## Metric spaces

Last update: 23 July 2014

## Metric spaces

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

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

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

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

The metric space uniformity on $X$ is the uniformity generated by the sets $Bε= {(x,y)∈X×X | d(x,y)<ε}$ for $\epsilon \in {ℝ}_{>0}\text{.}$

The metric space topology on $X$ is the topology generated by the sets $Bε(x)= {y∈X | d(x,y)<ε}$ for $x\in X$ and $\epsilon \in {ℝ}_{>0}\text{.}$

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

Homework: Let $\left(X,d\right)$ and $\left(Y,\rho \right)$ 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 $d\left(x,y\right)<\delta$ then $\rho \left(f\left(x\right),f\left(y\right)\right)<\epsilon \text{.}$

Homework: Let $\left(X,d\right)$ and $\left(Y,\rho \right)$ 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 $d\left(x,y\right)<\delta$ then $\rho \left(f\left(x\right),f\left(y\right)\right)<\epsilon \text{.}$

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 $\left(X,d\right)$ 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 $N\in {ℤ}_{>0}$ and ${x}_{1},{x}_{2},\dots ,{x}_{N}\in X$ such that $A⊆ B(x1,ε)∪ B(x2,ε)∪⋯∪ B(xN,ε).$ 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\left(x,y\right)

Let $\left(X,d\right)$ 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 $\left\{B\left(x,\epsilon \right) | x\in X\right\}$ is a cover of $A\text{.}$ Since $A$ is compact there exists $N\in {ℤ}_{>0}$ and ${x}_{1},{x}_{2},\dots ,{x}_{N}\in X$ such that ${ B(x1,ε),…, B(xN,ε) }$ is a finite cover of $A\text{.}$ So $A$ is totally bounded. (b) Assume $A\subseteq X$ is totally bounded. Let $N\in {ℤ}_{>0}$ and ${x}_{1},{x}_{2},\dots ,{x}_{N}\in X$ such that ${ B(x1,1),…, B(xN,1) }$ is a cover of $A\text{.}$ Let $C=3+\text{max}\left\{d\left({x}_{k},{x}_{\ell }\right) | k,\ell \in \left\{1,2,\dots ,N\right\}\right\}\text{.}$ Let $x,y\in A\text{.}$ Let $i,j\in \left\{1,2,\dots ,N\right\}$ such that $x∈B(xi,1)and y∈B(xj,1).$ Then $d(x,y) ≤ d(x,xi)+ d(xi,xj)+ d(xj,y) ≤ 1+max { d(xk,xℓ) | k,ℓ∈ {1,2,…,N} } +1 < C.$ So $A$ is bounded. $\square$

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

Homework: Let $A=\left(0,1\right)\subseteq ℝ$ where $ℝ$ has the standard metric $d(x,y)=|x-y|.$ 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. $⇐$: 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 $𝒩= { Np | p∈K} ,$ of $K$ has no finite subcover. $⇒$: 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 $𝒞=\left\{{C}_{1},{C}_{2},\dots \right\}$. If $𝒞$ does not have a finite subcover then, for each $n$, $(C≤n) c ≠∅ but ∩n C≤nc =∅.$ 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.