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 17: Compactness

Compactness

Let $(X,d)$ be a metric space. Let $A\subseteq X\text{.}$

The set $A$ is sequentially compact if every sequence in $A$ has a cluster point in $A\text{.}$

The set $A$ is Cauchy compact, or complete, if every Cauchy sequence in $A$ has a cluster point in $A\text{.}$

The set $A$ is ball compact if $A$ satisfies if $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ and $x_1,x_2,\dots ,x_N\in X$ such that $$A\subseteq B(x_1,\epsilon )\cup B(x_2,\epsilon )\cup \cdots \cup B(x_N,\epsilon )\text{.}$$

The set $A$ is cover compact if $A$ satisfies if $𝒮\subseteq 𝒯$ and $A\subseteq \bigcup _{U\in 𝒮}U$ then there exists $N\in {\mathbb{Z}}_{>0}$ and $U_1,U_2,\dots ,U_n\in 𝒮$ such that $$A\subseteq U_1\cup U_2\cup \cdots \cup U_n\text{.}$$ (in English: every open cover has a finite subcover).

Synonyms for ball compact are precompact and totally bounded.

Let $(X,d)$ be a metric space and let $A\subseteq X\text{.}$

(a)	If $A$ is cover compact then $A$ is sequentially compact.
(b)	If $A$ is sequentially compact then $A$ is cover compact.
(c)	If $A$ is sequentially compact then $A$ is Cauchy compact.
(d)	If $A$ is cover compact then $A$ is ball compact.
(e)	If $A$ is ball compact then $A$ is bounded.
(f)	If $A$ is Cauchy compact then $A$ is closed.

$$\begin{array}{ccccc}\text{cover compact}& \stackrel{\text{(d)}}{⟹}& \text{ball compact}& \stackrel{\text{(e)}}{⟹}& \text{bounded}\\ \text{(a)}\Downarrow \Uparrow \text{(b)}\\ \text{sequentially compact}& \stackrel{\text{(c)}}{⟹}& \text{Cauchy compact}& \stackrel{\text{(f)}}{⟹}& \text{closed}\end{array}$$

HW: Let $A=(0,1)\subseteq ℝ$ with the standard metric on $ℝ\text{.}$ Show that $A$ is ball compact and not closed but not cover compact (and not Cauchy compact).

HW: Let $X=ℝ$ with metric given by $d(x,y)=\text{min}\{\mid x-y\mid ,1\}\text{.}$ Show that $X$ is bounded but not ball compact.

HW: Let $A={ℝ}_{(0,1)}$ with metric $d(x,y)=\mid x-y\mid \text{.}$ Show that $A$ is closed but $A$ is not Cauchy compact.

HW: Show that $ℝ$ with the standard metric is Cauchy compact and not bounded but not cover compact (and not ball compact).

Let $(X,d)$ be a metric space and let $A\subseteq X\text{.}$ If $A$ is ball compact and Cauchy compact then $A$ is cover compact.

Let $A\subseteq {ℝ}^n$ with the standard metric on ${ℝ}^n\text{.}$ If $A$ is closed and bounded then $A$ is cover compact.

		Proof.
	Assume $A\subseteq {ℝ}^n$ and $A$ is closed and bounded. (a) Since ${ℝ}^n$ is complete, $A\subseteq {ℝ}^n$ and $A$ is closed then $A$ is complete. So $A$ is Cauchy compact. (b) Since $A\subseteq {ℝ}^n$ and $A$ is bounded then $A$ is ball compact. Since $A$ is Cauchy compact and ball compact. So $A$ is cover compact. (Note: (b) follows since $A\subseteq B(0,c)$ for some $c\in {ℝ}_{>0}$ and $B(0,c)$ is ball compact in ${ℝ}^n$ (since $B(0,c)$ has finite volume).) $\square$

HW: Show that a closed subset of a Cauchy compact set is Cauchy compact.

HW: Show that a bounded subset of a ball compact set is ball compact.

HW: Show that a closed subset of a cover compact set is cover compact.

${\ell }^2=\{\stackrel{\rightharpoonup }{x}=(x_1,x_2,\dots )\hspace{0.17em}|\hspace{0.17em}{x}_i\in ℝ\hspace{0.17em}\text{with}\hspace{0.17em}{\Vert \stackrel{\rightharpoonup }{x}\Vert }_2<\infty \}$ where $${\Vert \stackrel{\rightharpoonup }{x}\Vert }_2={\left(\sum _{i\in {\mathbb{Z}}_{>0}}{x}_i^2\right)}^{\frac{1}{2}}\text{for}\stackrel{\rightharpoonup }{x}=(x_1,x_2,\dots )\text{.}$$ Then let $$\begin{array}{rcl}{\displaystyle e_1}& =& {\displaystyle (1,0,0,\dots ),}\\ {\displaystyle e_2}& =& {\displaystyle (0,1,0,\dots ),}\\ {\displaystyle e_3}& =& {\displaystyle (0,0,1,0,\dots ),}\\ {\displaystyle }& ⋮& {\displaystyle }\end{array}$$ Then $e_1,e_2,e_3,\dots$ is a sequence in ${\ell }^2$ $\{e_1,e_2,e_3,\dots \}$ is bounded since $d(e_i,e_j)=\sqrt{1+1}=\sqrt{2}\text{.}$ The set $\{e_1,e_2,e_3,\dots \}$ is not ball compact.

Notes and References

These are a typed copy of Lecture 17 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 26, 2014.

page history