21 - Sequential Compactness

Sequential Compactness

Let $X$ be a metric space then

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

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

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

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 $\sigma$

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

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

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

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