Every sequence of $n^2+1$ distinct real numbers contains a subsequence of length $n + 1$ that is either strictly increasing or strictly decreasing.

Question

I have been studying Pigeonhole Principle and came across this theorem that is mentioned in the Discrete Mathematics book by Rosen. I understand the theorem and understand the example presented along with the same. But I am slightly confused by the what I understand to be the limitations in applicability of this theorem.

For example what about sequences that are not $n^2+1$ elements long, say sequences of length 11? While the theorem can be applied to sequences of length 10 by taking $n$ as 3 ($3^2+1=10$), I cannot find any $n$ such that $n^2+1=11$. Does that mean the theorem is not applicable in such cases? Does that mean there is no strictly increasing (or decreasing) subsequence of any particular length for any such sequences?

I read the theorem over and over again but it does seem like that the theorem is applicable to a very limited set of sequences i.e. sequences whose length can be written in form $n^2+1$. I was wondering if there was any possible relation between sequences of lengths that are not $n^2+1$ (say 7, 8, 9, 11 or 12) and the length of strictly increasing/decreasing sub-sequences that can be observed in such sequences.

Answer 1

Take a sequence of length $11$ and remove one element of your choice. The theorem guarantees that the resulting sequence contains a sequence of $4$ either strictly increasing or strictly decreasing numbers. This subsequence is obviously present also in the original sequence of length $11$.

In general, find the largest $n$ such that $n^2+1$ is less than or equal to the length of your sequence. Then you can apply the pigeonholing theorem.

Answer 2

For the future readers, here's a proof of the theorem with the Pigeonhole principle.

Let $\{a_1, a_2, \ldots, a_{n^2+1}\}$ be a sequence of distinct real numbers. Given every $a_k$, define a tuple $(i_k, d_k)$ where $i_k$ is the length of a longest increasing subsequence starting at $a_k$, and $d_k$ is the length of a longest decreasing subsequence starting at $a_k$.

If either $i_k \geq n+1$ or $d_k \geq n+1$ for some $k$, then we are done (as we would get a subsequence of length $n+1$ which is either strictly increasing or strictly decreasing. Note that the strictness on monotonicity is because the elements are distinct).

Otherwise, suppose $i_k \leq n$ and $d_k \leq n, \; \forall \; 1 \leq k \leq n^2+1$.

Now consider the set $\mathcal{A} = \{(i_k, d_k): 1 \leq k \leq n^2+1\}$

For $i_k$, we have less than or equal to $n$ choices (as $i_k \leq n$ by assumption), and same for $d_k$. So, the cardinality of $\mathcal{A}$ is at most $n^2$. That is, $|\mathcal{A}| \leq n^2$.

However, the set $\{(i_1, d_1), (i_2, d_2), \ldots, (i_{n^2+1}, d_{n^2+1})\}$ has $n^2+1$ elements.

Hence, by the pigeonhole principle, there exists $t,s$ such that $(i_t,d_t) = (i_s,d_s)$ for $ t\neq s$. Assume without loss of generality that $s < t$.

As $a_s \neq a_t$, we have that either $a_s < a_t$ or $a_s > a_t$.

• If $a_s < a_t$, then $i_s = i_t + 1$ ($a_s$ comes before $a_t$ and is smaller, so the length of the longest increasing subsequence starting from $a_s$ will be one more than that starting from $a_t$, since the former includes $a_s$. But this contradicts that $i_s = i_t)$.

• If $a_s > a_t$, then $d_s = d_t + 1$ (which contradicts $d_s = d_t$).

So, either $i_k \geq n+1$ or $d_k \geq n+1$ for some $k \in \{1, \ldots, n^2+1\}$. $\; \blacksquare$