By Odell E.

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**Additional resources for On the structure of separable infinite dimensional Banach spaces**

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Assume (xn ) is normalized in (X, · ) and admits a spreading x1 + x ˜2 | = 2|˜ x1 |. We obtain that model over X for | · | and every | · |c , c ∈ C. Suppose |˜ ˜2 c = 2 x ˜1 c for all c ∈ C. This enables us to deduce, after some work, things like x ˜1 + x lim lim y + β1 xm + β2 xn = lim y + (β1 + β2 )xm m n if β1 , β2 ≥ 0, and ultimately get m 1 in X. 42 The argument above was actually constructed to solve a problem of Milman of which the theorem was a by-product. 10. [OS4] X is reﬂexive (if and) only if there exists an equivalent norm | · | on X satisfying for any bounded (xn ) ⊆ X: If lim lim |xm + xn | = 2 lim |xn | then (xn ) is norm convergent.

Ei )n1 ∈ {X}n if for all ε > 0, (V ) has a winning strategy for choosing (xi )n1 with db ((xi )n1 , (ei )n1 ) < 1 + ε. 49 Or one can describe {X}n as the smallest compact subset of Mn so that, given ε > 0, (S) has a winning strategy to force (V ) to select (xi )n1 with db ((xi )n1 , {X}n ) < 1 + ε . These interpretations are discussed in [MMT], and easily lead to {X}n = ∅ for all n (alternately, we could use spreading models and Rosenthal’s 1 theorem to deduce this). , [(xi )i≥k ]. Also, if (xi ) is a shrinking basis for X it is not hard to see that {X, (xi )}tn = {X}n for all n.

Hence there exists U, V ⊆ Z so that for all (u, v) ∈ U × V , (x + u, y + v) ∈ / A. Let u = v = 0 to get x−y ≥ε x+y . Now if (x, y) rejects Z then for all W ⊆ Z there exists W ⊆ W so that for all w ∈ W , (x + w , y) rejects Z. Otherwise there exists W ⊆ Z so that for all U ⊆ W there exists u0 ∈ U so that (x + u0 , y) accepts Z. Thus for all V ⊆ W there exists (u1 , v) ∈ U × V so that (x + u0 + u1 , y + v) ∈ A. Hence (x, y) accepts W which contradicts (x, y) rejects Z. Assume (zi )n1 are chosen so that all reasonable pairs formed from them reject Z.