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

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.

Choose the sequenceEvery even index is a peak of a sequence $(a_n)$, since its value is $1$ and all later values are at most $1$. Every odd index is a peak of a sequence $(-a_n)$ for the same reason. Both sets of indices are infinite.

Choose the sequenceAfter every index there are positive terms arbitrarily large and negative terms arbitrarily small. Thus every term has both a larger later term and a smaller later term. The former excludes a peak of a sequence $(b_n)$, and the latter excludes a peak of a sequence $(-b_n)$.

Choose $N$ beyond every peak of a sequence of both $(a_n)$ and $(-a_n)$. Every term with index at least $N$ has a strictly larger later term and a strictly smaller later term.

Suppose for a contradiction that $a_n\to L$. Starting at $a_N$, recursively select a strictly increasing subsequence and a strictly decreasing subsequence. Both must have limit of a sequence $L$. If their second terms are $u>a_N$ and $v<a_N$, respectively, passage to their limits of sequences givesa contradiction. Hence $(a_n)$ does not converge. This argument includes the possibility that either set of peaks of a sequence is empty.

Enumerate both sets of peaks of a sequence in increasing order of index. If $a_n\to L$, the two selected subsequences both tend to $L$.

For the converse, suppose $a_{k_r}\to L$ and $a_{m_r}\to L$. Given $\epsilon >0$, choose $r$ large enough that both selected terms differ from $L$ by less than $\epsilon$. For every $n>\max \{k_r,m_r\}$, the two peak of a sequence conditions implyThe lower bound comes from the peak of a sequence of $(-a_n)$ at $m_r$. This is precisely the definition of $a_n\to L$. Thus convergence is equivalent to a common finite limit of a sequence for the upper and lower peak subsequences.