Metric spaces

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

Last update: 23 July 2014

Metric spaces

A metric space is a set $X$ with a function $d:X\times X\to {ℝ}_{\ge 0}$ such that

(a)	if $x\in X$ then $d(x,x)=0,$
(b)	if $x,y\in X$ and $d(x,y)=0$ then $x=y,$
(c)	if $x,y\in X$ then $d(x,y)=d(y,x),$
(d)	if $x,y,z\in X$ then $d(x,y)\le d(x,z)+d(z,y)\text{.}$

Let $(X,d)$ be a metric space. Let $x\in X$ and $\epsilon \in $ {ℝ}_{>0}\text{.}$$ The ball of radius $\epsilon$ at $x$ is the set $$B_{\epsilon }\left(x\right)=\{y\in X\hspace{0.17em}|\hspace{0.17em}d(x,y)<\epsilon \}\text{.}$$

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

The metric space uniformity on $X$ is the uniformity generated by the sets $$B_{\epsilon }=\{(x,y)\in X\times X\hspace{0.17em}|\hspace{0.17em}d(x,y)<\epsilon \}$$ for $\epsilon \in {ℝ}_{>0}\text{.}$

The metric space topology on $X$ is the topology generated by the sets $$B_{\epsilon }\left(x\right)=\{y\in X\hspace{0.17em}|\hspace{0.17em}d(x,y)<\epsilon \}$$ for $x\in X$ and $\epsilon \in {ℝ}_{>0}\text{.}$

Homework: Let $(X,d)$ be a metric space. Show that $X$ is Hausdorff.

Homework: Let $(X,d)$ and $(Y,\rho )$ be metric spaces and let $f:X\to Y$ be a function. Show that $f$ is uniformly continuous if and only if $f$ satisfies if $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $x,y\in X$ and $d(x,y)<\delta$ then $\rho (f\left(x\right),f\left(y\right))<\epsilon \text{.}$

Homework: Let $(X,d)$ and $(Y,\rho )$ be metric spaces and let $f:X\to Y$ be a function. Show that $f$ is continuous if and only if $f$ satisfies if $\epsilon \in {ℝ}_{>0}$ and $x\in X$ then there exists $\delta \in {ℝ}_{>0}$ such that if $y\in X$ and $d(x,y)<\delta$ then $\rho (f\left(x\right),f\left(y\right))<\epsilon \text{.}$

Homework: Show that the function $f:ℝ\to ℝ$ given by $f\left(x\right)=x^2$ is continuous but not uniformly continuous.

Compactness in metric spaces

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

A totally bounded subset of $X$ is a subset $A\subseteq X$ such that 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 $$A\subseteq B(x_1,\epsilon )\cup B(x_2,\epsilon )\cup \cdots \cup B(x_N,\epsilon )\text{.}$$ A bounded subset of $X$ is a subset $A\subseteq X$ such that there exists $C\in {ℝ}_{>0}$ such that if $x,y\in A$ then $d(x,y)<C\text{.}$

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

(a)	If $A$ is compact then $A$ is totally bounded.
(b)	If $A$ is totally bounded then $A$ is bounded.

		Proof.
	(a) Assume $A\subseteq X$ is compact. Let $\epsilon \in {ℝ}_{>0}\text{.}$ Then $\left\{B(x,\epsilon )\hspace{0.17em}\right|\hspace{0.17em}x\in X\}$ is a cover of $A\text{.}$ Since $A$ is compact there exists $N\in {\mathbb{Z}}_{>0}$ and $x_1,x_2,\dots ,x_N\in X$ such that $$\{B(x_1,\epsilon ),\dots ,B(x_N,\epsilon )\}$$ is a finite cover of $A\text{.}$ So $A$ is totally bounded. (b) Assume $A\subseteq X$ is totally bounded. Let $N\in {\mathbb{Z}}_{>0}$ and $x_1,x_2,\dots ,x_N\in X$ such that $$\{B(x_1,1),\dots ,B(x_N,1)\}$$ is a cover of $A\text{.}$ Let $C=3+\text{max}\left\{d(x_k,x_{\ell })\hspace{0.17em}\right|\hspace{0.17em}k,\ell \in \{1,2,\dots ,N\}\}\text{.}$ Let $x,y\in A\text{.}$ Let $i,j\in \{1,2,\dots ,N\}$ such that $$x\in B(x_i,1)\text{and}y\in B(x_j,1)\text{.}$$ Then $$\begin{array}{ccc}{\displaystyle d(x,y)}& \le & {\displaystyle d(x,x_i)+d(x_i,x_j)+d(x_j,y)}\\ & \le & {\displaystyle 1+\text{max}\left\{d(x_k,x_{\ell })\hspace{0.17em}\right|\hspace{0.17em}k,\ell \in \{1,2,\dots ,N\}\}+1}\\ & <& {\displaystyle C\text{.}}\end{array}$$ So $A$ is bounded. $\square$

Homework: Let $X=ℝ$ with metric $d:X\times X\to {ℝ}_{>0}$ given by $$d(x,y)=\text{min}\{|x-y|,1\}\text{.}$$ Show that $X$ is bounded but not totally bounded.

Homework: Let $A=(0,1)\subseteq ℝ$ where $ℝ$ has the standard metric $$d(x,y)=|x-y|\text{.}$$ Show that $A$ is totally bounded but not compact.

(This is [BR, Theorem 2.37]) Let $X$ be a metric space and let $E$ be a subset of $X$. The set $E$ is compact if and only if every infinite subset of $E$ has a close point in $E$.

		Proof.
	$\Leftarrow$: Let $K$ be a compact set and let $E$ be an infinite subset of $K$. If there is no close point of $E$ in $K$ then for each $p\in K$ there is a neighborhood $N_p$ of $p$ which contains no other element of $E$. Then the open cover $$𝒩=\left\{N_p\right|p\in K\}\text{,}$$ of $K$ has no finite subcover. $\Rightarrow$: Let $S$ be an infinite subset of $E$. The metric space $E$ has a countable base. So every open cover of $E$ has a countable subcover $𝒞=\{C_1,C_2,\dots \}$. If $𝒞$ does not have a finite subcover then, for each $n$, $${\left(C_{\le n}\right)}^c\ne \varnothing \text{but}\underset{n}{\cap }{C}_{\le n}^c=\varnothing .$$ Let $S$ be a set which contains a point from each $C_{\le n}^c$. Then $S$ has a limit point. But this is a contradiction. $\square$

Let $X$ be a metric space and let $E$ be a compact subset of $X$. Then $E$ is closed and bounded.

1. A k-cell is compact. [DEFINE K-CELL?]
2. Let $E$ be a subset of ${ℝ}^k$. If $E$ is closed and bounded then $E$ is compact.

		Proof.
	If $E$ is closed and bounded then $E$ is a closed subset of a $k$-cell. Since closed subsets of compact sets are compact $E$ is compact. $\square$

Notes and References

These are a typed copy of handwritten notes from the pdf 140721UniformSpacesscanned140721.pdf.

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