Erdős–Szekeres theorem

One can interpret the positions of the numbers in a sequence as x-coordinates of points in the Euclidean plane, and the numbers themselves as y-coordinates; conversely, for any point set in the plane, the y-coordinates of the points, ordered by their x-coordinates, forms a sequence of numbers (unless two of the points have equal x-coordinates). With this translation between sequences and point sets, the Erdős–Szekeres theorem can be interpreted as stating that in any set of at least rs − r − s + 2 points we can find a polygonal path of either r − 1 positive-slope edges or s − 1 negative-slope edges. In particular (taking r = s), in any set of at least n points we can find a polygonal path of at least ⌊√n-1⌋ edges with same-sign slopes. For instance, taking r = s = 5, any set of at least 17 points has a four-edge path in which all slopes have the same sign.

An example of rs − r − s + 1 points without such a path, showing that this bound is tight, can be formed by applying a small rotation to an (r − 1) by (s − 1) grid.

The Erdős–Szekeres theorem may also be interpreted in the language of permutation patterns as stating that every permutation of length at least rs + 1 must contain either the pattern 1, 2, 3, ..., r + 1 or the pattern s + 1, s, ..., 2, 1.

Given a sequence of length (r − 1)(s − 1) + 1, label each number n_i in the sequence with the pair (a_i, b_i), where a_i is the length of the longest monotonically increasing subsequence ending with n_i and b_i is the length of the longest monotonically decreasing subsequence ending with n_i. Each two numbers in the sequence are labeled with a different pair: if i < j and n_i ≤ n_j then a_i < a_j, and on the other hand if n_i ≥ n_j then b_i < b_j. But there are only (r − 1)(s − 1) possible labels if a_i is at most r − 1 and b_i is at most s − 1, so by the pigeonhole principle there must exist a value of i for which a_i or b_i is outside this range. If a_i is out of range then n_i is part of an increasing sequence of length at least r, and if b_i is out of range then n_i is part of a decreasing sequence of length at least s.

Steele (1995) credits this proof to the one-page paper of Seidenberg (1959) and calls it "the slickest and most systematic" of the proofs he surveys.[2][6]

Another of the proofs uses Dilworth's theorem on chain decompositions in partial orders, or its simpler dual (Mirsky's theorem).

To prove the theorem, define a partial ordering on the members of the sequence, in which x is less than or equal to y in the partial order if x ≤ y as numbers and x is not later than y in the sequence. A chain in this partial order is a monotonically increasing subsequence, and an antichain is a monotonically decreasing subsequence. By Mirsky's theorem, either there is a chain of length r, or the sequence can be partitioned into at most r − 1 antichains; but in that case the largest of the antichains must form a decreasing subsequence with length at least

${\displaystyle \left\lceil {\frac {rs-r-s+2}{r-1}}\right\rceil =s.}$

Alternatively, by Dilworth's theorem itself, either there is an antichain of length s, or the sequence can be partitioned into at most s − 1 chains, the longest of which must have length at least r.

The result can also be obtained as a corollary of the Robinson–Schensted correspondence.

Recall that the Robinson–Schensted correspondence associates to each sequence a Young tableau P whose entries are the values of the sequence. The tableau P has the following properties:

• The length of the longest increasing subsequence is equal to the length of the first row of P.
• The length of the longest decreasing subsequence is equal to the length of the first column of P.

Now, it is not possible to fit (r − 1)(s − 1) + 1 entries in a square box of size (r − 1)(s − 1), so that either the first row is of length at least r or the last row is of length at least s.