TheoremBase

The Subsequence Criterion for Convergence in a Metric Space

A subsequence of a subsequence is a subsequence. If every subsequence of a sequence in a metric space has in turn a subsequence converging to a fixed point, then the whole sequence converges to that point.

Statement

Let (X,d)(X,d) be a metric space, let N\mathbb{N} be the set of natural numbers carrying the addition and the order relations << and ≤\le of those definitions, let (xm)m∈N(x_{m})_{m\in\mathbb{N}} be a sequence in XX and let x∈Xx\in X. That a sequence (nk)k∈N(n_{k})_{k\in\mathbb{N}} in N\mathbb{N} is strictly increasing, and that (xnk)k∈N(x_{n_{k}})_{k\in\mathbb{N}} is then a subsequence of (xm)m∈N(x_{m})_{m\in\mathbb{N}}, are as defined there. Convergence in XX is convergence in (X,d)(X,d).

Then the following hold.

1. (A subsequence of a subsequence) Let (nk)k∈N(n_{k})_{k\in\mathbb{N}} be a strictly increasing sequence in N\mathbb{N}. Then nk<nln_{k}<n_{l} for all k,l∈Nk,l\in\mathbb{N} with k<lk<l. Consequently, if (kl)l∈N(k_{l})_{l\in\mathbb{N}} is a strictly increasing sequence in N\mathbb{N}, then (nkl)l∈N(n_{k_{l}})_{l\in\mathbb{N}} is strictly increasing, so every subsequence of a subsequence of (xm)m∈N(x_{m})_{m\in\mathbb{N}} is itself a subsequence of (xm)m∈N(x_{m})_{m\in\mathbb{N}}.

2. (The subsequence criterion) Suppose that for every strictly increasing sequence (nk)k∈N(n_{k})_{k\in\mathbb{N}} in N\mathbb{N} there is a strictly increasing sequence (kl)l∈N(k_{l})_{l\in\mathbb{N}} in N\mathbb{N} such that (xnkl)l∈N(x_{n_{k_{l}}})_{l\in\mathbb{N}} converges to xx in XX; that is, every subsequence of (xm)m∈N(x_{m})_{m\in\mathbb{N}} has in turn a subsequence converging to xx. Then (xm)m∈N(x_{m})_{m\in\mathbb{N}} converges to xx in XX.

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