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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.

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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.

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First I think that in the definition of a peak the inequality is strict (you will see why we need this in the proof). But even if it isn't here are some examples:

Let $a_n=\frac 1n$. This sequence has infinitely many peaks (at $1,2,...$). The monotone subsequence here is any subsequence of $a_n$. Another example (supposing the inequality is indeed non strict as you say): Let $a_n=(-1)^n$. This sequence has peaks at $0,2,...$. The monotone subsequence here is $a_{2n}=(-1)^{2n}=1$.

Let $b_n=\frac 1n+2013$, $n\le 7$ and $b_n=(1+\frac 1n)^n$, $n>7$. The peaks are $n=1,2,3,4,5,6,7$ and the monotone subsequence is $b_{n+7}=(1+\frac 1{n+7})^{n+7}$.

All the sequences above are bounded. Providing examples of peaks for unbounded sequences is equally simple

The notion of a peak is used only in the proof of the Bolzano Weierstrass Theorem (I myself hasn't seen it used anywhere else, but if anyone has, please comment on this). That's what makes the proof so magical. I don't understand why you say "Further by explaining using peaks, did we not already assume the sequence as monotone decreasing"?

For completeness here is the proof:

Suppose that $\left(a_n\right)$ has an infinite number of peaks $k_0<k_1<k_2<...<k_n<...$ and consider the subsequence $(a_{k_n})$. Then, $(a_{k_n})$ is strictly decreasing since $k_n>k_m\Rightarrow a_{k_n}<a_{k_m}$ and thus $(a_{k_n})$ is monotone.

Suppose that $\left(a_n\right)$ has an finite number of peaks and let $N\in \mathbb{N}$ be the last (greatest) peak. Then $k_{0}=N+1$ is not a peak and so \begin{equation}\exists k_1>k_0\in \mathbb{N}:a_{k_{1}}>a_{k_{0}}\end{equation} Having defined $k_n$ such as that $k_n>k_{n-1}>N$ then \begin{equation}\exists k_{n+1}>k_n\in \mathbb{N}:a_{k_{n+1}}>a_{k_{n}}\end{equation} The subsequence $(a_{k_n})$ is obviously increasing and so it is monotone

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The inequality is not strict. If it were, $a_n=(-1)^n$ would be a counterexample to the theorem. –  Brian M. Scott Sep 26 '13 at 19:32

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