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I am trying to prove this theorem "every sequence has a monotone subsequence"

I found this proof

Proof: Let us call a positive integer $n$ a peak of the sequence if $m > n \implies x_n > x_m$  i.e., if  $x_n$ is greater than every subsequent term in the sequence.

Suppose first that the sequence has infinitely many peaks, $n_1 < n_2 < n_3 < … < n_j < …$. Then the subsequence $\{x_{n_j}\}_j$  corresponding to these peaks is monotonically decreasing, and we are done.

So suppose now that there are only finitely many peaks, let $N$ be the last peak and set $n_1 = N + 1$.

Then $n_1$ is not a peak, since $n_1 > N$, which implies the existence of an $n_2 > n_1$ with $x_{n_2} \geq x_{n_1}.$  Again, as $n_2 > N$ it is not a peak, hence there is $n_3 > n_2$ with $x_{n_3} \geq x_{n_2}.$  Repeating this process leads to an infinite non-decreasing subsequence $x_{n_1} \leq x_{n_2} \leq x_{n_3} \leq \ldots$ as desired.

My question is about this part "So suppose now that there are only finitely many peaks"

is $0$ peak valid? for example the sequence $\{\dfrac{n}{n+1}\}$ has no peaks.

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  • $\begingroup$ Thank you, the first term is 1/2 , the second term is 2/3, therefore $x_n$ < $x_m$, the first term is not peak $\endgroup$ – Jose Vega Mar 20 '16 at 20:45
  • $\begingroup$ What does it mean for a sequence to have no peaks? $\endgroup$ – Friedrich Philipp Mar 20 '16 at 20:48
  • $\begingroup$ I understand that: The sequence is a increasing sequence $\endgroup$ – Jose Vega Mar 20 '16 at 20:50
  • $\begingroup$ No, you don't because it's wrong. ;-) $\endgroup$ – Friedrich Philipp Mar 20 '16 at 20:52
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Zero peak is valid and should be addressed. If a sequence has no peaks, then for every $n \in \mathbb{N}$, there exists $m > n$ such that $x_n \leq x_m$. Applying this to $n=1$ gives $n_2 > 1$, such that $x_{n_2} \geq x_1$. Applying to $n=n_2$ gives $n_3 > n_2$ such that $x_{n_3} \geq x_{n_2}$. In this way we get an increasing sequence $\left\{ n_k \right\}$ (here $n_1 = 1$), such that $\left\{ x_{n_k} \right\}$ is monotone.


EDIT: It looks like this is equivalent to setting $n_1=1$ if there are no peaks, and applying the same logic as the proof you wrote.

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    $\begingroup$ It doesn't need to be dealt with separately, you can just set $N=-1$ (assuming your sequences start with $n=0$) and the proof works just fine. $\endgroup$ – Najib Idrissi Mar 20 '16 at 20:59
  • $\begingroup$ Yes, thank you :) $\endgroup$ – Michael Harrison Mar 20 '16 at 21:00
  • $\begingroup$ @NajibIdrissi But if we set N = -1, wouldn't this mean that the last peak is -1? How does this make sense? $\endgroup$ – user825007 Sep 16 '20 at 2:07
  • $\begingroup$ Instead of thinking that the value of N means "the last peak occurs here", think that it means "no peaks occur after this index". Setting the value N=-1 when there are no peaks is just a convenient way to avoid splitting the case of zero peaks into a separate thing. $\endgroup$ – Michael Harrison Sep 16 '20 at 2:48
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Consider any sequence $(x_n).$

A "peak" element is an element such that all elements $x_i$ that succeed it in the sequence are no greater than it. In other words, an element $x_p$ is a "peak" element if for all $i > p_1,$ it is the case that $x_{p_1} \geq x_i.$

In the sequence $(x_n),$ either there are finitely many peaks, or there are infinitely many.

If there are finitely many, then there is a peak element $x_p$ that appears last. Then, for all $i \in \mathbb{N}$ such that $i > p,$ there exists some $M \in \mathbb{N}$ such that $M > i$ and $x_M > x_i.$ In such a case, there clearly exists a monotone increasing subsequence $(x_{n_i}).$

If there are infinitely many peaks, then there clearly exists a monotone decreasing subsequence,

$$x_{p_1}, x_{p_2}, x_{p_3}, \ldots$$

where $x_{p_1} \geq x_{p_2} \geq x_{p_3} \geq \cdots$ are all peak elements.

Hence, any sequence has some subsequence that is monotone, whether increasing or decreasing.

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