On Sequences, Sub-Sequences, and Sub-Sub-Sequences
Convergence criteria
There is a simple result about sequences that is both easy to state and surprisingly powerful in proofs.
Let \((x_n)\) be a sequence in a metric space, or more generally in any first-countable topological space. Then for a given point \(x\), \[ \begin{aligned} x_n \to x \Longleftrightarrow & \ \text{every subsequence } (x_{n_k}) \text{ has} \\ &\ \text{a further subsequence } (x_{n_{k_j}}) \to x. \end{aligned} \] The forward implication is immediate. If the whole sequence converges to \(x\), then every subsequence also converges to \(x\), and therefore every sub-subsequence does too.
The converse is the useful part. It says that to prove convergence of the full sequence, we do not need to show that every subsequence already converges. It is enough to show that every subsequence contains some further sub-subsequence that converges to the same limit \(x\). In practice, that is exactly what compactness arguments often give us.
Why the limit must be the same \(x\)
The phrase “to the same limit \(x\)” is the key. If we weaken the statement to \[ \text{every subsequence has a further convergent sub-subsequence,} \] then the conclusion is false. The alternating sequence \[ x_n = (-1)^n \] is the standard counterexample. Every subsequence contains a further constant-sign sub-subsequence, hence a convergent one, but the full sequence does not converge.
The weaker statement, without “the same \(x\)”, expresses a compactness-type property, not convergence. It says that subsequences cannot escape all accumulation points, but it does not say that there is only one accumulation point.
The stronger formulation, with the same \(x\), forces all roads back to the same \(x\), and that is exactly what turns compactness information into actual convergence.
Why “every subsequence converges to \(x\)” is too strong
There is a different statement one might first guess: \[ \text{every subsequence } (x_{n_k}) \to x. \] That certainly implies \(x_n \to x\), but almost vacuously so, because the original sequence is a subsequence of itself. Even if we insist on proper subsequences only, the statement still implies convergence: if the original sequence failed to converge to \(x\), then infinitely many terms would stay outside some neighborhood of \(x\), and those terms would form a proper subsequence that also fails to converge to \(x\).
The point of the theorem is precisely that the weaker-looking condition with sub-subsequences is already enough. That is why the formulation matters.
Proof
The proof of the nontrivial implication is short and easy.
Assume that every subsequence \((x_{n_k})\) has a further subsequence \((x_{n_{k_j}})\) converging to \(x\). We want to prove \(x_n \to x\).
Suppose not. Then there is some neighborhood \(U\) of \(x\) such that infinitely many terms of the sequence lie outside \(U\). Choose those terms to form a subsequence \((x_{n_k})\) with \(x_{n_k} \notin U\) for every \(k\). But then no further subsequence of \((x_{n_k})\) can converge to \(x\), because any sequence converging to \(x\) must eventually lie in every neighborhood of \(x\), in particular in \(U\). That contradiction proves \(x_n \to x\).
The proof shows the mechanism very clearly: failure of convergence produces a bad subsequence that stays away from \(x\), and such a subsequence cannot contain any sub-subsequence converging to \(x\).
Applications
compactness plus uniqueness of subsequential limits gives convergence of the full sequence.
The theorem is commonly used in proofs by:
- start with an arbitrary subsequence \((x_{n_k})\);
- use compactness, boundedness, or some extraction principle to obtain a further convergent sub-subsequence \((x_{n_{k_j}})\);
- prove that any such convergent sub-subsequence must converge to \(x\);
- conclude that the original sequence \((x_n)\) converges to \(x\).
That pattern appears everywhere: weak convergence, compactness arguments in PDE, probabilistic tightness arguments, and many lower-semicontinuity proofs.
When sequences are not enough
In metric spaces, sequences do a remarkable amount of work. They characterize convergence, and they also detect closure: a point \(x\) lies in the closure of a set \(A\) if and only if there is a sequence in \(A\) converging to \(x\).
That fact fails in general topological spaces. There are spaces where \(x \in \overline{A}\), but no sequence in \(A\) converges to \(x\). The problem is that a sequence has only countably many stages, while the neighborhood structure around a point may be too large or too branched to be captured by a countable progression.
Nets are introduced to solve these problems. Nets are discussed in a companion post.