Counting composants of bR₊: 1, 2 or 2^c?
by
Taras Banakh
Nipissing University (North Bay, Canada) and Lviv University (Lviv, Ukraine)
Coauthors: Andreas Blass (University of Michigan)

The famous Mazurkiewicz Theorem asserts that the number of composants of a metrizable indecomposable continuum K is uncountable. Moreover, K contains a copy of Cantor set meeting each composant in at most one point. This result is not true is for non-metric continua. A counterexample is given by the Stone-Cech compcatification bR₊ of the half-line R₊=[0, ¥), the most famous non-metrizable indecomposable continuum. Surprisingly enough, but composants of bR₊ cannot be counted in ZFC alone: bR₊ has 1 composant under NCF, 2 composants under (r < s) and 2^c composants under (r ³ d).

By an old result of Mioduszewski, the number of composants of bR₊ equals the number of coherence classes of ultrafilters on w = {0, 1, 2, ...}, where two ultrafilters F, U are coherent if f(F)=f(U) for some finite-to-one map f : w® w.

Our principal result is the Finite-2^c Dichotomy, asserting that the number of coherence classes of ultrafilters is either finite or 2^c. In the latter case bw contains a closed infinite subset meeting each coherence class in at most one point. This result is proved by two different methods depending on relationship between the small cardinals u and d.

Applying the Finite-2^c Dichotomy to bR₊, we conclude that the number of composants of bR₊ is either finite or 2^c. In the latter case, bR₊ contains a topological copy of bw, which meets each composant of bR₊ in a most one point. In a sense, this is a counterpart of the Mazurkiewicz Theorem for bR₊.

Unfortunately, the Finite-2^c Dichotomy does answer the following Open Problem: Which finite numbers greater than 2 can be consistently equal to the number of composants of bR₊?

Date received: February 23, 2005