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.
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.
Definition: a complete graph KN is a simple undirected graph in which
every pair of distinct vertices is connected by a unique edge.
It is easy to see that KN has
edges.
We can now use Ramsey theorem, with p=r=2, where
is interpreted as
the edge set of the complete graph KN on N vertices. Ramsey’s theorem
asserts that if each of the edges of KN is colored red or blue, we can always
find a complete subgraph on n vertices with all edges red or all edges blue.
N=6
n=3
N=5
A set X  R d is convexindependent if for every x X, we have x  conv(X\ {x}).
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∊S
A∉conv(S\{A})
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.
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.
For every natural number k, there exists a number N(k) such that any N(k)-point set
X⊂R2 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.
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
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):
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).
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
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|<f(k-1,l) ->
|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.
Let X be a set in R d . A k - pointset Y  X is called a k - holein X if
Y is convexindependent and conv(Y) X  Y.
Example:
Y determines a convex k-gon with no point of X inside.
X
X
Y
Y
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 sufficiently large point set contains
a convex 6-hole. This proof is much more complex,
And we will not show that.
•
•
•
•
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.
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
H
I
No
points
•
•
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:
H
I
The point x lies in one of these quadrilaterals, and the vertices of the other
quadrilateral together with x form a 5-hole.
•
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.
H
x
I
y
r
* 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.
•
If |r∩H|≥3, we get a 5-hole formed by x,y and 3 points of r∩H.
H
x
.
I
y
r

For |r∩H|≤2, we have one of the two cases:
(2)
(1)
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.
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:
(i)
No line determined by two points of X∪Y is vertical.
(ii)
Each line determined by two points of X lies above all points of Y.
(iii)
Each line determined by two points of Y lies below all the points of X.
X
Y
For a set X={x1,…,xn} with no two points having equal x-coordinates and with
notation chosen so that the x-coordinates of the xi increase with I, we define the
sets X0={x2,x4,…} and X1={x1,x3,…}.
A finite set H⊂R2 is a Horton set if
•
|H|≤1
•
|H|≥1 and both H0 and H1 are Horton sets. H1 lies high above H0 or H0 lies
high above H1.
H0
x2
x4
x6
x8
Example: H(3)
H1
x1
x3 x x7
5
For every n≥1, an n-point Horton set exists.
Note: in the proof, we will construct a Horton set of size 2k, but we can produce a
smaller Horton set from a larger one by deleting points from the right.
By induction.
We define H (0) as thepoint(0,0).Suppose that wecan constructa Hortonset H (k) with
2 k pointswhose x - coordinateare 0,1,...2k -1.
Let A  2H(k) , and B  A  (1,h k ), where h k is a sufficiently large number.
H(k)
A
B
We set H(k+1) = A∪B. It easily seen that if
Hk is large enough, B lies above A, and so H(k+1) is Horton as well.
B
H(k+1) = A∪B
b=(1+xA,h(k)+yA)
A
A set X in R2 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.
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
H0 lies deep below H1 (the other possible case is analogous). Let C⊆H be a 4-cup.
If C  H 0 or C  H1 , thena pointclosing C fromaboveexistsby theinductivehypothesis
(H0 and H1 are also Hortonsets).
T hus,let C  H 0    C  H1.
T hecup C may haveat most 2 pointsin H1 :
If therewere 3 points,say a, b, c, thenH 0 lies below thelines ab and bc, and
so theremainingpointof C, which wassupposed to lie in H 0 , can't forma cup with {a, b, c}.
C
This means that C has at least 2 points, a and b, in the lower part H0.
Since the points of H0 and H1 alternate along the x-axis, there is a point c∊H1
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
H
c
H1
H0
a
b
No Horton set contains a 7-hole.
For contradiction, suppose there is a 7-hole X in the considered Horton set H. If
X⊆H0 or X⊆H1, we use induction (H0 and H1 are both Horton sets).
Otherwise, we select the part (H0 or H1) containing the larger portion of X;
If this part, is say, H0, and it lies deep below H1, H0 has at least 4 points of X.
H
X
H1
H0
X
These 4 points must form a cup in H0, for if 3 of them were a cap, no point of
H1 could complete them to a convex independent set.
H1
H
H0
By the previous lemma, H0 contains a point closing the 4-cup from above
(Because H0 is a Horton set). Such a point must be contained in the convex
hull of the 7-hole X. A contradiction.
H1
H
H0