Suslin Sets

Proof. Suppose z is in the RHS. Then, for some f ∈ Y X , z ∈ Sxfx for all x ∈ X. Hence for each x ∈ X there is y = fx so that z ∈ Sxy, so z is in the LHS. Suppose z is in the LHS. Then for each x ∈ X there is some y ∈ Y so that z ∈ Sxy. Hence for each x ∈ X the set Tx = { y : z ∈ Sxy } is not empty. By AC there is a function f ∈ Y x so that, for all x ∈ X, fx ∈ Tx. Hence, for all x ∈ X, z ∈ Sxfx , so z is in the RHS. We write P for the set of positive integers. On P we define a binary operator ∗ by k ∗ j = 2k−1(2j − 1). Since every positive integer can be factored uniquely as a product of a power of 2 and an odd integer we have: 2.2. (k, j) 7→ k ∗ j is a bijection of P onto P. By a word of length n over P we mean an element w of P, written as w = 〈w1, w2, . . . , wn〉. The set of all words of positive length is written P. 2.3. s = 〈s1, s2, . . . , sk〉 7→ k ∗ s1 ∗ s2 ∗ · · · ∗ sk defines a bijection of P onto P (where ∗ is considered to be right associative).