My favorite example of: the pigeonhole principle

This is part of a new sequence of posts titled,

My favorite example of: {x} ,

for different values of {x} . Today, {x} is the pigeonhole principle.

The Erdös-Szekeres Theorem: Consider any sequence of distinct numbers. There must exist a subsequence of ${\sqrt{n}}$ numbers such that is either entirely increasing or entirely decreasing.

The Erdös-Szekeres Theorem is a classic result in permutation combinatorics. Although the theorem was first presented in 1935, the proof I’m going to describe appears in the 1959 paper “A simple proof of a theorem of Erdös and Szekeres”, written by Abraham Seidenberg. Seidenberg’s proof is arguably one of the slickest applications of the pigeonhole principle ever.

The Proof.

Consider a sequence ${x = x_1, x_2, \ldots, x_n}$ of {n} distinct numbers. For each number ${x_i}$ , define

$\displaystyle a_i = \text{ length of longest increasing subsequence ending in }x_i$

and define

$\displaystyle b_i = \text{ length of longest decreasing subsequence ending in }x_i.$

For example, if {x = 1, 4, 5, 3, 2} , then ${a_5 = 2}$ is the length of the longest increasing subsequence ending in ${x_5 = 2}$ and ${b_5 = 3}$ is the length of the longest decreasing subsequence ending in ${x_5 = 2}$ .

In order to prove the theorem, Seidenberg made one crucial observation:

Observation: If {i < j} then either ${a_i \neq a_j}$ or ${b_i \neq b_j}$ .

To see why the observation holds, consider the case where ${x_i < x_j}$ . Then for any increasing subsequence {S} that ends in ${x_i}$ , the longer subsequence ${S \cup \{x_j\}}$ is an increasing subsequence that ends in ${x_j}$ . This implies that ${a_j \ge a_i + 1}$ and thus that ${a_i \neq a_j}$ . A similar argument can be used in the case where ${x_i > x_j}$ to show that ${b_j \ge b_i + 1}$ , and thus that ${b_i \neq b_j}$ . In both cases, the observation holds.

In order to prove the theorem, our goal is to show that there is some {i} for which either ${a_i \ge \sqrt{n}}$ or ${b_i \ge \sqrt{n}}$ . Seidenberg realized that he could prove this with a simple application of the pigeonhole principle.

Suppose for conradiction that ${a_i < \sqrt{n}}$ and ${b_i < \sqrt{n}}$ for all {i} . Then there are fewer than ${\sqrt{n} \cdot \sqrt{n} = n}$ possibilities for each pair ${(a_i, b_i)}$ . It follows by the pigeonhole principle that at least two of the pairs

$\displaystyle (a_1, b_1), (a_2, b_2), (a_3, b_3), \ldots, (a_n, b_n)$

are equal to one-another. That is, there exists $i \neq j$ such that (a_i, b_i) = (a_j, b_j) . But this contradicts Seidenberg’s observation, and thus completes the proof.

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