# Higher order Tsirelson Spaces and their modified versions are isomorphic

Hùng Việt Chu  Department of Mathematics, Texas A&M University, College
Station, TX 77843, USA<EMAIL_ADDRESS>and  Thomas Schlumprecht  Department
of Mathematics, Texas A&M University, College Station, TX 77843, USA and
Faculty of Electrical Engineering, Czech Technical University in Prague,
Zikova 4, 166 27, Prague, Czech Republic<EMAIL_ADDRESS>
###### Abstract.

We prove that for every countable ordinal $\xi$, the Tsirelson’s space
$T_{\xi}$ of order $\xi$, is naturally, i.e., via the identity, $3$-isomorphc
to its modified version. For the first step, we prove that the Schreier family
$\mathcal{S}_{\xi}$ is the same as its modified version
$\mathcal{S}^{M}_{\xi}$, thus answering a question by Argyros and Tolias. As
an application, we show that the algebra of linear bounded operators on
$T_{\xi}$ has $2^{\mathfrak{c}}$ closed ideals.

###### Key words and phrases:

Tsirelson spaces of higher order, modified Tsirelson spaces, closed operator
ideals

###### 1991 Mathematics Subject Classification:

Primary: 46B20. Secondary: 46B06, 46B25.

The work was supported in part by the grant DMS-2054443 from the National
Science Foundation.

###### Contents

  1. 1 Introduction
  2. 2 Schreier families of higher order
  3. 3 Equality of the families $\mathcal{S}_{\xi}$ and $\mathcal{S}^{M}_{\xi}$, $\xi<\omega_{1}$
  4. 4 Modified Tsirelson spaces of higher order
  5. 5 Application: the cardinality of the closed ideals of $\mathcal{L}(T[\mathcal{S}_{\xi},\theta])$

##  1\. Introduction

In 1974, Tsirelson [17] constructed the first known space that does not
contain any isomorphic copies of $\ell_{p}$, $1\leqslant p<\infty$, or
$c_{0}$. Soon later, Figiel and Johnson [8] introduced the dual of Tsirelson’s
original space, and showed that the $p$-convexification of it is a uniformly
convex Banach space, not containing any $\ell_{q}$, $1<q<\infty$. Nowadays the
space defined by Figiel and Johnson is called Tsirelson space, and denoted by
$T$, while Tsirelson’s original space is often denoted by $T^{*}$.

In 1976, Johnson [11] defined a modified version $T^{M}$ of $T$, nowadays
called the modified Tsirelson space, which also does not contain $\ell_{p}$,
$1\leqslant p<\infty$, or $c_{0}$, but has the additional property that for
any $m\in\mathbb{N}$, there is a co-finite dimensional subspace $Y$ in which
every $m$-dimensional subspace is 2-isomorphic to a subspace of a finite-
dimensional subspace $F$ of $Y$ which is $2$-isomorphic to
$\ell_{1}^{\mbox{dim}(F)}$. Later, Casazza and Odell [7] showed that the norms
on $T^{M}$ and $T$ are actually equivalent.

The space $T$ is a reflexive space with a $1$-unconditional basis $(e_{i})$,
which has the property that the averages of any normalized block basis are not
converging in norm to zero. It is therefore a reflexive example of a space in
which a weakly null sequence may not have Cesáro summing subsequences. In [1],
Alspach and Argyros introduced iterated averages of order $\xi$, $\xi$ being a
countable ordinal, and constructed reflexive spaces $T_{\xi}$ for which the
iterated averages of order $\xi$ do not converge to $0$. We refer to these
spaces as higher order Tsirelson spaces. As in the base case, we can also
define their modified version. The main goal of this paper is to show that
Tsirelson spaces of any order $\xi<\omega_{1}$ are isomorphic (via the
identity) to their modified version. This extends a result by Manoussakis [14]
who proved this result for finite ordinals $\xi$.

Our paper is organized as follows: in Section 2, we recall Schreier families
$\mathcal{S}_{\xi}$ of order $\xi<\omega_{1}$, as well as their modified
version $\mathcal{S}^{M}_{\xi}$, $\omega_{1}$ denoting the first uncountable
ordinal. The main goal of Section 3, is to show the equality of
$\mathcal{S}_{\xi}$ and $\mathcal{S}^{M}_{\xi}$. In Section 4, we recall the
definition of Tsirelson spaces $T_{\xi}$ of order $\xi<\omega_{1}$, and their
modified versions $T^{M}_{\xi}$, and it is shown that their norms are
equivalent. Section 5 is concerned with the cardinality of the closed ideals
of the algebra of bounded linear operators on $T_{\xi}$.

##  2\. Schreier families of higher order

In this section, we recall the definition of Schreier families of order $\xi$,
where $\xi$ is a countable ordinal, and their modified version. We introduce
some notation: $[\mathbb{N}]$, $[\mathbb{N}]^{<\omega}$, and
$[\mathbb{N}]^{\omega}$ denote the subsets, the finite subsets, and the
infinite subsets of the natural numbers $\mathbb{N}$. For
$A\in[\mathbb{N}]^{<\omega}$ and $B\in[\mathbb{N}]$, we write $A<B$, if $a<b$
for all $a\in A$ and $b\in B$. For $A\in[\mathbb{N}]$ and $k\in\mathbb{N}$, we
let $A_{>k}=A\cap\\{k+1,k+2,\ldots\\}$, $A_{\geqslant
k}=A\cap\\{k,k+1,k+2,\ldots\\}$, $A_{<k}=A\cap\\{1,2,\ldots,k-1\\}$, and
$A_{\leqslant k}=A\cap\\{1,2,\ldots,k\\}$. As a matter of convention, we put
$\max\emptyset=0$ and $\min\emptyset=\infty$, and thus, $A<\emptyset$ and
$A>\emptyset$ for all $A\in[\mathbb{N}]^{<\omega}$. For $m\in\mathbb{N}$, we
write $m\leqslant A$ (or $m<A$, respectively), if $m\leqslant\min A$ (or
$m<\min A$, respectively).

Identifying elements of $[\mathbb{N}]$ in the canonical way with
$\\{0,1\\}$-valued functions on $\mathbb{N}$ and thus elements of
$\\{0,1\\}^{\mathbb{N}}$, we equip $[\mathbb{N}]$ with the product topology of
the discrete topology on $\\{0,1\\}$. We say that
$\mathcal{F}\subset[\mathbb{N}]^{<\omega}$ is compact if it is compact in that
product topology. We say that $\mathcal{F}\subset[\mathbb{N}]^{<\omega}$ is
hereditary if $A\in\mathcal{F}$ and $B\subset A$ imply that $B\in\mathcal{F}$.
Given $a_{1}<\cdots<a_{n},b_{1}<\cdots<b_{n}$ in $\mathbb{N}$, we say that
$\\{b_{1},\dots,b_{n}\\}$ is a spread of $\\{a_{1},\ldots,a_{n}\\}$ if
$a_{i}\leqslant b_{i}$ for all $i\in\\{1,\ldots,n\\}$. A family
$\mathcal{F}\subset[\mathbb{N}]^{<\omega}$ is called spreading if every spread
of every element of $\mathcal{F}$ is also in $\mathcal{F}$. A family that is
compact, hereditary, and spreading is said to be regular.

We now define the Schreier family
$\mathcal{S}_{\xi}\subset[\mathbb{N}]^{<\omega}$ for each $\xi<\omega_{1}$ by
transfinite induction. Particularly,

$\mathcal{S}_{0}\ :=\ \\{\\{n\\}\,:\,n\in\mathbb{N}\\}\cup\\{\emptyset\\}.$

Assuming that $\mathcal{S}_{\gamma}$ has been defined for all $\gamma<\xi$. If
$\xi=\gamma+1$, we let

$\mathcal{S}_{\xi}\ =\ \left\\{\bigcup_{i=1}^{n}E_{i}\,:\,n\leqslant
E_{1}<E_{2}<\cdots<E_{n},E_{i}\in\mathcal{S}_{\gamma},i=1,2,\ldots,n\right\\}.$
(2.1)

If $\xi$ is a limit ordinal, we choose a fixed sequence of successor ordinals
$(\alpha(\xi,n))_{n=1}^{\infty}\subset[1,\xi)$, which increases to $\xi$ and
put

$\mathcal{S}_{\xi}\ =\ \\{E:\exists n\ \leqslant\ E\mbox{ with
}E\in\mathcal{S}_{\alpha(\xi,n)}\\}.$ (2.2)

The sequence $(\alpha(\xi,n))_{n=1}^{\infty}$ is called a $\xi$-approximating
sequence. As needed later in Section 5, it follows by transfinite induction
that $\mathcal{S}_{1}\subset\mathcal{S}_{\xi}$ for all $\xi<\omega_{1}$.
Furthermore, we can and will later put additional conditions on
$(\alpha(\xi,n))_{n=1}^{\infty}$.

The standard transfinite induction gives that the families
$\mathcal{S}_{\xi}$, for $\xi<\omega_{1}$, are regular, and they are almost
increasing in the following sense: for all ordinals $\gamma<\xi$,

$\mbox{ there is an }m\in\mathbb{N}\mbox{ so that if
}A\in\mathcal{S}_{\gamma}\mbox{ and }\min A\geqslant m,\mbox{ then
}A\in\mathcal{S}_{\xi}.$ (2.3)

In the case of a limit ordinal $\xi$, Property (2.3) enables us to require the
following additional property on a $\xi$-approximating sequence
$(\alpha(\xi,n))_{n=1}^{\infty}$:

$\displaystyle\alpha(\xi,1)=1,$ $\displaystyle\text{and for all
}n\in\mathbb{N},\text{ if }A,B\in\mathcal{S}_{\alpha(\xi,n)}\mbox{ with
}1<A<B,\mbox{ then }A\cup B\in\mathcal{S}_{\alpha(\xi,n+1)}.$ (2.4)

Indeed, suppose that we have a limit ordinal $\xi$ and have chosen
approximating sequences $(\alpha(\gamma,n))_{n=1}^{\infty}$ for all limit
ordinals $\gamma<\xi$. We first choose an arbitrary $\xi$-approximating
sequence $(\alpha^{\prime}(\xi,n))_{n=1}^{\infty}$. We shall build a new
$\xi$-approximating sequence $(\alpha(\xi,n))_{n=1}^{\infty}$ from
$(\alpha^{\prime}(\xi,n))_{n=1}^{\infty}$ be choosing
$\ell_{1}<\ell_{2}<\ell_{3}<\cdots$ recursively so that

$\\{A\cup B:A,B\in\mathcal{S}_{\alpha^{\prime}(\xi,n)+\ell_{n}},1<A<B\\}\
\subset\ \mathcal{S}_{\alpha^{\prime}(\xi,n+1)+\ell_{n+1}},$

and then putting $\alpha(\xi,n)=\alpha^{\prime}(\xi,n)+\ell_{1}$.

Set $\ell_{1}=0$. Assume that $\ell_{n}$ has been defined. Consider
$\alpha^{\prime}(\xi,n+1)+\ell_{n}>\alpha^{\prime}(\xi,n)+\ell_{n}=\alpha(\xi,n)$.
By (2.3), there exists $k_{n}\in\mathbb{N}$ such that $A_{\geqslant
k_{n}}\in\mathcal{S}_{\alpha^{\prime}(\xi,n+1)+\ell_{n}}$, for any
$A\in\mathcal{S}_{\alpha(\xi,n)}$. Thus for any two sets $1<A<B$ in
$\mathcal{S}_{\alpha(\xi,n)}=\mathcal{S}_{\alpha^{\prime}(\xi,n)+\ell_{n}}$ it
follows that $U:=A_{\geqslant k_{n}}\cup B_{\geqslant
k_{n}}\in\mathcal{S}_{\alpha^{\prime}(\xi,n+1)+\ell_{n}+1}$. By (2.1),
$\\{2,3,\ldots,k_{n}-1\\}\in\mathcal{S}_{\alpha^{\prime}(\xi,n+1)+k_{n}-3}$;
hence, $V:=A_{<k_{n}}\cup
B_{<k_{n}}\in\mathcal{S}_{\alpha^{\prime}(\xi,n+1)+k_{n}-3}$. Since $V<U$ are
in $\mathcal{S}_{\alpha^{\prime}(\xi,n+1)+\max\\{k_{n}-3,\ell_{n}+1\\}}$, we
know that $U\cup
V\in\mathcal{S}_{\alpha^{\prime}(\xi,n+1)+\max\\{k_{n}-2,\ell_{n}+2\\}}$.
Therefore, if we set $\ell_{n+1}=\max\\{k_{n}-2,\ell_{n}+1\\}+1$ and
$\alpha(\xi,n+1)=\alpha^{\prime}(\xi,n+1)+\ell_{n+1}$. The new
$\xi$-approximating sequence $(\alpha(\xi,n))_{n=1}^{\infty}$ increases to
$\xi$ and satisfies $\eqref{e11}$.

As a consequence of (2), for a $\xi$-approximating sequence
$(\alpha(\xi,n))_{n=1}^{\infty}$, we have
${\mathcal{S}}_{\alpha(\xi,m)}\subset{\mathcal{S}}_{\alpha(\xi,n)}$ whenever
$m\leqslant n$, which, in combination with (2.2), gives a neat way to write
$\mathcal{S}_{\xi}$ as

$\mathcal{S}_{\xi}\ =\ \\{E\,:\,\exists n\leqslant\min E\mbox{ with
}E\in\mathcal{S}_{\alpha(\xi,n)}\\}\ =\
\\{E\,:\,E\in\mathcal{S}_{\alpha(\xi,\min E)}\\}.$ (2.5)

For $\xi<\omega_{1}$, we denote by MAX$(\mathcal{S}_{\xi})$ the maximal
elements of $\mathcal{S}_{\xi}$. Since $\mathcal{S}_{\xi}$ is compact, every
element of $\mathcal{S}_{\xi}$ is a subset of some maximal element of
$\mathcal{S}_{\xi}$ (see [5, Section 2] for details).

######  Proposition 2.1.

If $0<\xi=\gamma+1<\omega_{1}$ and $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$,
then there are unique elements $A_{i}\in\text{\rm
MAX}(\mathcal{S}_{\gamma}),i=1,2,\ldots,\min A$, so that

$A_{1}\ <\ A_{2}\ <\ \cdots\ <\ A_{\min A}\mbox{ and }A\ =\ \cup_{i=1}^{\min
A}A_{i}.$

###### Proof.

Pick $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$. By definition, we can write
$A=\cup_{i=1}^{n}A_{i}$, for $n\leqslant A_{1}<\cdots<A_{n}$, and $A_{i}$ are
nonempty sets in $\mathcal{S}_{\gamma}$. We shall show that $n=\min A$, and
each $A_{i}$ is in $\text{\rm MAX}(\mathcal{S}_{\gamma})$.

Observe that $\min A_{1}=\min A$, so $n\leqslant\min A$. If $n<\min A$, then
find some $m>A_{n}$ to have

$n+1\ \leqslant\ A_{1}<\cdots<A_{n}<A_{n+1}\ :=\ \\{m\\}$

By definition, the set $A^{\prime}=\cup_{i=1}^{n+1}A_{i}\in\mathcal{S}_{\xi}$
and $A\subsetneq A^{\prime}$, which contradicts the maximality of $A$. Hence,
$n=\min A$.

We claim that every $A_{i}$ is in $\text{\rm MAX}(\mathcal{S}_{\gamma})$.
Otherwise, let $j$ be the smallest such that $A_{j}\notin\text{\rm
MAX}(\mathcal{S}_{\gamma})$. If $j=\min A$, then form $A^{\prime}_{\min
A}:=A_{\min A}\cup\\{m\\}\in\mathcal{S}_{\gamma}$, for some $m>A$. Then the
set $\cup_{i=1}^{j-1}A_{i}\cup A^{\prime}_{\min A}$ is in $\mathcal{S}_{\xi}$,
again contradicting the maximality of $A$. If $j<\min A$, we form

$\displaystyle A^{\prime}_{j}\ :=\ A_{j}\cup\\{\min A_{j+1}\\}$ $\displaystyle
A^{\prime}_{i}\ :=\ (A_{i}\backslash\\{\min A_{i}\\})\cup\\{\min
A_{i+1}\\}\mbox{ for }j\ <\ i\ \leqslant\ \min A-1,\mbox{ and }$
$\displaystyle A^{\prime}_{\min A}\ :=\ (A_{\min A}\backslash\\{\min A_{\min
A}\\})\cup\\{m\\},\mbox{ for some }m>A.$

Clearly, $A^{\prime}_{i}\in\mathcal{S}_{\gamma}$ for $i=j,\ldots,\min A$.
Hence,

$\left(\bigcup_{i=1}^{j-1}A_{i}\right)\cup\left(\bigcup_{i=j}^{\min
A}A^{\prime}_{i}\right)\ \in\ \mathcal{S}_{\xi},$

contradicting the maximality of $A$.

Finally, the uniqueness of the decomposition is obvious. Indeed, suppose that
$A=\cup_{i=1}^{\min A}A_{i}=\cup_{i=1}^{\min A}A^{\prime}_{i}$, where
$A_{1}<\cdots<A_{\min A}\in\text{\rm MAX}(\mathcal{S}_{\gamma})$ and
$A^{\prime}_{1}<\cdots<A^{\prime}_{\min A}\in\text{\rm
MAX}(\mathcal{S}_{\gamma})$. Let $j$ be the smallest (if exists) such that
$A_{j}\neq A^{\prime}_{j}$. Then either $A_{j}\subsetneq A^{\prime}_{j}$ or
$A^{\prime}_{j}\subsetneq A_{j}$. Either way violates that both $A_{j}$ and
$A_{j}^{\prime}$ are in MAX$(\mathcal{S}_{\gamma})$. Therefore,
$A_{j}=A_{j}^{\prime}$ for $j=1,\ldots,\min A$. ∎

######  Proposition 2.2.

If $\xi<\omega_{1}$ is a limit ordinal, then $A\in\text{\rm
MAX}(\mathcal{S}_{\xi})$ if and only if $A\in\text{\rm
MAX}(\mathcal{S}_{\alpha(\xi,\min A)})$.

###### Proof.

Let $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$. By (2.5),
$A\in\mathcal{S}_{\alpha(\xi,\min A)}$. If $A\notin\text{\rm
MAX}(\mathcal{S}_{\alpha(\xi,\min A)})$, then form $A^{\prime}:=A\cup\\{\max
A+1\\}$, which must still be in $\mathcal{S}_{\alpha(\xi,\min
A)}=\mathcal{S}_{\alpha(\xi,\min A^{\prime})}$. Hence,
$A^{\prime}\in\mathcal{S}_{\xi}$, contradicting the maximality of $A$.

Conversely, let $A\in\text{\rm MAX}(\mathcal{S}_{\alpha(\xi,\min A)})$. Then
$A\in\mathcal{S}_{\xi}$. If $A\notin\text{\rm MAX}(\mathcal{S}_{\xi})$, then
$A^{\prime}:=A\cup\\{\max A+1\\}\in\mathcal{S}_{\xi}$. By the definition of
$\mathcal{S}_{\xi}$, $A^{\prime}\in\mathcal{S}_{\alpha(\xi,\min
A^{\prime})}=\mathcal{S}_{\alpha(\xi,\min A)}$, thus contradicting the
maximality of $A$. ∎

It is convenient to define a $\xi$-approximating sequence
$(\alpha(\xi,n))_{n=1}^{\infty}$ also for a successor ordinal $\xi<\omega_{1}$
by simply letting $\alpha(\xi,n)=\xi$ for all $n\in\mathbb{N}$. Thus, it
follows that $A\in\mathcal{S}_{\alpha(\xi,\min A)}$ for
$A\in\mathcal{S}_{\xi}$ (whether $\xi$ is a limit ordinal or not), and we have
the following unified recursive definition of $\mathcal{S}_{\xi}$:

$\mathcal{S}_{\xi}\ =\ \left\\{\bigcup_{i=1}^{\min
A_{1}}A_{i}\,:\,A_{1}<\cdots<A_{\min A_{1}}\mbox{ and
}A_{i}\in\mathcal{S}_{\alpha(\xi,\min A_{1})-1},i=1,\ldots,\min
A_{1}\right\\},$

where $\gamma-1$ is the predecessor for a sucessor ordinal $\gamma$. Moreover,
if $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$, then this decomposition into
increasing sets $A_{i}\in\text{\rm MAX}(\mathcal{S}_{\alpha(\xi,\min
A_{1})-1})$ is unique.

We conclude this section by introducing the modified Schreier families, which
are defined using transfinite induction for all $\xi<\omega_{1}$, as follows

$\mathcal{S}_{0}^{M}\ =\ \mathcal{S}_{0}\ =\
\\{\\{n\\}\,:\,n\in\mathbb{N}\\}\cup\\{\emptyset\\}.$

Assuming that $\mathcal{S}_{\gamma}^{M}$ has been defined for all
$\gamma<\xi$. If $\xi=\gamma+1$, we let

$\mathcal{S}_{\xi}^{M}=\left\\{\bigcup_{i=1}^{n}E_{i}\,:\,\mbox{sets
}E_{i}\mbox{'s are pairwise disjoint, }E_{i}\in\mathcal{S}_{\gamma}^{M},\mbox{
and }E_{i}\geqslant n,i=1,\ldots,n\right\\}.$

If $\xi$ is a limit ordinal, we choose a fixed sequence
$(\alpha(\xi,n):n\in\mathbb{N})\subset[1,\xi)$ which increases to
$\varepsilon$ and put

$\mathcal{S}^{M}_{\xi}=\\{E\,:\,\exists n\ \leqslant\ \min E\mbox{ with
}E\in\mathcal{S}^{M}_{\alpha(\xi,n)}\\}.$

##  3\. Equality of the families $\mathcal{S}_{\xi}$ and
$\mathcal{S}^{M}_{\xi}$, $\xi<\omega_{1}$

In this section, we answer a question stated in [3, pp. 16] by showing that
for any $\xi<\omega_{1}$, the Schreier family $\mathcal{S}_{\xi}$ of order
$\xi<\omega_{1}$ coincides with its modified version $\mathcal{S}^{M}_{\xi}$.
This result was independently obtained in [6]. We would like to give here a
different, self-contained proof and lay the foundation for the next section,
where we prove that the spaces $T_{\xi}$ and $T^{M}_{\xi}$ are naturally (via
the formal identity) isomorphic. More precisely, we will prove Theorem 3.4,
from which, it will easily follow that
$\mathcal{S}_{\xi}=\mathcal{S}_{\xi}^{M}$, for $\xi<\omega_{1}$. Theorem 3.4
will also be used in Section 4.

######  Definition 3.1 (The $\xi$-analysis of maximal $\mathcal{S}_{\xi}$
sets).

Given $\xi<\omega_{1}$, we define recursively the $\xi$-analysis $T(\xi,A)$ of
a maximal $\mathcal{S}_{\xi}$ set $A$ or the $\xi$-analysis tree of $A$, for
short. We also define for any $D\in T(\xi,A)$, the $\xi$-order of $D$ inside
$A$, denoted by $\mbox{order}_{\xi,A}(D)$, which is an ordinal not larger than
$\xi$.

If $\xi=0$ and $A=\\{a\\}\in\text{\rm MAX}(\mathcal{S}_{0})$, put
$T(0,A)=\\{A\\}$ and order${}_{\xi,A}(A)=0$. Assume that $T(\gamma,A)$ has
been defined for all $\gamma<\xi$, and $A\in\text{\rm
MAX}(\mathcal{S}_{\gamma})$, and for all $D\in T(\gamma,A)$, we have defined
the ordinal $\mbox{order}_{\gamma,A}(D)\leqslant\gamma$. We define $T(\xi,A)$
for $A\in\mathcal{S}_{\xi}$.

Case 1: $\xi$ is a successor ordinal. Write $\xi=\gamma+1$. For $A\in\text{\rm
MAX}(\mathcal{S}_{\xi})$, write $A=\cup_{i=1}^{\min A}A_{i}$, where
$A_{1}<\cdots<A_{\min A}$ and $A_{i}\in\text{\rm MAX}(\mathcal{S}_{\gamma})$.
Then

$T(\xi,A)\ =\ \\{A\\}\cup\bigcup_{i=1}^{\min A}T(\gamma,A_{i}),\mbox{ and }$
$\mbox{order}_{\xi,A}(A)\ =\ \xi,\mbox{ and }\mbox{order}_{\xi,A}(D)\ =\
\mbox{order}_{\gamma,A_{i}}(D),$

for $D\in T(\gamma,A_{i})$ and $i=1,\ldots,\min A$.

Case 2: $\xi$ is a limit ordinal. For $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$,
let $T(\xi,A)=T(\alpha(\xi,\min A),A)$, and
$\mbox{order}_{\xi,A}(D)=\mbox{order}_{\alpha(\xi,\min A),A}(D)$ for all $D\in
T(\xi,A)$.

######  Remark 3.2.

For $\xi<\omega_{1}$ and $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$, we make the
following observations:

  1. (1)

If $\xi$ is a successor ordinal or $0$, we have $\mbox{order}_{\xi,A}(A)=\xi$;
if $\xi$ is a limit ordinal, we have $\mbox{order}_{\xi,A}(A)=\alpha(\xi,\min
A)$, and $T(\xi,A)=T(\alpha(\xi,\min A),A)$.

  2. (2)

For every $D\in T(\xi,A)$, $\mbox{order}_{\xi,A}(D)$ is a successor ordinal,
and furthermore, $D\in\text{\rm MAX}(\mathcal{S}_{\mbox{order}_{\xi,A}(D)})$.

  3. (3)

If $D,E\in T(\xi,A)$ and $D\subsetneq E$, then
$\mbox{order}_{\xi,A}(D)<\mbox{order}_{\xi,A}(E)$.

  4. (4)

$T(\xi,A)$ is a tree with respect to inclusion, i.e., $D$ is a successor of
$E$ if $D\subsetneq E$. In this tree, $A$ is the root, while the leaves (or
terminal nodes) consist of all the singletons $\\{a\\}$, for $a\in A$.

  5. (5)

Let $D\in T(\xi,A)$ be not a terminal node. Denote the set of immediate
successors of $D$ in $T(\xi,A)$ by $\mbox{succ}_{\xi}(D)$, i.e.,
$\mbox{succ}_{\xi}(D)=$

$\\{E\in T(\xi,A)\,:\,E\subsetneq D\mbox{ and }\not\exists E^{\prime}\in
T(\xi,A)\mbox{ s.t. }E\subsetneq E^{\prime}\subsetneq D\\}.$

Then $D=\cup_{E\in\mbox{succ}(D)}E$. If we write
$\mbox{order}_{\xi,A}(D)=\gamma+1$, then $\mbox{succ}(D)$ consists of $\min D$
elements $D_{1}<D_{2}<\cdots<D_{\min D}$, where $D_{i}\in\text{\rm
MAX}(\mathcal{S}_{\gamma})$. Then there are two cases:

     * •

either $\gamma$ is a successor ordinal, in which case
$\mbox{order}_{\xi,A}(D_{i})=\gamma$.

     * •

or $\gamma$ is a limit ordinal, in which case, we have $D_{i}\in\text{\rm
MAX}(\mathcal{S}_{\alpha(\gamma,\min D_{i})})$ and
$\mbox{order}_{\xi,A}(D_{i})\ =\ \alpha(\gamma,\min D_{i})$ for
$i=1,2,\ldots,\min D$.

  6. (6)

For $D\in T(\xi,A)$, $T(\mbox{order}_{\xi,A}(D),D)$ is a subtree of $A$, and

$T(\mbox{order}_{\xi,A}(D),D)\ =\ \\{E\in T(\xi,A)\,:\,E\subset D\\}.$

  7. (7)

Let $D,E\in T(\xi,A)$, $D\neq E$, then there are only four possibilities:

     * •

$D\subsetneq E$, and order${}_{\xi,A}(D)<\mbox{order}_{\xi,A}(E)$,

     * •

$E\subsetneq D$, and order${}_{\xi,A}(E)<\mbox{order}_{\xi,A}(D)$,

     * •

$E<D$,

     * •

$D<E$.

We define for $\xi<\omega_{1}$, $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$, and
$a\in A$, the $\xi$-order of $a$ inside $A$:

$\mbox{order}_{\xi,A}(a)\ =\ \max\\{\mbox{order}_{\xi,A}(D)\,:\,D\in
T(\xi,A)\mbox{ and }a\ =\ \min D\\}.$

Roughly speaking, to find $\mbox{order}_{\xi,A}(a)$, we look for the set $D$
closest to the root and whose minimum is $a$, then set
$\mbox{order}_{\xi,A}(a)=\mbox{order}_{\xi,A}(D)$. Note that for every $a\in
A$, we have $\\{a\\}\in T(\xi,A)$, so $\mbox{order}_{\xi,A}(a)$ is well-
defined, and $\mbox{order}_{\xi,A}(a)=0$ if and only if $a$ is not the minimum
of any other node of $T(\xi,A)$. For $a\in A$, let $D_{\xi,A}(a)$ be the
unique set $D$ closest to the root (possibly the root itself) and whose
minimum is $a$. If $E\in T(\xi,A)$ and $a\in E$ with $a>\min E$, then
$D_{\xi,A}(a)=D_{\mbox{order}_{\xi,A}(E),E}(a)$ and
$\mbox{order}_{\xi,A}(a)=\mbox{order}_{\mbox{order}_{\xi,A}(E),E}(a)$.

######  Lemma 3.3 (The replacement lemma).

Let $\xi<\omega_{1}$, $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$, and $a\in A$
with $a>\min A$. Put

$\displaystyle m_{1}$ $\displaystyle\ =\ \max\\{\max D\,:\,D\in
T(\xi,A),D<D_{\xi,A}(a)\\},\mbox{ and }$ $\displaystyle m_{2}$ $\displaystyle\
=\ \min\\{\min D\,:\,D\in T(\xi,A),D>D_{\xi,A}(a)\\},$

where $\min\emptyset=\infty$ in the case there is no node after
$D_{\xi,A}(a)$. Let $E$ be the immediate predecessor of $D_{\xi,A}(a)$ in
$T(\xi,A)$. Write $\mbox{order}_{\xi,A}(E)=\gamma+1$. Then for every
$\widetilde{D}\in\text{\rm MAX}(\mathcal{S}_{\gamma})$ with
$\widetilde{D}\subset(m_{1},m_{2})$, we have

$\widetilde{A}\ :=\ (A\backslash D_{\xi,A}(a))\cup\widetilde{D}\ \in\
\text{\rm MAX}(\mathcal{S}_{\xi}).$

###### Proof.

We prove the lemma by induction on $n$, the number of predecessors of
$D_{\xi,A}(a)$.

Base case: $n=1$. Let $\xi<\omega_{1}$, $A\in\text{\rm
MAX}(\mathcal{S}_{\xi})$, and $a\in A$ with $a>\min A$. Since $n=1$, $A$ is
the immediate predecessor of $D_{\xi,A}(a)$ in $T(\xi,A)$. Write
$A=\cup_{i=1}^{\min A}A_{i}$, where $\min A\leqslant
A_{1}<A_{2}<\cdots<A_{\min A}$ and $A_{i}\in\text{\rm
MAX}(\mathcal{S}_{\gamma})$. For some $i_{0}=2,3,\ldots,\min A$, we have
$A_{i_{0}}=D_{\xi,A}(a)$. Note that $i_{0}\neq 1$ because $a>\min A$. We have

$A_{1}\ <\ A_{2}\ <\ \cdots\ <\ A_{i_{0}-1}\ <\ \widetilde{D}\ <\ A_{i_{0}+1}\
<\ \cdots\ <\ A_{\min A}.$

It may be possible that in the above chain, $\widetilde{D}$ is the last set,
which happens exactly when $i_{0}=\min A$. By definition,

$\widetilde{A}\ =\ (A\backslash D_{\xi,A}(a))\cup\widetilde{D}\ =\
\bigcup_{i=1,i\neq i_{0}}^{\min A}A_{i}\cup\widetilde{D}\ \in\ \text{\rm
MAX}(\mathcal{S}_{\xi}).$

Inductive hypothesis: assume that the lemma is true for $n$ predecessors and
all $\xi<\omega_{1}$.

Let us fix $\xi<\omega_{1}$, $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$, and
$a\in A$ with $a>\min A$. Suppose that $D_{\xi,A}(a)$ has $n+1$ predecessors
$(E_{i})_{i=1}^{n+1}$ in $T(\xi,A)$ with

$D_{\xi,A}(a)\ =\ E_{0}\ \subsetneq\ E_{1}\ \subsetneq\ E_{2}\ \subsetneq\
\cdots\ \subsetneq\ E_{n}\ \subsetneq\ E_{n+1}\ =\ A.$

Put

$\displaystyle m_{1}$ $\displaystyle\ =\ \max\\{\max D\,:\,D\in
T(\xi,A),D<D_{\xi,A}(a)\\},\mbox{ and }$ $\displaystyle m_{2}$ $\displaystyle\
=\ \min\\{\min D\,:\,D\in T(\xi,A),D>D_{\xi,A}(a)\\}.$

Choose $\widetilde{D}\subset(m_{1},m_{2})$ such that
$\widetilde{D}\in\text{\rm MAX}(\mathcal{S}_{\mbox{order}_{\xi,A}(a)})$.

Let $\xi=\gamma+1$ and consider the tree $T(\gamma,E_{n})$. Put

$\displaystyle m^{\prime}_{1}$ $\displaystyle\ =\ \max\\{\max D\,:\,D\in
T(\gamma,E_{n}),D<D_{\gamma,E_{n}}(a)\\},\mbox{ and }$ $\displaystyle
m^{\prime}_{2}$ $\displaystyle\ =\ \min\\{\min D\,:\,D\in
T(\gamma,E_{n}),D>D_{\gamma,E_{n}}(a)\\}.$

Since $D_{\gamma,E_{n}}(a)=D_{\xi,A}(a)$ and $T(\gamma,E_{n})$ is a subtree of
$T(\xi,A)$, we know that

$m^{\prime}_{1}\ \leqslant\ m_{1}<m_{2}\ \leqslant\ m_{2}^{\prime}.$

Furthermore, $a>\min E_{n}$ because $D_{\xi,A}(a)$ has at least two
predecessors. Applying the inductive hypothesis to the tree $T(\gamma,E_{n})$
and $a\in E_{n}$, we have

$(E_{n}\backslash D_{\xi,A}(a))\cup\widetilde{D}\ =\ (E_{n}\backslash
D_{\gamma,E_{n}}(a))\cup\widetilde{D}\ \in\ \text{\rm
MAX}(\mathcal{S}_{\gamma}).$

It is worth noting that due to how $m_{1}^{\prime}$ and $\widetilde{D}$ are
chosen, $(E_{n}\backslash D_{\xi,A}(a))\cup\widetilde{D}$ and $E_{n}$ have the
same minimum. Write $A=\cup_{i=1}^{\min A}A_{i}$, for $\min A\leqslant
A_{1}<\cdots<A_{\min A}$ and $A_{i}\in\text{\rm MAX}(\mathcal{S}_{\gamma})$.
Let $A_{i_{0}}=E_{n}$. We have

$A_{1}\ <\ \cdots\ <\ A_{i_{0}-1}\ <\ (E_{n}\backslash
D_{\xi,A}(a))\cup\widetilde{D}\ <\ A_{i_{0}+1}\ <\ \cdots<A_{\min A}.$

These sets are in $\text{\rm MAX}(\mathcal{S}_{\gamma})$, and the
transformation does not change the minimum of $A_{i_{0}}$. Therefore,

$(A\backslash D_{\xi,A}(a))\cup\widetilde{D}\ =\ \left(\cup_{i=1,i\neq
i_{0}}^{\min A}A_{i}\right)\cup((E_{n}\backslash
D_{\xi,A}(a))\cup\widetilde{D})\ \in\ \text{\rm MAX}(\mathcal{S}_{\xi}).$

This completes our proof. ∎

######  Theorem 3.4.

Let $\xi<\omega_{1}$. If $A_{1},\ldots,A_{n}$, $n\in\mathbb{N}$, are pairwise
disjoint, nonempty sets in $\mathcal{S}_{\xi}$ with $\min A_{1}<\cdots<\min
A_{n}$, then there exist nonempty sets $A_{1}^{\prime}<\cdots<A_{n}^{\prime}$
in $\mathcal{S}_{\xi}$ such that
$\cup_{i=1}^{n}A_{i}=\cup_{i=1}^{n}A_{i}^{\prime}$ and $\min
A_{i}\leqslant\min A_{i}^{\prime}$, for $i=1,\ldots,n$.

######  Remark 3.5.

Let us first note that if $\xi<\omega_{1}$ and $A\in\mathcal{S}_{\xi}$ with
$1\in A$, then $A=\\{1\\}$. Therefore, we can assume in the statement of
Theorem 3.4 that $1\notin A_{i}$ for $i=1,2,\ldots,n$.

###### Proof of Theorem 3.4.

We will prove the theorem by transfinite induction for all $\xi<\omega_{1}$.
For the base case $\xi=0$, the theorem is trivially true.

Inductive hypothesis: suppose that the theorem is true for all $\eta<\xi$, for
some $\xi\geqslant 1$. We shall show that the theorem is true for the ordinal
$\xi$. We do so in three steps:

  * •

Step 1: we prove the theorem for a successor ordinal $\xi$ and $n=2$.

  * •

Step 2: we prove the theorem for a limit ordinal $\xi$ and $n=2$.

  * •

Step 3: we show that the theorem holds for an arbitrary $n\in\mathbb{N}$.

Step 1: Write $\xi=\eta+1$. Let $A,B\in\mathcal{S}_{\xi}$ be nonempty. Without
loss of generality, we can assume that $2\leqslant\min A<\min B<\max A$. By
enlarging $A$, if necessary, we also assume that $A\in\text{\rm
MAX}(\mathcal{S}_{\xi})$.

We decompose $A$ as $A=\cup_{i=1}^{\min A}A_{i}$ with
$A_{i}\in\mathcal{S}_{\eta}\backslash\\{\emptyset\\}$ and $\min A\leqslant
A_{1}<\cdots<A_{\min A}$. Also decompose $B=\cup_{i=1}^{\ell}B_{i}$ with
$B_{i}\in\mathcal{S}_{\eta}\backslash\\{\emptyset\\}$ and $\ell\leqslant
B_{1}<\cdots<B_{\ell}$. Then we order the sets $\\{A_{i}:i\leqslant\min
A\\}\cup\\{B_{i}:i\leqslant\ell\\}$ into sets $(D_{i})_{i=1}^{\min A+\ell}$
and apply the inductive hypothesis to obtain nonempty sets
$(\widetilde{D}_{i})_{i=1}^{\min A+\ell}\subset\mathcal{S}_{\eta}$ such that

$\widetilde{D}_{1}\ <\ \cdots\ <\ \widetilde{D}_{\min A+\ell}\mbox{ and
}\bigcup_{i=1}^{\min A+\ell}D_{i}\ =\ \bigcup_{i=1}^{\min
A+\ell}\widetilde{D}_{i}.$

We form $D^{\prime}_{1}$ by unioning $\widetilde{D}_{1}$ with the smallest
elements of $\cup_{i=2}^{\min A+\ell}\widetilde{D}_{i}$, if necessary, until
we obtain a maximal $\mathcal{S}_{\eta}$ set. Then form $D^{\prime}_{2}$ from
the smallest elements in $(\widetilde{D}_{i})_{i=2}^{\min A+\ell}\backslash
D^{\prime}_{1}$ so that $D^{\prime}_{2}$ is a maximal $\mathcal{S}_{\eta}$
set. In general, we form $D^{\prime}_{j}$, for $j=2,\ldots,\min A$, from the
smallest elements in $(\widetilde{D}_{i})_{i=j}^{\min
A+\ell}\backslash\left(\cup_{i=1}^{j-1}D^{\prime}_{i}\right)$. Then the sets
$(D^{\prime}_{i})_{i=1}^{\min A}$ are maximal $\mathcal{S}_{\eta}$. Let
$\widetilde{A}=\cup_{i=1}^{\min A}D^{\prime}_{i}$.

After the above process, some of the sets
$(\widetilde{D}_{i}\backslash\widetilde{A})_{i=\min A+1}^{\min A+\ell}$ are
possibly empty. Discard the empty sets and rename the nonempty ones as
$(D^{\prime}_{i})_{i=\min A+1}^{\min A+\ell^{\prime}}$. Here
$\ell^{\prime}\leqslant\ell$, and there must exist some nonempty sets;
otherwise, we have a contradiction with the maximality of $A$. Let
$\widetilde{B}=\cup_{i=\min A+1}^{\min A+\ell^{\prime}}D^{\prime}_{i}$.

Since all the sets $(B_{i})_{i=1}^{\ell}$ have $\min B_{i}\geqslant\min
B_{1}$, there are at least $\ell$ sets $\widetilde{D}_{i}$’s whose minimum is
at least $\min B_{1}$. Hence, we have

$\displaystyle\min\widetilde{A}\ =\ \min D^{\prime}_{1}\ =\
\min\widetilde{D}_{1}\ =\ \min A,$ $\displaystyle\min\widetilde{B}\ =\ \min
D^{\prime}_{\min A+1}\ \geqslant\ \min\widetilde{D}_{\min A+1}\ \geqslant\
\min B_{1}\ \geqslant\ \ell\ \geqslant\ \ell^{\prime}.$

Hence, both $\widetilde{A}$ and $\widetilde{B}$ are in $\mathcal{S}_{\xi}$,
$\widetilde{A}<\widetilde{B}$, and $\min A=\min\widetilde{A}$ and $\min
B\leqslant\min\widetilde{B}$, as desired. Moreover, if $A$ is
$\mathcal{S}_{\xi}$ maximal, then $\widetilde{A}$ is $\mathcal{S}_{\xi}$
maximal as well.

Step 2: Let $\xi$ be a limit ordinal and $n=2$. Let $A,B\in\mathcal{S}_{\xi}$
be nonempty. Without loss of generality, we can assume that $\min A<\min
B<\max A$ and $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$. Let $a=\min\\{x\in
A:x>\min B\\}$ and let $\gamma+1=\mbox{order}_{\xi,A}(E)$, where $E$ is the
immediate predecessor of $D_{\xi,A}(a)$ in the tree $T(\xi,A)$.

Put $F=\\{x\in A\cup B:x\geqslant\min B\\}$ and write
$F=\\{f_{1},f_{2},\ldots,f_{\ell}\\}$, with $\min
B=f_{1}<f_{2}<\cdots<f_{\ell}$. Then $D_{\xi,A}(a)\cup B\subset F$. Let
$m<\ell$ be such that
$\widetilde{D}=\\{f_{1},f_{2},\ldots,f_{m}\\}\in\text{\rm
MAX}(\mathcal{S}_{\gamma})$. Such an $m$ must exist because $F$ contains a
maximal $\mathcal{S}_{\gamma}$ set, namely $D_{\xi,A}(a)$. Both
$\widetilde{D}$ and
$D_{\xi,A}(a)=\\{f_{1}^{\prime},f_{2}^{\prime},\ldots,f_{k}^{\prime}\\}$ are
subsets of $F$, and because $\widetilde{D}$ consists of the smallest elements
of $F$, we know that $f_{i}<f_{i}^{\prime}$ for $i=1,\ldots,\min\\{m,k\\}$. It
must be that $k\geqslant m$; otherwise, the spreading property and
$\widetilde{D}\in\mathcal{S}_{\gamma}$ would imply that
$D_{\xi,A}(a)\cup\\{a^{\prime}\\}\in\mathcal{S}_{\gamma}$, for every
$a^{\prime}>\max A\cup B$, is in $\text{\rm MAX}(\mathcal{S}_{\gamma})$. This
contradicts the maximality of $D_{\xi,A}(a)$. Therefore,
$\max\widetilde{D}=f_{m}<f^{\prime}_{k}=\max D_{\xi,A}(a)$. We obtain

$\max\widetilde{D}\ <\ \max D_{\xi,A}(a)\ <\ \min\\{\min D\,:\,D\in
T(\xi,A)\mbox{ and }D>D_{\xi,A}(a)\\}\ =\ m_{2}.$

Furthermore,

$\min\widetilde{D}\ =\ \min B\ >\ \max\\{\max D\,:\,D\in T(\xi,A)\mbox{ and
}D<D_{\xi,A}(a)\\}\ =\ m_{1}.$

It follows from Lemma 3.3 that $\widetilde{A}=(A\backslash
D_{\xi,A}(a))\cup\widetilde{D}\in\text{\rm MAX}(\mathcal{S}_{\xi})$, and $\min
A=\min\widetilde{A}$. Define
$\widetilde{B}=(B\backslash\widetilde{D})\cup(D_{\xi,A}(a)\backslash\widetilde{D})$.
It is easy to verify that

$\displaystyle\widetilde{A}\cup\widetilde{B}\ =\ A\cup B\mbox{ and }\min B\ <\
\min\widetilde{B},$ $\displaystyle B\backslash\widetilde{D}\ \in\
\mathcal{S}_{\alpha(\xi,\min B)},\mbox{ and}$ $\displaystyle
D_{\xi,A}(a)\backslash\widetilde{D}\ \in\ \mathcal{S}_{\alpha(\xi,\min A)}\
\subset\ \mathcal{S}_{\alpha(\xi,\min B)}.$

Applying the inductive hypothesis to $\widetilde{B}$, we can find two sets
$\widetilde{B}_{1},\widetilde{B}_{2}\in\mathcal{S}_{\alpha(\xi,\min B)}$ with
$\widetilde{B}_{1}<\widetilde{B}_{2}$. By (2), we know that
$\widetilde{B}=\widetilde{B}_{1}\cup\widetilde{B}_{2}\in\mathcal{S}_{\alpha(\xi,\min
B+1)}\subset\mathcal{S}_{\alpha(\xi,\min\widetilde{B})}$. Hence,
$\widetilde{B}\in\mathcal{S}_{\xi}$.

Either we are now in the situation that $\widetilde{A}<\widetilde{B}$ in which
case, we are done, or $\max\widetilde{A}>\min\widetilde{B}$, and we would
repeat the above process. However, we can only iterate the process finitely
many times because, after each iteration, we increase the minimum of a subset
of $A\cup B$ by $1$. This completes our Step 2. As in Step 1, our new set
$\widetilde{A}$ is can be chosen to be in $\text{\rm MAX}(\mathcal{S}_{\xi})$
if $A\in\text{\rm MAX}(\mathcal{S}_{\xi})$.

Step 3: We show by induction on $n$ that if $A_{1},A_{2},\ldots,A_{n}$ are
disjoint, nonempty sets in $\mathcal{S}_{\xi}$, then there are nonempty sets
$\widetilde{A}_{1},\widetilde{A}_{2},\ldots,\widetilde{A}_{n}$ in
$\mathcal{S}_{\xi}$ so that
$\widetilde{A}_{1}<\widetilde{A}_{2}<\cdots<\widetilde{A}_{n}$ and
$\cup_{j=1}^{n}A_{j}=\cup_{j=1}^{n}\widetilde{A}_{j}$. Assume that $\min
A_{1}<\min A_{2}<\cdots<\min A_{n+1}$. By possibly enlarging the sets, we can
also assume that $A_{1},A_{2},\ldots,A_{n+1}\in\text{\rm
MAX}(\mathcal{S}_{\xi})$.

We can recursively find $n$ pairs of nonempty sets
$(\widetilde{A}_{1}^{(j)},A_{j}^{\prime})\in\text{\rm
MAX}(\mathcal{S}_{\xi})\times\mathcal{S}_{\xi}$, where $j=2,\ldots,n+1$, so
that

$\displaystyle\widetilde{A}_{1}^{(j)}\ <\ A_{i}^{\prime},\quad
i=2,\ldots,j,\mbox{ and}$ (3.1)
$\displaystyle\widetilde{A}_{1}^{(j)}\cup\bigcup_{i=2}^{j}A_{i}^{\prime}\ =\
\bigcup_{i=1}^{j}A_{i}.$ (3.2)

Indeed, the case $j=2$ follows from Step 1 and Step 2. Suppose we have chosen
$\widetilde{A}_{1}^{(j)}$ and $A^{\prime}_{i}$ for $j\leqslant n$ and
$i=2,\ldots,j$. From $\widetilde{A}_{1}^{(j)}$ and $A_{j+1}$, we obtain
$\widetilde{A}_{1}^{(j+1)}\in\text{\rm MAX}(\mathcal{S}_{\xi})$ and
$A^{\prime}_{j+1}\in\mathcal{S}_{\xi}$ so that
$\widetilde{A}_{1}^{(j+1)}<A^{\prime}_{j+1}$ and
$\widetilde{A}_{1}^{(j+1)}\cup A^{\prime}_{j+1}=\widetilde{A}_{1}^{(j)}\cup
A_{j+1}$. Note that
$\max\widetilde{A}_{1}^{(j+1)}\leqslant\max\widetilde{A}_{1}^{(j)}$ because
otherwise,
$\widetilde{A}_{1}^{(j)}\subsetneq\widetilde{A}_{1}^{(j+1)}\in\mathcal{S}_{\xi}$,
which contradicts the $\mathcal{S}_{\xi}$-maximality of
$\widetilde{A}_{1}^{(j)}$. This together with the supposition that (3.1) holds
for $j$ guarantees that (3.1) holds for $j+1$. Using
$\widetilde{A}_{1}^{(j+1)}\cup A^{\prime}_{j+1}=\widetilde{A}_{1}^{(j)}\cup
A_{j+1}$ and the supposition that (3.2) holds for $j$, we can easily verify
that (3.2) holds for $j+1$. This shows the existence of
$\widetilde{A}_{1}^{(j)}$ and $A_{j}^{\prime}$, $j=2,3,\ldots,n+1$.

Put $\widetilde{A}_{1}=\widetilde{A}_{1}^{(n+1)}$. Then
$\widetilde{A}_{1}<A_{i}^{\prime}$, for $i=1,2,\ldots,n+1$. Apply the
inductive hypothesis to the $n$ sets
$(A_{i}^{\prime})_{i=2}^{n+1}\subset\mathcal{S}_{\xi}$ to have
$(\widetilde{A}_{i})_{i=2}^{n+1}\subset\mathcal{S}_{\xi}$ satisfying

$\widetilde{A}_{2}\ <\ \cdots\ <\ \widetilde{A}_{n+1}\quad\mbox{ and
}\quad\bigcup_{i=2}^{n+1}\widetilde{A}_{i}\ =\
\bigcup_{i=2}^{n+1}A^{\prime}_{i}.$

It follows from (3.2) that

$\widetilde{A}_{1}\ <\ \widetilde{A}_{2}\ <\ \cdots\ <\
\widetilde{A}_{n+1}\quad\mbox{ and }\quad\bigcup_{i=1}^{n+1}\widetilde{A}_{i}\
=\ \widetilde{A}_{1}^{(n+1)}\cup\bigcup_{i=2}^{n+1}A^{\prime}_{i}\ =\
\bigcup_{i=1}^{n+1}A_{i}.$

This completes our proof. ∎

######  Corollary 3.6.

For all countable ordinal $\xi\in[1,\omega_{1})$, we have
$\mathcal{S}_{\xi}=\mathcal{S}^{M}_{\xi}$.

###### Proof.

We proceed by transfinite induction. The case $\xi=1$ is trivial. Assume that
the corollary holds for all $\gamma<\xi$ for some $\xi\geqslant 2$.

Case 1: if $\xi$ is a successor ordinal, then write $\xi=\gamma+1$. Let
$F\in\mathcal{S}^{M}_{\xi}$. By definition, $F$ can be written as
$\cup_{i=1}^{k}F_{i}$, for $(F_{i})_{i=1}^{k}\subset\mathcal{S}^{M}_{\gamma}$,
being pairwise disjoint, and $(\min F_{i})_{i=1}^{k}\in\mathcal{S}_{1}$. By
the inductive hypothesis, $\mathcal{S}^{M}_{\gamma}=\mathcal{S}_{\gamma}$;
hence, $(F_{i})_{i=1}^{k}\subset\mathcal{S}_{\gamma}$. By Theorem 3.4, we can
find $(F_{i}^{\prime})_{i=1}^{k}\subset\mathcal{S}_{\gamma}$ such that

$F^{\prime}_{1}\ <\ F^{\prime}_{2}\ <\ \cdots\ <\ F^{\prime}_{k},\\{\min
F^{\prime}_{i}\,:\,1\leqslant i\leqslant k\\}\in\mathcal{S}_{1},\mbox{ and
}\cup_{i=1}^{k}F^{\prime}_{i}\ =\ \cup_{i=1}^{k}F_{i}.$

Therefore, $F$ is in $\mathcal{S}_{\xi}$.

Case 2: $\xi$ is a limit ordinal. Pick $F\in\mathcal{S}^{M}_{\xi}$, then
$F\in\mathcal{S}^{M}_{\alpha(\xi,k)}$ for some $k\leqslant\min F$. By the
inductive hypothesis,
$F\in\mathcal{S}_{\alpha(\xi,k)}\subset\mathcal{S}_{\alpha(\xi,\min F)}$,
which implies that $F\in\mathcal{S}_{\xi}$. ∎

##  4\. Modified Tsirelson spaces of higher order

Manoussakis [14] proved that the Tsirelson space $T[\mathcal{S}_{\xi},\theta]$
is naturally isomorphic to its modified version
$T_{M}[\mathcal{S}_{\xi},\theta]$ for all finite ordinals $\xi$. To do so, he
compared $T[\mathcal{S}_{\xi},\theta]$ to an auxiliary space $Y$ then compared
$T_{M}[\mathcal{S}_{\xi},\theta]$ to $Y$, where the latter step used a
technique from Bellenot [4]. In this section, we extended Manoussakis’ result
by proving the isomorphism between $T[\mathcal{S}_{\xi},\theta]$ and
$T_{M}[\mathcal{S}_{\xi},\theta]$ for all ordinals $\xi<\omega_{1}$. To
compare $T[\mathcal{S}_{\xi},\theta]$ to $Y$, we borrow Manoussakis’ argument
and adapt it to handle limit ordinals (see Lemma 4.3.) To compare
$T_{M}[\mathcal{S}_{\xi},\theta]$ to $Y$, we present a different proof than
Manoussakis’ and Bellenot’s.

Let $c_{00}$ be the vector space of all sequences $x=(x_{n})$ in $\mathbb{R}$
for which the support, $\text{supp }x=\\{n\in\mathbb{N}:x_{n}\not=0\\}$, is
finite. We denote by $(e_{n})$ the unit vector basis of $c_{00}$ and by
$(e^{*}_{n})$ its coordinate functionals. All our Banach spaces will be
completions of $c_{00}$ under some norm, for which $(e_{n})$ is a normalized
and unconditional basis.

Let $\mathcal{F}$ denote a regular family of finite subsets of $\mathbb{N}$. A
sequence of finite sets $(E_{i})_{i=1}^{k}$ of positive integers is said to be
$\mathcal{F}$-admissible if

$E_{1}\ <\ E_{2}\ <\ \cdots\ <\ E_{k}\mbox{ and }\\{\min E_{i}\,:\,1\leqslant
i\leqslant k\\}\ \in\ \mathcal{F}.$

A sequence of finite sets $(E_{i})_{i=1}^{k}$ of positive integers is said to
be $\mathcal{F}$-allowable if

$(E_{i})_{i=1}^{k}\mbox{ are pairwise disjoint and }\\{\min
E_{i}\,:\,1\leqslant i\leqslant k\\}\ \in\ \mathcal{F}.$

Let $\theta$ be a number with $0<\theta<1$. The Tsirelson space
$T[\mathcal{F},\theta]$ is the completion of $c_{00}$ under the implicit norm

$\|x\|_{\mathcal{F},\theta}\ =\
\max\left\\{\|x\|_{\infty},\theta\sup\left\\{\sum_{i=1}^{k}\|E_{i}x\|\,:\,(E_{i})_{i=1}^{k}\mbox{
is }\mathcal{F}\mbox{-admissible}\right\\}\right\\}.$

The modified Tsirelson space $T_{M}[\mathcal{F},\theta]$ is the completion of
$c_{00}$ under the implicit norm

$\|x\|_{\mathcal{F},\theta,M}\ =\
\max\left\\{\|x\|_{\infty},\theta\sup\left\\{\sum_{i=1}^{k}\|E_{i}x\|\,:\,(E_{i})_{i=1}^{k}\mbox{
is }\mathcal{F}\mbox{-allowable}\right\\}\right\\}.$

It was shown in [1], where the space $T[\mathcal{S}_{\xi},\theta]$ was
introduced, that $T[\mathcal{S}_{\xi},\theta]$ is reflexive, and, thus, in
particular, the unit basis cannot have a block basis equivalent to the unit
vector basis of $\ell_{1}$. It follows from [2, Theorem 1.27] that
$T_{M}[\mathcal{S}_{\xi},\theta]$ is also reflexive.

Our goal in this section is to show that $T[\mathcal{S}_{\xi},\theta]$ is
$3$-isomorphic to $T_{M}[\mathcal{S}_{\xi},\theta]$ for every countable
ordinal $\xi$ and for every $0<\theta<1$. To do so, we shall use the auxiliary
space $T[\mathcal{S}_{\xi}[\mathcal{A}_{2}],\theta]$, where
$\mathcal{A}_{n}=\\{E\subset\mathbb{N}:|E|\leqslant n\\}$. It suffices to
prove the following two theorems.

######  Theorem 4.1.

For every $x\in c_{00}$, we have

$\|x\|_{\mathcal{S}_{\xi}[\mathcal{A}_{2}],\theta}\ \leqslant\
3\|x\|_{\mathcal{S}_{\xi},\theta},$

######  Theorem 4.2.

For every $x\in c_{00}$, we have

$\|x\|_{\mathcal{S}_{\xi},\theta,M}\ \leqslant\
\|x\|_{\mathcal{S}_{\xi}[\mathcal{A}_{2}],\theta}$

and thus, by Theorem 4.1,

$\|x\|_{\mathcal{S}_{\xi},\theta}\ \leqslant\
\|x\|_{\mathcal{S}_{\xi},\theta,M}\ \leqslant\
3\|x\|_{\mathcal{S}_{\xi},\theta}.$

To prove Theorem 4.1, we recall some notation for the statement of the key
Lemma 4.3. Given two families $\mathcal{M}$ and $\mathcal{N}$ of subsets of
$\mathbb{N}$, let

$\mathcal{M}[\mathcal{N}]\ =\
\left\\{\bigcup_{i=1}^{k}F_{i}\,:\,F_{i}\in\mathcal{N}\mbox{ and
}(F_{i})_{i=1}^{k}\mbox{ is }\mathcal{M}\mbox{-admissible}\right\\},\mbox{ and
}$ $(\mathcal{M})^{3}\ =\ \mathcal{A}_{3}[\mathcal{M}]\ =\ \left\\{M_{1}\cup
M_{2}\cup M_{3}\,:\,M_{1}<M_{2}<M_{3}\mbox{ and
}M_{i}\in\mathcal{M}\right\\}.$

Below are several useful facts:

  1. (1)

For families $\mathcal{L},\mathcal{M}$, and $\mathcal{N}$, it holds that

$(\mathcal{L}[\mathcal{M}])[\mathcal{N}]\ =\
\mathcal{L}[\mathcal{M}[\mathcal{N}]].$ (4.1)

###### Proof.

Let $F_{1}<\cdots<F_{k}$, where $F_{i}\in\mathcal{N}$ and $(\min
F_{i})_{i=1}^{k}\in\mathcal{L}[\mathcal{M}]$. Setting $j_{0}=0$, we can find
$\ell$ and $j_{1}<\cdots<j_{\ell}$ such that $\\{\min
F_{j_{i-1}+1},\ldots,\min F_{j_{i}}\\}\in\mathcal{M}$ for $i=1,\ldots,\ell$,
and $(\min F_{j_{i}+1})_{i=0}^{\ell-1}\in\mathcal{L}$. Now let
$E_{n}=\cup_{i=j_{n-1}+1}^{j_{n}}F_{i}$, $n=1,2,\ldots,\ell$. Clearly,
$E_{n}\in\mathcal{M}[\mathcal{N}]$. Therefore,
$\cup_{i=1}^{k}F_{i}=\cup_{n=1}^{\ell}E_{n}\in\mathcal{L}[\mathcal{M}[\mathcal{N}]]$.

Conversely, we take $F_{1}<\cdots<F_{k}$, where each
$F_{i}\in\mathcal{M}[\mathcal{N}]$ and the set $\\{\min F_{1},\ldots,\min
F_{k}\\}$ is in $\mathcal{L}$. Write $F_{i}=\cup_{j=1}^{\ell_{i}}E^{i}_{j}$,
for $E^{i}_{j}\in\mathcal{N}$, $(\min
E^{i}_{j})_{j=1}^{\ell_{i}}\in\mathcal{M}$, and
$E^{i}_{1}<\cdots<E^{i}_{\ell_{i}}$. Note that $\\{\min E^{i}_{j}:1\leqslant
i\leqslant k,1\leqslant j\leqslant\ell_{i}\\}$ is in
$\mathcal{L}[\mathcal{M}]$. Hence,
$\cup_{i=1}^{k}\left(\cup_{j=1}^{\ell_{i}}E^{i}_{j}\right)\in(\mathcal{L}[\mathcal{M}])[\mathcal{N}]$.

This completes our proof. ∎

  2. (2)

From (4.1), we deduce that for a family $\mathcal{F}$ of subsets of
$\mathbb{N}$ and for all $\xi<\omega_{1}$,

$\mathcal{S}_{\xi+1}[\mathcal{F}]\ =\
\mathcal{S}_{1}[\mathcal{S}_{\xi}[\mathcal{F}]].$ (4.2)

  3. (3)

For all $\xi<\omega_{1}$,

$\mathcal{S}_{1}[(\mathcal{S}_{\xi})^{3}]\ \subset\
(\mathcal{S}_{\xi+1})^{3}.$ (4.3)

###### Proof.

Take $F_{1}<\cdots<F_{k}$, $F_{i}\in(\mathcal{S}_{\xi})^{3}$, and $\\{\min
F_{1},\ldots,\min F_{k}\\}\in\mathcal{S}_{1}$. Write $F_{i}=F^{i}_{1}\cup
F^{i}_{2}\cup F^{i}_{3}$, for $F^{i}_{j}\in\mathcal{S}_{\xi}$ and
$F^{i}_{1}<F^{i}_{2}<F^{i}_{3}$. Reindex the sets $\\{F^{i}_{j}:1\leqslant
i\leqslant k,1\leqslant j\leqslant 3\\}$ into $(E_{i})_{i=1}^{3k}$ satisfying
$k\ \leqslant\ E_{1}\ <\ E_{2}\ <\ \cdots\ <\ E_{3k}$. Since $k\leqslant
E_{1}<E_{2}<\cdots<E_{k}$ and $E_{i}\in\mathcal{S}_{\xi}$, we know that
$\cup_{i=1}^{k}E_{i}\in\mathcal{S}_{\xi+1}$. Next, $2k\leqslant
E_{k+1}<E_{k+2}<\cdots<E_{3k}$ and $E_{i}\in\mathcal{S}_{\xi}$; hence,
$\cup_{i=k+1}^{3k}E_{i}\in\mathcal{S}_{\xi+1}$. We conclude that

$\bigcup_{i=1}^{k}F_{i}\ =\
\bigcup_{i=1}^{k}E_{i}\cup\bigcup_{i=k+1}^{3k}E_{i}\ \in\
(\mathcal{S}_{\xi+1})^{2}.$

∎

Using the above facts, we can prove the following lemma.

######  Lemma 4.3.

Let $1\leqslant\xi<\omega_{1}$. It holds that

$(\mathcal{S}_{\xi}[\mathcal{A}_{2}])[\mathcal{A}_{3}]\ \subset\
(\mathcal{S}_{\xi})^{3}.$ (4.4)

###### Proof.

We proceed by transfinite induction on $\xi$. When $\xi=1$, (4.1) gives

$(\mathcal{S}_{1}[\mathcal{A}_{2}])[\mathcal{A}_{3}]\ =\
\mathcal{S}_{1}[\mathcal{A}_{2}[\mathcal{A}_{3}]]\ =\
\mathcal{S}_{1}[\mathcal{A}_{6}].$

Let $F\in\mathcal{S}_{1}[\mathcal{A}_{6}]$, then $F=\cup_{i=1}^{k}F_{i}$,
where $k\leqslant F_{1}<F_{2}<\cdots<F_{k}$ and $|F_{i}|\leqslant 6$. Write
$F=\\{m_{1},m_{2},\ldots,m_{\ell}\\}$ with $k\leqslant m_{1}\leqslant
m_{2}\leqslant\cdots\leqslant m_{\ell}$ and $\ell\leqslant 6k$. The same
reasoning as in the proof of (4.3) guarantees that
$F\in(\mathcal{S}_{1})^{3}$.

Inductive hypothesis: Suppose that (4.4) holds for all $\gamma<\xi$ for some
$\xi\geqslant 2$. We shall show that it is true for $\xi$.

Case 1: $\xi$ is a successor ordinal. Write $\xi=\gamma+1$. By (4.1), the
inductive hypothesis, and (4.3), we have

$\displaystyle(\mathcal{S}_{\gamma+1}[\mathcal{A}_{2}])[\mathcal{A}_{3}]$
$\displaystyle\ =\
((\mathcal{S}_{1}[\mathcal{S}_{\gamma}])[\mathcal{A}_{2}])[\mathcal{A}_{3}]\
=\ (\mathcal{S}_{1}[\mathcal{S}_{\gamma}[\mathcal{A}_{2}]])[\mathcal{A}_{3}]$
$\displaystyle\ =\
\mathcal{S}_{1}[(\mathcal{S}_{\gamma}[\mathcal{A}_{2}])[\mathcal{A}_{3}]]\
\subset\ \mathcal{S}_{1}[(\mathcal{S}_{\gamma})^{3}]\ \subset\
(\mathcal{S}_{\xi})^{3}.$

Case 2: $\xi$ is a limit ordinal. Take
$F\in(\mathcal{S}_{\xi}[\mathcal{A}_{2}])[\mathcal{A}_{3}]$. Then
$F=\cup_{i=1}^{k}F_{i}$, for $|F_{i}|\leqslant 3$ and $(\min
F_{i})_{i=1}^{k}\in\mathcal{S}_{\xi}[\mathcal{A}_{2}]$. Hence, $\\{\min
F_{i}:i=1,3,\ldots,2\lfloor(k-1)/2\rfloor+1\\}\in\mathcal{S}_{\xi}$. By (2.5),
$\\{\min
F_{i}:i=1,3,\ldots,2\lfloor(k-1)/2\rfloor+1\\}\in\mathcal{S}_{\alpha(\xi,\min
F)}$. By the inductive hypothesis, $F\in(\mathcal{S}_{\alpha(\xi,\min
F)}[\mathcal{A}_{2}])[\mathcal{A}_{3}]\subset(\mathcal{S}_{\alpha(\xi,\min
F)})^{3}$. Hence, $F=E_{1}\cup E_{2}\cup E_{3}$, for $E_{1}<E_{2}<E_{3}$ and
$E_{i}\in\mathcal{S}_{\alpha(\xi,\min F)}$. Since
$E_{i}\in\mathcal{S}_{\alpha(\xi,\min F)}$ and $\min E_{i}\geqslant\min F$, it
follows from (2.5) that $E_{i}\in\mathcal{S}_{\xi}$, for $i=1,2,3$. Therefore,
$F\in(\mathcal{S}_{\xi})^{3}$. ∎

For a regular family $\mathcal{F}$ and $0<\theta<1$, we now define $1$-norming
sets $K=K(\mathcal{F},\theta)$ and $K^{M}=K^{M}(\mathcal{F},\theta)$ for
$T[\mathcal{F},\theta]$ and $T_{M}[\mathcal{F},\theta]$, respectively. Let
$K_{0}=K_{0}(\mathcal{F},\theta)=\\{0\\}\cup\\{\pm
e^{*}_{n}:n\in\mathbb{N}\\}$. Assuming that $K_{n}=K_{n}(\mathcal{F})$ has
been defined, we define $K_{n+1}=K_{n+1}(\mathcal{F})$ by

$K_{n+1}\ =\
K_{n}\cup\\{\theta(f^{*}_{1}+\cdots+f^{*}_{d})\,:\,d\in\mathbb{N},f_{i}^{*}\in
K_{n},\mbox{ and }(\operatorname{supp}f^{*}_{i})_{i=1}^{d}\mbox{ is
}\mathcal{F}\mbox{-admissible}\\}.$

Let $K=\cup_{n=0}^{\infty}K_{n}$. Then $K$ is a $1$-norming set for
$T[\mathcal{F},\theta]$, i.e., for every $x\in T[\mathcal{F},\theta]$,

$\|x\|_{\mathcal{F},\theta}\ =\ \sup_{f^{*}\in K}\langle x,f^{*}\rangle.$

Similarly, we define a norming set $K^{M}$ for $T_{M}[\mathcal{F},\theta]$.
Let $K_{0}^{M}=K_{0}^{M}(\mathcal{F})=\\{0\\}\cup\\{\pm
e^{*}_{n}:n\in\mathbb{N}\\}$. Assuming that $K_{n}^{M}=K_{n}^{M}(\mathcal{F})$
has been defined, we define $K^{M}_{n+1}=K^{M}_{n+1}(\mathcal{F})$ by

$K^{M}_{n+1}\ =\
K^{M}_{n}\cup\\{\theta(f^{*}_{1}+\cdots+f^{*}_{d})\,:\,d\in\mathbb{N},f_{i}^{*}\in
K^{M}_{n},(\operatorname{supp}f^{*}_{i})_{i=1}^{d}\mbox{ is
}\mathcal{F}\mbox{-allowable}\\}.$

From now on, by the norming set of $T[\mathcal{F},\theta]$ and
$T_{M}[\mathcal{F},\theta]$, we are referring to either $K$ and $K^{M}$,
respectively.

We observe the following. Let $f^{*}\in K^{M}$, and we write it as
$f^{*}=\sum_{j=1}^{\infty}a_{j}e^{*}_{j}$, for some $(a_{j})\in c_{00}$, and
if $N\subset\mathbb{N}$ then $P_{N}(f^{*})=\sum_{j\in N}a_{j}e^{*}_{j}\in
K^{M}$; moreover, $P_{N}(f^{*})\in K$ if $f^{*}\in K$. We call $P_{N}(f^{*})$
the projection of $f^{*}$ onto $N$.

###### Proof of Theorem 4.1.

Let $x\in c_{00}$ and $f^{*}$ be arbitrary in the norming set
$K=\cup_{n=0}^{\infty}K_{n}$ of
$T[\mathcal{S}_{\xi}[\mathcal{A}_{2}],\theta]$. To verify our claim, it
suffices to show by induction on $n$ that there exist $h^{*}_{1},h^{*}_{2}$,
and $h^{*}_{3}$ in the norming set of $T[\mathcal{S}_{\xi},\theta]$ such that

$f^{*}(x)\ \leqslant\
\sum_{i=1}^{3}f^{*}_{i}(x),\bigcup_{i=1}^{3}\operatorname{supp}f^{*}_{i}\subset\operatorname{supp}f^{*},\mbox{
and
}\operatorname{supp}f^{*}_{1}\\!<\\!\operatorname{supp}f^{*}_{2}\\!<\\!\operatorname{supp}f^{*}_{3}.$
(4.5)

If $f^{*}=\pm e^{*}_{m}$ for some $m\in\mathbb{N}$, there is nothing to prove.
Suppose that (4.5) holds for every $f^{*}\in K_{n}$ for some $n\geqslant 0$.
We show that it holds for $f^{*}\in K_{n+1}\backslash K_{n}$. Write
$f^{*}=\theta\sum_{i=1}^{d}f^{*}_{i}$, for $f^{*}_{i}\in K_{n}$ and
$(\operatorname{supp}f^{*}_{i})_{i=1}^{d}$ is
$\mathcal{S}_{\xi}[\mathcal{A}_{2}]$-admissible. By the inductive hypothesis,
for $i=1,2,\ldots,d$, we can find three functionals $g^{*}_{i,1},g^{*}_{i,2}$,
and $g^{*}_{i,3}$ in the norming set of $T[\mathcal{S}_{\xi},\theta]$
satisfying

$f_{i}^{*}(x)\ \leqslant\
\sum_{j=1}^{3}g^{*}_{i,j}(x),\bigcup_{j=1}^{3}\operatorname{supp}g^{*}_{i,j}\
\subset\ \operatorname{supp}f^{*}_{i},\mbox{ and
}\operatorname{supp}g^{*}_{i,1}\ <\ \operatorname{supp}g^{*}_{i,2}\ <\
\operatorname{supp}g^{*}_{i,3}.$

Lemma 4.3 and
$(\min\operatorname{supp}f^{*}_{i})_{i=1}^{d}\in\mathcal{S}_{\xi}[\mathcal{A}_{2}]$
imply that

$\\{\min\operatorname{supp}g^{*}_{i,j}\,:\,1\leqslant i\leqslant d,1\leqslant
j\leqslant 3\\}\ \in\ (\mathcal{S}_{\xi}[\mathcal{A}_{2}])[\mathcal{A}_{3}]\
\subset\ (\mathcal{S}_{\xi})^{3}.$

Reindex $(g^{*}_{i,j})_{i,j}$ into the sequence $(g^{*}_{i})_{i=1}^{3d}$ with
consecutive supports and find $j_{1}$ and $j_{2}$ such that

$\displaystyle\\{\min\operatorname{supp}g^{*}_{1},\ldots,\min\operatorname{supp}g^{*}_{j_{1}}\\}\
\in\ \mathcal{S}_{\xi},$
$\displaystyle\\{\min\operatorname{supp}g^{*}_{j_{1}+1},\ldots,\min\operatorname{supp}g^{*}_{j_{2}}\\}\
\in\ \mathcal{S}_{\xi},\mbox{ and }$
$\displaystyle\\{\min\operatorname{supp}g^{*}_{j_{2}+1},\ldots,\min\operatorname{supp}g^{*}_{3d}\\}\
\in\ \mathcal{S}_{\xi}.$

Then $h_{1}^{*}=\theta\sum_{i=1}^{j_{1}}g^{*}_{i}$,
$h_{2}^{*}=\theta\sum_{i=j_{1}+1}^{j_{2}}g^{*}_{i}$, and
$h_{3}^{*}=\theta\sum_{i=j_{2}+1}^{3d}g^{*}_{i}$ are in the norming set of
$T[\mathcal{S}_{\xi},\theta]$, satisfying
$\operatorname{supp}h_{1}^{*}<\operatorname{supp}h_{2}^{*}<\operatorname{supp}h_{3}^{*}$
and
$\cup_{i=1}^{3}\operatorname{supp}h_{i}^{*}\subset\operatorname{supp}f^{*}$.
Furthermore,

$f^{*}(x)\ =\ \theta\sum_{i=1}^{d}f^{*}_{i}(x)\ \leqslant\
\theta\sum_{i=1}^{d}\sum_{j=1}^{3}g^{*}_{i,j}(x)\ =\
\theta\sum_{i=1}^{3d}g^{*}_{i}(x)\ =\ \sum_{i=1}^{3}h_{i}^{*}(x).$

This completes our proof. ∎

In order to prove Theorem 4.2, we shall verify

$K^{M}(\mathcal{S}_{\xi},\theta)\ \subset\
K(\mathcal{S}_{\xi}[\mathcal{A}_{2}],\theta),\text{ for all
$\xi<\omega_{1}$,}$ (4.6)

which implies that

$\|x\|_{\mathcal{S}_{\xi},\theta,M}\ \leqslant\
\|x\|_{\mathcal{S}_{\xi}[\mathcal{A}_{2}],\theta},\text{ for all }x\in
c_{00}.$

Note that, for a given regular regular family
$\mathcal{F}\subset[\mathbb{N}]^{<\omega}$, the sets $K(\mathcal{F})$ and
$K^{M}(\mathcal{F})$ are invariant under sign changes of the coefficients,
i.e., for all $k\in\mathbb{N}$, $(a_{i})_{i=1}^{k}\subset\mathbb{R}$,
$n_{1}<n_{2}<\cdots<n_{k}$ in $\mathbb{N}$, and
$(\varepsilon_{j})_{j=1}^{k}\subset\\{\pm 1\\}$, we have

$\displaystyle\sum_{j=1}^{k}a_{j}e^{*}_{n_{j}}\ \in\ K(\mathcal{F})$
$\displaystyle\iff\sum_{j=1}^{k}\varepsilon_{j}a_{j}e^{*}_{n_{j}}\ \in\
K(\mathcal{F}),\mbox{ and }$ $\displaystyle\sum_{j=1}^{k}a_{j}e^{*}_{n_{j}}\
\in\ K^{M}(\mathcal{F})$
$\displaystyle\iff\sum_{j=1}^{k}\varepsilon_{j}a_{j}e^{*}_{n_{j}}\ \in\
K^{M}(\mathcal{F}).$

It is therefore enough to show that

$P^{M}(\mathcal{S}_{\xi},\theta)\ \subset\
P(\mathcal{S}_{\xi}[\mathcal{A}_{2}],\theta),$ (4.7)

where for a regular family $\mathcal{F}$, we denote by $P^{M}(\mathcal{F})$
and $P(\mathcal{F})$ the elements of $K^{M}(\mathcal{F})$ and
$K(\mathcal{F})$, respectively, with nonnegative coefficients.

We fix a regular family $\mathcal{F}$ and abbreviate $P=P(\mathcal{F})$ and
$P^{M}=P^{M}(\mathcal{F})$; for $n\in\mathbb{N}_{0}$, we let
$P_{n}=P(\mathcal{F})\cap K_{n}(\mathcal{F})$ and
$P^{M}_{n}=P^{M}(\mathcal{F})\cap K^{M}_{n}(\mathcal{F})$.

  1. (1)

If $f^{*}\in P_{0}^{M}=P_{0}\setminus\\{0\\}$, then $f^{*}=e^{*}_{s}$, for
some $s\in\mathbb{N}$.

  2. (2)

If $\ell\in\mathbb{N}$ and $f^{*}\in P^{M}_{\ell}\backslash P_{0}^{M}$, then
the coefficients of $f^{*}$ are of the form $\theta^{j}$, with
$j\in\\{1,2,\ldots,\ell\\}$. We define the level sets  of $f^{*}$ to be

$L_{j}(f^{*})\ =\ \\{s\in\mathbb{N}\,:\,f^{*}(e_{s})=\theta^{j}\\},\text{ for
$j=1,2,\ldots,\ell$.}$

It follows that the sets $L_{j}(f^{*})$ are pairwise disjoint and that

$f^{*}\ =\ \sum_{j=1}^{\ell}\theta^{j}\sum_{s\in L_{j}(f^{*})}e^{*}_{s}.$

For $A\in[\mathbb{N}]^{<\omega}$, we write $1^{*}_{A}$ for the functional
$\sum_{s\in A}e^{*}_{s}$. We will now discuss conditions under which a
pairwise disjoint sequence $(A_{j})_{j=1}^{\ell}$ in $[\mathbb{N}]^{<\omega}$
has the property that $\sum_{j=1}^{\ell}\theta^{j}1^{*}_{A_{j}}\in
P_{\ell}^{M}$. If that is the case, and if $A_{\ell}\not=\emptyset$ (and thus,
$\sum_{j=1}^{\ell}\theta^{j}1^{*}_{A_{j}}\in P_{\ell}^{M}\setminus
P_{\ell-1}^{M}$), we call $(A_{j})_{j=1}^{\ell}$ an $\mathcal{F}$-allowable
sequence of length $\ell$.

######  Proposition 4.4.

Assume that $(A_{j})_{j=1}^{\ell}$ is an $\mathcal{F}$-allowable sequence of
length $\ell$ and that

$A^{\prime}_{\ell}\mbox{ is a spread of }A_{\ell}\mbox{ and is disjoint from
}\bigcup_{j=1}^{\ell-1}A_{j}.$ (4.8)

Then $(A_{1},A_{2},\ldots,A_{\ell-1},A^{\prime}_{\ell})$ is also
$\mathcal{F}$-allowable of length $\ell$.

###### Proof.

The case for $\ell=1$ follows from the assumption that $\mathcal{F}$ is
spreading. Assume that the claim is true for some $\ell\in\mathbb{N}$, and let
$(A_{j})_{j=1}^{\ell+1}$ be an $\mathcal{F}$-allowable sequence of length
$\ell+1$, and let $A^{\prime}_{\ell+1}$ be a spread of $A_{\ell+1}$. We have
$f^{*}=\sum_{j=1}^{\ell+1}\theta^{j}1^{*}_{A_{j}}\in P_{\ell+1}^{M}\setminus
P_{\ell}^{M},$ and thus we can write $f^{*}$ as
$f^{*}=\theta\sum_{i=1}^{k}f^{*}_{i}$ with $k\in\mathbb{N}$, $f^{*}_{i}\in
P_{\ell}^{M}$, for $i=1,2,\ldots,k$, and $(\mbox{supp}(f^{*}_{i}))_{i=1}^{k}$
being $\mathcal{F}$-allowable. It follows that

$\displaystyle A_{1}$ $\displaystyle\ =\ L_{1}(f^{*})\ =\
\\{s\in\mathbb{N}\,:\,\exists i=1,2,\ldots,k\mbox{ with
}f^{*}_{i}=e^{*}_{s}\\},$ and for $j=2,3,\ldots,\ell+1$, we have
$\displaystyle A_{j}$ $\displaystyle\ =\ L_{j}(f^{*})\ =\
\bigcup_{i=1}^{k}L_{j-1}(f^{*}_{i}).$

Since $A^{\prime}_{\ell+1}$ is a spread of $A_{\ell+1}$, there is a bijection
$\rho:A_{\ell+1}\to A^{\prime}_{\ell+1}$, with $\rho(a)\geqslant a$ for $a\in
A_{\ell+1}$. For $i=1,2,\ldots,k$, put
$A^{\prime}_{\ell+1,i}=\rho(L_{\ell}(f^{*}_{i}))$, then
$A^{\prime}_{\ell+1,i}$ is a spread of $L_{\ell}(f^{*}_{i})$, or
$L_{\ell}(f^{*}_{i})=\emptyset$. For $i=1,\ldots,k$, let

$g^{*}_{i}\ =\ \begin{cases}f^{*}_{i},&\mbox{ if
}L_{\ell}(f^{*}_{i})=\emptyset,\\\
\sum_{j=1}^{\ell-1}\theta^{j}1^{*}_{L_{j}(f^{*}_{i})}+\theta^{\ell}1_{A^{\prime}_{\ell+1,i}}^{*},&\mbox{
otherwise}.\end{cases}$

By the inductive hypothesis, we have $g^{*}_{i}\in P^{M}_{\ell}$. Note that
(4.8) and the fact that $(\operatorname{supp}(f^{*}_{i}))_{i=1}^{k}$ is
$\mathcal{F}$-allowable implies that $(\mbox{supp}(g^{*}_{i}))_{i=1}^{k}$ is
$\mathcal{F}$-allowable. Therefore, we deduce that

$\sum_{j=1}^{\ell}\theta^{j}1^{*}_{A_{j}}+\theta^{\ell+1}1^{*}_{A^{\prime}_{\ell+1}}\
=\ \theta\sum_{i=1}^{k}g^{*}_{i}\ \in\ P^{M}_{\ell+1}.$

This completes our proof. ∎

We introduce a coding for elements in $P$ and $P^{M}$ using certain finite
trees. For $n\in\mathbb{N}$, we let $[[\mathbb{N}]]^{n}$ be the set of non
strictly increasing sequences in $\mathbb{N}$ of length $n$, and put
$[[\mathbb{N}]]^{<\omega}=\bigcup_{n=1}^{\infty}[[\mathbb{N}]]^{n}$. For
$s=(s_{1},s_{2},\ldots,s_{m})$ and $t=(t_{1},t_{2},\ldots,t_{n})$, we say that
$t$ extends $s$ and write $s\prec t$ if $m<n$, $s_{i}=t_{i}$, for
$i=1,2,\ldots,m$, and we write $s\preceq t$ if $s\prec t$ or $s=t$. For
$t=(t_{1},t_{2},\ldots,t_{n})=[[\mathbb{N}]]^{<\omega}$, we call $n$ the
length of $t$, denoted by $|t|$; for $j=1,2,\ldots,|t|$, we denote the $j$th-
element of $t$ by $t(j)$. If $s\in[[\mathbb{N}]]^{m}$ and
$t\in[[\mathbb{N}]]^{n}$, for some $m,n\in\mathbb{N}$, and $s_{m}\leqslant
t_{1}$, we put

$(s,t)\ =\ (s_{1},s_{2},\ldots,s_{m},t_{1},t_{2},\ldots,t_{n})\ \in\
[[\mathbb{N}]]^{m+n}.$

We also define $(\emptyset,s)=(s,\emptyset)=s$.

We introduce the lexicographical order on $[[\mathbb{N}]]^{<\omega}$: for
$s=(s_{1},s_{2},\ldots,s_{m})$ and $t=(t_{1},t_{2},\ldots,t_{n})$ in
$[[\mathbb{N}]]^{<\omega}$, we let

$s\ \prec_{\text{lex}}\ t\iff\exists 1\ \leqslant\ i_{0}\ <\ \min(m,n)\mbox{
s.t. }s_{i}\ =\ t_{i}\text{ for $1\leqslant i<i_{0}$ and }s_{i_{0}}\ <\
t_{i_{0}}.$

Note that if $s,t\in[[\mathbb{N}]]^{<\omega}$ are not comparable with each
other with respect to “$\preceq$”, then either $s\prec_{\text{lex}}t$ or
$t\prec_{\text{lex}}s$.

A finite subset $T$ of $[[\mathbb{N}]]^{<\omega}$ is called an
$\mathcal{F}$-allowable tree if

  1. (T1)

If $t\in T$ and $s\preceq t$, then $s\in T$.

  2. (T2)

There is a $t_{0}\in T$ with $|t_{0}|=1$, which has the property that
$t_{0}\preceq t$ for all $t\in T$.

From (T1) and (T2), it follows that $T$ is a tree, for which $t_{0}$ is the
root. For $t\in T\setminus\\{t_{0}\\}$, we let $t^{-}$ be the unique
predecessor of $t$ in $T$. For $t\in T$, we call $s\in T$ an immediate
successor of $t$ if $t\prec s$ and $|s|=|t|+1$ and denote the set of immediate
successors of $t$ by $S^{T}_{t}$. A node $t\in T$ is called a terminal node if
$S_{t}^{T}=\emptyset$.

  1. (T3)

If $t\in T$ is not a terminal node, then
$(t,t(n))=(t(1),t(2),\ldots,t(n),t(n))\in T$, and $\\{j\in\mathbb{N}:(t,j)\in
T\\}\in\mathcal{F}$.

To state the next condition, we define the support of a node $s\in T$ as

$\mbox{supp}(s)\ =\ \\{t(|t|)\,:\,t\succeq s,\mbox{ and }t\text{ is terminal
in $T$}\\}.$

  1. (T4)

If $s,t\in T$ with $s\prec_{\text{lex}}t$, then $\operatorname{supp}(s)$ and
$\operatorname{supp}(t)$ are disjoint.

We call $T$ an $\mathcal{F}$-admissible tree if in addition to Conditions (T1)
– (T3), the following condition holds

  1. (T5)

If $s,t\in T$ with $s\prec_{\text{lex}}t$, then
$\operatorname{supp}(s)<\operatorname{supp}(t)$.

For an $\mathcal{F}$-allowable tree $T$, we call

$|T|\ =\ \max\\{|t|\,:\,t\in T\\}\ =\ \max\\{|t|\,:\,t\text{ is terminal node
of $T$}\\}$

the length of $T$. Let $\mathcal{T}^{(\ell)}$ and $\mathcal{T}^{M,(\ell)}$ be
the the collection of all $\mathcal{F}$-admissible and $\mathcal{F}$-allowable
trees of length $\ell$, respectively. We identify
$\mathcal{T}^{(1)}=\mathcal{T}^{M,(1)}$ with $\mathbb{N}$ and put
$\mathcal{T}=\bigcup_{\ell=1}^{\infty}\mathcal{T}^{(\ell)}$ and
$\mathcal{T}^{M}=\bigcup_{\ell=1}^{\infty}\mathcal{T}^{M,(\ell)}$. The
following easy proposition exhibits two ways to reduce trees to trees of
smaller lengths.

######  Proposition 4.5.

Let $\ell\in\mathbb{N}_{\geqslant 2}$ and $T\in\mathcal{T}^{M,(\ell)}$.

  1. (1)

$T^{\prime}=\\{t\in T\,:\,|t|\leqslant\ell-1\\}$ is in
$\mathcal{T}^{M,(\ell-1)}$.

  2. (2)

Let $n_{1}$ be the root of $T$ and let $n_{1}<n_{2}<\cdots<n_{k}$ be chosen in
$\mathbb{N}$, so that

$\\{(n_{1},n_{i})\,:\,i=1,2,\ldots,k\\}\ =\ \\{t\in T\,:\,|t|=2\\}.$

Then for $i=1,2,\ldots,k$,

$T^{(i)}\ =\
\\{(n_{i},s)\,:\,s\in\\{\emptyset\\}\cup[[\mathbb{N}]]^{<\omega}\text{ with
}(n_{1},n_{i},s)\in T\\}\ \in\ \bigcup_{j=1}^{\ell-1}\mathcal{T}^{M,(j)},$

and at least for some $i=1,2,\ldots,k$, we have
$T^{(i)}\in\mathcal{T}^{M,(\ell-1)}$.

Moreover, if $T$ is in $\mathcal{T}^{(\ell)}$, then $T^{\prime}$ as defined in
(1) is in $\mathcal{T}^{(\ell-1)}$, and $T^{(i)}$, $i=1,2,\ldots,k$, as
defined in (2), are in $\cup_{j=1}^{\ell-1}\mathcal{T}^{(j)}$.

We define now for each $\mathcal{F}$-allowable tree $T$ a functional $\Psi(T)$
as follows.

######  Proposition 4.6.

For $T\in\mathcal{T}^{M}$, we define

$\Psi(T)\ =\ \begin{cases}e^{*}_{n}\quad&\text{if
$T=\\{n\\}\in\mathcal{T}^{M,(1)}$,}\\\
\sum_{j=2}^{\ell}\theta^{j-1}1^{*}_{\\{t(j):t\in T\text{is terminal node, and
$|t|=j$}\\}}\quad&\text{if $T\in\mathcal{T}^{M,(\ell)}$, $\ell\geqslant
2$.}\end{cases}$

Then $\Psi$ is a surjective map from $\mathcal{T}^{M,(1)}$ onto
$P^{M}_{0}\setminus\\{0\\}$, and for $\ell\geqslant 2$, $\Psi$ is a surjective
map from $\mathcal{T}^{M,(\ell)}$ onto $P^{M}_{\ell-1}\setminus
P^{M}_{\ell-2}$.

Moreover, if $T$ is a surjective map from $\mathcal{T}^{(1)}$ onto
$P_{0}\setminus\\{0\\}$, and for $\ell\geqslant 2$, $\Psi$ is a surjective map
from $\mathcal{T}^{(\ell)}$ onto $P_{\ell-1}\setminus P_{\ell-2}$.

###### Proof.

We prove the claim by induction on $\ell\in\mathbb{N}$. For $\ell=1$, $\Psi$
is clearly a bijective map between $\mathcal{T}^{M,(1)}$ and
$P_{0}^{M}\setminus\\{0\\}$.

Assume that there is $\ell\geqslant 2$ such that the claim is true for all
$\ell^{\prime}<\ell$ and let $T\in\mathcal{T}^{M,(\ell)}$. Choose
$k\in\mathbb{N}$ and $T^{(i)}\in\bigcup_{j=1}^{\ell-1}\mathcal{T}^{M,(j)}$,
for $i=1,2,\ldots,k$, as in Proposition 4.5 item (2). Denote, for each
$i=1,2,\ldots,k$, the root of $T^{(i)}$ by $n_{i}$ such that
$n_{1}<n_{2}<\cdots<n_{k}$, and $n_{1}$ is the root of $T$. By Condition (T3),
we have $\\{n_{i}:i=1,2,\ldots,k\\}\in\mathcal{F}$. Let
$A_{1}=\\{i=1,2,\ldots,k:|T^{(i)}|=1\\}$ and
$A_{2}=\\{i=1,2,\ldots,k:|T^{(i)}|>1\\}$. We note that

$\displaystyle\Psi(T)$ $\displaystyle\ =\
\sum_{j=2}^{\ell}\theta^{j-1}1^{*}_{\\{t(j)\,:\,t\in T\text{ is a terminal
node, and $|t|=j$}\\}}$ $\displaystyle\ =\
\theta\sum_{j=2}^{\ell}\theta^{j-2}1^{*}_{\\{t(j)\,:\,t\in T\text{ is a
terminal node, and $|t|=j$}\\}}$ $\displaystyle\ =\ \theta\Big{(}\sum_{i\in
A_{1}}e^{*}_{n_{i}}+\sum_{i\in
A_{2}}\sum_{j=2}^{\ell-1}\theta^{j-1}1^{*}_{\\{t(j)\,:\,t\in T^{(i)}\text{ is
a terminal node and $|t|=j$}\\}}\Big{)}\ =\
\theta\sum_{i=1}^{k}\Psi(T^{(i)}).$

It follows from the induction hypothesis that $\Psi(T^{(i)})\in
P_{\ell-2}^{M}$, for $i=1,2,\ldots,k$, and since
$(\min\operatorname{supp}(\Psi(T^{(i)}))=\\{n_{1},n_{2},\ldots,n_{k}\\}\in\mathcal{F}$,
we know that $\Psi(T)\in P^{M}_{\ell-1}$. Since for at least one
$i=1,2,\ldots$, we have $T^{(i)}\in\mathcal{T}^{M,(\ell-1)}$, it follows that
$\Psi(T)\in P^{M}_{\ell-1}\setminus P^{M}_{\ell-2}$.

In order to verify that $\Psi:\mathcal{T}^{M,(\ell)}\rightarrow
P^{M}_{\ell-1}\setminus P^{M}_{\ell-2}$ is surjective, let $f^{*}\in
P^{M}_{\ell-1}\setminus P^{M}_{\ell-2}$ and write
$f^{*}=\theta\sum_{i=1}^{k}f^{*}_{i}$, with $f^{*}_{i}\in P^{M}_{\ell-2}$ and
$\\{\mbox{supp}(f^{*}_{i}):i=1,2,\ldots,k\\}$ is $\mathcal{F}$-allowable. Note
that in the case $\ell>2$, for at least one $i=1,2,\ldots,k$ , we have that
$f^{*}_{i}\not\in P^{M}_{\ell-3}$. By the inductive hypothesis, we can write
$f^{*}_{i}=\Psi(T^{(i)})$ for $i=1,2,\ldots,k$, where
$T^{(i)}\in\mathcal{\cup}_{j=1}^{\ell-1}\mathcal{T}^{M,(j)}$. We can assume
that $n_{1}<n_{2}<\cdots<n_{k}$, where $n_{i}$ is the root of $T^{(i)}$. We
put

$T\ =\
\\{n_{1}\\}\cup\\{(n_{1},n_{i},s)\,:\,i=1,2,\ldots,k,s\in[[\mathbb{N}]]^{<\omega}\cup\\{\emptyset\\}\text{
with }(n_{i},s)\in T^{(i)}\\}$

and note that $T\in\mathcal{T}^{M,(\ell)}$ and $\Psi(T)=f^{*}$.

The moreover part follows from the moreover part in Proposition 4.5. ∎

Finally, we state our key lemma which we shall prove by induction. We assume
that $\mathcal{F}$ has the following property (by Theorem 3.4,
$\mathcal{S}_{\xi}$, $\xi<\omega_{1}$ has this property):

  1. ($*$)

If $A_{1},\ldots,A_{n}$, $n\in\mathbb{N}$, are nonempty sets in $\mathcal{F}$
with $\min A_{1}<\cdots<\min A_{n}$, then there exist nonempty sets
$A_{1}^{\prime}<\cdots<A_{n}^{\prime}$ in $\mathcal{F}$ such that
$\cup_{i=1}^{n}A_{i}=\cup_{i=1}^{n}A_{i}^{\prime}$ and $\min
A_{i}\leqslant\min A_{i}^{\prime}$, for $i=1,\ldots,n$.

We denote the root of a tree $T$ by $r(T)$ and denote
$\operatorname{supp}(r(T))$ by $\operatorname{supp}T$. Recall that

$(\mathcal{F})^{s}\ :=\
\left\\{\cup_{i=1}^{s}E_{i}\,:\,E_{1}<\cdots<E_{s},E_{i}\in\mathcal{S}\right\\},$

which, by Property ($*$), equals to the family

$(\mathcal{F})_{s}\ :=\
\left\\{\cup_{i=1}^{s}E_{i}\,:\,E_{1},\ldots,E_{s}\mbox{ are pairwise
disjoint},E_{i}\in\mathcal{S}\right\\}.$

######  Lemma 4.7.

(The key lemma) Let $T_{1},T_{2},\ldots,T_{\ell}$ be $\mathcal{F}$-allowable
trees, having pairwise disjoint supports, with
$\\{r(T_{i}):i=1,2,\ldots,\ell\\}\in(\mathcal{F})^{s}$ for some given
$s\in\mathbb{N}$. Then there exist $\mathcal{F}[\mathcal{A}_{2}]$-admissible
trees $T_{1}^{\prime},T_{2}^{\prime},\ldots,T^{\prime}_{\ell^{\prime}}$, such
that

  1. (1)

$\operatorname{supp}T_{1}^{\prime}<\operatorname{supp}T_{2}^{\prime}<\cdots<\operatorname{supp}T^{\prime}_{\ell^{\prime}}$.

  2. (2)

$\sum_{i=1}^{\ell}\Psi(T_{i})=\sum_{i=1}^{\ell^{\prime}}\Psi(T^{\prime}_{i})$.

  3. (3)

$\ell^{\prime}\leqslant\ell+k$, where $k=\\#\\{T_{i}:|T_{i}|=1\\}$.

  4. (4)

$\\{r(T^{\prime}_{i}):i=1,2,\ldots,\ell^{\prime}\\}\in(\mathcal{F}[\mathcal{A}_{2}])^{s}$.
More precisely, if $\\{r(T_{i}):i=1,\ldots,\ell\\}$ is decomposed as the union
of $\mathcal{F}$ sets $B_{1}<B_{2}<\cdots<B_{s}$, the set
$\\{r(T^{\prime}_{i}):i=1,\ldots,\ell^{\prime}\\}$ can be written as the union
of consecutive $\mathcal{F}[\mathcal{A}_{2}]$ sets
$B^{\prime}_{1}<B^{\prime}_{2}<\cdots<B^{\prime}_{s}$ such that $\min
B_{i}\leqslant\min B_{i}^{\prime}$ for $i=1,\ldots,s$.

To see that Lemma 4.7 implies Theorem 4.2, we simply let $\ell=1$ and $T_{1}$
be an $\mathcal{F}$-allowable tree. If $|T_{1}|=1$, then there is nothing to
prove. If $|T_{1}|\geqslant 2$, the lemma gives exactly one
$\mathcal{F}[\mathcal{A}_{2}]$-admissible tree $T_{1}^{\prime}$ such that
$\Psi(T_{1}^{\prime})=\Psi(T_{1})$, and we are done. To prove Lemma 4.7, we
need the following easy observation.

######  Proposition 4.8.

Let $k,m\in\mathbb{N}$, $a_{1}<a_{2}<\cdots<a_{k}$ in $\mathbb{N}$,
$B_{1}<B_{2}<\cdots<B_{m}$ in $[\mathbb{N}]^{<\omega}$, have the following
properties:
$\\{a_{1},a_{2},\ldots,a_{k}\\}\cap\bigcup_{j=1}^{m}B_{j}=\emptyset$ and
$\\{a_{1},a_{2},\ldots,a_{k}\\}\cup\left\\{\min
B_{j}:j=1,2,\ldots,m\right\\}\in(\mathcal{F})^{s}$, for some $s\in\mathbb{N}$.
Then the family

$\mathcal{D}\ =\ \big{(}\\{(a_{i},a_{i+1})\cap B_{j}\,:\,0\leqslant i\leqslant
k,1\leqslant j\leqslant
m\\}\setminus\\{\emptyset\\}\big{)}\cup\\{\\{a_{i}\\}\,:\,1\leqslant
i\leqslant k\\},$

where $a_{0}=0$, and $a_{k+1}=\infty$ can be ordered into
$(D_{j})_{j=1}^{\ell}$ with $D_{1}<D_{2}<\cdots<D_{\ell}$, $\ell\leqslant
m+2k$, and

$\left\\{\min D_{j}:j=1,2,\ldots,\ell\right\\}\ \in\
(\mathcal{F}[{\mathcal{A}}_{2}])^{s}.$

###### Proof.

Ordering: Let $D=(a_{i},a_{i+1})\cap B_{j}\not=\emptyset$ and
$D^{\prime}=(a_{i^{\prime}},a_{i^{\prime}+1})\cap
B_{j^{\prime}}\not=\emptyset$. Then if $j<j^{\prime}$, it follows that
$D<D^{\prime}$, and if $j=j^{\prime}$, and $i<i^{\prime}$, then also
$D<D^{\prime}$. Thus we can order $\mathcal{D}$ into
$D_{1}<D_{2}<\ldots<D_{l}$.

Size of $\mathcal{D}:$ By construction, it follows for each $D\in\mathcal{D}$,
that either $D=\\{a_{i}\\}$, for some $i=1,2\ldots k$, or $\min D=\min B_{j}$,
for some $j=1,2,\ldots,m$, or there is an $i=1,\ldots k$, and a $j=1,2,\ldots
m$, for which $\min B_{j}<a_{i}<\min D\leqslant\max B_{j}$. Since there can at
most be $k$ elements in $\mathcal{D}$ of the third kind, it follows that
$\ell\leqslant 2k+m$.

From $\mathcal{F}$ to $\mathcal{F}[\mathcal{A}_{2}]$: Note that the elements
of $\\{\min D:D\in\mathcal{D}\\}$, which are of the third kind, form a subset
of a spread of $\\{a_{i}:i=1,2,\ldots,k\\}$, which implies that $\\{\min
D:D\in\mathcal{D}\\}\in(\mathcal{F}[\mathcal{A}_{2}])^{s}$. ∎

Let us give a short overview of the proof: We will prove the lemma by
induction on

$\max\text{length }\mathcal{C}\ :=\ \max\\{|T|\,:\,T\in\mathcal{C}\\}.$

In the induction step, we decompose each $T_{i}$, which are not singletons,
into trees of smaller lengths, using Proposition 4.5. We apply the induction
hypothesis to this, possibly much larger, collection of trees of smaller
lengths. Then we “glue” the obtained new collection of
$\mathcal{F}[\mathcal{A}]_{2})$-admissible trees appropriately together. The
possible existence of singletons among the family $(T_{i})_{i=1}^{\ell}$ may
make it necessary to increase the number of trees so that the roots of
$(T_{i}^{\prime})_{i=1}^{\ell^{\prime}}$ may only be in
$(\mathcal{F}([\mathcal{A}]_{2}))^{s}$ instead of $(\mathcal{F})^{s}$.

###### Proof of Lemma 4.7.

Let $\mathcal{C}$ be a finite collection of trees. We prove our claim by
induction on the maximum length of trees in our collection.

Base case: $\max\text{length }\mathcal{C}=1$. For $\ell,s\in\mathbb{N}$,
suppose that $\mathcal{C}$ contains $\ell$ singleton trees
$T_{1},T_{2},\ldots,T_{\ell}$ and
$(r(T_{i}))_{i=1}^{\ell}\in(\mathcal{F})^{s}$. We simply set
$T_{i}^{\prime}=T_{i}$, for $i=1,\ldots,\ell$.

Inductive hypothesis: suppose that the lemma holds for all finite collections
of trees $\mathcal{C}$ with $\max\text{length }\mathcal{C}\leqslant u$ for
some $u\geqslant 1$. We shall show that it holds for all finite collections of
trees $\mathcal{C}$ with $\max\text{length }\mathcal{C}=u+1$.

Pick $\ell$ and $s\in\mathbb{N}$. Let $\mathcal{C}$ consist of $\ell$ disjoint
$\mathcal{F}$-allowable trees $T_{1},\ldots,T_{\ell}$ with $\max_{1\leqslant
i\leqslant\ell}|T_{i}|=u+1$ and $(r(T_{i}))_{i=1}^{s}\in(\mathcal{F})^{s}$.
Let $A_{1}=\\{n:|T_{n}|=1\\}$ and $A_{2}=\\{n:|T_{n}|>1\\}$, and order $A_{2}$
into $a_{1}<a_{2}<\cdots<a_{m}$. For each $n\in A_{2}$, let $T^{(i)}_{n}$,
$i=1,\ldots,k_{n}$, be the trees defined as in Proposition 4.5 item (2) for
$T_{n}$. For the collection of disjoint $\mathcal{F}$-allowable trees

$\mathcal{C}^{\prime}\ =\ \\{T^{(i)}_{n}\,:\,n\in A_{2},i=1,\ldots,k_{n}\\},$

it follows that $\max\text{length }\mathcal{C}^{\prime}\leqslant u$ and
$\\{r(T):T\in\mathcal{C}^{\prime}\\}\in(\mathcal{F})_{m}$. By Property ($*$),
$\\{r(T):T\in\mathcal{C}^{\prime}\\}\in(\mathcal{F})^{m}$, and we can write

$\\{r(T):T\in\mathcal{C}^{\prime}\\}\ =\ \bigcup_{i\in A_{2}}B_{i},\quad
B_{a_{1}}\ <\ \cdots\ <\ B_{a_{m}},B_{i}\ \in\ \mathcal{F},\min B_{i}\
\geqslant\ r\left(T_{i}\right).$

Applying the inductive hypothesis to $\mathcal{C}^{\prime}$, we obtain
$\mathcal{F}[\mathcal{A}_{2}]$-admissible trees
$\widetilde{T}_{1},\widetilde{T}_{2},\ldots,\widetilde{T}_{v}$ with
$\operatorname{supp}\widetilde{T}_{1}<\operatorname{supp}\widetilde{T}_{2}<\cdots<\operatorname{supp}\widetilde{T}_{v}$,
$(r(\widetilde{T}_{i}))_{i=1}^{v}\in(\mathcal{F}[\mathcal{A}_{2}])^{m}$, and
we can write

$(r(\widetilde{T}_{i}))_{i=1}^{v}\ =\ \bigcup_{i\in A_{2}}B^{\prime}_{i},\quad
B^{\prime}_{a_{1}}\ <\ \cdots\ <\ B^{\prime}_{a_{m}},B^{\prime}_{i}\ \in\
\mathcal{F}[\mathcal{A}_{2}],\min B^{\prime}_{i}\ \geqslant\ \min B_{i}.$

For $n\in A_{2}$, define the following
$\mathcal{F}[\mathcal{A}_{2}]$-admissible trees:

$\widehat{T}_{n}\ =\ \\{(\min B^{\prime}_{n}),(\min
B^{\prime}_{n},t):t\in\widetilde{T}_{i},r(\widetilde{T}_{i})\in
B^{\prime}_{n}\\},$ (4.9)

and for $n\in A_{1}$, put $\widehat{T}_{n}=T_{n}$.

Denote the new collection $\\{\widehat{T}_{n}:n=1,2,\ldots,\ell\\}$ of trees
by $\mathcal{C}^{\prime\prime}$. From above, $\min B^{\prime}_{n}\geqslant
r(T_{n})$ for each $n\in A_{2}$; hence, the roots of the new trees in (4.9)
form a spread of $\\{r(T_{n}):n\in A_{2}\\}$. Therefore,
$\\{r(T):T\in\mathcal{C}^{\prime\prime}\\}$ is a spread of
$\\{r(T):T\in\mathcal{C}\\}$ and thus, is in $(\mathcal{F})^{s}$. Note that
the set of singletons in $\mathcal{C}^{\prime\prime}$ is the same as in
$\mathcal{C}^{\prime}$. By our construction, if
$\widehat{T}_{i},\widehat{T}_{j}$ are not singletons with
$r(\widehat{T}_{i})<r(\widehat{T}_{j})$, then
$\operatorname{supp}\widehat{T}_{i}<\operatorname{supp}\widehat{T}_{j}$.
Furthermore, since
$\sum_{T\in\mathcal{C}^{\prime}}\Psi(T)=\sum_{i=1}^{v}\Psi(\widetilde{T}_{i})$,
we know that

$\sum_{i=1}^{\ell}\Psi(\widehat{T}_{i})\ =\ \sum_{i=1}^{\ell}\Psi(T_{i}).$

We now apply Proposition 4.8 to the sequence
$(r(\widehat{T}_{i}))_{|\widehat{T}_{i}|=1}$ and the collection of consecutive
sets $\\{\operatorname{supp}\widehat{T}_{i}:1\leqslant
i\leqslant\ell,|\widehat{T}_{i}|>1\\}$ to obtain sets $(D_{i})_{i=1}^{p}$
satisfying $D_{1}<D_{2}<\cdots<D_{p}$, $(\min
D_{i})_{i=1}^{p}\in(\mathcal{F}[\mathcal{A}_{2}])^{s}$, and

$p\ \leqslant\
\\#\\{\widehat{T}_{i}\,:\,|\widehat{T}_{i}|>1\\}+2\cdot\\#\\{\widehat{T}_{i}\,:\,|\widehat{T}_{i}|=1\\}\
=\ \ell+|A_{1}|\ =\ \ell+k.$

Now, we define $(T^{\prime}_{i})_{i=1}^{p}$ based on the following cases for
each $D_{i}$, $1\leqslant i\leqslant p$. Order the set
$\\{r(\widehat{T}_{i}):|\widehat{T}_{i}|=1\\}$ into
$\widehat{a}_{1}<\widehat{a}_{2}<\cdots<\widehat{a}_{k}$. For convenience, let
$\widehat{a}_{0}=0$ and $\widehat{a}_{k+1}=\infty$.

Case 1: if $D_{i}=r(\widehat{T}_{j})$ for some $|\widehat{T}_{j}|=1$, we
choose $T^{\prime}_{i}=r(\widehat{T}_{j})$.

Case 2: if
$D_{i}=(\widehat{a}_{j},\widehat{a}_{j+1})\cap\operatorname{supp}\widehat{T}_{q}$
for some $0\leqslant j\leqslant k$ and for some $q$ with
$|\widehat{T}_{q}|>1$, we choose an $\mathcal{F}[\mathcal{A}_{2}]$-admissible
tree $T^{\prime}_{i}$ for which $\Psi(T^{\prime}_{i})$ is the projection of
$\Psi(\widehat{T}_{q})$ onto $(\widehat{a}_{j},\widehat{a}_{j+1})$. This can
be done thanks to Proposition 4.6.

The trees $(T^{\prime}_{i})_{i=1}^{p}$ are
$\mathcal{F}[\mathcal{A}_{2}]$-admissible and obviously satisfy Properties
(1), (2), and (3) of our claim. In order to verify that
$(T^{\prime}_{i})_{i=1}^{p}$ satisfies (4), we write

$\\{r(T_{i}:i=1,\ldots,\ell\\}\ =\ \bigcup_{i=1}^{s}E_{i},\quad E_{1}\ <\
\cdots\ <\ E_{s},E_{i}\ \in\ \mathcal{F}.$

Since $r(\widehat{T}_{i})\geqslant r(T_{i})$, if we write

$\\{r(\widehat{T}_{i}):i=1,\ldots,\ell\\}\ =\
\bigcup_{i=1}^{s}E^{\prime}_{i},\quad E^{\prime}_{1}\ <\ \cdots\ <\
E^{\prime}_{s},|E^{\prime}_{i}|=|E_{i}|,$

then $E^{\prime}_{i}$ is a spread of $E_{i}$, and
$E^{\prime}_{i}\in\mathcal{F}$. As in the proof of Proposition 4.8, the set
$(r(T^{\prime}_{i}))_{i=1}^{p}=(\min D_{i})_{i=1}^{p}$ is formed by adding to
$(r(\widehat{T}_{i}))_{i=1}^{\ell}$ at most one integer between two
consecutive integers in $(r(\widehat{T}_{i}))_{i=1}^{\ell}$. Therefore, if we
write

$\\{r(T^{\prime}_{i}):i=1,\ldots,\ell\\}\ =\
\bigcup_{i=1}^{s}E^{\prime\prime}_{i},\quad E^{\prime\prime}_{1}\ <\ \cdots\
<\ E^{\prime\prime}_{s},$

where

$E^{\prime\prime}_{i}\ =\ [\min E^{\prime}_{i},\min
E^{\prime}_{i+1}-1]\cap\\{r(T^{\prime}_{i})\,:\,1\leqslant i\leqslant
p\\}\mbox{ and }\min E^{\prime}_{s+1}=\infty,$

then $E^{\prime\prime}_{i}$ is in $\mathcal{F}[\mathcal{A}_{2}]$, and $\min
E^{\prime\prime}_{i}=\min E^{\prime}_{i}\geqslant\min E_{i}$. This completes
our proof. ∎

From Lemma 4.7 and Theorem 4.1, we obtain

######  Corollary 4.9.

For any $\xi<\omega_{1}$, $0<\theta<1$, and $x\in c_{00}$, it follows that

$\|x\|_{\mathcal{S}_{\xi},\theta}\ \leqslant\
\|x\|_{\mathcal{S}_{\xi},\theta,M}\ \leqslant\
\|x\|_{\mathcal{S}_{\xi},\theta}.$

The following observation will be important in the next section.

######  Corollary 4.10.

Let $\xi<\omega_{1}$. Any collection of normalized vectors $(y_{j})_{j=1}^{k}$
in $T[\mathcal{S}_{\xi},\theta]$ for which
$\big{(}\operatorname{supp}(y_{j})\big{)}_{j=1}^{k}$ is
$\mathcal{S}_{\xi}$-allowable, is $\frac{\theta}{3}$-equivalent to the
$\ell_{1}^{k}$\- unit vector basis.

###### Proof.

For $(a_{j})_{j=1}^{k}\subset\mathbb{R}$, we have

$\left\|\sum_{j=1}^{k}a_{j}y_{j}\right\|_{\mathcal{S}_{\xi},\theta}\
\geqslant\
\frac{1}{3}\left\|\sum_{j=1}^{k}a_{j}y_{j}\right\|_{\mathcal{S}_{\xi},\theta,M}\
\geqslant\
\frac{1}{3}\sum_{j=1}^{k}\|a_{j}y_{j}\|_{\mathcal{S}_{\xi},\theta,M}\ =\
\frac{1}{3}\sum_{j=1}^{k}|a_{j}|.$

∎

##  5\. Application: the cardinality of the closed ideals of
$\mathcal{L}(T[\mathcal{S}_{\xi},\theta])$

We recall some notation. For a Banach space $X$, $\mathcal{L}(X)$ denotes the
algebra of bounded linear operators on $X$. A closed ideal in $\mathcal{L}(X)$
is a closed subspace $\mathcal{J}$ of $\mathcal{L}(X)$ (in the strong operator
norm) which is closed under multiplication from the right and left by elements
of $\mathcal{L}(X)$. For $\mathcal{T}\subset\mathcal{L}(X)$, let
$\mathcal{J}^{\mathcal{T}}$ denote the closed ideal generated by
$\mathcal{T}$, i.e.,

$\mathcal{J}^{\mathcal{T}}=\overline{\left\\{\sum_{j=1}^{n}A_{j}\circ
T_{j}\circ
B_{j}:n\in\mathbb{N},\,T_{j}\in\mathcal{T},\,A_{j},B_{j}\in\mathcal{L}(X),j=1,2,\ldots,n\right\\}}.$

For $T\in\mathcal{L}(X)$, we write $\mathcal{J}^{T}=\mathcal{J}^{\\{T\\}}$ and
call $\mathcal{J}^{T}$ a singly generated closed ideal.

In [16], it was shown that ${\mathcal{L}}(\ell_{p}\oplus\ell_{q})$,
$1<p<q<\infty$, has $\mathfrak{c}$ many singly generated closed subideals,
which is, for trivial set theoretical reasons, the largest possible. In [9],
this result was extended to the spaces $\ell_{p}\oplus c_{0}$,
$\ell_{p}\oplus\ell_{\infty}$, and $\ell_{1}\oplus\ell_{p}$, $1<p<\infty$. It
might be worth mentioning here that the case $\ell_{1}\oplus c_{0}$ is still
open. It is not even known whether or not $\mathcal{L}(\ell_{1}\oplus c_{0})$
has infinite many closed subideals. Johnson and Schechtman [13] proved that
the space $\mathcal{L}(L_{p})$, $1<p<\infty$, has even $2^{\mathfrak{c}}$
closed subideals, which is also, for set-theoretical reason, the largest
possible. Using an argument from [13], it was also possible to show that
$\mathcal{L}(X)$ has $2^{\mathfrak{c}}$ many closed subideals, where
$X=\ell_{p}\oplus\ell_{q},\ell_{p}\oplus c_{0},\ell_{1}\oplus\ell_{p}$,
$\ell_{p}\oplus\ell_{\infty}$, for $1<p<q<\infty$ [10]. The argument from
[13], which we will also use, is as follows:

######  Proposition 5.1.

Assume that $(T_{N}:N\in[\mathbb{N}]^{\omega})$ is a uniformly bounded family
of operators on a Banach space $X$ which satisfies the following tree
condition for some $c>0$:

$T_{M}\ \in\ \mathcal{J}^{T_{N}}\text{ if $M\subset N$, and
}\mbox{dist}(T_{M},\mathcal{J}^{T_{N}})\ \geqslant\ c\text{ if }M\setminus
N\in[\mathbb{N}]^{\omega}.$ (T)

Let $\mathcal{C}\subset[\mathbb{N}]^{\omega}$ be of cardinality
$\mathfrak{c}$, consisting of pairwise almost disjoint sets (meaning that
$M\cap N$ is finite, for all $M,N\in\mathcal{C}$, $M\not=N$). Then
$(\mathcal{J}^{\\{T_{N}:N\in\mathcal{A}\\}}:\mathcal{A}\subset\mathcal{C})$ is
a family of pairwise distinct closed subideals.

In unpublished work, Johnson [12] showed that the space of operators on
$T_{1}=T[\mathcal{S}_{1},\theta]$ has $2^{\mathfrak{c}}$ closed ideals. We
will here reproduce his arguments, and based on the results shown in the
previous sections, we will deduce that his arguments extend to
$T[\mathcal{S}_{\xi},\theta]$, for all $\xi<\omega_{1}$. We start by recalling
a result by Johnson, which is based on [15, Theorem A].

######  Proposition 5.2.

[11, Proposition 1] For each $m\in\mathbb{N}$, there is a number
$L(m)\in\mathbb{N}$ (the proof shows that one can take $L(m)=2^{2m^{6}}$) so
that for every Banach space $X$ with a $1$ unconditional basis $(e_{j})$, the
following holds:

For each $m$-dimensional subspace $E$ of $X$, there exists an automorphism
$S:X\to X$, with $\|S\|\cdot\|S^{-1}\|\leqslant 2$, and pairwise disjoint
supported vectors $(x_{j})_{j=1}^{L(m)}$ in $X$ so that
$S(E)\subset\text{span}(x_{j}:j=1,2,\ldots,L(m))$.

######  Corollary 5.3.

Let $\xi<\omega_{1}$, $0<\theta<1$, $X=T[\mathcal{S}_{\xi},\theta]$, and
$(e_{j})$ be its unit basis. For any $m\in\mathbb{N}$ and any $m$-dimensional
subspace $E$ of $X_{L(m)}=\overline{\text{span}(e_{j}:j\geqslant L(m))}$
(where $L(m)$ is as in Proposition 5.2), there exists an $L(m)$-dimensional
subspace $F$ of $X_{L(m)}$ generated by a sequence $(x_{j})_{j=1}^{L(m)}$ of
disjointly supported vectors, which is $\frac{\theta}{3}$ equivalent to the
$\ell_{1}^{L(m)}$-vector basis, and there exists an automorphism
$S:X_{L(m)}\to X_{L(m)}$, with $\|S^{-1}\|\cdot\|S\|\leqslant 2$ and
$S(E)\subset F$.

###### Proof.

We apply Proposition 5.2 to the space $X_{L(m)}$. Then our claim follows from
Corollary 4.10 and our previous observation that
$\mathcal{S}_{1}\subset\mathcal{S_{\xi}}$. ∎

We are now ready to state the main result of this section.

######  Theorem 5.4.

[12] Assume that $X$ is a Banach space with a $1$-unconditional basis
$(e_{j})$ which is not equivalent to the unit vector basis of $\ell_{1}$ and
satisfies the following condition for some $c>0$ and some sequence
$(L(m))_{m=1}^{\infty}\subset\mathbb{N}$:

For any $m\in\mathbb{N}$ and any $m$-dimensional subspace $E$ of
$X_{L(m)}=\overline{\text{span}(e_{j}:j\geqslant L(m))}$, there exists a
finite-dimensional subspace $F$ of $X_{L(m)}$ generated by a basis
$(f_{j})_{j=1}^{L(m)}$, which is $c$-equivalent to the $\ell_{1}^{L(m)}$-unit
basis, and there exists an automorphism $S:X_{L(m)}\to X_{L(m)}$, with
$\|S^{-1}\|\cdot\|S\|\leqslant 2$ and $S(E)\subset F$.

Then there exists a family $(T_{N}:N\in[\mathbb{N}]^{\omega})$ of spatial
projections ($T_{N}$ is a projection on some subspace of $X$ generated by a
subsequence of $(e_{j})$), which satisfies (T).

###### Proof.

Using on the one hand that $(e_{j})$ is not equivalent to the unit vector
basis of $\ell_{1}$, on the other hand the sequence $(L(m))$, we choose
recursively natural numbers $m_{1}<n_{1}<m_{2}<n_{2}<\cdots$ as follows:

We let $m_{1}=1$, and assuming that $m_{k}\in\mathbb{N}$ has been chosen, we
choose $n_{k}>m_{k}$ so that there exist numbers
$(p_{j}^{(k)})_{j=m_{k}}^{n_{k}}\subset[0,1]$ so that
$\sum_{j=m_{k}}^{n_{k}}p_{j}^{(k)}=1$ and

$\left\|\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}e_{j}\right\|\ <\ \frac{1}{km_{k}},$
(5.1)

and assuming $n_{k}\in\mathbb{N}$ has been chosen, we choose
$m_{k+1}=L(n_{k})$.

For $N\in[\mathbb{N}]^{\omega}$, we define $T_{N}$ to be the projection onto
the subspace of $X$ generated by $(e_{i}:i\in\bigcup_{j\in N}[m_{j},n_{j}])$
and claim that the so defined family $(T_{N}:N\in[\mathbb{N}]^{\omega})$
satisfies (T). The first condition in (T) is trivially satisfied. To verify
the second condition, we define for $k\in\mathbb{N}$ the following functional
$\Psi_{k}\in\mathcal{L}^{*}(X)$:

$\Psi_{k}(S)\ =\ \sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}e^{*}_{j}(S(e_{j})),\text{
for }S\in\mathcal{L}(X).$

Clearly, $\|\Psi_{k}\|_{\mathcal{L}^{*}(X)}=1$, and if
$N\in[\mathbb{N}]^{\omega}$ and $k\in N$, then $\Psi_{k}(T_{N})=1$. It is
therefore enough to show that for $A,B\in\mathcal{L}(X)$,
$\|A\|,\|B\|\leqslant 1$, and $M,N\in[\mathbb{N}]^{\omega}$, with $M\setminus
N\in[\mathbb{N}]^{\omega}$, we have

$\lim_{k\to\infty,k\in M\setminus N}\Psi_{k}(A\circ T_{N}\circ B)\ =\ 0$ (5.2)

We compute for $k\not\in N$ that

$\displaystyle\big{|}\Psi_{k}(A\circ T_{N}\circ B)\big{|}$ $\displaystyle\
\leqslant\ \sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}|e^{*}_{j}(AT_{N}B)(e_{j})|$
$\displaystyle\ \leqslant\ \sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}\|T_{N}B(e_{j})\|$
$\displaystyle\ \leqslant\
\underbrace{\sum_{s=1}^{n_{k-1}}\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}|B_{s}(e_{j})|}_{=\
\Sigma_{1}}+\underbrace{\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}\|P\circ
B(e_{j})\|}_{=\ \Sigma_{2}},$

where $B_{s}=e^{*}_{s}\circ B\in X^{*}$, with $\|B_{s}\|_{X^{*}}\leqslant 1$,
for $s\in\mathbb{N}$, and $P:X\to X_{m_{k+1}}$ is the basis projection onto
$X_{m_{k+1}}=\overline{\mbox{span}(e_{j}:j\geqslant m_{k+1})}$.

To estimate $\Sigma_{1}$, we observe that

$\displaystyle\Sigma_{1}\ =\
\sum_{s=1}^{n_{k-1}}B_{s}\left(\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}\mbox{sign}(B_{s}(e_{i}))e_{j}\right)\
\leqslant\
\sum_{s=1}^{n_{k-1}}\left\|\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}\mbox{sign}(B_{s}(e_{i}))e_{j}\right\|\
\leqslant\ \frac{1}{k}.$ (5.3)

To estimate $\Sigma_{2}$, we consider the space $E=P\circ
B(\mbox{span}(e_{j}:m_{k}\leqslant j\leqslant n_{k}))$ which is a subspace of
$X_{m_{k+1}}=X_{L(n_{k})}$ and has dimension not larger than $n_{k}$.
Therefore, we can find a finite-dimensional subspace $F$ of $X_{L(n_{k})}$,
with a basis $(f_{j})_{j=1}^{\mbox{dim}F}$ which is $c$-equivalent to the
$\ell_{1}^{\mbox{dim}F}$-unit basis, and an automorphism $S$ on $X_{m_{k+1}}$,
for which $S(E)\subset F$, and $\|S\|\cdot\|S^{-1}\|\leqslant 2$. We have

$\displaystyle\Sigma_{2}$ $\displaystyle\ \leqslant\
2\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}\|SPB(e_{j})\|$ $\displaystyle\ \leqslant\
2\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}\sum_{s=1}^{\dim(F)}|\widetilde{B}_{s}(e_{j})|$
where $\widetilde{B}_{s}(x)=f^{*}_{s}(SPB(x))$, for
$s=1,2,\ldots,\mbox{dim}F$, $x\in X$, $(f^{*}_{j})_{j=1}^{\mbox{dim}F}$ being
the coordinate functionals in $F$, $\displaystyle\ =\
2\sum_{s=1}^{\dim(F)}\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}|\widetilde{B}_{s}(e_{j})|$
$\displaystyle\ =\
2\sum_{s=1}^{\dim(F)}\widetilde{B}_{s}\left(\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}\mbox{sign}(\widetilde{B}_{s}(e_{j}))e_{j}\right)$
$\displaystyle\ \leqslant\
2c\left\|\sum_{s=1}^{\dim(F)}\widetilde{B}_{s}\left(\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}\mbox{sign}(\widetilde{B}_{s}(e_{j}))e_{j}\right)f_{s}\right\|_{F}$
$\displaystyle\ =\
2c\left\|SPB\left(\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}\mbox{sign}(\widetilde{B}_{s}(e_{j}))e_{j}\right)\right\|_{F}$
$\displaystyle\ \leqslant\
4c\left\|\sum_{j=m_{k}}^{n_{k}}p^{(k)}_{j}e_{j}\right\|\ \leqslant\
\frac{4c}{k}.$

This, together with (5.3), implies our claim (5.2). ∎

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