Complete Metric Spaces

Recall 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 $M$:

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

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

${\begin{aligned}\quad d(x_{m},x_{n})=\left|\left(1-{\frac {1}{m}}\right)-\left(1-{\frac {1}{n}}\right)\right|=\left|{\frac {1}{n}}-{\frac {1}{m}}\right|\leq \left|{\frac {1}{n}}\right|+\left|{\frac {1}{m}}\right|={\frac {1}{n}}+{\frac {1}{m}}\end{aligned}}$

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

${\begin{aligned}\quad m\geq N>{\frac {2}{\epsilon }}\quad \mathrm {so} \quad {\frac {1}{m}}\leq {\frac {1}{N}}<{\frac {\epsilon }{2}}\\\quad n\geq N>{\frac {2}{\epsilon }}\quad \mathrm {so} \quad {\frac {1}{n}}\leq {\frac {1}{N}}<{\frac {\epsilon }{2}}\end{aligned}}$

Hence for all $m,n\geq N$ we have that:

${\begin{aligned}\quad d(x_{m},x_{n})\leq {\frac {1}{n}}+{\frac {1}{m}}<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

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.

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Content obtained and/or adapted from:

• Complete Metric Spaces, mathonline.wikidot.com under a CC BY-SA license