Klee's Trigonometry Problem

Abstract

ment of F is continuous. Note as well that F is contained in xFLK D(0, r(x)), where D(0, r (x)) is the closed disk in the complex plane with center 0 and radius r (x). Let F1 denote F with the topology of uniform convergence, and let F2 denote F with the topology it inherits as a subspace of xFxK D(0, r (x)). We will show that F2 is a closed subspace of -xK D(0, r (x)) and that the identity map from F2 to F1 is continuous. It will then follow from the Tychonoff theorem that F2 (and hence FI) is compact. The fact that F2 is a closed subset of IXEK D(0, r(x)) is clear: if OEA is a net in F2 converging to f in xEK D(0, r (x)) then f,(x) -f(x) for each x in K, so f belongs to F2. The proof that the identity mapping from F2 to Fl is continuous is proved as in the traditional proof of the Ascoli-Arzelt Theorem: Suppose that E > 0. The fact that K is compact shows that there exist finitely many points xl, ... x, of K such that K = U=l (xj, E). Now if aEA is a net converging to f in F2 then there exists ao in A such that If,(xj) f(xj)l ao. Now for any such a and any x in K we may choose j in { 1,..., n} such that x belongs to co (xj, E). It follows that