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The following question was asked in my combinatorics exam -

“Let $n$ be a positive integer. Exhibit an arrangement of integers between $1$ to $n^2$ which has no increasing or decreasing subsequence of length $n + 1$.”

This question has a very similar structure to the Rosen’s subsequence theorem, and I am aware of its proof using the pigeonhole principle, as given in the link above by @Rajdeep.

But i am not really sure on how to provide the requisite construction for my question above — the pigeonhole principle fails here as we have $n^2$ numbers and $n^2$ pairs, if we proceed in a similar fashion as done in the proof linked above.

So could somebody provide me with some hints or solutions ? I may be completely wrong as well in the using pigeonhole principle for this problem, so different solutions are also welcomed.

Vinay Karthik
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    See https://math.stackexchange.com/questions/1040400/a-sequence-of-n2-real-numbers-which-contains-no-monotonic-subsequence-of-more also https://math.stackexchange.com/questions/1144232/proving-erd%c5%91s-szekeres-bound-with-counterexample and several other questions linked there. – Gerry Myerson Mar 10 '24 at 11:02
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    It's weird to call it "Rosen's subsequence theorem", when the result is not due to Rosen but to Erdős and Szekeres. This is even mentoned in the comments of the question linked to here. – Misha Lavrov Mar 10 '24 at 20:50
  • @MishaLavrov, i was not aware of this fact, our professor taught it as Rosen’s subsequence theorem.I thank you for informing me . – Vinay Karthik Mar 12 '24 at 05:25

1 Answers1

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Split all the numbers in groups of $n$ numbers and then make the following sequence (each string is one group):

$$n^2-n+1, n^2-n+2,…,n^2,$$ $$…,$$ $$2n+1,2n+2,…,3n,$$ $$n+1,n+2,…,2n,$$ $$1,2,…,n.$$

Every strictly increasing subsequence is of maximal length $n$, since for every number in the sequence all the numbers that are greater than it are in the same group. If you want to make a strictly decreasing subsequence, you have to take at most one number from each group, so again the length is at most $n$.

Aig
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