The Boundedness of Convergent Sequences in Metric Spaces

# The Boundedness of Convergent Sequences in Metric Spaces

Recall from The Uniqueness of Limits of Sequences in Metric Spaces page that if $(M, d)$ is a metric space and $(x_n)_{n=1}^{\infty}$ is a sequence in $M$ that is convergent then the limit of this sequence $p \in M$ is unique.

We will now look at another rather nice theorem which states that if $(x_n)_{n=1}^{\infty}$ is convergent then it is also bounded.

 Theorem 1: Let $(M, d)$ be a metric space and let $(x_n)_{n=1}^{\infty}$ be a sequence in $M$. If $(x_n)_{n=1}^{\infty}$ is convergent then the set $\{ x_1, x_2, ..., x_n, ... \}$ is bounded.
• Proof: Let $(M, d)$ be a metric space and let $(x_n)_{n=1}^{\infty}$ be a sequence in $M$ that converges to $p \in M$, i.e., $\lim_{n \to \infty} x_n = p$. Then $\lim_{n \to \infty} d(x_n, p) = 0$. So for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $d(x_n, p) < \epsilon$. So for $\epsilon_0 = 1 > 0$ there exists an $N(\epsilon_0) \in \mathbb{N}$ such that if $n \geq N(\epsilon_0)$ then:
(1)
\begin{align} \quad d(x_n, p) < \epsilon_0 = 1 \end{align}
• Now consider the elements $x_1, x_2, ..., x_{N(\epsilon_0)}$. This is a finite set of elements and furthermore the set of distances from these elements to $p$ is finite:
(2)
\begin{align} \quad \{ d(x_1, p), d(x_2, p), ..., d(x_{N(\epsilon_0) - 1}, p) \} \end{align}
• Define $M ^*$ to be the maximum of these distances:
(3)
\begin{align} \quad M^* = \max \{ d(x_1, p), d(x_2, p), ..., d(x_{N(\epsilon_0) - 1}, p) \} \end{align}
• So if $1 \leq n < N(\epsilon_0)$ we have that $d(x_n, p) \leq M^*$ and if $n \geq N(\epsilon_0)$ then $d(x_n, p) < 1$. Let $M = \max \{ M^*, 1 \}$. Then for all $n \in \mathbb{N}$, $d(x_n, p) \leq M$. So consider the open ball $B(p, M + 1)$. Then $x_n \in B(p, M + 1)$ for all $n \in \{1, 2, ... \}$ so:
(4)
\begin{align} \quad \{ x_1, x_2, ..., x_n, ... \} \subseteq B(p, M+1) \end{align}
• Therefore $\{ x_1, x_2, ..., x_n, ... \}$ is a bounded set in $M$. $\blacksquare$