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.
Best Answer
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.