$\gdef\ucal{\mathcal U}$

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)

1. Compactness is an absolute (or intrinsic) property of a metric space. If $X$ is a metric space, and $K \In X$ is a subset, when we say $K$ is compact, we mean $K$ as a 'stand-alone' metric space (totally forgetting about $X$, but only using the distance function inherited from $X$) is compact.

2. We proved last time: If $K \In X$ is (open cover) compact, then $K$ is closed and bounded. (Discussion: if you replace open cover compact by sequential compactness, can you prove the two conclusions directly (without using the equivalence of the two definitions)?)

3. We know that $[0,1]$ is sequentially compact (by Heine-Borel theorem), can you show that $[0,1]^2$ is sequentially compact? (Hint: given a sequence $(p_n)$ in $[0,1]^2$, first show that you can pass to subsequence to make the sequence of the first coordinate converge, then …)

Have you wondered, what subset of a metric space is both open and closed?

We say a metric space $X$ is connected, if $X$ cannot be written as disjoint union of two non-emtpy open subset. In other word, the only subsets in $X$ that is both open and closed are $X$ and $\emptyset$.

For example, $\Q$ is not connected, since $(-\infty, \sqrt{5})$ is both open and closed in $Q$. (why?)

For example, $X=\{1,2,3\}$ (with induced metric from $\R$) is not connected, since $\{1\}$ is both open and closed in $X$. (Discussion: Equip $X$ with the induced metric, can you show that $\{1\}$ is both open and closed? Equip $X$ with the induced topology, can you show that $\{1\}$ is both open and closed? )

Theorem: a subset $E \In \R$ is connected, if and only if, for any $x,y \in E$, we have $[x,y] \In E$.
Proof: next time.