The dimension of remainders of rim-compact spaces

Abstract

Answering a question of Isbell we show that there exists a rim-compact space X such that every compactification Y of X has dim(Y \X) ≥ 1. It is known that a space X is rim-compact if and only if X has a compactification Y such that Y X is zero-dimensionally embedded in Y , i.e. Y has a base for the open sets whose boundaries are disjoint from Y X. Note that this implies that ind(Y \X) ≤ 0. On the other hand, if there is a compactification Y of the space X such that ind(Y \X) ≤ 0 then X need not be rim-compact; an example was given by Smirnov in [1958]. If X is Lindelof at infinity, i.e. each compact subset of X is contained in a compact set with a countable base for its neighbourhoods, then X is rimcompact if and only if it has a compactification Y such that ind(Y \X) ≤ 0; such a compactification then also has dim(Y X) ≤ 0. This suggests the question whether every rim-compact space X has a compactification Y such that dim(Y \X) ≤ 0 (see Isbell [1964] and Aarts and Nishiura [1993]; see also Diamond, Hatzenbuhler and Mattson [1988] for related problems). We shall show that the answer is negative. We construct a strongly zero-dimensional spaceX such that for every compactification Y ofX we have dim(Y \X) ≥ 1. Our space X will be such that βX \X is metrizable, zero-dimensional but not strongly zero-dimensional. Originally, the remainder was Roy’s space ∆; however, by using Kulesza’s space we were able to obtain an example of the smallest possible weight ω1. Convention. We identify an ordinal with its set of predecessors so that for example ω1 + 1 = ω1 ∪ {ω1}. All ordinals under consideration carry the order topology. 1991 Mathematics Subject Classification: Primary 54F45; Secondary 54D35. 288 J. M. Aarts and E. Coplakova The construction. Let K be Kulesza’s space, i.e. K is a (completely) metrizable subspace of ω 1 of weight w(K) = ω1 which is dense in ω ω 1 and satisfies indK = 0 and IndK = 1 (see Kulesza [1990]). Consider now Z = (ω1 + 1) = (ω1 + 1) × (ω1 + 1) and let X = Z (K × {ω1}) . Then (ω1+1)×ω1 ⊆ X ⊆ Z. As (ω1+1) is compact we may conclude that Z = (ω1 +1)×(ω1 +1) is the Cech–Stone compactification of (ω1 +1)×ω1 (see Engelking [1989, Problem 3.12.20(c)]). Hence βX = Z as well. Note that βX is a product of compact zero-dimensional spaces hence βX is also compact and zero-dimensional and therefore strongly zero-dimensional. It follows that X itself is strongly zero-dimensional, hence zero-dimensional and a fortiori rim-compact. It is also easily seen that w(X) = ω1. It remains to show that dim(αX \X) ≥ 1 for every compactification αX of X. Let αX be a compactification of X. Consider f :βX → αX, the extension of the natural embedding idX :X → αX over βX. As βX and αX are compact, the mapping f is perfect. Now, βX \X = f−1[αX \X] so f βX \X is also perfect. But βX \X = K × {ω1} and K is metrizable. To finish our argument we need the following theorem, due to Morita and Nagami (see Engelking [1989]). Theorem. If f :X → Y is a closed mapping of a metrizable space X to a metrizable space Y and for every y ∈ Y , Ind f−1[{y}] ≤ k for k ≥ 0, then IndX ≤ IndY + k. Now, K×{ω1} and αX \X are metrizable, the mapping f :K×{ω1} → αX \X is perfect and for each x ∈ αX \X the fiber f−1[{x}] is compact and zero-dimensional, hence strongly zero-dimensional. Note that this means that Ind(f−1[{x}]) ≤ 0 for each x ∈ αX \X, so by the theorem Ind(K × {ω1}) ≤ Ind(αX \X) + 0 , hence dim(αX \X) = Ind(αX \X) ≥ Ind(K × {ω1}) = 1 .