By Bruck R.H.

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**Example text**

4 The Class NCX 23 is Borel. To see this notice that m k (yn ) ∼ (hn ) ⇔ ∀m ∀a0 , . . , am ∈ Q ∀l m an hn (l) ≤ k n=0 and ∀p ∃i 1 k m an y n − n=0 1 ≤ p+1 an y n n=0 m an hn (i) . n=0 The sequence (fn ) consists of Borel functions. Therefore, the relation Ik in C(2N )N × S deﬁned by (yn ), X ∈ Ik ⇔ (yn ), (fn (X)) ∈ Ek is Borel. Finally, by property (P5) in Sect. 1, the relation R in SB×C(2N )N deﬁned by Y, (yn ) ∈ R ⇔ span{yn : n ∈ N} = Y is Borel. Now let A∗ = {Y ∈ SB : ∃X ∈ A with Y ∼ = X ∗ } be the dual class of A.

Of N such that one of the following (mutually exclusive) cases must occur: Case 1. The set {tn : n ∈ L} is an antichain. Our hypothesis in this case implies that for every n, m ∈ L with n = m the segments sn and sm are 0 incomparable. We deﬁne z = yl0 + . . + ylk0 . As the family (sli )ki=0 consists of pairwise incomparable segments of T , we get that k0 z ≥ Psli (z) i=0 2 1/2 (c) k0 = Psli (yli ) 2 1/2 (b) ≥ C k0 + 1. 15) i=0 Now we set w = z/ z ∈ Y . 14), for every segment s of T we have Ps (w) ≤ C 1 + k0 ε √ < .

P} with the property that there exists t ∈ Tjσ with t σ. For this particular jσ we have the trivial estimate Pσ (ujnσ ) ≤ 1 for every n ∈ N. Now ﬁx i ∈ {0, . . , p} with i = jσ . Then every node t in Ti is not an initial segment of σ. We deﬁne si = {σ|k : k ≥ 1} \ TGi = {σ|k : k ≥ 1} \ {τ |n : n ≥ 1 and τ ∈ Gi }. Notice, ﬁrst, that si is a ﬁnal segment of σ. Also notice that si is nonempty. Indeed, our assumption that every node t in Ti is not an initial segment of σ, simply reduces to the fact that si contains the ﬁnal segment {σ|k : k > l0 }.