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13. 13. Let A ⊆ S × H be as in the statement of the proposition. 14, the map ΦA : S → K(H) defined by ΦA (s) = As is Borel, and so, the set {As : s ∈ S} is an analytic subset of K(H). Let n ∈ N. 10, the set ΩDn = {K ∈ K(H) : Dn∞ (K) = ∅} is Π11 and the map K → |K|Dn is a Π11 -rank on ΩDn . 9, we see that for every s ∈ S and for every ξ < ω1 if Dnξ (As ) = ∅, then Dnξ+1 (As ) Dnξ (As ). It follows that the sequence {Dnξ (As ) : ξ < ω1 } of iterated derivatives of As must be stabilized at ∅. Hence {As : s ∈ S} ⊆ ΩDn .

Then the following are satisfied: (i) If Y is X-compact, then Y is X-singular. (ii) If Y is X-singular, then Y is weakly X-singular. Proof. Part (i) is straightforward. To see part (ii) let Y be an X-singular subspace of T2X and let A be an arbitrary finite subset of [T ]. Consider the B-tree TA generated by A. Notice that there exist final segments s0 , . . , sk of T and a finite-dimensional subspace F of T2X such that XTA ∼ = F ⊕ k ⊕X . 6, if the operator P : Y → X were an sn TA TA n=0 40 3 The 2 Baire Sum isomorphic embedding, then there would existed some σ ∈ A and a subspace Y of Y such that the operator Pσ : Y → Xσ is an isomorphic embedding too.

Then for every ε > 0 there exists a normalized block sequence (yn ) is Y such that lim sup Pσ (yn ) < ε for every σ ∈ [T ]. Proof. The proof is a quest of a contradiction. So, suppose that there exist a block weakly X-singular subspace Y of T2X and ε > 0 such that for every normalized block sequence (yn ) in Y there exists σ ∈ [T ] such that lim sup Pσ (yn ) ≥ ε. Let p ∈ N and r > 0 to be determined later. We start with a normalized block sequence (yn0 ) in Y . By our assumptions, there exist σ0 ∈ [T ] and L0 ∈ [N] such that Pσ0 (yn0 ) > ε/2 for every n ∈ L0 .

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