Monotone subsequence theorem

Every real sequence has a monotone subsequence, without a boundedness assumption. Call an index a peak when its term is at least every later term. Infinitely many peaks give a nonincreasing subsequence; beyond finitely many peaks each term has a larger later term, which recursively gives an increasing subsequence.

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.

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.

Suppose $k_1<k_2<\cdots$ are peaks of a sequence $(a_n)$ and $m_1<m_2<\cdots$ are peaks of a sequence $(-a_n)$. Then $(a_n)$ converges precisely when these two subsequences have a common finite limit of a sequence $L$.

Necessity follows because every subsequence inherits the limit of a sequence of a convergent sequence. For sufficiency, choose one selected value from each subsequence within $\epsilon$ of $L$. Every term after both selected indices lies between the two selected values:This is exactly $a_n\to L$.

Suppose both $(a_n)$ and $(-a_n)$ have only finitely many peaks of a sequence. Beyond some index $N$, every term has a strictly larger later term and a strictly smaller later term. Starting from $a_N$, construct one strictly increasing and one strictly decreasing monotone subsequence.

If the original sequence had limit of a sequence $L$, both selected subsequences would tend to $L$. The increasing one has a second term strictly above $a_N$, so $L>a_N$; the decreasing one similarly forces $L<a_N$. This contradiction proves nonconvergence. For example, $a_n=(-1{)}^{n}n$ has no peaks of a sequence in either sign and diverges.