The Boundedness of Cauchy Sequences in Metric Spaces

# The Boundedness of Cauchy Sequences in Metric Spaces

Recall from the Cauchy Sequences in Metric Spaces page that if $(M, d)$ is a metric space then a sequence $(x_n)_{n=1}^{\infty}$ is said to be a Cauchy sequence if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $d(x_m, x_n) < \epsilon$.

We have already seen that if $(x_n)_{n=1}^{\infty}$ is a convergent sequence in $M$ then it is also Cauchy. We will now look at an important result which will say that if $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence then it is bounded.

 Theorem 1: Let $(M, d)$ be a metric space. If $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence then $(x_n)_{n=1}^{\infty}$ is also bounded.
• Proof: Let $(x_n)_{n=1}^{\infty}$ be a Cauchy sequence. Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $d(x_m, x_n) < \epsilon$
• Then for $\epsilon_1 = 1 > 0$ we have that there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $d(x_m, x_n) < \epsilon_1 = 1$.
• Suppose that $m, n < N$. Then the set $\{ d(x_m, x_n) : 1 \leq m \leq N, 1 \leq n \leq N \}$ is a finite set with $N^2$ elements and hence a maximum exists. Let $M = \max \{ d(x_m, x_n) : 1 \leq m \leq N, 1 \leq n < N \}$. Now if $m < N$ and $n \geq N$ then we have by the Triangle inequality that:
(1)
\begin{align} \quad d(x_m, x_n) \leq d(x_m, x_{N}) + d(x_{N}, x_n) < M + 1 \end{align}
• Hence we see that for all cases, if $m, n \in \mathbb{N}$ then $d(x_m, x_n) < M + 1$, so $(x_n)_{n=1}^{\infty}$
• So the ball $B(x_m, M+1)$ encapsulates all terms in the sequence $(x_n)_{n=1}^{\infty}$, so $(x_n)_{n=1}^{\infty}$ is bounded. $\blacksquare$