While proving every sequence of real numbers has a monotone subsequence, we take two cases, either there are infinitely many "peaks" or else "finitely many" peaks. However, I am unable to grasp from mere definition what "peaks" mean. Though I know that they are defined as $a_i$ is peak if $a_i\le a_j$ for all $i> j$.
Is it related to infimum? It does not make sense to have infinitely or even finitely many "peaks". Further by explaining using peaks, did we not already assume the sequence as monotone decreasing?
I would appreciate if you can illustrate sequences where there are finitely many peaks and infinitely many peaks. and illustrate with them how they are monotone subsequence.
Best Answer
You got the definiion of peak slightly wrong. We say that the sequence $(a_n)_{n\in\mathbb N}$ has a peak at index $i\in\mathbb N$, if none of the subsequent members exceeds the value there, i.e. if $a_j\le a_i$ for all $j>i$. We do not exclude, however, that earlier members exceed it.
With this definition it should be clear that infinitely many peaks form a (not necessarily strictly) decreasing subsequence; and if there are only finitely many peaks one can always find (for all positions after the last peak) a next sequence member that exceeds the current one, i.e. one finds a (strictly) increasing subsequence.
An example with infinitely many peaks: $$ 1, -1, \frac12, -\frac12, \frac13,-\frac13,\frac14,-\frac14,\ldots$$ Here all positive membes are in fact peaks because all later members are smaller.
An example with finitely many peaks: $$ 42, 17, 33, 0, -\frac12,-\frac13,-\frac14,-\frac15,-\frac15,-\frac16,\ldots$$ Here, the $42$, the $33$ and the $0$ are peaks.