In the previous example, N(k)=5 and k=4. We will see 2 different proofs.

9. Color a 4-tuple T ⊂ X red if its four points are convex independent and blue otherwise. If N is sufficiently large, We can use Ramsey’s theorem: p=4 (4-tuple) r=2 (red,blue) n=k (subset in X, |Y|) Ramsey’s theorem provides a k-point subset Y ⊂ X such that all 4-tuples from Y have the same color. From the previous claim we get that for k≥5, this color can’t be blue, because any 5 points determine at least one red 4-tuple. Y is convex independent, since every 4 of it’s points are (Otherwise, there exists a triangle with a point inside-4 points that aren’t convex independent, Contradiction). For the second proof, we will need a few definitions

10. Let X be a finite point set in the plane in general position. We call X a cup if X is convex independent and its convex hull is bounded from above by a single edge (if the points of X lie on the graph of a convex function). We define a cap with a single edge bounding the convex hull from below by a single edge (if the points of X lie on the graph of a concave function). A k-cap is a cap with k points, and similarly for an l-cup. For example (k=5, l=4):

11. We define f(k,l) as the smallest number N such that any N-point set in general position contains a k-cup or an l-cap. By induction on k and l, we will prove: Erdos-Szekeres theorem will follows from this with N(k) ≤ f(k,k) (this means there is a k-cup or a k-cap, and k-cap and k-cup are both convex independent sets).

12. By induction on l and k. Basis: 1-cap or 1-cup are just single points, and every 2 points form are both 2-cap and 2-cup. So the claim holds for k,l≤2. We now assume k,l≥3. For the induction step: We will prove that f(k,l)≤f(k-1,l)+f(k,l-1)-1 and the claim will follow from that: Pascal’s triangle with n=k+l-4 and m=k-2

13. Let k,l≥3, and consider a set X in general position with N=f(k-1,k)+f(k,l-1)-1 points. We prove that it contains a k-cup or an l-cap. This will establish the inequality f(k,l)≤f(k-1,l)+f(k,l-1)-1. Suppose that there is no l-cap in X. Let E ⊆ X be the set of points p ∊ X such that X contains a (k-1) cup ending with p. Because X\E contains no (k-1) cup then |X\E| |E|≥|X|-f(k-1,l)=N-f(k-1,l) +1 =f(k,l-1). Either the set E contains a k-cup, and then we are done, or there is an (l-1) cap. The first point p of such an (l-1) cap is, by the definition of E, the last point of some (k-1) cup in X, and in this situation, either the cup or the cap can be extended by one point.

14. Example: Y determines a convex k-gon with no point of X inside. X Y X Y

15. Is it true that for every k there exists an n(k) such that any n(k)-point set in the plane in general position has a k-hole? A construction due to Horton shows that this is false for k≥7, but true for k≤5. We will see the proofs for that. What about k=6? It took almost a quarter of a century after Horton’s construction to answer the existence question for 6-holes. In 2007/08 Nicol´as and independently Gerken proved that every su ffi ciently large point set contains a convex 6-hole. This proof is much more complex, And we will not show that.

16. Prove the existence of a 5-hole. Define and construct Horton sets. Prove a lemma regarding Horton sets property. Use the lemma to show that no Horton set contains a 7-hole.

17. Every sufficiently large planar point set in general position contains a 5-hole. By the Erdos-Szekres theorem, we may assume that there exists a 6-point convex independent subset of our set X. Define H ⊆ X as a 6-point convex independent subset with the smallest possible |X∩conv(H)|. Let I=conv(H)∩(X\H) be the points inside the convex hull of H. X H I

18. If I is empty, then we have a 6-hole (from Erdos-Szekers theorem). If there is one point x in I, we consider a diagonal that partitions the hexagon into two quadrilaterals: The point x lies in one of these quadrilaterals, and the vertices of the other quadrilateral together with x form a 5-hole. H H I I No point s

19. If |I|≥2, we choose an edge xy of conv(I). Let r be an open half-plane bounded by the line xy and containing no points of I. *Half plane is a planar region consisting of all points on one side of an infinite straight line, and no points on the other side. If the points on the line are not included, then it is called an open half-plane. x y H r I

20. If |r∩H|≥3, we get a 5-hole formed by x,y and 3 points of r∩H.. x y H r I

21.  For |r∩H|≤2, we have one of the two cases: By replacing u and v by x and y in (2), or u by x in (1), we obtain a 6-point convex independent set having fewer points inside H, which is a contradiction. (1) (2)

22. Let X and Y be finite sets in the plane. We say that X is high above Y (and that is Y is deep below X) if the following hold:  No line determined by two points of X ∪ Y is vertical.  Each line determined by two points of X lies above all points of Y.  Each line determined by two points of Y lies below all the points of X. For a set X={x 1,…,x n } with no two points having equal x-coordinates and with notation chosen so that the x-coordinates of the x i increase with I, we define the sets X 0 ={x 2,x 4,…} and X 1 ={x 1,x 3,…}. X Y

23. A finite set H ⊂ R2 is a Horton set if |H|≤1 |H|≥1 and both H 0 and H 1 are Horton sets. H 1 lies high above H 0 or H 0 lies high above H 1. H0H0 H1H1 x1x1 x2x2 x4x4 x6x6 x8x8 x3x3 x5x5 x7x7 Example: H (3)

24. For every n≥1, an n-point Horton set exists. Note: in the proof, we will construct a Horton set of size 2 k, but we can produce a smaller Horton set from a larger one by deleting points from the right. H (k) A B

25. We set H (k+1) = A ∪ B. It easily seen that if H k is large enough, B lies above A, and so H (k+1) is Horton as well. H (k+1) = A∪B B A b=(1+x A,h (k) +y A )

26. A set X in R 2 is r-closed from above, if for any r-cup in X there exists a point in X lying above the r-cup. Similarly, we define a set r-closed from below using r-caps.

27. Every Horton set is 4-closed from above and 4-closed from below. By induction on the size of the Horton set. Let H be a Horton set, and assume that H 0 lies deep below H 1 (the other possible case is analogous). Let C ⊆ H be a 4-cup. C

28. This means that C has at least 2 points, a and b, in the lower part H 0. Since the points of H 0 and H 1 alternate along the x-axis, there is a point c ∊ H 1 between a and b in the ordering by x-coordinates. This c is above the segment ab, and so it closes the cup C from above. We argue similarly for a 4-cup. C H0H0 H1H1 b c a H

29. No Horton set contains a 7-hole. For contradiction, suppose there is a 7-hole X in the considered Horton set H. If X ⊆ H 0 or X ⊆ H 1, we use induction (H 0 and H 1 are both Horton sets). Otherwise, we select the part (H 0 or H 1 ) containing the larger portion of X; If this part, is say, H 0, and it lies deep below H 1, H 0 has at least 4 points of X. H1H1 H0H0 H X X

30. These 4 points must form a cup in H 0, for if 3 of them were a cap, no point of H 1 could complete them to a convex independent set. By the previous lemma, H 0 contains a point closing the 4-cup from above (Because H 0 is a Horton set). Such a point must be contained in the convex hull of the 7-hole X. A contradiction. H1H1 H0H0 H H1H1 H0H0 H