Complete Metric Spaces

# Complete Metric Spaces

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

Consider any metric space $(M, d)$. If $(M, d)$ is such that every Cauchy sequence converges in $M$, then we give this metric space a special name.

 Definition: Let $(M, d)$ be a metric space. Then $(M, d)$ is said to be Complete if every Cauchy sequence converges in $M$.

In general, it is much easier to show that a metric space is not complete by finding a Cauchy sequence that does not converge in the space. For example, consider the set $M = [0, 1)$ with the standard Euclidean metric $d(x, y) = \mid x - y \mid$ defined for all $x, y \in M$. Then $(M, d)$ is a metric space. Now, consider the following sequence in $M$:

(1)
\begin{align} \quad (x_n)_{n=1}^{\infty} = \left ( 1 - \frac{1}{n} \right )_{n=1}^{\infty} = \left ( 0, \frac{1}{2}, \frac{2}{3}, ... \right ) \end{align}

We claim that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence. Let's prove this. Let $m, n \in \mathbb{N}$, and consider:

(2)
\begin{align} \quad d(x_m, x_n) = \biggr \lvert \left ( 1 - \frac{1}{m} \right ) - \left ( 1 - \frac{1}{n} \right ) \biggr \rvert = \biggr \lvert \frac{1}{n} - \frac{1}{m} \biggr \rvert \leq \biggr \lvert \frac{1}{n} \biggr \rvert + \biggr \lvert \frac{1}{m} \biggr \rvert = \frac{1}{n} + \frac{1}{m} \end{align}

Choose $N$ such that $N > \frac{2}{\epsilon}$. Then if $m, n \in \mathbb{N}$ are such that $m, n, \geq N$ then:

(3)
Hence for all $m, n \geq N$ we have that:
So $(x_n)_{n=1}^{\infty}$ is indeed a Cauchy sequence. However, it should be intuitively clear that $\lim_{n \to \infty} x_n = 1$, but $1 \not \in [0, 1)$! Therefore $(x_n)_{n=1}^{\infty}$ does not converge in $M$ and hence $(M, d)$ is not a complete metric space.