Convergent Sequences and Subsequences in Metric Spaces

# Convergent Sequences and Subsequences in Metric Spaces

We will now look at some nice criterion which tells us that in a metric space $(M, d)$, a sequence $(x_n)_{n=1}^{\infty}$ converges to $p$ if and only if every subsequence $(x_{n_k})_{k=1}^{\infty}$ converges to $p$. This is analogous to the similar result when we looked at convergent sequences of real numbers.

 Theorem 1: Let $(M, d)$ be a metric space, let $(x_n)_{n=1}^{\infty}$ be a sequence in $M$, and let $p \in M$. Then $(x_n)_{n=1}^{\infty}$ converges to $p$ if and only if every subsequence $(x_{n_k})_{k=1}^{\infty}$ of $(x_n)_{n=1}^{\infty}$ converges to $p$.
• Proof: $\Rightarrow$ Let $(x_n)_{n=1}^{\infty}$ be a convergent sequence in $M$ that converges to $p$. Then $\lim_{n \to \infty} x_n = p$. So $\lim_{n \to \infty} d(x_n, p) = 0$. Therefore, for all $\epsilon > 0$ there exists an $N = N(\epsilon) \in \mathbb{N}$ such that if $n \geq N$ then:
(1)
\begin{align} \quad d(x_n, p) < \epsilon \end{align}
• Let $(x_{n_k})_{k=1}^{\infty}$ be any subsequence of $(x_n)_{n=1}^{\infty}$. For each $\epsilon > 0$ we have that for some $K \in \mathbb{N}$ that $n_K \geq N(\epsilon)$ and so for all $k \geq K$ we have that $n_k \geq N(\epsilon)$ so by the convergence of $(x_n)_{n=1}^{\infty}$ we have that:
(2)
\begin{align} \quad d(x_{n_k}, p) < \epsilon \end{align}
• So for each $\epsilon > 0$ there exists a $K \in \mathbb{N}$ such that if $k \geq K$ then $d(x_{n_k}, p) < \epsilon$, therefore, the subsequence $(x_{n_k})_{k=1}^{\infty}$ converges to $p$.
• $\Leftarrow$ Suppose that every subsequence $(x_{n_k})_{k=1}^{\infty}$ of $(x_n)_{n=1}^{\infty}$ converges to $p$. Then $(x_n)_{n=1}^{\infty}$ is a subsequence of itself and converges to $p$. $\blacksquare$