4.7: More on Compactness

Another useful approach to compactness is based on the notion of a covering of a set (already encountered in Problem 10 in §6). We say that a set $F$ is covered by a family of sets $G_{i}(i \in I)$ iff

$$F \subseteq \bigcup_{i \in I} G_{i}.$$

If this is the case, $\left\{G_{i}\right\}$ is called a covering of $F .$ If the sets $G_{i}$ are open, we call the set family $\left\{G_{i}\right\}$ an open covering. The covering $\left\{G_{i}\right\}$ is said to be finite (infinite, countable, etc.) iff the number of the sets $G_{i}$ is.

If $\left\{G_{i}\right\}$ is an open covering of $F,$ then each point $x \in F$ is in some $G_{i}$ and is its interior point $\left(\text { for } G_{i} \text { is open }\right),$ so there is a globe $G_{x}\left(\varepsilon_{x}\right) \subseteq G_{i} .$ In general, the radii $\varepsilon_{x}$ of these globes depend on $x,$ i.e., are different for different points $x \in F .$ If, however, they can be chosen all equal to some $\varepsilon$, then this $\varepsilon$ is called a Lebesgue number for the covering $\left\{G_{i}\right\}$ (so named after Henri Lebesgue). Thus $\varepsilon$ is a Lebesgue number iff for every $x \in F,$ the globe $G_{x}(\varepsilon)$ is contained in some $G_{i} .$ We now obtain the following theorem.

Theorem $\PageIndex{1}$

(Lebesgue). Every open covering $\left\{G_{j}\right\}$ of a sequentially compact set $F \subseteq(S, \rho)$ has at least one Lebesgue number $\varepsilon .$ In symbols,

$$(\exists \varepsilon>0)(\forall x \in F)(\exists i) \quad G_{x}(\varepsilon) \subseteq G_{i}.$$

Proof

Seeking a contradiction, assume that $(1)$ fails, i.e., its negation holds. As was explained in Chapter 1, §§1-3, this negation is

$$(\forall \varepsilon>0)\left(\exists x_{\varepsilon} \in F\right)(\forall i) \quad G_{x_{\varepsilon}}(\varepsilon) \nsubseteq G_{i}$$

(where we write $x_{\varepsilon}$ for $x$ since here $x$ may depend on $\varepsilon ) .$ As this is supposed to hold for all $\varepsilon>0,$ we take successively

$$\varepsilon=1, \frac{1}{2}, \ldots, \frac{1}{n}, \dots$$

Then, replacing $" x_{\varepsilon} "$ by $" x_{n} "$ for convenience, we obtain

$$(\forall n)\left(\exists x_{n} \in F\right)(\forall i) \quad G_{x_{n}}\left(\frac{1}{n}\right) \nsubseteq G_{i}.$$

Thus for each $n,$ there is some $x_{n} \in F$ such that the globe $G_{x_{n}}\left(\frac{1}{n}\right)$ is not contained in any $G_{i} .$ We fix such an $x_{n} \in F$ for each $n,$ thus obtaining a sequence $\left\{x_{n}\right\} \subseteq F .$ As $F$ is compact (by assumption), this sequence clusters at some $p \in F .$

The point $p,$ being in $F,$ must be in some $G_{i}(\text { call it } G),$ together with some globe $G_{p}(r) \subseteq G .$ As $p$ is a cluster point, even the smaller globe $G_{p}\left(\frac{r}{2}\right)$ contains infinitely many $x_{n} .$ Thus we may choose $n$ so large that $\frac{1}{n}<\frac{r}{2}$ and $x_{n} \in G_{p}\left(\frac{r}{2}\right)$. For that $n, G_{x_{n}}\left(\frac{1}{n}\right) \subseteq G_{p}(r)$ because

$$\left(\forall x \in G_{x_{n}}\left(\frac{1}{n}\right)\right) \quad \rho(x, p) \leq \rho\left(x, x_{n}\right)+\rho\left(x_{n}, p\right)<\frac{1}{n}+\frac{r}{2}<\frac{r}{2}+\frac{r}{2}=r.$$

As $G_{p}(r) \subseteq G$ (by construction), we certainly have

$$G_{x_{n}}\left(\frac{1}{n}\right) \subseteq G_{p}(r) \subseteq G.$$

However, this is impossible since by $(2)$ no $G_{x_{n}}\left(\frac{1}{n}\right)$ is contained in any $G_{i} .$ This contradiction completes the proof. $\square$

Our next theorem might serve as an alternative definition of compactness. In fact, in topology (which studies more general than metric spaces), this $i s$ is the basic definition of compactness. It generalizes Problem 10 in §6.

Theorem $\PageIndex{2}$

(generalized Heine-Borel theorem). A set $F \subseteq(S, \rho)$ is compact iff every open covering of $F$ has a finite subcovering.

That is, whenever $F$ is covered by a family of open sets $G_{i}(i \in I), F$ can also be covered by a finite number of these $G_{i}$.

Proof

Let $F$ be sequentially compact, and let $F \subseteq \cup G_{i},$ all $G_{i}$ open. We have to show that $\left\{G_{i}\right\}$ reduces to a finite subcovering.

By Theorem $1,\left\{G_{i}\right\}$ has a Lebesgue number $\varepsilon$ satisfying $(1) .$ We fix this $\varepsilon>0 .$ Now by Note 1 in §6, we can cover $F$ by a finite number of $\varepsilon$ -globes,

$$F \subseteq \bigcup_{k=1}^{n} G_{x_{k}}(\varepsilon), \quad x_{k} \in F.$$

Also by $(1),$ each $G_{x_{k}}(\varepsilon)$ is contained in some $G_{i} ;$ call it $G_{i_{k}} .$ With the $G_{i_{k}}$ so fixed, we have

$$F \subseteq \bigcup_{k=1}^{n} G_{x_{k}}(\varepsilon) \subseteq \bigcup_{k=1}^{n} G_{i_{k}}.$$

Thus the sets $G_{i_{k}}$ constitute the desired finite subcovering, and the "only if' in the theorem is proved.

Conversely, assume the condition stated in the theorem. We have to show that $F$ is sequentially compact, i.e., that every sequence $\left\{x_{m}\right\} \subseteq F$ clusters at some $p \in F .$

Seeking a contradiction, suppose $F$ contains $n o$ cluster points of $\left\{x_{m}\right\} .$ Then by definition, each point $x \in F$ is in some globe $G_{x}$ containing at most finitely many $x_{m} .$ The set $F$ is covered by these open globes, hence also by finitely many of them (by our assumption). Then, however, $F$ contains at most finitely many $x_{m}$ (namely, those contained in the so-selected globes), whereas the sequence $\left\{x_{m}\right\} \subseteq F$ was assumed infinite. This contradiction completes the proof. $\square$