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 15: Completion of a metric space

$(\hat{X},\hat{d})$ 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) $\overline{\phi \left(X\right)}=\hat{X}\text{.}$ (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 $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})\le \hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{z})+\hat{d}(\stackrel{\rightharpoonup }{z},\stackrel{\rightharpoonup }{y})\text{.}$ (ab) To show: If $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}\in \hat{X}$ then $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})=\hat{d}(\stackrel{\rightharpoonup }{y},\stackrel{\rightharpoonup }{x})\text{.}$ Assume $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}\in \hat{X}$ with $\stackrel{\rightharpoonup }{x}=(x_1,x_2,\dots )$ and $\stackrel{\rightharpoonup }{y}=(y_1,y_2,\dots )\text{.}$ Since $d(x_n,y_n)=d(y_n,x_n),$ $$\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})=\underset{n\to \infty }{\text{lim}}d(x_n,y_n)=\underset{n\to \infty }{\text{lim}}d(y_n,x_n)=\hat{d}(\stackrel{\rightharpoonup }{y},\stackrel{\rightharpoonup }{x})\text{.}$$ (ac) To show: If $\stackrel{\rightharpoonup }{x}\in \hat{X}$ then $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{x})=0\text{.}$ Assume $\stackrel{\rightharpoonup }{x}\in \hat{X}\text{.}$ To show: $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{x})=0\text{.}$ Since $d(x_n,x_n)=0,$ $$\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{x})=\underset{n\to \infty }{\text{lim}}d(x_n,x_n)=\underset{n\to \infty }{\text{lim}}0=0\text{.}$$ (ad) To show: If $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}\in \hat{X}$ and $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})=0$ then $\stackrel{\rightharpoonup }{x}=\stackrel{\rightharpoonup }{y}\text{.}$ Assume $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y}\in \hat{X}$ and $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})=0\text{.}$ To show: $\stackrel{\rightharpoonup }{x}=\stackrel{\rightharpoonup }{y}\text{.}$ This is a consequence of the definition of $\hat{X}$ which requires that $\stackrel{\rightharpoonup }{x}=\stackrel{\rightharpoonup }{y}$ if $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})=0\text{.}$ (ae) If $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y},\stackrel{\rightharpoonup }{z}\in \hat{X}$ then $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})\le \hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{z})+\hat{d}(\stackrel{\rightharpoonup }{z},\stackrel{\rightharpoonup }{y})\text{.}$ Assume $\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y},\stackrel{\rightharpoonup }{z}\in \hat{X}\text{.}$ To show: $\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})\le \hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{z})+\hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})\text{.}$ $$\begin{array}{rcl}{\displaystyle \hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{y})}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}d(x_n,y_n)}\\ & \le & {\displaystyle \underset{n\to \infty }{\text{lim}}(d(x_n,z_n)+d(z_n,y_n))}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}d(x_n,z_n)+\underset{n\to \infty }{\text{lim}}d(z_n,y_n)}\\ & =& {\displaystyle \hat{d}(\stackrel{\rightharpoonup }{x},\stackrel{\rightharpoonup }{z})+\hat{d}(\stackrel{\rightharpoonup }{z},\stackrel{\rightharpoonup }{y}),}\end{array}$$ where the next to last equality follows from the continuity of addition in ${ℝ}_{\ge 0}\text{.}$ (d) To show: $\overline{\phi \left(X\right)}=\hat{X}\text{.}$ To show: If $\stackrel{\rightharpoonup }{z}\in \hat{X}$ then there exists a sequence ${\stackrel{\rightharpoonup }{x}}_1,{\stackrel{\rightharpoonup }{x}}_2,\dots$ in $\phi \left(X\right)$ such that $\underset{n\to \infty }{\text{lim}}{\stackrel{\rightharpoonup }{x}}_n=\stackrel{\rightharpoonup }{z}\text{.}$ Let $\stackrel{\rightharpoonup }{z}=(z_1,z_2,\dots )\in \hat{X}\text{.}$ To show: There exists ${\stackrel{\rightharpoonup }{x}}_1,{\stackrel{\rightharpoonup }{x}}_2,\dots$ in $\phi \left(X\right)$ with $\underset{n\to \infty }{\text{lim}}{\stackrel{\rightharpoonup }{x}}_n=\stackrel{\rightharpoonup }{z}\text{.}$ Let $$\begin{array}{rcl}{\displaystyle {\stackrel{\rightharpoonup }{x}}_1}& =& {\displaystyle (z_1,z_1,z_1,z_1,\dots )=\phi \left(z_1\right),}\\ {\displaystyle {\stackrel{\rightharpoonup }{x}}_2}& =& {\displaystyle (z_2,z_2,z_2,z_2,\dots )=\phi \left(z_2\right),}\\ {\displaystyle {\stackrel{\rightharpoonup }{x}}_3}& =& {\displaystyle (z_3,z_3,z_3,z_3,\dots )=\phi \left(z_3\right),\dots }\end{array}$$ Then ${\stackrel{\rightharpoonup }{x}}_1,{\stackrel{\rightharpoonup }{x}}_2,\dots$ is the sequence $\phi \left(z_1\right),\phi \left(z_2\right),\dots$ in $\phi \left(X\right)\text{.}$ To show: $\underset{n\to \infty }{\text{lim}}{\stackrel{\rightharpoonup }{x}}_n=\stackrel{\rightharpoonup }{z}\text{.}$ To show: $\underset{n\to \infty }{\text{lim}}\hat{d}({\stackrel{\rightharpoonup }{x}}_n,\stackrel{\rightharpoonup }{z})=0\text{.}$ To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then $\hat{d}({\stackrel{\rightharpoonup }{x}}_n,\stackrel{\rightharpoonup }{z})<\epsilon \text{.}$ Assume $\epsilon \in {ℝ}_{>0}\text{.}$ Let $N\in {\mathbb{Z}}_{>0}$ be such that if $r,s\in {\mathbb{Z}}_{>0}$ and $r>N$ and $s>N$ then $d(z_r,z_s)<{\epsilon }_2\text{.}$ To show: If $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then $\hat{d}\left({\stackrel{\rightharpoonup }{x}}_n\stackrel{\rightharpoonup }{z}\right)<\epsilon \text{.}$ Assume $n\in {\mathbb{Z}}_{>0}$ and $n>N\text{.}$ To show: $\hat{d}({\stackrel{\rightharpoonup }{x}}_n,\stackrel{\rightharpoonup }{z})<\epsilon \text{.}$ To show: $\underset{k\to \infty }{\text{lim}}d(z_n,z_k)<\epsilon \text{.}$ $$\underset{k\to \infty }{\text{lim}}d(z_n,z_k)\le \frac{\epsilon }{2}<\epsilon ,$$ since $d(z_n,z_k)<\frac{\epsilon }{2}$ for $k>N\text{.}$ So $\underset{n\to \infty }{\text{lim}}{\stackrel{\rightharpoonup }{x}}_n=\stackrel{\rightharpoonup }{z}\text{.}$ So $\overline{\phi \left(X\right)}=\hat{X}\text{.}$ (b) To show: $(\hat{X},\hat{d})$ is complete. To show: If ${\stackrel{\rightharpoonup }{x}}_1,{\stackrel{\rightharpoonup }{x}}_2,\dots$ is a Cauchy sequence in $\hat{X}$ then ${\stackrel{\rightharpoonup }{x}}_1,{\stackrel{\rightharpoonup }{x}}_2,\dots$ converges. Assume $$\begin{array}{rcl}{\displaystyle {\stackrel{\rightharpoonup }{x}}_1}& =& {\displaystyle (x_{11},x_{12},x_{13},\dots ),}\\ {\displaystyle {\stackrel{\rightharpoonup }{x}}_2}& =& {\displaystyle (x_{21},x_{22},x_{23},\dots ),}\\ {\displaystyle {\stackrel{\rightharpoonup }{x}}_3}& =& {\displaystyle (x_{31},x_{32},x_{33},\dots ),\dots }\end{array}$$ is a Cauchy sequence in $\hat{X}\text{.}$ To show: There exists $\stackrel{\rightharpoonup }{z}=(z_1,z_2,\dots )$ in $\hat{X}$ such that $\underset{n\to \infty }{\text{lim}}{\stackrel{\rightharpoonup }{x}}_n=\stackrel{\rightharpoonup }{z}\text{.}$ Using that $\overline{\phi \left(X\right)}$ in $\hat{X},$ for $k\in {\mathbb{Z}}_{>0}$ let $z_k\in X$ be such that $\hat{d}(\phi \left(z_k\right),{\stackrel{\rightharpoonup }{x}}_k)<\frac{1}{k}\text{.}$ $$\begin{array}{ccc}\begin{array}{rcl}{\displaystyle {\stackrel{\rightharpoonup }{x}}_1}& =& {\displaystyle (x_{11},x_{12},x_{13},\dots ),}\\ {\displaystyle {\stackrel{\rightharpoonup }{x}}_2}& =& {\displaystyle (x_{21},x_{22},x_{23},\dots ),}\\ {\displaystyle {\stackrel{\rightharpoonup }{x}}_3}& =& {\displaystyle (x_{31},x_{32},x_{33},\dots ),}\\ {\displaystyle }& ⋮& {\displaystyle }\end{array}& & \begin{array}{rcl}{\displaystyle \phi \left(z_1\right)}& =& {\displaystyle (z_1,z_1,z_1,\dots ),}\\ {\displaystyle \phi \left(z_2\right)}& =& {\displaystyle (z_2,z_2,z_2,\dots ),}\\ {\displaystyle \phi \left(z_3\right)}& =& {\displaystyle (z_3,z_3,z_3,\dots ),}\\ {\displaystyle }& ⋮& {\displaystyle }\end{array}\end{array}$$ To show: (ba) $\stackrel{\rightharpoonup }{z}=(z_1,x_2,\dots )$ is a Cauchy sequence. (bb) $\underset{n\to \infty }{\text{lim}}{\stackrel{\rightharpoonup }{x}}_n=\stackrel{\rightharpoonup }{z}\text{.}$ (ba) To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $r,s\in {\mathbb{Z}}_{>0}$ and $r>N$ and $s>N$ then $d(z_r,z_s)<\epsilon \text{.}$ Assume $\epsilon \in {ℝ}_{>0}\text{.}$ To show: There exists $N\in {\mathbb{Z}}_{>0}$ such that if $r,s\in {\mathbb{Z}}_{>0}$ and $r>N$ and $s>N$ then $d(z_r,z_s)<\epsilon \text{.}$ Let $N_1=\lceil \frac{3}{\epsilon }\rceil +1,$ so that $\frac{1}{N_1}<\frac{\epsilon }{3}\text{.}$ Let $N_2\in {\mathbb{Z}}_{>0}$ such that if $r,s\in {\mathbb{Z}}_{>0}$ and $r>N_2$ and $s>N_2$ then $\hat{d}({\stackrel{\rightharpoonup }{x}}_r,{\stackrel{\rightharpoonup }{x}}_s)<\frac{\epsilon }{3}\text{.}$ Let $N=\text{max}(N_1,N_2)\text{.}$ To show: If $r,s\in {\mathbb{Z}}_{>0}$ and $r>N$ and $s>N$ then $d(z_r,z_s)<\epsilon \text{.}$ Assume $r,s\in {\mathbb{Z}}_{>0}$ and $r>N$ and $s>N\text{.}$ To show: $d(z_r,z_s)<\epsilon \text{.}$ $$\begin{array}{rcl}{\displaystyle d(z_r,z_s)}& =& {\displaystyle \hat{d}(\phi \left(z_r\right),\phi \left(z_s\right))}\\ & \le & {\displaystyle \hat{d}(\phi \left(z_r\right),{\stackrel{\rightharpoonup }{x}}_r)+\hat{d}({\stackrel{\rightharpoonup }{x}}_r,{\stackrel{\rightharpoonup }{x}}_s)+\hat{d}({\stackrel{\rightharpoonup }{x}}_s,\phi \left(z_s\right))}\\ & <& {\displaystyle \frac{1}{r}+\frac{\epsilon }{3}+\frac{1}{s}}\\ & <& {\displaystyle \frac{1}{N_1}+\frac{\epsilon }{3}+\frac{1}{N_1}}\\ & =& {\displaystyle \frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}}\\ & =& {\displaystyle \epsilon \text{.}}\end{array}$$ So $\stackrel{\rightharpoonup }{z}$ is Cauchy. (bb) To show: $\underset{n\to \infty }{\text{lim}}\hat{d}({\stackrel{\rightharpoonup }{x}}_n,\stackrel{\rightharpoonup }{z})=0\text{.}$ $$\begin{array}{rcl}{\displaystyle \underset{n\to \infty }{\text{lim}}\hat{d}({\stackrel{\rightharpoonup }{x}}_n,\stackrel{\rightharpoonup }{z})}& \le & {\displaystyle \underset{n\to \infty }{\text{lim}}(\hat{d}({\stackrel{\rightharpoonup }{x}}_n,\phi \left(z_n\right))+\hat{d}\left(\phi (z_n,\stackrel{\rightharpoonup }{z})\right))}\\ & \le & {\displaystyle \underset{n\to \infty }{\text{lim}}(\frac{1}{n}+\hat{d}(\phi \left(z_n\right),\stackrel{\rightharpoonup }{z}))}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{1}{n}+\underset{n\to \infty }{\text{lim}}\hat{d}(\phi \left(z_n\right),\stackrel{\rightharpoonup }{z})}\\ & =& {\displaystyle 0+0}\\ & =& {\displaystyle 0\text{.}}\end{array}$$ So $(\hat{X},\hat{d})$ is complete. So $(\hat{X},\hat{d})$ with $\phi$ is a completion. $\square$

Notes and References

These are a typed copy of Lecture 15 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 21, 2014.

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