Metric Spaces

$$\newcommand{\ms}{(X, d)} \newcommand{\Rn}{\mathbb R^n} \newcommand{\nat}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\bd}{\boldsymbol} \newcommand{\ltwo}{\mathcal l^2} \newcommand{\clo}{\overline} \newcommand{\O}{\mathcal O} $$

1. Metric Space

Definition. A pair $(X, d)$ is called a metric space if $X$, as a set, is equipped with a mapping

$$d:X\times X\to\mathbb{R}_{\geq0} $$

that satisfies: $\forall x,y,z\in X$

1. $d(x,x)=0$
2. $d(x,y)=d(y,x)$
3. $d(x,y)\leq d(x,z)+d(z,y)$

This mapping $d$ is called a metric of $X$.

Definition. Neighborhood

Given a metric space $(X, d)$, a set $N_r(p)$ is called a neighborhood of a point $p\in X$ with radius $r$ if

$$N_r(p)=\{x\mid d(x,p)<r\} $$

Definition. Interior Point

Given a metric space $(X, d)$, a subset $E\subset X$, a point $p\in E$ is called an interior point if

$$\exists \epsilon>0. N_\epsilon(p)\subset E $$

The set of all interior points of $E$ is denoted by $E^\circ$

Definition. Openness

Given a metric space $(X, d)$, a subset $E\subset X$ is called open if $E=E^\circ$

Definition. Limit Point

Given a metric space $(X, d)$ with a subset $E\subset X$, a point $p$ is called a limit point of $E$ if

$$\forall \epsilon>0.N_\epsilon(p)\cap E\setminus\{p\}\neq \emptyset $$

The set of all limit points of $E$ is denoted by $E'$, called the derivative set of $E$

Definition. Closedness

Given a metric space $(X, d)$, a subset $E\subset X$ is called closed if $E'\subset E$

Theorem. A set $E$ is open in $X$ iff $E^c$ is closed.

Pf. The proof instantly comes from the definition of closedness and opensess

Theorem. Openness/Closedness is independent of metric

Let $d_1$, $d_2$ be two metrics on $X$, then TTFE.

1. $U\subset X$ is $d_1$-open iff it’s $d_2$-open
2. $C\subset X$ is $d_1$-closed iff it’s $d_2$-closed.
3. a sequence in $X$ is $d_1$-convergent iff it’s $d_2$-convergent.

Pf. Instantly comes from definition.

The above theorem shows that openess/closedness is independent of the choice of metric, this is in fact because these two properties are topological properties of a set, as we’ll see soon in this post

Definition. Boundedness

Given a metric space $(X, d)$, a subset $E\subset X$ is called bounded if

$$\exists p\in X.\exists r>0.\forall x\in E.d(x,p)<r $$

Definition. Distance

Given a metric space $(X, d)$, a subset $E\subset X$, the distance between point $x\in X$ to the subset $E$ is defined as

$$d(x,E):=\inf\{d(x,p)\mid p\in E\} $$

Definition. $\epsilon$-dense, $\epsilon$-net

Given a metric space $(X, d)$, a subset $E\subset X$. A set $N=N(\epsilon)$ is called $\epsilon$-dense in $E$ if

$$\forall x\in E.d(x,N)\leq \epsilon $$

Then we call $N$ an $\epsilon$-net of $E$.

Definition. Totally boundedness (Precompactness)

Given a metric space $(X, d)$, a subset $E\subset X$ is called totally bounded if $\forall \epsilon>0$. there exists an finite $\epsilon$-net of $E$.

Lemma. Closure of precompact set if compact

If $Y\subset \ms$ is precompact, then $\clo Y$ is compact

sometimes this is the definition, while the above definition will be a property as a result.

Lemma. Totally boundedness $\implies$ boundedness

Definition. Completeness

A metric space $\ms$ is called complete if every Cauchy sequence in $X$ has a convergent subsequence.

Theorem. Rudin Def 3.12

Closed subsets of complete set is complete

Definition. Open cover

Given a metric space $\ms$, a collection $\mathcal O=\{O_\alpha\}_{\alpha\in\Lambda}\subset X$ is called an open cover of a subset $U\subset X$ if

1. $\forall \alpha\in\Lambda.O_\alpha$ is open
2. $U\subset \bigcup_{\alpha\in\Lambda}O_\alpha$

Definition. Compactness

Given a metric space $\ms$, a set $K\subset X$ is called compact if any open cover of $K$ can be reduced to a finite subcover.

Theorem. Rudin Thm 2.35

Closed subsets of a compact set is compact

Example.

1. $\{\frac 1 n\mid n\in \nat\}$ is not compact in $\mathbb R$.
2. $\{0\}\cup \{\frac 1 n\mid n\in \nat\}$ is compact in $\mathbb R$.

Definition. Sequentially Compactness

A metric space $\ms$ is called sequentially compact if every sequence in $X$ has a convergent subsequence.

2. Compactness

We are going to talk about several properties of compact metric spaces here.

Theorem. Bolzano-Weierstrass Theorem

Every closed and bounded subset of $\Rn$ is sequentially compact.

Theorem. Lebesgue Number Lemma

$\ms$ is sequentially compact, $\mathcal O=\{O_\alpha\mid \alpha\in \Lambda\}$ is an open covering of $X$. Then $\exists \delta>0$, s.t.

$$\forall p\in X.\exists \alpha \in \Lambda.N_{\delta}(p)\subset O_{\alpha} $$

Such $\delta$ is called a Lebesgue number of $\mathcal O$

Pf. A detailed proof can be found on wikipedia.

LN Lemma describes how effective an open cover is.

BTW this theorem looks like axiom of choice so much, and in fact one of its proof is shown using axiom of choice.

Theorem. Heine-Borel Theorem

Let $\ms$ be a metric space, then TTFE

1. $X$ is compact
2. $X$ is sequentially compact
3. $X$ is complete and totally bounded

Pf. I’ll prove by showing $1\iff 2\iff 3$
($1\implies 2$)
Let $(X, d)$ be a compact metric space. Let $S=\{x_n\}_{n=1}^\infty$ be a Cauchy sequence in $X$ but does not converge. Then by definition,

$$\forall x\in X. \exists r=r(x) >0.N_r(x)\cap S\setminus \{x\} $$

Since $\{N_{r_x}(x)\mid x\in X\}$ is an open cover of $X$, by compactness there exists a finite subcover, denoted by $\{N_{r_i}(p_i)\}_{i=1}^n= X$. Note that

$$\forall i\in \{1, 2,\ldots, n\}.N_{r_i}(p_i)\cap S\setminus \{p_i\} =\emptyset $$

so

$$\begin{aligned} & &\forall i\in \{1, 2,\ldots, n\}.N_{r_i}(p_i)\cap S &\subset \{p_i\}&\\ & \implies &(\bigcup_{i=1}^n N_{r_i}(p_i))\cap S &\subset \{p_i\}_{i=1}^n&\\ & \iff & S &\subset\{p_i\}_{i=1}^n& \end{aligned} $$

So the Cauchy sequence that we constructed is finite (not even Cauchy), contradiction.

($2\implies 3$)
It’s obvious that if $X$ is sequentially compact, then $X$ is complete, i.e. any Cauchy sequence must converge because if $n$ is sufficiently large, $x_n$ are sufficiently closed to each other. So we’ll only prove $X$ is totally bounded in this case. If $X$ is not totally compact, i.e. $\exists \epsilon>0$ such that $X$ has no $\epsilon$-nets, then by definition we can pick a sequence $\{p_n\}_{n=1}^\infty$ such that

$$\begin{aligned} p_1&\in X& & N_{\epsilon}(p_1)\subsetneq X\\ p_2&\in (X\setminus N_{\epsilon}(p_1))& & \bigcup_{i=1}^nN_\epsilon (p_i)\subsetneq X\\ &\vdots& & \vdots\\ p_{n+1}&\in(X\setminus \bigcup_{i=1}^nN_{\epsilon}(p_i))&&\bigcup_{i=1}^{n}N_{\epsilon}(p_i)\subsetneq X\\ &\vdots&&\vdots \end{aligned} $$

By the way we construct the sequence, it’s obvious that it has no convergent subsequence because any two elements in the sequence have distance no smaller than $\epsilon$. By contrapositive, the proof is done.

($3\implies 2$)
A detailed proof could be found here. This is the most non-trivial part of this post.

($2\implies 1$)
If $X$ is sequentially compact. Take $\O=\{O_\alpha\mid \alpha\in \Lambda\}$ be an open cover of $X$. By Lebesgue Number Lemma, a sequentially compact metric space has its Lebesgue number, denoted by $\delta$. Since we’ve shown ($2\iff 3$), we know $X$ is also totally bounded, so there exists a finite $\delta$-net

$$\{x_n\}_{n=1}^N $$

such that

$$X=\bigcup_{n=1}^NN_{\delta}(x_n) $$

By Lebesgue’s Number Lemma, we know $X$ is an (open) cover of $X$, and the proof is done

Note that the completeness in the third statement shouldn’t be substituted with “closed”, considering the following example

$X=(0,1]\subset \R$ as a metric space, then

• $(0, \frac 1 2]$ is closed $X$ because it contains all its limit points
• $(0, \frac 1 2]$ is totally bounded because $\forall \epsilon>0$, one can separate the set into no more than $\lfloor{\frac 1 {2\epsilon}}\rfloor + 1$ segments and construct such an $\epsilon$-net
• but $(0,\frac 1 2]$ is not compact, considering such an open cover $\{(\frac 1 n, \frac 1 2]\mid n\in\nat \}$

Such a space is called a Ethanian Space, where Ethan is my friend lol.

Corollary. Heine-Borel Theorem in $\Rn$

Let $K\subset \Rn$, then TTFE.

1. $K$ is compact
2. $K$ is sequentially compact
3. $K$ is closed and bounded.

If not treated as a corollary, the formal definition of the above theorem could be found at Baby Rudin Thm 2.41

In fact, Heine-Borel is first introduced in $\Rn$, and what we proved above in an arbitrary space is usually called Extended Heine-Borel Theorem.

A metric space $\ms$ is said to have Heine-Borel property if any closed and bounded set in $X$ is compact. Bolzano-Weierstrass theorem shows that $\Rn$ has Heine-Borel property.

So far the following map can precisely describe relations between all these properties of metric spaces that we’ve learn.

$$\require{AMScd} \begin{CD} \text{sequentially compact}\\ @|\\ \text{compact} \\ @|\\ \text{complete and totally bounded}@>>>\text{complete}@>>> \text{closed}\\ @VVV\\ \text{totally bounded}\\ @VVV\\ \text{bounded} \end{CD} $$

Counterexamples

1. Closed and bounded $\not\Rightarrow$ sequentially compact

In $\ltwo$-space where $\ltwo:=\{\bd x\in\R^\omega\mid \sum_{i=1}^\infty x_i^2<\infty\}$ and $d_{\ltwo}:=(x, y)\mapsto \sum_{i=1}^\infty (x_i-y_i)^2$, consider $\{e_n\}_{n\in\nat}\subset \ltwo$

$$e_n:=\alpha\mapsto \begin{cases} 1 & \alpha = n\\ 0 & \text{else} \end{cases} $$

we have

• $\{e_n\}$ is closed because it has no limit points
• $\{e_n\}$ is bounded by definition

obviously has no convergent subsequence because it’s discrete.

3. Several applications

Definition. Given a metric space $\ms$, a pair $(Y, d)$ is called a subspace of $\ms$ if $Y\subset X$.

Theorem. Completeness of subspace

$(Y, d)$ is a subspace of a complete metric space $\ms$, then $Y$ is complete iff $Y$ is closed.

Pf.

$(\Leftarrow)$ Let $\{x_n\}$ be a Cauchy sequence in $Y$, then by definition $\{x_n\}$ is Cauchy in $X$. Since $X$ is complete, suppose

$$\lim_{n\to\infty}x_n=x_0 $$

Since $Y$ is closed, $x_0\in Y$. Since the choice of sequence is arbitrary, we have $\{x_n\}$ converges in $Y$, then $Y$ is complete.

$(\Rightarrow)$ Let $x_0\in Y'$. By definition there exists a sequence $\{x_n\}\in Y$ such that

$$\lim_{n\to\infty}x_n=x_0 $$

Since $\{x_n\}$ converges in $X$, it is then Cauchy in $X$, by definition it is then Cauchy in $Y$. Since $Y$ is complete, we have $x_0\in Y$. Since the choice of $x_0$ is arbitrary, we have $Y'\subset Y$, indicating that $Y$ is closed.

Definition. Product Space

$(X, d_X)$ and $(Y, d_Y)$ are two metric spaces, then the set $(X\times Y, d)$ is called a product space of $X$ and $Y$ with the metric space $d:(X\times Y)^2\to\R_{\ge 0}$ defined by

$$((x_1, y_1), (x_2, y_2))\mapsto \sqrt{d_X^2(x_1, x_2)+d_Y^2(y_1, y_2)} $$

One can easily check that this metric is well-defined. The non-trivial part is triangle property, which could be shown using Cauchy-Schwarz inequality.

Property. Let $X\times Y$ be the product space of two metric spaces $X$ and $Y$, then the following are true

1. $\forall r>0.\forall x\in X,y\in Y.N_r(x)\times N_r(y)$ is open in $X\times Y$
2. If $X$ and $Y$ are both complete, then $X\times Y$ is complete
3. if $X$ and $Y$ are both sequentially compact, then $X\times Y$ is sequentially compact.