A solution to the Blumberg problem

Abstract

The question arose: Which Baire spaces are Blumberg? and, in particular, the Blumberg problem: Must every compact Hausdorff space be Blumberg? Partial answers were given by R. Levy [5], [6] and H. E. White [8]. In fact, there is a non-Blumberg compact Hausdorff space. An unusual feature of our example is that it is the disjoint union of two spaces, one or the other of which fails to be Blumberg, depending on whether or not the continuum hypothesis holds. Let B be the Boolean algebra of Lebesgue measurable subsets of [0, 1], and let / be the ideal of null sets of B. The reduced measure algebra is the complete Boolean algebra B/I. Let St(B/I) be the Stone space of the reduced measure algebra. As is shown in [3], St(B/I) is a compact Hausdorff, extremally disconnected space in which first category sets are nowhere dense. Also, it has weight 2 ° and satisfies the countable chain condition. It can be shown that St(B/I) is the union of 2 o nowhere dense sets. From these facts, we obtain the following theorem.