Spaces for which the Stone-Weierstrass theorem holds

Abstract

If X is a topological space, a subset A of C{X), the set of bounded continuous real functions on X, is said to separate the points of X if for every pair x, y of distinct points of X there is a function/in A with f{x)^f{ y).A space X is called completely Hausdorff if C{X) separates the points of X.The Stone-Weierstrass theorem states : If X is a compact Hausdorff space, and if A is a subalgebra of C{X) which (i) separates the points of X and (ii) contains the constants, then A is uniformly dense in C{X){2).In the following, we shall say the Stone-Weierstrass theorem holds for a space X provided that X is completely Hausdorff, and that every subalgebra of C{X) which satisfies (i) and (ii) is uniformly dense in C{X).An extension space of a topological space X is a pair ( Y, h), where F is a topological space, n is a homeomorphism of X into Y, and h{X) is dense in Y; if h is the identity map, the reference to n is omitted, and Y itself is called an extension space of X; {Y, h) is called proper if h{X) is a proper subset of Y.We shall call a completely Hausdorff space X completely Hausdorff-complete if and only if X has no proper extension space ( Y, h) such that F is a completely Hausdorff space.A filter on a space X is called completely regular provided that it has a base SS of open sets such that for each B in â §, there is a set £'<=£ in 38 and a function fe C{X) which maps X into [0, 1], is 0 on £', and is 1 on X-B.In [3] Banaschewski proved that the Stone-Weierstrass theorem holds for a completely Hausdorff space X if and only if every completely regular filter Jon I has the property that C~\{F\Fe&}i=0.Using this result, we shall prove the following:Theorem 1.The Stone-Weierstrass theorem holds for a completely Hausdorff space X if and only if X is completely Hausdorff-complete.Proof.Suppose that X has a proper extension space ( Y, h) such that F is a completely Hausdorff space.Fix a point v in Y-h{X), and letZ={/£ C{h{X)) | /= g | h{X), where g is in C{Y), and g{y)=0}.For each/£Z and number 0<r, define W{f, t)={z\ -t<f{z)<t}, and let & be the filter on h{X) generated by the collection of all finite intersections of elements of {W{f t) \ fe Z, 0 < r}.It is not difficult to see that IF is a completely regular filter on h{X): take (~){W{fi,ti)\i=l,...,n}eßr;for each integer i, l&itiM, choose a number