The Subsequence Criterion for Convergence in a Metric Space

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

Then the following hold.

1. (A subsequence of a subsequence)¶ Let $(n_{k})_{k\in\mathbb{N}}$ be a strictly increasing sequence in $\mathbb{N}$. Then $n_{k}<n_{l}$ for all $k,l\in\mathbb{N}$ with $k<l$. Consequently, if $(k_{l})_{l\in\mathbb{N}}$ is a strictly increasing sequence in $\mathbb{N}$, then $(n_{k_{l}})_{l\in\mathbb{N}}$ is strictly increasing, so every subsequence of a subsequence of $(x_{m})_{m\in\mathbb{N}}$ is itself a subsequence of $(x_{m})_{m\in\mathbb{N}}$.

2. (The subsequence criterion)¶ Suppose that for every strictly increasing sequence $(n_{k})_{k\in\mathbb{N}}$ in $\mathbb{N}$ there is a strictly increasing sequence $(k_{l})_{l\in\mathbb{N}}$ in $\mathbb{N}$ such that $(x_{n_{k_{l}}})_{l\in\mathbb{N}}$ converges to $x$ in $X$; that is, every subsequence of $(x_{m})_{m\in\mathbb{N}}$ has in turn a subsequence converging to $x$. Then $(x_{m})_{m\in\mathbb{N}}$ converges to $x$ in $X$.