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A monotone subsequence is a subsequence which is a monotone sequence. The indices always increase strictly, but equal selected values are permitted. The monotone subsequence theorem guarantees one for every sequence of real numbers.

For a sequence of real numbers, a common finite limit of a sequence for all monotone subsequences forces the whole sequence to have that limit of a sequence. There is no need to assume a bounded sequence if the hypothesis already includes convergence of every monotone subsequence.

If $a_n$ fails to tend to $L$, select a subsequence with $|a_{n_j}-L|\ge \epsilon$ for some fixed $\epsilon >0$. The monotone subsequence theorem gives a monotone subsequence of this subsequence, which contradicts the common limit of a sequence $L$. Conversely, every subsequence of a convergent sequence has the original limit of a sequence.

For a bounded sequence, the bounded monotone sequence theorem already ensures that all monotone subsequences converge. Only agreement of their limits of sequences remains to be checked.

Call an index $n$ a peak if $a_n\ge a_m$ for every $m>n$. If there are infinitely many peaks, list their indices increasingly. Each selected value is at least every later selected value, so this gives a nonincreasing subsequence.

If there are only finitely many peaks, choose an index beyond the last one. Every index from then onward has a later index with strictly larger value. Select these recursively: $n_{j+1}>n_j$ with $a_{n_{j+1}}>a_{n_j}$. This gives an increasing subsequence. Neither complex argument requires the original sequence to be bounded. Every real sequence therefore has a monotone subsequence, as asserted by the monotone subsequence theorem.

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

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.

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.

A peak of a sequence of real numbers is an index $k$ whose value is at least every later value. This is a condition on the entire tail, stronger than comparing only adjacent terms. Equality is allowed: every index of a constant sequence is a peak of a sequence.

Infinitely many peaks of a sequence, listed in increasing order of index, give a nonincreasing monotone subsequence. If only finitely many exist, every index beyond the last has a larger later term; repeatedly choosing one gives a strictly increasing monotone subsequence. This proves the monotone subsequence theorem.

A peak of a sequence of $(-a_n)$ supplies a lower bound for all subsequent terms of $(a_n)$. If both $(a_n)$ and $(-a_n)$ have a peak of a sequence, the two selected values bound a common tail. Adding bounds for the finite initial segment proves that $(a_n)$ is a bounded sequence.