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Claim: every sequence of $m^2 + 1$ distinct numbers has an increasing or decreasing subsequence of length $m+1$.

Proof:

Idea

Associate every position $ 1 <= l <= m^2 +1 $ with a pair of lengths of longest increasing (decreasing) subsequences $(i_{l}, d_{l})$ beginning at position $l$.

==> Assume claim is false.

==> There are at most $ m^2 $ pairs $(i_{l}, d_{l})$ that can occur

==> $m^2+1 - m^2 = 1$ ==> $1$ pair two indices $s < t$ with an equal $(i,d)$ pair.

==> Since numbers are distinct. 2 cases If $a_s < a_t$: Then longer subsequence $ i_{t} $ can be built with sequence up to $s + a_s + a_t$ sequence contradicting that pairs are equal.

==> Analogous for $a_t < a_s$

==> Claim is true

My problem is seeing how one goes from Claim is false to There are at most $m^2$ pairs that can occur.

Prajwal Kansakar
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Mathflunkie
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1 Answers1

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This actually follows from a more general result by the great Erdos: https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Szekeres_theorem.

But I'll just answer the part you're confused about in the question. So, we assume the claim is false. What's the claim? That there is an increasing or decreasing sequence of size $m+1$. Equivalently, the claim says that there is an increasing or decreasing subsequence of at least length $m+1$, for if there were a longer one, we could always make one of exactly length $m+1$ by taking a subsequence of that. This in turn relies on the fact that the subsequence relation is transitive i.e. if $S$ is a susequence of $R$ which is a subsequence of $U$ then $S$ is a subsequence of $U$.

So, negating the claim says that every subsequence is of length less than $m+1$, which equivalent to saying each such subsequence is of length less than or equal $m$. Now, this implies that the longest subsequence ending at any point is of length at most $m$, i.e. its length is in the set $\{1, 2, \ldots, m\}$, of which there are $m$ elements. Note, a maximal sequence of length $0$ is impossible because we can always simply choose the number itself as the singleton sequence, so our maximal sequences are all of length at least $1$.

Now, since at each point we have pairs of maximal increasing and decreasing subsequences up to and including that point, and since there are $m$ choices for each, there are at most $m^2$ combinations of these:

$$(1, 1), (1, 2), \ldots, (1, m) \ldots (2, 1), (2, 2), (2, m) \ldots (m, 1), (m, 2), \ldots (m, m)$$

You could also draw an $m$ by $m$ grid to represent all the possible points. Every number in the main sequence will have a pair of points associated to it that will be in this grid, assuming the claim is false. The rest of the proof I assume you can follow? The main other point is to show all pairs must be distinct, which you can prove in a separate lemma. Then combining this with the "fact" that your points must all be in the grid, under the assumption that the claim is false, leads to a contradiction via application of the pigeonhole principle. You then conclude the claim must be true.

Colm Bhandal
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