4.7: More on Compactness
This page is a draft and is under active development.
( \newcommand{\kernel}{\mathrm{null}\,}\)
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 Gi(i∈I) iff
F⊆⋃i∈IGi.
If this is the case, {Gi} is called a covering of F. If the sets Gi are open, we call the set family {Gi} an open covering. The covering {Gi} is said to be finite (infinite, countable, etc.) iff the number of the sets Gi is.
If {Gi} is an open covering of F, then each point x∈F is in some Gi and is its interior point ( for Gi is open ), so there is a globe Gx(εx)⊆Gi. In general, the radii εx of these globes depend on x, i.e., are different for different points x∈F. If, however, they can be chosen all equal to some ε, then this ε is called a Lebesgue number for the covering {Gi} (so named after Henri Lebesgue). Thus ε is a Lebesgue number iff for every x∈F, the globe Gx(ε) is contained in some Gi. We now obtain the following theorem.
(Lebesgue). Every open covering {Gj} of a sequentially compact set F⊆(S,ρ) has at least one Lebesgue number ε. In symbols,
(∃ε>0)(∀x∈F)(∃i)Gx(ε)⊆Gi.
- Proof
-
Seeking a contradiction, assume that (1) fails, i.e., its negation holds. As was explained in Chapter 1, §§1-3, this negation is
(∀ε>0)(∃xε∈F)(∀i)Gxε(ε)⊈Gi
(where we write xε for x since here x may depend on ε). As this is supposed to hold for all ε>0, we take successively
ε=1,12,…,1n,…
Then, replacing "xε" by "xn" for convenience, we obtain
(∀n)(∃xn∈F)(∀i)Gxn(1n)⊈Gi.
Thus for each n, there is some xn∈F such that the globe Gxn(1n) is not contained in any Gi. We fix such an xn∈F for each n, thus obtaining a sequence {xn}⊆F. As F is compact (by assumption), this sequence clusters at some p∈F.
The point p, being in F, must be in some Gi( call it G), together with some globe Gp(r)⊆G. As p is a cluster point, even the smaller globe Gp(r2) contains infinitely many xn. Thus we may choose n so large that 1n<r2 and xn∈Gp(r2). For that n,Gxn(1n)⊆Gp(r) because
(∀x∈Gxn(1n))ρ(x,p)≤ρ(x,xn)+ρ(xn,p)<1n+r2<r2+r2=r.
As Gp(r)⊆G (by construction), we certainly have
Gxn(1n)⊆Gp(r)⊆G.
However, this is impossible since by (2) no Gxn(1n) is contained in any Gi. This contradiction completes the proof. ◻
Our next theorem might serve as an alternative definition of compactness. In fact, in topology (which studies more general than metric spaces), this is is the basic definition of compactness. It generalizes Problem 10 in §6.
(generalized Heine-Borel theorem). A set F⊆(S,ρ) is compact iff every open covering of F has a finite subcovering.
That is, whenever F is covered by a family of open sets Gi(i∈I),F can also be covered by a finite number of these Gi.
- Proof
-
Let F be sequentially compact, and let F⊆∪Gi, all Gi open. We have to show that {Gi} reduces to a finite subcovering.
By Theorem 1,{Gi} has a Lebesgue number ε satisfying (1). We fix this ε>0. Now by Note 1 in §6, we can cover F by a finite number of ε -globes,
F⊆n⋃k=1Gxk(ε),xk∈F.
Also by (1), each Gxk(ε) is contained in some Gi; call it Gik. With the Gik so fixed, we have
F⊆n⋃k=1Gxk(ε)⊆n⋃k=1Gik.
Thus the sets Gik 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 {xm}⊆F clusters at some p∈F.
Seeking a contradiction, suppose F contains no cluster points of {xm}. Then by definition, each point x∈F is in some globe Gx containing at most finitely many xm. 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 xm (namely, those contained in the so-selected globes), whereas the sequence {xm}⊆F was assumed infinite. This contradiction completes the proof. ◻