On Intervals of Ordered Sets

Abstract

The present discussion stems from the following specific problem in the theory of (linearly) ordered sets: let m be any infinite cardinal; is it true that every ordered set of power m has a family of mutually disjoint intervals which is also of power m? This problem is solved, completely or partially, for all except those m > No which are strongly inaccessible.2 The solution is negative for all regular cardinals which are not strongly inaccessible, and is positive for some singular cardinals. Whether it is positive for all singular cardinals we do not know; the statement that it is (proposition Q) proves to be equivalent to a certain hypothesis from the domain of cardinal arithmetic (proposition P) which somewhat resembles, but is weaker than, the generalized hypothesis of the continuum. In particular the solution is negative for every cardinal which is a power of 2. For strongly inaccessible cardinals > No the problem remains open. These results are presented in ?2. The results in this section have been obtained jointly with Alfred Tarski and are included here with his kind permission. Some related problems, and some questions concerning the inaccessible numbers, are discussed in ?3. ?4 is devoted to a few somewhat less closely related theorems on decompositions of sets. 1. In this section we recall some definitions and notation and make a few preliminary remarks.3 The cardinal number of an arbitrary set M is denoted by M; the cardinal number of the set of all ordinals less than any fixed ordinal 4 is given the special symbol A. For every ordinal a, ;f(a) denotes the least cardinal p such that Na can be expressed as the sum of p cardinals each K,,, holds for every a. It follows that if ,, = no (for any n and d) then d < cf(a) (since (01)I n=b). In particular, as deduced by J. Konig a half-century ago, we have that 2NO K, (since cf(cw) = 0). But it is