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We can rewrite N (ν) as the union, over all n ∈ N and over all sequences ∞ {pi }i=1 ⊆ N , of the sets q1 Mn11 q2 Mn11 ,n2 .. .. qk .. ,q 2 ∩ · · · ∩ Mn11 ,n22 .. qk ,q k ∩ ··· ∩ ··· . . ,nk ∩ · · · .. .. . 33) is indexed by k subscripts and superscripts. The choices of the subscripts and superscripts are constrained by the following requirements:   in any row, the list of subscripts is constant,  in any row, the list of superscripts grows by concatenation,  in any column, the list of subscripts grows by concatenation.

11 Let X and Y be topological spaces and let f : X → Y be continuous. If X is sequentially compact, X ⊇ C1 ⊇ C2 ⊇ · · ·, and if each ∞ Ci is a closed subset of X, then f ( ∞ i=1 Ci ) = i=1 f (Ci ). ∞ Proof. 10, we need only show ∞ i=1 f (Ci ) ⊆ f ( i=1 Ci ), so ∞ suppose y ∈ i=1 f (Ci ). For each i, there is xi ∈ Ci with f (xi ) = y, and because the sets Ci are decreasing, we have xj ∈ Ci whenever j ≥ i. Set x0,j = xj for j = 1, 2, . .. Since C1 is sequentially compact, there is a ∞ convergent subsequence {x1,j }∞ j=1 of {x0,j }j=1 .

Consequently, we also have Π01 ⊆ Π02. Since Σ01 ⊆ Π02 ⊆ Σ0α holds whenever 2 < α and since Π01 ⊆ Σ02 holds, we have Σ01 ⊆ Σ0α and Π01 ⊆ Π0α for all 1 < α < ω1 . Next consider 1 ≤ β < α < ω1 . Suppose Σ0γ ⊆ Σ0α and Π0γ ⊆ Π0α hold whenever γ < β. Any set A ∈ Σ0β must be of the form A = ∪∞ i=1 Ai with 0 each Ai ∈ Πγ for some γ < β. Then since β < α we see that A ∈ Σ0α . Thus Σ0β ⊆ Σ0α . Similarly, we have Π0β ⊆ Π0α . 3 We have Σ0α = α<ω1 Π0α . 32) is the σ-algebra of Borel subsets of RN . Proof. 32).

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