Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Lecture 14: Examples of complete spaces

Completions

Let $(X,d)$ be a metric space.

The completion of $(X,d)$ is a metric space $(\hat{X},\hat{d})$ with an isometry $\phi :X\to \hat{X}$ such that $(\hat{X},\hat{d})$ is complete and $\overline{\phi \left(X\right)}=\hat{X}\text{.}$

HW: (Uniqueness of completions). If $({\hat{X}}_1,{\hat{d}}_1)$ with ${\phi }_1:X\to {\hat{X}}_1$ and $({\hat{X}}_2,{\hat{d}}_2)$ with ${\phi }_2:X\to {\hat{X}}_2$ are completions of $(X,d)$ then there exists $f:{\hat{X}}_1\to {\hat{X}}_2$ such that

(a)	$f$ is an isometry,
(b)	$f$ is a bijection,
(c)	$f\circ {\phi }_1={\phi }_2\text{.}$

$$\begin{array}{ccc}& & {\hat{X}}_1\\ & {}^{{\phi }_1}\nearrow & \downarrow f\\ X& \\ & \underset{{\phi }_2}{\searrow }\\ & & {\hat{X}}_2\end{array}$$

Existence of completions

Let $(X,d)$ be a metric space.

Let $\hat{X}$ be the set of Cauchy sequences $\stackrel{\rightharpoonup }{x}:{\mathbb{Z}}_{>0}\to X$ with $\stackrel{\rightharpoonup }{x}=\stackrel{\rightharpoonup }{y}$ if $\underset{n\to \infty }{\text{lim}}d(x_n,y_n)=0,$ where $$\begin{array}{cccc}\stackrel{\rightharpoonup }{x}:& {\mathbb{Z}}_{>0}& ⟶& X\\ & n& ⟼& x_n\end{array}$$ and $$\begin{array}{cccc}\stackrel{\rightharpoonup }{y}:& {\mathbb{Z}}_{>0}& ⟶& X\\ & n& ⟼& y_n\end{array}\text{.}$$

Define $\hat{d}:\hat{X}\times \hat{X}\to {ℝ}_{\ge 0}$ by $$d(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})=\underset{n\to \infty }{\text{lim}}d(x_n,y_n),$$ where $$\begin{array}{cccc}\stackrel{\rightharpoonup }{x}:& {\mathbb{Z}}_{>0}& ⟶& X\\ & n& ⟼& x_n\end{array}$$ and $$\begin{array}{cccc}\stackrel{\rightharpoonup }{y}:& {\mathbb{Z}}_{>0}& ⟶& X\\ & n& ⟼& y_n\end{array}\text{.}$$

Define $\phi :X\to \hat{X}$ by $\phi \left(x\right)=(x,x,\dots ),$ i.e. $$\begin{array}{cccc}\phi \left(x\right):& {\mathbb{Z}}_{>0}& ⟶& X\\ & n& ⟼& x\end{array}\text{.}$$

$(\hat{X},\hat{d})$ with $\phi :X\to \hat{X}$ is a completion of $(X,d)\text{.}$

		Proof.
	To show: (a) $(\hat{X},\hat{d})$ is a metric space. (b) $(\hat{X},\hat{d})$ is complete. (c) $\phi :X\to \hat{X}$ is an isometry. (d) $\overline{\phi \left(X\right)}=\hat{X}\text{.}$ (c) To show: If $x,y\in X$ then $d(\phi \left(x\right),\phi \left(y\right))=d(x,y)\text{.}$ Assume $x,y\in X\text{.}$ To show: $d(\phi \left(x\right),\phi \left(y\right))=d(x,y)\text{.}$ $$\begin{array}{rcl}{\displaystyle d(\phi \left(x\right),\phi \left(y\right))}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}d(\phi {\left(x\right)}_n,\phi {\left(y\right)}_n)}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}d(x,y)}\\ & =& {\displaystyle d(x,y)\text{.}}\end{array}$$ So $\phi$ is an isometry. (a) To show: $(\hat{X},\hat{d})$ is a metric space. To show: (aa) $\hat{d}:\hat{X}\times \hat{X}\to {ℝ}_{\ge 0}$ given by $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})=\underset{n\to \infty }{\text{lim}}d(x_n,y_n),$ is a function. (ab) If $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}\in \hat{X}$ then $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})=\hat{d}(\stackrel{\rightharpoonup }{y},\stackrel{\rightharpoonup }{x})\text{.}$ (ac) If $\stackrel{\rightharpoonup }{x}\in \hat{X}$ then $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{x})=0\text{.}$ (ad) If $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}\in \hat{X}$ and $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})=0$ then $\stackrel{\rightharpoonup }{x}=\stackrel{\rightharpoonup }{y}\text{.}$ (ae) If $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y},\stackrel{\rightharpoonup }{z}\in \hat{X}$ then $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})\le \hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{z})+\hat{d}(\stackrel{\rightharpoonup }{z},\stackrel{\rightharpoonup }{y})\text{.}$ (aa) To show: If $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}\in \hat{X}$ then there exists a unique $z\in {ℝ}_{\ge 0}$ such that $z=\underset{n\to \infty }{\text{lim}}d(x_n,y_n)\text{.}$ Assume $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}\in \hat{X}$ with $\stackrel{\rightharpoonup }{x}=(x_1,x_2,\dots )$ and $\stackrel{\rightharpoonup }{y}=(y_1,y_2,\dots )\text{.}$ Let $d_1,d_2,\dots$ be the sequence in ${ℝ}_{\ge 0}$ given by $$d_n=d(x_n,y_n)\text{.}$$ To show: There exists $z\in {ℝ}_{\ge 0}$ such that $z=\underset{n\to \infty }{\text{lim}}{d}_n\text{.}$ Since limits in metric spaces are unique when they exist, $z$ will be unique if it exists, see Rubinstein Notes Proposition 2.9. To show: $d_1,d_2,\dots$ is a Cauchy sequence in ${ℝ}_{\ge 0}\text{.}$ This will show that $z$ exists since Cauchy sequences in ${ℝ}_{\ge 0}$ converge, since ${ℝ}_{\ge 0}$ is complete, see Rubinstein notes Theorem 4.6. To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $n>N$ then $\mid d_m-d_n\mid <\epsilon \text{.}$ Assume $\epsilon \in {ℝ}_{>0}\text{.}$ Let $N=\text{max}(N_1,N_2),$ where $N_1$ is such that if $n,m>N_1$ then $d(x_m,x_n)<\frac{\epsilon }{2},$ $N_2$ is such that if $n,m>N_2$ then $d(y_m,y_n)<\frac{\epsilon }{2}\text{.}$ $\text{(}{N}_1$ and $N_2$ exist since $\stackrel{\rightharpoonup }{x}$ and $\stackrel{\rightharpoonup }{y}$ are Cauchy sequences). To show: If $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $N>N$ then $\mid d_m-d_n\mid <\epsilon \text{.}$ Assume $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $n>N\text{.}$ To show: $\mid d_m-d_n\mid <\epsilon \text{.}$ $$\begin{array}{rcl}{\displaystyle \mid d_m-d_n\mid }& =& {\displaystyle \mid d(x_m,y_m)-d(x_n,y_n)\mid }\\ & <& {\displaystyle \mid d(x_n,x_m)+d(y_n,y_m)\mid ,}\end{array}$$ since $d(x_n,y_n)\le d(x_n,x_m)+d(x_m,y_m)+d(y_n,y_m)\text{.}$ So $$\mid d_m-d_n\mid <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \text{.}$$ So $d_1,d_2,\dots$ is a Cauchy sequence in ${ℝ}_{\ge 0}\text{.}$ So $z=\underset{n\to \infty }{\text{lim}}{d}_n$ exists in ${ℝ}_{\ge 0}\text{.}$ $\square$

Notes and References

These are a typed copy of Lecture 14 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 20, 2014.

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