Definable Combinatorics of Some Borel Equivalence Relations

Abstract

If X is a set, E is an equivalence relation on X, and n ∈ ω, then define [X]nE = \(x0, ..., xn - 1) ∈ nX : (∀ i,j)(i ≠ j ⇒ (xi \ E \ xj))\. For n ∈ ω, a set X has the n-J\'onsson property if and only if for every function f : [X]n= → X, there exists some Y ⊂eq X with X and Y in bijection so that f[[Y]n=] ≠ X. A set X has the J\'onsson property if and only for every function f : (n ∈ ω[X]n=) → X, there exists some Y ⊂eq X with X and Y in bijection so that f[n ∈ ω [Y]n=] ≠ X. Let n ∈ ω, X be a Polish space, and E be an equivalence relation on X. E has the n-Mycielski property if and only if for all comeager C ⊂eq nX, there is some 11 A ⊂eq X so that E ≤_11 E A and [A]nE ⊂eq C. The following equivalence relations will be considered: E0 is defined on ω2 by x \ E0 \ y if and only if (∃ n)(∀ k > n)(x(k) = y(k)). E1 is defined on ω(ω2) by x \ E1 \ y if and only if (∃ n)(∀ k > n)(x(k) = y(k)). E2 is defined on ω2 by x \ E2 \ y if and only if Σ\1n + 1 : n ∈ x \ \ y\ < ∞, where denotes the symmetric difference. E3 is defined on ω(ω2) by x \ E3 \ y if and only if (∀ n)(x(n) \ E0 \ y(n)). Holshouser and Jackson have shown that R is J\'onsson under AD. It will be shown that E0 does not have the 3-Mycielski property and that E1, E2, and E3 do not have the 2-Mycielski property. Under ZF + AD, ω 2 / E0 does not have the 3-J\'onsson property.

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