No abstract

Definition 1. For a set S and a cardinal κ,In particular, 2ω denotes the power set of the natural numbers and not the cardinal 2ℵ0. We regard 2ω as a topological space with the usual product topology.Definition 2. A set S ⊆ 2ω is Ramsey if there is an M ∈ [ω]ω such that either [M]ω ⊆ S or else [M]ω ⊆ 2ω − S.Erdös and Rado [3, Example 1, p. 434] showed that not every S ⊆ 2ω is Ramsey. In view of the nonconstructive character of the counterexample, one might expect (as Dana Scott has suggested) that all sufficiently definable sets are Ramsey. In fact, our main result (Theorem 2) is that all Borei sets are Ramsey. Soare [10] has applied this result to some problems in recursion theory.The first positive result on Scott's problem was Ramsey's theorem [8, Theorem A]. The next advance was Nash-Williams' generalization of Ramsey's theorem (Corollary 2), which can be interpreted as saying: If S1 and S2 are disjoint open subsets of 2ω, there is an M ∈ [ω]ω such that either [M]ω ⋂ S1 = ∅ or [M]ω ∩ S2 = ⊆. (This is halfway between “clopen sets are Ramsey” and “open sets are Ramsey.”) Then Galvin [4] stated a generalization of Nash-Williams' theorem (Corollary 1) which says, in effect, that open sets are Ramsey; this was discovered independently by Andrzej Ehrenfeucht, Paul Cohen, and probably many others, but no proof has been published.

No abstract

Let us consider the following infinite game between two players, Empty and Nonempty. We are given a large set S. Empty opens the game by choosing a large subset S0 of S; then Nonempty chooses a large set S1 ⊆ S0; then Empty chooses large S2 ⊆ S, etc. The game is over after ω moves. If ⋂n=0xSn is empty then Empty wins, and if ⋂n=0∞Sn is nonempty then Nonempty wins.If “large” means “infinite”, then Empty can beat Nonempty rather easily: he chooses So countable, S0 = {a0, a1,…, an,…}, and then he chooses S2 such that a0 ∉ S2, S4 such that a1, ∉ S4 and so on.Next we assume that S is a set of uncountable cardinality, and that “large” means “of cardinality ∣S∣”. Then still Empty can win, but his winning strategy is somewhat more sophisticated: Let us identify S with a cardinal number κ. Thus each subset of S of size κ is a set of ordinals below κ. For each X ⊆ κ of size κ, let fx be the unique order-preserving mapping of X onto κ, and let F(X) = {x ϵ X: f(x) is a successor ordinal}. Empty's strategy is to play S0 = F(K), and when Nonempty plays S2k − 1, let S2k = F(S2k − 1).

The purpose of this note is to characterize those distributive lattices that can be isomorphically embedded in free lattices. If it is known (cf. (2)) that in a free lattice every element is either additively or multiplicatively irreducible, and consequently every sublattice of a free lattice must also have this property. We therefore begin by studying the class of all those distributive lattices in which this condition is satisfied.