Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics II / 1 / a

Call $k$ a peak of a sequence when $a_k\ge a_m$ for every $m>k$. If there are infinitely many peaks of a sequence, list their indices in increasing order, $k_1<k_2<\cdots$. Since $k_{r+1}>k_r$,This is a decreasing subsequence; throughout this argument, a decreasing monotone sequence may have equal successive terms.

If there are only finitely many peaks of a sequence, choose $N$ beyond all their indices; take $N=1$ if there are none. Every $n\ge N$ fails to be a peak of a sequence, so some $m>n$ satisfies $a_m>a_n$. Starting with $n_1=N$, choose such an $n_{r+1}>n_r$ recursively. This givesThe two cases prove the monotone subsequence theorem without any boundedness hypothesis.

The Bolzano-Weierstrass theorem states: every bounded sequence of real numbers has a convergent subsequence.

By the monotone subsequence theorem proved in the preceding part, a bounded sequence $(a_n)$ has a monotone subsequence $(b_r)$. It inherits the bounds of $(a_n)$. For completeness, the bounded monotone sequence theorem follows directly from completeness of the real numbers. If $(b_r)$ is increasing, let $L=\sup \{b_r:r\ge 1\}$. For every $\epsilon >0$, the definition of supremum supplies $r_0$ with $L-\epsilon <b_{r_0}\le L$. Thenso $b_r\to L$. If $(b_r)$ is decreasing, apply the increasing case to $(-b_r)$, or use its infimum. Thus in either case

It is useful to prove the contrapositive. Suppose $k$ is a peak of a sequence $(a_n)$ and $m$ is a peak of a sequence $(-a_n)$. The second condition says $a_j\ge a_m$ for every $j>m$. HenceOnly finitely many earlier terms remain, and their absolute values have a finite maximum. Together these give a bound for every term, so $(a_n)$ is a bounded sequence. Consequently,

If $a_n\to L$, every subsequence also has limit of a sequence $L$: its increasing indices eventually exceed any prescribed threshold. This applies in particular to every monotone subsequence.

Conversely, suppose $(a_n)$ is a bounded sequence and all its monotone subsequences converge to the same $L$. At least one exists by the monotone subsequence theorem. If $a_n$ did not tend to $L$, there would be ${\epsilon }_0>0$ and a subsequence $(a_{n_j})$ such thatApply the monotone subsequence theorem to $(a_{n_j})$. The resulting further subsequence is also a monotone subsequence of $(a_n)$, so it tends to $L$, contradicting the displayed bound. ThereforeThe bounded monotone sequence theorem ensures that these subsequences have finite limits of sequences; the hypothesis requires those values to agree.