4.7: More on Compactness
- Page ID
- 20964
This page is a draft and is under active development.
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.
(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
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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.
(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
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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. \(\square\)