$\gdef\ucal{\mathcal U}$

Last time we discussed two versions of compactness: the sequential compactness and the open cover compactness.

Theorem: let $X$ be a metric space. Then, $X$ is sequentially compact, if and only if $X$ is open cover compact.

Last time we have shown open cover compactness implies sequential compactness, today we show the converse.

Suppose a metric space $(X,d)$ is sequentially compact, namely, any sequence $(p_n)$ in $X$ has a convergent subsequence. How to show that any open cover $\ucal$ of $X$ has a finite sub-cover? We first introduce

Definition (A Lebesgue number of a cover $\ucal$) we say $\lambda$ is a Lebesgue number of the covering $\ucal$, if for any $p \in X$, there is some $U \in \ucal$, such that $B_\lambda(p) \In U$.

Lemma Every open cover $\ucal$ of a sequentially compact space $X$ has a Lebesgue number $\lambda>0$.
Proof: suppose not, then for every $\lambda$, there exists a point $p$, such that $B_\lambda(p)$ is not contained in any $U \in \ucal$. Let $\lambda$ take values $1/n$ for $n=1,2,\cdots$, and we get a sequence $p_n$ for $\lambda=1/n$. By sequential compactness, this sequence has a convergent subsequence, say converge to $p \in X$. However, $p$ is contained in some open set $U_0 \in \ucal$, thus there is $r>0$ that $B_r(p) \In U_0$. Thus, let $\epsilon = r/3$, and $N$ large enough such that $1/N < r/3$, since $p_n$ sub-converge to $p$, there exists $n > N$ with $d(p_n, p) < \epsilon$. Thus, we have $$ B_{1/n}(p_n) \In B_{r/3}(p_n) \In B_r(p) \In U_0 $$ which contradict with $B_{1/n}(p_n)$ is not contained in any $U \in \ucal$. This proves the lemma.

Lemma If $X$ is a sequentially compact space, then for any $r>0$, $X$ can be covered by finitely many open balls of radius $r$. Proof: We claim that $X$ can only contain finitely many disjoint open balls with radius $r/2$. Then, take such a maximal $r/2$-radius ball packing, and replace the radius $r/2$ balls by radius $r$ balls, claim the bigger balls cover $X$. (Discussion: prove the two claims)

Given the above two lemma, one can prove that any open cover of a sequentially compact space $X$ admits a finite subcover. (Discussion: prove it)