Subsequences

A subsequence of a sequence $\{a_n\}_{n=0}^{\infty}$ is a sequence formed by deleting elements of the $a_n$ to produce a new $a_n$. This subsequence is usually written as $a_{n_k}$, where $\{n_k\}_{k=0}^{\infty}$ is an increasing sequence of positive integers.

Because all the terms of a subsequence are also terms of the original sequence, the properties of the subsequence are closely tied to the properties of the sequence, and so a great deal of information about one can be determined by studying the other. Sequences are objects of central importance in analysis, but they can be difficult to work with, and their simpler subsequences provide a useful tool for understanding them.

The convergence of a sequence can be characterized in terms of the convergence of its subsequences.

A sequence converges to a limit $x$ if and only if every subsequence also converges to the limit $x$.

For one direction, suppose that $a_n\to x$, and consider some subsequence $a_{n_k}$. Because $a_n\to x$, for every $\epsilon>0$, there is $N>0$ such that $$n>N\implies |a_n-x|<\epsilon.$$ Then, for $a_{n_k}$, there must be some $K$ such that $k>K\implies n_k>N,$ because the $n_k$ are increasing. But if $n_k>N,$ then $|a_{n_k}-x|<\epsilon,$ so $a_{n_k}\to x$ as well. Thus any subsequence converges as well.

For the other direction, suppose that every subsequence of $a_{n_k}$ converges, and suppose for the sake of contradiction that $a_n$ does not converge. Then, for every $i>0$, there is an arbitrarily large $a_n$ such that $|a_n-x|>\frac1i$ (if this were not true, then the sequence would converge). Then, taking the $a_n$ that satisfy this for each $i$, choosing them to be increasing, gives a subsequence that does not converge; contradiction. $_\square$

Does the sequence $a_n=\frac{1}{3n+1}$ converge?

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Yes. This is a subsequence of the convergent sequence $a_n=\frac{1}{n}$, so it must converge. $_\square$

A very important theorem regarding the convergence of subsequences of sequences in $\mathbb{R}$ is the Bolzano-Weierstrass theorem.

Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

We claim that every sequence in $\mathbb{R}$ has a monotone subsequence. Because every bounded monotone sequence converges, the result follows.

Consider a sequence $a_n$, and call a term $a_k$ a peak if it is larger than every subsequent term in the sequence. Now, if a sequence has infinitely many peaks, the subsequence corresponding to the peaks is monotonically decreasing. If there are only finitely many peaks, suppose that the last peak is at $N$. Then, the $N+1$ term is not a peak, so there is a term after it larger than it. That term is also not a peak, so there is a term after that larger than it. In this way, a monotonically increasing subsequence can be constructed.

Either way, the sequence has a monotone subsequence, and this subsequence will converge, proving the result. $_\square$

The Bolzano-Weierstrass theorem can be used to give proofs of the Ascoli-Arzela theorem, as well as be applied to various problems in economics.