By R. B. Burckel

Publication via Burckel, R. B.

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**Sample text**

Therefore A Hence is uniformly closed A::> A]R + iA]R C]R(X) +iC]R(X) =C(X). A. Browder gives another nice proof of this on pp. 88-89 of [8]. Also Arenson [1] gives a proof. 4 If X lying in is a compact Hausdorff space, A C(X) which contains the constants and separates the points of implies ~ A~ must have Re A C~(X) uniformly closed in C ~(X). : a Banach algebra A~ and C ~ (X). is uniformly dense in Re A Since is assumed to be uniformly closed, we Re A = C ~(X). : and the injection is continuous.

Is uniformly dense in Re A Since is assumed to be uniformly closed, we Re A = C ~(X). : and the injection is continuous. -= t=( IN,F). F is complete and E C F E C F. F If then E M such that is bounded in F so and evidently Now suppose E is complete and show first that for some (2) is dense in E = {1,2, •••• } Then By hypothesis there is a constant Therefore any bounded sequence in E are (real or complex) IN,E), the bounded functions from IN normed as usual, and in addition E,F x E F & E is dense in F.

KEC That is, f( P a minimal one E. By hypothesis on not perfect and so has an isolated point is closed and so by minima1ity of E f{x O ) is compact,so is is isolated in f pre-images X, the set xO. Then we must have P = f{E), that is, P\{f{XO)} = f{E)\{f{X O )} E\{xO } n K) = P. KEC There is therefore in the family of all compact of e), = E is E\{XO} f{E\{X O }) f{E\{X O}). F As f{E\(X O }) = P\{f{XO)}' and so P, our final contradiction. 1. 10, is from the paper of Bade and Curtis in [8J, pp. 90-92: they 24 attribute the proof to Katznelson.