Complete Metric Space

Cauchy Sequence

Cauchy Sequence

Let $(X,d)$ be a metric space. A sequence $(x_n)$ in $X$ is called a Cauchy sequence if for any $\epsilon >0$ there exists $N$ such that $d(x_n,x_m)<\epsilon$ for all $n,m>N$.

Theorem

Convergent sequences in metric spaces are Cauchy sequences.

Proof Suppose $(x_n)$ is convergent with $x_n\to x$. Then for any $\epsilon >0$, there exists $N$ such that $d(x_n,x)<\epsilon /2$ for all $n>N$. Therefore, for all $n,m>N$, we have $d(x_n,x_m)\le d(x_n,x)+d(x,x_m)<\epsilon /2+\epsilon /2=\epsilon$Thus $(x_n)$ is a Cauchy sequence. $\square$

Proposition

Cauchy sequences in metric spaces are bounded.

Proof Let $(x_n)$ be a Cauchy sequence. Then there exists $N$ such that $d(x_n,x_N)<1$ for all $n\ge N$. Let $R=\max \left\{d(x_1,x_N),\cdots ,d(x_{N-1},x_N)\right\}+1$Then $d(x_n,x_N)<R$ for all n. Therefore $x_n\in B_R(x_N)$ for all $n$, i.e. $(x_n)$ is bounded. $\square$

Theorem

Let $(X,d)$ be a metric space and let $(x_n)$ be a Cauchy sequence in $X$. If $(x_n)$ has a convergent subsequence, then $(x_n)$ is convergent.

Proof Let $(x_n)$ be a Cauchy sequence and let $(x_{n_k})$ be a convergent subsequence of $(x_n)$ with $x_{n_k}\to x$. We have $d(x_n,x)\le d(x_n,x_{n_k})+d(x_{n_k},x)\le {\sup }_{m\ge n}d(x_n,x_m)+d(x_{n_k},x)$for sufficiently large $k$. By taking $k\to \infty$ and using $d(x_{n_k},x)\to 0$, we obtain $d(x_n,x)\le {\sup }_{m\ge n}d(x_n,x_m)$Since $(x_n)$ is Cauchy, the right hand side goes to $0$ as $n\to \infty$. Therefore $d(x_n,x)\to 0$ as $n\to \infty$, that is $x_n\to x$. $\square$

Theorem

Every Cauchy sequence in ${\mathbb{R}}^k$ is convergent.

Proof Let $(x_n)$ be a Cauchy sequence in ${\mathbb{R}}^k$. Then $(x_n)$ is bounded. Thus, it follows from Bolzano-Weierstrass theorem that $(x_n)$ has a convergent subsequence. Consequently, by invoking previous theorem, $(x_n)$ is convergent.

Completeness

Complete Metric Space

A metric space $(X,d)$ is called complete if every Cauchy sequence in $X$ is convergent.

e.g.

• ${\mathbb{R}}^k$ with the Euclidean metric is complete.
• $(0,1)$ with the induced Euclidean metric is not complete.
• $C[-1,1]$ is complete under the metric $d_{\infty }(f,g)={\max }_{x\in [-1,1]}|f(x)-g(x)|$

Theorem

Let $(X,d)$ be a complete metric space and $S\subset X$. Then $(S,d)$ is complete iff $S$ is closed in $X$.

Proof Let $(x_n)$ be a sequence in $S$ with $x_n\to x\in X$. Then $(x_n)$ is a Cauchy sequence in $X$ and hence is a Cauchy sequence in $S$. Since $S$ is complete, $x_n\to \tilde{x}$ for some $\tilde{x}\in S$. By uniqueness of limit, we have $x=\tilde{x}$ and thus $x\in S$ . This shows that $S$ is closed by corollary. Conversely, let $(x_n)$ be a Cauchy sequence in $(S,d)$. Then $(x_n)$ is also a Cauchy sequence in $(X,d)$. Since $(X,d)$ is complete, $x_n\to x$ for some $x\in X$. Since $S$ is closed, $x\in S$. Therefore $(S,d)$ is complete. $\square$

Theorem

Every metric space $(X,d)$ has a completion, i.e. there exists a complete metric space $(\tilde{X},\tilde{d})$ and an injection $f:X\to \tilde{X}$ such that $d(x,y)=d(f(x),f(y)),\forall x,y\in X$and $\overline{f(X)}=\tilde{X}$.

Contraction

Fixed Point

Let $(X,d)$ be a metric space and $f:X\to X$ a map. A point $x\in X$ is called a fixed point of $f$ if $f(x)=x$.

Contraction

Let $(X,d)$ be a metric space and $f:X\to X$ a map. $f$ is called a contraction if there exists $q\in (0,1)$ such that $d(f(x),f(y))\le qd(x,y),\forall x,y\in X$where $q$ is called the contraction rate of $f$.

Contraction Mapping Theorem

Let $(X,d)$ be a complete metric space. If $f:X\to X$ is a contraction, then it has a unique fixed point.