09. Sequential Compactness

9.1 Definitions

Sequential Compactness

Let $X$ be a metric space then

$X$ is sequentially compact if any sequence of elements in $X$ has a convergent subsequence

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9.2 Closure and Boundedness Properties

Closed and Boundedness of Sequentially Compact Subspaces lemma

Sequentially compact subspace of a metric space is closed and bounded

Proof

Let $X$ be the metric space and $Y\subseteq X$ be the sequentially compact subspace

Suppose that $Y$ is not closed then $\overline{Y}\setminus Y$ is non-empty
Let $a\in \overline{Y}\setminus Y$ so by Interior points and convergent sequences then

There exists sequence $(y_n{)}_{n=1}^{\infty }$ of elements of $Y$ with ${\lim }_{n\to \infty }{y}_n=a$ then
Any subsequence of $(y_n{)}_{n=1}^{\infty }$ also converges to $a$ and by uniqueness of limits means
The sequence doesn’t converge to an element of $Y$ hence $Y$ is not sequentially compact

Hence by contrapositive if $Y$ is sequentially compact it is closed

Suppose that $Y$ is not bounded then
Pick arbitrary point $a_0\in Y$ and sequence $(y_n{)}_{n=1}^{\infty }$ such that

$$d(y_0,y_n)\ge n\text{ for all }n\in \mathbb{N}$$

Suppose subsequence $(y_{n_k}{)}_{n=1}^{\infty }$ converges to $b$ then for large $k$ then $d(y_{n_k},b)<1$ so

$$d(y_0,b)\ge d(y_0,y_{n_k})-d(y_{n_k},b)\ge n_k-1$$

Since $n_k\to \infty$ as $k\to \infty$ then as $d(y_0,b$ is fixed then it is a contradiction

Hence $Y$ is bounded

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Property of Closed Subset of a Sequentially Compact Metric Space lemma

Closed subset of a sequentially compact metric space is sequentialy compact

Proof

Let $X$ be the metric space and $Y$ be the closed subspace
Consider sequence $(y_n{)}_{n=1}^{\infty }$ of elements of $Y$

It is also a sequence of elements of $X$ so by sequential compactness of $X$ then
$(y_n{)}_{n=1}^{\infty }$ has a convergent subsequence converging to $a\in X$

As $Y$ is closed then the limit of any convergent subsequence of elements of $Y$ lies in $Y$
Hence

$$a\in Y$$

So $Y$ is sequentially compact

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9.3 Continuous Functions on Sequentially Compact Space

Image of a Sequentially Compact Metric Space under a Continuous Map lemma

Image of a sequentially compact metric space under a continuous map is sequentially compact

Proof

Let $X$ be sequentially compact
Suppose that $f:X\to Y$ is continuous

Let $\sigma =(f(x_n){)}_{n=1}^{\infty }$ be a sequence of elements of $f(X)$

Sequence $(x_n)$ contains a convergent subsequence $(x_{n_k})$ with

$$x_{n_k}\to a\text{ as }k\to \infty$$

Since $f$ is continuous then

$$f(x_{n_k})\to f(a)$$

Hence ${\sigma }^{\prime }=(f(x_{n_k}){)}_{k=1}^{\infty }$ is a convergent subsequence of $\sigma$

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Property of Continuous Function from Sequentially Compact Metric Space to $\mathbb{R}$ lemma

Continuous function from a sequentially compact metric space to $\mathbb{R}$ is uniformly continuous

Proof

Let $X$ be a sequentially compact metric space
Suppose $f:X\to \mathbb{R}$ be continuous but not uniformly continuous

There exists $ϵ>0$ such that for each $n\in \mathbb{N}$
There is $a_n,b_n\in X$ such that

$$d(a_n,b_n)<\frac{1}{n}\text{ but }|f(a_n)-f(b_n)|\ge ϵ$$

Since $X$ is sequentially compact then $(a_n{)}_{n=1}^{\infty }$ has subsequence converging to point $l$

Consider corresponding subsequence $(b_{n_k}{)}_{k=1}^{\infty }$
Since $d(a_{n_k},b_{n_k})\le \frac{1}{n_k}\to 0$ then $b_{n_k}$ also converges to $l$ as $k\to \infty$

Now $f$ is continuous at $l$ so there is $\delta >0$ such that for all $x\in X$ with $d(l,x)<\delta$ then

$$\left|f(l)-f(x)\right|<\frac{ϵ}{2}$$

For large $n$ then $\frac{d}{(}l,a_n)<\delta$ and $d(l,b_n)<\delta$ hence

$$ϵ\le |f(a_n)-f(b_n)|\le |f(a_n)-f(l)|+|f(l)-f(b_n)|<\frac{ϵ}{2}+\frac{ϵ}{2}$$

which is a contradiction hence $f$ is uniformly continuous

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9.4 Product Spaces

Convergence in Product Space with Convergence in Metric Space lemma

Let $X,Y$ be metric spaces then

Sequence $((x_n,y_n){)}_{n=1}^{\infty }$ in $X\times Y$ converges if and only if

$$(x_n{)}_{n=1}^{\infty }\text{ converges in }X\text{ and }(y_n{)}_{n=1}^{\infty }\text{ converges in }Y$$

Proof

Projection Maps $p_X:X\times Y\to X$ and $p_Y:X\times Y\to Y$ are continuous
$p_X$ and $p_Y$ are also Lipschitz continuous with constant $1$

If ${\lim }_{n\to \infty }(x_n,y_n)=(a,b)$ then

$$\begin{array}{rl}\underset{n\to \infty }{\lim }{x}_n& =\underset{n\to \infty }{\lim }{p}_X(x_n,y_n)=p_X(a,b)=a\\ \underset{n\to \infty }{\lim }{y}_n& =\underset{n\to \infty }{\lim }{p}_Y(x_n,y_n)=p_Y(a,b)=b\end{array}$$

Conversely if $x_n\to a$ and $y_n\to b$ then

$$d_{X\times Y}((x_n,y_n),(a,b))=\sqrt{d_X(x_n,a{)}^2+d_Y(y_n,b{)}^2}\to 0$$

as $n\to \infty$ hence $(x_n,y_n)\to (a,b)$ as $n\to \infty$

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Property of Product of Two Sequentially Compact Metric Space

Product of two sequentially compact Metric Space is sequentially compact

Proof

Let $((x_n,y_n{)}_{n=1}^{\infty }$ be a sequence in $X\times Y$
As $X$ is sequentially compact then sequence $\sigma =(x_n{)}_{n=1}^{\infty }$ has a convergent subsequence

$${\sigma }^{\prime }=(x_{n_k}{)}_{k=1}^{\infty }\text{ with }{x}_{n_k}\to a\text{ as }k\to \infty$$

Consider sequence $(y_{n_k}{)}_{n=1}^{\infty }$ in $Y$
As $Y$ is sequentially compact then it has a convergent subsequence

$$(y_{n_{k_r}}{)}_{r=1}^{\infty }\text{ with }{y}_{n_{k_r}}\to b\text{ as }r\to \infty$$

Let ${\sigma }^{\prime \prime }=(x_{n_{k_r}}{)}_{n=1}^{\infty }$ is a subsequence of ${\sigma }^{\prime }$ then it converges to $a$

By Convergence in product space with convergence in metric space then

$$(x_{n_{k_r}},y_{n_{k_r}})\to (a,b)\text{ as }r\to \infty$$

Hence $X\times Y$ is sequentially compact

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Bolzano-Weierstrass corollary

Any closed and bounded subset of ${\mathbb{R}}^n$ is sequentially compact

Proof

Let $X\subseteq {\mathbb{R}}^n$ be the set
Since $X$ is bounded, it is contained in some cube $[-M,M{]}^n$

By the Bolzano-Weierstrass Theorem on $\mathbb{R}$ then $[-M,M]$ is sequentially compact
Therefore by Property of product of two sequentially compact metric space then

$$[-M,M{]}^n\text{ is sequentially compact}$$

Since $X$ is closed, then it is sequentially compact by Property of closed subset of a sequentially compact metric space

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9.5 Sequentially Compact equals Complete and Totally Bounded

Property of Sequentially Compact Metric Spaces

Let $X$ be a sequentially compact metric space then

$$X\text{ is complete and bounded}$$

with the converse not holding in general

Proof

Suppose $X$ is sequentially compact then
By Closed and boundedness of sequentially compact subspaces then it is bounded

Suppose $(x_n{)}_{n=1}^{\infty }$ is a Cauchy Sequence in $X$
Since $X$ is sequentially compact then $(x_n{)}_{n=1}^{\infty }$ has a convergent subsequence $(x_{n_k}{)}_{k=1}^{\infty }$

Suppose ${\lim }_{k\to \infty }{x}_{n_k}=a$
For $ϵ>0$ then as $(x_n{)}_{n=1}^{\infty }$ is Cauchy there exists $N$ such that for all $n,m\ge N$ then

$$d(x_n,x_m)<\frac{ϵ}{2}$$

Since ${\lim }_{k\to \infty }{x}_{n_k}=a$ then there exists $k$ such that $n_k\ge k\ge nN$ and

$$d(x_{n_k},a)<\frac{ϵ}{2}$$

Hence

$$d(x_n,a)\le d(x_n,x_{n_k})+d(x_{n_k},a)<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ$$

Hence ${\lim }_{n\to \infty }{x}_n=a$

Converse Example
Let $C_b(\mathbb{R})$ be the normed space of continuous bounded functions on the real line with

$$||\cdot |{|}_{\infty }-\text{norm}\text{ and the associated metric}$$

Let $X=\overline{B}(0,1)$ (functions having sup norm bounded by $1$)
Hence $X$ is closed and bounded

As $C_b(\mathbb{R})$ is complete by Completeness of the Space of Bounded Continuous Functions
then $X$ is complete

Define function on $\varphi :\mathbb{R}\to \mathbb{R}\varphi :\mathbb{R}\to \mathbb{R}$ by

$$\varphi (t)=\{\begin{array}{ll}2t+1& -\frac{1}{2}\le t\le 0\\ & \\ 1-2t& 0\le t\le \frac{1}{2}\end{array}$$

With $\varphi (t)=0$ for $t\notin \left[-\frac{1}{2},\frac{1}{2}\right]$

For each $n\in N$ define $f_n(t)=\varphi (t-n)$
So all functions $f_n$ lie in $X=\overline{B}(0,1)$

However if $n\ne m$ then $f_n(n)=n$ whilst $f_m(n)=0$ so

$$||f_n-f_m|{|}_{\infty }=1$$

Hence sequence $(f_n{)}_{n=1}^{\infty }$ has no Cauchy subsequence thus no convergent subsequence

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15 - Sequentially Compactness Characterisation Theorem

Sequentially Compactness Characterisation Theorem

Let $X$ be a metric space hten

$$X\text{ is sequentially compact }⟺X\text{ is complete and totally bounded}$$

Proof

Suppose that metric space $X$ is sequentially compact then
By Property of sequentially compact metric spaces then $X$ is complete

Suppose that $X$ is not totally bounded
Let $ϵ$ be such there is no way to cover $X$ by finitely many open balls of radius $ϵ$

Using a greedy algorithm, select infinite sequence $(x_n{)}_{n=1}^{\infty }$ of elements of $X$
where the elements are separated by at least $ϵ$ so that

$$d(x_i,x_j)\ge ϵ\text{ for }i\ne j$$

Suppose $x_1,\cdots ,x_n$ have already been selected
By assumption then balls $B(x_i,ϵ)$ do not cover $X$
So select point $x_{n+1}\in X$ that doesn’t lie in any balls thus

$$d(x_i,x_{n+1})\ge ϵ\text{ for }i=1,\cdots ,n$$

It is clear that the sequence has no convergent subsequence hence by contradiction then
$X$ is totally bounded

Suppose that metric space $X$ is complete and totally bounded
Let $\sigma$ be a sequence of elements of $X$

Using the total boundedness assumption for balls of radii $1,\frac{1}{2},\frac{1}{4},\cdots$
Hence for non-negative integer $m$ there is a finite collection of open balls

$$B_1^{(m)},\cdots ,B_{k_m}^{(m)}\text{ of radius }{2}^{-m}\text{ which cover }X$$

Start with balls $B_1^{(0)},\cdots ,B_{k_0}^0$ of radius $1$
Then one of these balls contain infinitely many elements of sequence of sequence $\sigma$
Let $B_0$ be the ball with that property
Let ${\sigma }^{(0)}$ be the infinite subsequence of $\sigma$ of elements contained in $B_0$

Looking at balls $B_1^{(1)},\cdots ,B_{k_1}^{(1)}$ of radius $\frac{1}{2}$
Similarly one of the balls contains many elements of new subsequence ${\sigma }^{(0)}$
Let $B_1$ be ball
Let ${\sigma }^{(1)}$ be the infinite subsequence of ${\sigma }^{(0)}$ of elements contained in $B_1$

Continuing to produce new subsequences ${\sigma }^{(2)},\cdots$ with ${\sigma }^{(r)}$ contained in $B_r$ and ${\sigma }^{(r-1)}$

Consider sequence ${\sigma }^{\ast }$ obtained by diagonal argument
So the $i$th element of ${\sigma }^{\ast }$ is the $i$th element of ${\sigma }^{(i)}$
Hence ${\sigma }^{\ast }$ is a subsequence of $\sigma$ so write

$${\sigma }^{\ast }=(x_n{)}_{n=1}^{\infty }\text{ with }{x}_n\in B_r\text{ for all }n\ge r$$

We have that ${\sigma }^{\ast }$ is a Cauchy sequence
Given $ϵ>0$ let $N$ be such that $2^{-N}<\frac{ϵ}{2}$ hence for $n,m\ge N$ then
$x_n,x_m$ both lie in $B_N$ which is a ball of radius $2^{-N}<ϵ$ hence

$$d(x_n,x_m)<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ$$

Thus as $X$ is complete then sequence ${\sigma }^{\ast }$ converges
Thus $X$ is sequentially compact as $\sigma$ was an arbitrary sequence in $X$

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