A Result in Dual Ramsey Theory

نویسندگان

  • Lorenz Halbeisen
  • Pierre Matet
چکیده

We present a result which is obtained by combining a result of Carlson with the Finitary Dual Ramsey Theorem of Graham-Rothschild. We start by introducing some notation. We conform to the usual practice of identifying the least infinite ordinal ω with the set of non-negative integers. Given α, β ≤ ω, a partition of α into β blocks is an onto function X : α→ β such that min ( X−1({n}) ) < min ( X−1({m}) ) whenever n < m < β. Thus, the blocks of X are ordered as their leaders (i.e., their least elements). The leader function ` : (α) × β → α is defined by `(X,m) := min ( X−1({m}) ) . Hence, the function m 7→ `(X,m) enumerates the leaders of X in increasing order. Given X ∈ (α) and Y ∈ (α), where α, β, γ ≤ ω, we let Y 6 X if Y is coarser than X, i.e., each block of Y is a union of blocks of X. Given α, β, γ ≤ ω and X ∈ (α), (X) := {Y ∈ (α) : Y 6 X}. Given α, β ≤ ω and k < ω, (α)βk denotes the set of all X ∈ (α) such that (a) X−1({n}) is finite if k ≤ n < β, and (b) max ( X−1({n}) ) < `(X,n+ 1) if k ≤ n and n+ 1 < β. Given α, β, γ ≤ ω, X ∈ (α) and k,m < ω such that k ≤ γ and m ≤ β, (k,m,X) is the set of all Y ∈ (X) such that { `(Y, i) : i < k } ⊆ { `(X, j) : j < m } . Note that (0,m,X) = (1,m,X) = (X) for all m ≤ β. The authors thank the referee for helpful comments. ∗The author wishes to thank the Swiss National Science Foundation for supporting him. 2000 Mathematics Subject Classification: 03E05 05D10 Key-words and phrases: partition, Dual Ellentuck Theorem, Finitary Dual Ramsey Theorem 1 2 The amalgamation function A is defined as follows: Given X ∈ (ω) and t ∈ (p), where 0 < m ≤ p < ω, A(t,X) is the partition of ω whose blocks are ⋃ i∈t−1({0}) X−1({i}) , . . . ⋃ i∈t−1({m−1}) X−1({i}) , X−1({p}) , X−1({p+ 1}) , . . . For t ∈ (p), where m ≤ p < ω, let Ot := {X ∈ (ω) : X p = t}. We topologize (ω) by taking as basic open sets ∅ and Ot for t ∈ ⋃ m≤p<ω (p). A function F : (ω) → r, where 1 ≤ r < ω, is clopen if F−1({i}) is a clopen subset of (ω) for each i < r. Our starting point is the following immediate consequence of the Dual Ellentuck Theorem (Theorem 4.1 in [1]) of Carlson-Simpson. Proposition 1. Given X ∈ (ω) and a clopen F : (ω) → r, where 1 ≤ r < ω, there is Y ∈ (X) such that F is constant on (Y ). Even if every block of X is finite, there may not be any homogeneous Y having infinitely many finite blocks. Proposition 2. There is a clopen F : (ω) → 2 with the property that there is no Y ∈ (ω) such that F is constant on (Y ) and Y has infinitely many finite blocks. Proof. Define F : (ω) → 2 by stipulating that F (X) = 0 if and only if X−1({1}) ∩ `(X, 3) ⊆ `(X, 2). Obviously, F is clopen. Now suppose that there is Y ∈ (ω) such that Y has infinitely many finite blocks and F is constant on (Y ). Pick Z ∈ (ω)1 with Z 6 Y . Then F is constant on (Z), which is clearly impossible. a Carlson established a “specialized” version (Theorem 6.9 of [1], which follows from Theorem 2 of [2]) of the Dual Ellentuck Theorem that deals with partitions of ω having finitely many infinite blocks. Carlson’s result immediately implies the following. Proposition 3. Given k < ω, X ∈ (ω)k and a clopen F : (ω) → r, where 1 ≤ r < ω, there is Y ∈ (ω)k ∩ (k, k,X) such that F is constant on (k, k, Y ). The purpose of this paper is to present the combinatorial result which is obtained by combining Proposition 3 with the Finitary Dual Ramsey theorem of GrahamRothschild [3]. This last reads as follows. Proposition 4. Suppose that 1 ≤ k ≤ m < ω and 1 ≤ r < ω. Then there is p < ω such that p ≥ m and the following holds: Given f : (p) → r, there is s ∈ (p) such that f is constant on (s).

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عنوان ژورنال:
  • J. Comb. Theory, Ser. A

دوره 100  شماره 

صفحات  -

تاریخ انتشار 2002