The descriptive complexity of the family of Banach spaces with the π-property

We show that the set of all separable Banach spaces that have the π-property is a Borel subset of the set of all closed subspaces of C(Δ), where Δ is the Cantor set, equipped with the standard Effros-Borel structure. We show that if α < ω1, the set of spaces with Szlenk index at most α which have a shrinking FDD is Borel.


Introduction
Let C( ) be the space of continuous functions on the Cantor space . It is well known that C( ) is isometrically universal for all separable Banach spaces. We denote SE the set of all closed subspaces of C( ) equipped with the standard Effros-Borel structure. In [1], Bossard considered the topological complexity of the isomorphism relation and of many subsets of SE. In addition, it has been shown that the set of all separable Banach spaces that have the bounded approximation property (BAP) is a Borel subset of SE, and that the set of all separable Banach spaces that have the metric approximation property (MAP) is also Borel [5].
We recall that the Banach space X has the π λ -property if there is a net of finite rank projections (S α ) on X converging strongly to the identity on X with limsup α S α λ (see [2]). We say that the Banach space X has the π-property if it has the π λ -property for some λ ≥ 1.
In this note we show that the set of all separable Banach spaces that have the π-property is a Borel subset of SE. This bears some consequences on the complexity of the class of spaces with finite-dimensional decompositions. For instance, we show that in the set of spaces whose Szlenk index is bounded by some countable ordinal, the subset consisting of spaces which have a shrinking finite-dimensional decomposition is Borel.

Main results
Here is the main technical lemma.
n=1 is a dense sequence in a Banach space X . Then, X has the π-property if and only if ∃λ > 1 ∀c ∈ (0, 1 4 where K , R, N vary over N and , λ , and λ vary over Q. Proof Indeed, suppose X has the π λ -property. Then there exists a sequence (P n ) of finite rank projections such that P n < λ, for all n and P n converge strongly to the identity. By perturbing P n , we may suppose that P n maps into the finite-dimensional subspace [x 1 , . . . , x R n ] for some R n in N but we still have (2.2) and P n < λ. Then, for every N , we may perturb P n slightly so that P n < λ and P n (x i ) belongs to the Q-linear span of the x j for all i N , such that (2.1), (2.2) and (2.3) still hold. Define x t , the three inequalities hold for all α 1 , . . . , α N ∈ Q, and i K .
Conversely, suppose that the above criterion holds and that > 0. Pick a rational 3λ > > 0 and a K . So let λ and R be given as above. Then for every N and i N , define where the σ i are given depending on N . We have that for all c ∈ (0, 1 4 ) Q. In particular, for every i, the sequences (y N i ) ∞ N =i , and (z N i ) ∞ N =i are contained in a bounded set in a finite-dimensional space. So by a diagonal procedure, we may find some subsequence (N l ) so that y i = lim l→∞ y N l i and z i = lim for all α i ∈ Q, for all c ∈ (0, 1 4 ) Q. Now, since the x i are dense in X , there are uniquely defined bounded linear operators T K , : is a dense sequence in X and the operators T K , are uniformly bounded, then 1 4 . Therefore, by [3,Theorem(3.7)], X has the π λ+1 -property as c → 0.

Theorem 2.2 The set of all separable Banach spaces that have the π-property is a Borel subset of SE.
Proof Let K , R, N vary over N and , λ vary over Q. Let also c ∈ (0, 1 4 ) Q, σ ∈ (Q R ) N , and α ∈ Q N . Then we consider the set E c,K , ,λ ,R,N ,σ,α ⊆ C( ) N such that: There is a Borel map d : SE −→ C( ) N such that d(X ) = X , by [8,Theorem (12.13)]. Moreover, the previous Lemma implies that X has the π-property ⇐⇒ d(X ) ∈ E.
Therefore, {X ∈ SE; X has the π-property} is a Borel subset of SE.
We will now prove that this result implies, with some work, that in some natural classes the existence of a finite-dimensional decomposition happens to be a Borel condition. We first consider the class of reflexive spaces. The commuting bounded approximation property (CBAP) implies the bounded approximation property (BAP) by the definition of the CBAP. By Grothendieck's theorem (see [ X has the MAP and the π-property} such that {X ∈ R; X has a FDD} = B ∩ R.
We will extend this simple observation to some classes of non-reflexive spaces. The following result has been proved in [7]. The proof below follows the lines of [6].

Proposition 2.3
Let X be a Banach space with separable dual. If X has the MAP for all equivalent norms, then X * has the MAP.
Proof Since X * is separable, there is an equivalent Fréchet differentiable norm on X . If . X is a Fréchet differentiable norm and x ∈ S X , there exists a unique x * ∈ S X * such that x * (x) = 1, and x * is a strongly exposed point of B X * . Since by assumption X equipped with this norm has the MAP, there exists an approximating sequence (T n ) with T n 1, and then for all x * ∈ X * we have T * n (x * ) w * → x * . For all x * ∈ X * which attains its norm, we have T * n (x * ) − x * X * → 0. Bishop-Phelps theorem yields that for all x * ∈ X * , T * n (x * ) − x * X * → 0. Therefore, X * has the MAP.
The set S D of all Banach spaces with separable dual spaces is coanalytic in SE and the Szlenk index Sz is a coanalytic rank on S D (see [1,Corollary (3.3) and Theorem (4.11)]). In particular, the set S α = {X ∈ SE; Sz(X ) α} is Borel in SE (see [8]). In this Borel set, the following holds.

Theorem 2.4 The set of all separable Banach spaces in S α that have a shrinking FDD is Borel in SE.
Proof Indeed, by [  Questions: As seen before, a separable Banach space has CBAP if and only if it has an equivalent norm for which it has MAP. It follows that the set {X ∈ SE; X has the CBAP} is analytic. It is not clear if it is Borel or not. Also, it is not known if there is a Borel subset B of SE, such that {X ∈ S D; X * has the AP} = B ∩ S D.
This would be an improvement of Theorem (2.3). Finally, what happens when we replace FDD by basis is not clear: for instance, the set of all spaces in S α which have a basis is clearly analytic. Is it Borel?