Yongheng Zhang
When we say a metric space X is compact, usually we mean any open covering of X has a finite subcovering. If we have a collection of open sets which covers X, no matter whether it is a covering of a subcovering, the collection of the complement of each open set has an empty intersection by De Morgen’s law and vice versa. So another formulation of compactness is that any collection of closed subsets of X with an empty intersection has a finite subcollection also with an empty intersection. Furthermore, taking the contrapositive we have an equivalent defintion of compactness: for any col- lection C of closed subsets of X, if every finite subcollection of C has nonempty intersection, then the intersection of the sets in C is nonempty. We say a collection of sets of X has the finite intersection property if every finite subcollection of X has nonempty intersetion. So a concise version of the previous statement is
A space X is compact if and only if every collection of closed subsets of X with the finite intersec- tion property has nonempty intersection.
Equipped with the above statement, we will find a fine connection between compactness and the Bolzano-Weierstrass property. Recall that when we say a space X has the Bolzano-Weierstrass prop- erty, we mean any sequence in X has a cluster point (Is the cluster point in X?). In fact, compactness implies the Bolzano-Weierstrass property. To see why, let < xn > be a sequence in a ∞ ∞ compact space X. Then we consider the collection of closed sets A = {Ak}k=1 where Ak = {xn}n=k. Why is Ak ∈ X? Clearly A has the finite intersection property. Because X is compact, we know T∞ that k=1 Ak is nonempty. Let x be in the intersection. Let Ox be an open set containing x. For any N, x ∈ AN = {xN , xN+1,...}. So there is n ≥ N such that xn ∈ Ox. This precisely says that x is a cluster point of < xn >.
Recall that a space X is called sequentially compact if every sequence in it has a convergent sub- sequence. It can be readily checked that sequential compactness is the same as having the Bolzano- Weierstrass property (In the backward direction, we construct a subsequence converging to the cluster point).
Now assume only that X is sequentially compact. Let f ∈ C(X). We will see that f assumes its minimum (and maximum) and use it to show the existence of Lebesgue number of a compact metric space. Let m = infx∈X {f(x)}. By successively shrinking > 0 in the defintion of infimum, we obtain a sequence < xn > in X such that limn→∞ f(xn) = m. Since X is sequentially compact, < xn > has a subsequence < xnk > converging to some point x ∈ X. Because f is continuous, we have f(x) = limk→∞ f(xnk ). Thus, m = f(x), as desired. Similarly the maximum is achieved.
Given an open covering U of a metric space X, if there is a number > 0 such that for any 0 < δ < and x ∈ X, there is O ∈ U such that the open ball Bδ(x) is contained in O, then we say that is the Lebesgue number of the open covering U. In fact, if a space X is sequentially compact, then its Lebesgue number exists.
Now we prove this. Let U be an open covering of sequentially compact X. If X ∈ U, then ... On the other hand, suppose X/∈ U. Define a function f by f(x) = sup{r : there is O ∈ U such that Br(x) ⊂
1 2
O}. Obviously, f > 0. (But why f < ∞?) Notice that for any x, y ∈ X, f(y) ≥ f(x) − ρ(x, y). Thus, f is continuous. By sequential compactness, f assumes its minimum, i.e., there is x ∈ X and > 0 such that infx∈X f(x) = > 0. Therefore, for any δ < and x ∈ X, there is O ∈ U such that Bδ(x) ⊂ O. So is the Lebesgue number. (if we increase , then we can not guarantee the uniform containment.)
A metric space X is called totally bounded if for any > 0 it can be covered by a finite number of open balls with radius . One of its key words is finiteness, so it is closely related to compactness. Later we will use it to show that being sequentially compact implies compactness. But first, we prove that a sequentially compact space is totally bounded. We prove the contrapositive. Suppose X is not totally bounded, i.e., there is > 0 such that for any finite collection of points {x1, . . . , xN } in X, there is a point x ∈ X such that ρ(xn, x) ≥ for all n = 1,...,N. Given this > 0, there is x1 ∈ X such that there is x2 ∈ X such that ρ(x2, x1) ≥ . Then, there is x3 ∈ X such that ρ(x3, xn) ≥ for n = 1, 2. Inductively, there is xN+1 ∈ X such that ρ(xN+1, xn) ≥ for all n = 1,...,N. Hence, we have a sequence < xn > such that any two terms in them are at a distance at least . Therefore, it is impossible for < xn > to have a convergent subsequence.
Given the existence of the Lebesgue number and the property of totally bounded of a sequentially compact space, we show that the space is compact. To see this, let U be an open covering of compact X; let > 0 be the Lebesgue number and let 0 < δ < . Because X is totally bounded, there is a N finite collection of points {x1, . . . , xN } in X such that {Bδ(xn)}n=1 covers X. But each Bδ(xn) is in N some open set Oxn ∈ U. Consequently, the finite collection {Oxn }n=1 covers X and we are done.