Peak Point Lemma

Theorem
Let $$\left \langle {x_n} \right \rangle$$ be a sequence in $\R$.

Then $$\left \langle {x_n} \right \rangle$$ has a subsequence which is monotone.

Proof

 * First, suppose that every set $$\left\{{x_n: n > N}\right\}$$ has a maximum.

If this is the case, we can find a sequence $$n_r \in \N$$ such that:


 * $$x_{n_1} = \max_{n > 1} \left({x_n}\right)$$
 * $$x_{n_2} = \max_{n > n_1} \left({x_n}\right)$$
 * $$x_{n_3} = \max_{n > n_2} \left({x_n}\right)$$

and so on.

Obviously, $$n_1 < n_2 < n_3 < \cdots$$, so at each stage we are taking the maximum of a smaller set than at the previous one. And at each stage, the previous maximum has already been taken as the previous element in the sequence.

Thus, $$\left \langle {x_{n_r}} \right \rangle$$ is a decreasing subsequence of $$\left \langle {x_n} \right \rangle$$.


 * Second, suppose it is not true that every set $$\left\{{x_n: n > N}\right\}$$ has a maximum.

Then there will exist some $$N_1$$ such that $$\left\{{x_n: n > N_1}\right\}$$ has no maximum.

So it follows that, given some $$x_m$$ with $$m > N_1$$, we can find an $$x_n$$ following that $$x_m$$ such that $$x_n > x_m$$.

(Otherwise, the biggest of $$x_{N_1+1}, \ldots, x_m$$ would be a maximum for $$\left\{{x_n: n > N_1}\right\}$$.)

So, we define $$x_{n_1} = x_{N_1+1}$$.

Then $$x_{n_2}$$ can be the first term after $$x_{n_1}$$ such that $$x_{n_2} > x_{n_1}$$.

Then $$x_{n_3}$$ can be the first term after $$x_{n_2}$$ such that $$x_{n_3} > x_{n_2}$$.

And so on.

Thus we get an increasing subsequence of $$\left \langle {x_n} \right \rangle$$.


 * There are only these two possibilities, and from both we get a subsequence that is monotone: one is decreasing and one is increasing.

We can of course choose to investigate whether every set $$\left\{{x_n: n > N}\right\}$$ has a minimum. The same conclusion will be reached.