Ph.D. Qualifying Exam – Spring 2004

Fourth problemLet X be a metric space.1. Prove that if X is countably compact, then X is compact.2. Prove that if every continuous function on X is bounded, then X is compact.Solution1 Suppose that X is countably compact, which means that every countable open cover ofX has a finite subcover. Then we can derive a property analogous to the finite intersectionproperty: if (F n ) n∈N is a (countable) collection of closed sets such that ⋂ F n = ∅, thereexists a finite subcollection F n1 , . . .,F nk such that F n1 ∩ · · · ∩ F nk = ∅.n∈NThis is proven exactly as the finite intersection property is proven: the (F c n) n∈Nform a countable open cover of X. Thus there is a finite subcover F c n 1∪ · · · ∪ F c n kof X, and we have F n1 ∩ · · ·F nk = ∅.Because X is a metric space, sequential compactness is equivalent to compactness. Solet (x n ) n∈N be a sequence of elements of X. The set of limit points of (x n ) n∈N isL = ⋂ {x k | k n}Suppose that L is empty. Then there exist integers n 1 . . . n j such that∅ = j ⋂n∈Nk=1{x k | k n j }But these closed sets form a decreasing sequence, therefore∅ = {x k | k n j }which is, of course, impossible. Hence L is not empty: (x n ) n∈N has convergent subsequences.2 Suppose that X is not compact. There exists a sequence (x n ) n∈N of elements of Xthat has no convergent subsequence. Denote by X the collection (x n ) n∈N . Then X isclosed: indeed, a convergent sequence of distinct points of X would allow us to constructa convergent subsequence of (x n ) n∈N .The same conclusion holds for X n = X \ {x n }, for every integer n: this set is closed.Since it does not contain x n , the distance dist(x n , X n ) is a positive real number r n .6