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# 4.7: More on Compactness

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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. $$\square$$