Erdős–Szekeres theorem

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The natural numbers from 1 to 101 are written in some order. Prove that this sequence contains either an increasing or a decreasing subsequence of length 11.

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Remark.
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A subsequence is what is obtained from a sequence by deleting some of its terms.

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(1)

Among any 5 points in general position in the plane there is a convex 4-gon.

(2)

There exist 8 points in general position in the plane among which there is no convex 5-gon.

(3)

Among any 9 points in general position in the plane there is a convex 5-gon.

(4)

Erdős–Szekeres Theorem. For some nn, among any nn points in general position in the plane there is a convex 10-gon. (Cf. Problem 4.4.4.)

The hypergraph Ramsey number Rl(m1,…,mk)R_l(m_1, \ldots , m_k), m1,…,mk⩾lm_1, \ldots , m_k \geqslant l, is the smallest positive integer xx such that for any coloring of all ll-element subsets of an xx-element set in kk colors there exist ii and a subset of size mim_i all of whose ll-element subsets are colored in the ii-th color. (Hypergraph Ramsey number'' is a single term defined above; knowing the term hypergraph'' is not needed to understand it.)

For example, it is obvious that R2(m1,…,mk)=R(m1,…,mk)R_2(m_1, \ldots , m_k) = R(m_1, \ldots , m_k) and R3(3,n)=nR_3(3, n) = n. Problem 4.3.1 asks to prove that R3(5,4)⩽8 000R_3(5,4) \leqslant 8\, 000. And solving Problem 4.3.2(4) requires proving that R3(10,10)R_3(10,10) or R4(5,10)R_4(5,10) exists.

An mm-cup (mm-cap) is a subset of mm points of the graph of a convex (concave) function. Denote by f(k,l)f(k, l) the minimal number nn such that among any nn points in the plane that have distinct abscissas, no three of which lie on a line, there is either a kk-cup or an ll-cap. (It is not obvious that such a number exists. Therefore a remark analogous to the one made in Problem 4.1.2(1) applies to the formulas in this problem.)

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(1)

Among any f(m,m)f(m, m) points in the plane there is a convex mm-gon. (Cf. Problem 4.3.2.)

(2)

f(k,l)⩽f(k−1,l)+f(k,l−1)−1f(k, l) \leqslant f(k-1, l) + f(k, l-1) - 1.

(3)

f(k,l)⩽(k+l−4k−2)+1f(k, l) \leqslant \dbinom {k+l-4}{k-2} + 1.

(4)

f(k,l)=(k+l−4k−2)+1f(k, l) = \dbinom {k+l-4}{k-2} + 1.

References

[ГДИ+16] Глибичук А.А., Дайняк А.Б., Ильинский Д.Г., Купавский А.Б., Райгородский А.М., Скопенков А.Б., Чернов А.А. Элементы дискретной математики в задачах. МЦНМО, 2016.