The Gelfand Subalgebra of Real or Nonarchimedean Valued Continuous Functions

Abstract

Let L be either the field of real numbers or a nonarchimedean rank-one valued field. For topological space T we study the Gelfand subalgebra CO(T, L) of the algebra of all L-valued continuous functions C(T, L). The main result is that if T is a paracompact locally compact Hausdorff space, which is ultraregular if L is nonarchimedean, then for f E C(T, L) the following statements are equivalent: (1) There exists a compact set K C T such that f(T K) is finite, (2) f has finite range on every discrete closed subset of T, and (3) f E CO(T, L). Throughout this paper L will stand for either the valued field R or a nonarchimedean rank-one valued field, and T for a Hausdorff topological space. T will be assumed completely regular in the real case and ultraregular in the nonarchimedean case. We will denote by C(T, L), or simply by C if there is no confusion, the algebra over L consisting of all L-valued continuous functions on T, and by CK the ideal of those continuous f E C with compact support. Let 9DT be the set of maximal ideals of C, and denote by C0 the Gelfand subalgebra of C, consisting of all f e C with the property that, for each M E 'DT, there exists X E L such that (f X) E M. (The concept of Gelfand subalgebras of general algebras has been introduced by N. Shell -formerly N. Shilkret-in [6].) We will denote by CF the subalgebra of C consisting of those f E C for which there exists a compact set K C T such that f(T K) is finite. PROPOSITION 1. For any f E CO, f(T) is compact. PROOF. First, we prove that f(T) is a precompact set. Real case. It suffices to see that f is bounded, and this follows from [5, 5.7(b)]. Nonarchimedean case. Take > 0. Note that any two (closed-open) --radius spheres B(a) = ( LI IE A a I < e} are equal or disjoint. Choose an indexed set (aci),I inf(T) such that B(aci),EI is disjoint and covers f(T). We claim I is finite. In fact, assume, to the contrary, that I is infinite. Let Ai = U,#,f'-(B(a,)). Since the family of closed-open sets (A,)iEI has the finite intersection property, there exists M E 91T such that Ai E Z[M] for any i E I. On the other hand, since f E C0 there exists X E L such that (fX) C M and hence Z(fx) n A, # 0 for any i E I, which is a contradiction. Thus I is finite and so f(T) is precompact. Received by the editors April 20, 1982 and, in revised form, February 1, 1983. 1980 Mathematics Subject Classification. Primary 54C40.