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 $X$