# Banach spaces of $\mathcal{I}$-convergent sequences

Michael A. Rincón-Villamizar, Carlos Uzcátegui Aylwin

###### Abstract.

We study the space $c_{0,\mathcal{I}}$ of all bounded sequences $(x_{n})$ that
$\mathcal{I}$-converge to $0$, endowed with the sup norm, where $\mathcal{I}$
is an ideal of subsets of $\mathbb{N}$. We show that two such spaces,
$c_{0,\mathcal{I}}$ and $c_{0,\mathcal{J}}$, are isometric exactly when the
ideals $\mathcal{I}$ and $\mathcal{J}$ are isomorphic. Additionally, we
analyze the connection of the well-known Katětov pre-order $\leq_{K}$ on
ideals with some properties of the space $c_{0,\mathcal{I}}$. For instance, we
show that $\mathcal{I}\leq_{K}\mathcal{J}$ exactly when there is a (not
necessarily onto) Banach lattice isometry from $c_{0,\mathcal{I}}$ to
$c_{0,\mathcal{J}}$, satisfying some additional conditions.

We present some lattice-theoretic properties of $c_{0,\mathcal{I}}$,
particularly demonstrating that every closed ideal of $\ell_{\infty}$ is equal
to $c_{0,\mathcal{I}}$ for some ideal $\mathcal{I}$ on $\mathbb{N}$. We also
show that certain classical Banach spaces are isometric to $c_{0,\mathcal{I}}$
for some ideal $\mathcal{I}$, such as the spaces $\ell_{\infty}(c_{0})$ and
$c_{0}(\ell_{\infty})$. Finally, we provide several examples of ideals for
which $c_{0,\mathcal{I}}$ is not a Grothendieck space.

##  1\. Introduction

An ideal on $\mathbb{N}$ is a collection $\mathcal{I}$ of subsets of
$\mathbb{N}$ closed under finite unions and taking subsets of its elements. A
sequence $(x_{n})$ in $\mathbb{R}$ is said to be $\mathcal{I}$-convergent to
$x\in\mathbb{R}$, denoted as $\mathcal{I}$-$\lim x_{n}=x$, if for each
$\varepsilon>0$, the set
$\\{n\in\mathbb{N}\,\colon\,|x_{n}-x|\geq\varepsilon\\}$ belongs to
$\mathcal{I}$. When $\mathcal{I}$ is ${\sf Fin}$, the ideal of finite subsets
of $\mathbb{N}$, we have the classical convergence in $\mathbb{R}$. The
$\mathcal{I}$-convergence was introduced in [17], although many authors had
already studied this concept in particular cases and in different contexts
(see, for instance, [2, 8, 9, 18]). The main goal of this paper is to study
the following space:

$\displaystyle
c_{0,\mathcal{I}}=\\{(x_{n})\in\ell_{\infty}\,\colon\,\mathcal{I}-\lim
x_{n}=0\\}.$

This space has recently received some attention (see, for instance, [16, 19]).
It is known that $c_{0,\mathcal{I}}$ is a closed subspace (also, an ideal) of
$\ell_{\infty}$ and it is isometric to $C_{0}(U_{\mathcal{I}})$ for some open
set $U_{\mathcal{I}}$ of $\beta\mathbb{N}$ (see [16] or Proposition 3.1). Two
extreme examples are worth keeping in mind. On the one hand, as we mentioned
before, $c_{0,{\sf Fin}}$ is exactly $c_{0}$. On the other hand, if
$\mathcal{I}$ is the trivial ideal $\mathcal{P}(\mathbb{N})$, we obtain the
whole space $\ell_{\infty}$.

A natural question is to determine when two such spaces are isomorphic
(isometric). As a consequence of the Banach-Stone theorem, $c_{0,\mathcal{I}}$
and $c_{0,\mathcal{J}}$ are isometric exactly when $\mathcal{I}$ and
$\mathcal{J}$ are isomorphic as ideals. However, the same does not hold for
isomorphism. Indeed, it is known that if $\mathcal{I}$ is a maximal ideal,
then $c_{0,\mathcal{I}}$ is complemented in $\ell_{\infty}$ and thus
isomorphic to $\ell_{\infty}$ (see [19]); but $\ell_{\infty}$ is equal to
$c_{0,\mathcal{P}(\mathbb{N})}$ and a maximal ideal is not isomorphic to
$\mathcal{P}(\mathbb{N})$.

Ideals on countable sets have been studied for a long time, and several ways
of comparing them have been investigated (see the surveys [14, 24]). We are
particularly interested in the Katětov pre-order (see, for instance, [15]).
Given two ideals $\mathcal{I}$ and $\mathcal{J}$ on two countable sets $X$ and
$Y$, respectively, we say that $\mathcal{I}$ is Katětov below $\mathcal{J}$,
denoted as $\mathcal{I}\leq_{K}\mathcal{J}$, if there is $f:Y\to X$ such that
$f^{-1}(A)\in\mathcal{J}$ for all $A\in\mathcal{I}$. Following the ideas
behind a proof of Holsztynki’s theorem [22], we show that
$\mathcal{I}\leq_{K}\mathcal{J}$ exactly when there is a (non-necessarily
onto) Banach lattice isometry from $c_{0,\mathcal{I}}$ to $c_{0,\mathcal{J}}$
satisfying some additional conditions (see Theorem 3.7).

We study some lattice-theoretic properties of the spaces $c_{0,\mathcal{I}}$
and, in particular, we show that any closed ideal of $\ell_{\infty}$ is equal
to $c_{0,\mathcal{I}}$ for some ideal $\mathcal{I}$ on $\mathbb{N}$. We found
an interesting connection between the ideal theoretic notion of orthogonal
ideals [12, 23] and the notion of $c_{0}$-disjoint subspaces of
$\ell_{\infty}$ (see section 4).

We show how to represent some classical Banach spaces as a space of the form
$c_{0,\mathcal{I}}$ for some ideal $\mathcal{I}$ on $\mathbb{N}$. For example,
we found ideals providing an isometric representation of the spaces
$\ell_{\infty}(c_{0})$ and $c_{0}(\ell_{\infty})$. These two spaces have been
recently shown to be non-isomorphic (see [5]). We present a different argument
to show that they are not isometric. We also found an uncountable collection
of pairwise non-isometric spaces of the form $c_{0,\mathcal{I}}$. In [11,
Problem 9], it was asked about ideals $\mathcal{I}$ such that
$c_{0,\mathcal{I}}$ is Grothendieck; we present several non-Grothendieck
spaces.

##  2\. Preliminaries

###  2.1. Banach spaces and Banach lattices.

We will use standard terminology and notation for Banach lattices and Banach
space theory. For unexplained definitions and notations, we refer to [1, 21].
All Banach lattices analyzed here are assumed to be real. $B_{X}$ stands for
the closed unit ball of $X$. The positive cone of a Banach lattice $X$ is
denoted by $X^{+}$. A sublattice $Y$ of a Banach lattice $X$ is an ideal if
$x\in Y$ whenever $|x|\leq|y|$ for some $y\in Y$. The closed ideal generated
by a subset $A$ of $X$ is denoted by $\langle A\rangle$. If $X$ and $Y$ are
isomorphic Banach spaces, we write $X\sim Y$.

If $X$ and $Y$ are Banach lattices, a linear operator $T\colon X\to Y$ is
called a Banach lattice isomorphism if $T$ is a Banach isomorphism such that
$T(x\wedge y)=Tx\wedge Ty$ for all $x,y\in X$. Furthermore, if $T$ is an
isometry, we say that $T$ is a Banach lattice isometry. Finally, when $T$ is
not necessarily onto we will called it an into Banach lattice isomorphism
(isometry, respectively).

Now, we introduce some notation. Let $(X_{j})_{j\in\mathbb{N}}$ be a family of
Banach spaces.

  1. (1)

$\ell_{\infty}((X_{j})_{j\in\mathbb{N}})$ denotes the $\ell_{\infty}$-sum of
$(X_{j})_{j\in\mathbb{N}}$, that is, the Banach space of all bounded sequences
$\displaystyle(x_{j})\in\prod_{j}X_{j}$ endowed with the norm $\|\cdot\|$
given by $\displaystyle\|(x_{j})\|=\sup_{j}\|x_{j}\|$.

  2. (2)

$c_{0}((X_{j})_{j\in\mathbb{N}})$ is the $c_{0}$-sum of
$(X_{j})_{j\in\mathbb{N}}$, that is, the closed subspace of
$\ell_{\infty}((X_{j})_{j\in\mathbb{N}})$ consisting of all null sequences,
i.e, $\displaystyle\lim_{j}\|x_{j}\|=0$.

When $X=X_{j}$ for all $j\in\mathbb{N}$, these spaces are denoted by
$\ell_{\infty}(X)$ and $c_{0}(X)$, respectively. Finally, for $X=\mathbb{R}$,
$\ell_{\infty}(X)$ and $c_{0}(X)$ correspond to $\ell_{\infty}$ and $c_{0}$,
respectively.

######  Remark 2.1.

It is easy to see that if each $X_{j}$ is isometric (isomorphic) to $X$, then
$\ell_{\infty}((X_{j})_{j\in\mathbb{N}})$ ($c_{0}((X_{j})_{j\in\mathbb{N}})$)
is isometric (isomorphic, respectively) to $\ell_{\infty}(X)$ ($c_{0}(X)$,
respectively).

###  2.2. Ideals.

Recall that an ideal $\mathcal{I}$ on a set $X$ is a collection of subsets of
$X$ satisfying:

  1. (1)

$\emptyset\in\mathcal{I}$;

  2. (2)

If $A\subset B$ and $B\in\mathcal{I}$, then $A\in\mathcal{I}$;

  3. (3)

If $A,B\in\mathcal{I}$, then $A\cup B\in\mathcal{I}$.

We always assume that every finite subset of $X$ belongs to $\mathcal{I}$. The
dual filter of an ideal $\mathcal{I}$ is denoted by $\mathcal{I}^{*}$ and
consists of all sets of the form $X\setminus A$ for some $A\in\mathcal{I}$. If
$\mathcal{A}$ is a family of subsets of $X$, $\mathcal{I}(\mathcal{A})$
denotes the ideal generated by $\mathcal{A}$ which consists of all subsets of
finite unions of sets from $\mathcal{A}$. If $A\subset X$ and $\mathcal{I}$ is
an ideal on $X$, we denote the restriction of $\mathcal{I}$ to $A$ by
$\mathcal{I}\restriction A=\\{A\cap B\,\colon B\in\mathcal{I}\\}$ which is an
ideal on $A$. If $\mathcal{I}$ and $\mathcal{J}$ are ideals on $X$, the ideal
$\mathcal{I}\sqcup\mathcal{J}$ is defined as
$\mathcal{I}\sqcup\mathcal{J}:=\\{A\cup
B\,\colon\,A\in\mathcal{I},B\in\mathcal{J}\\}.$

Two ideals $\mathcal{I}$ and $\mathcal{J}$ on $\mathbb{N}$ are isomorphic if
there is a bijection (called isomorphism) $f\colon\mathbb{N}\to\mathbb{N}$
such that $A\in\mathcal{J}$ iff $f^{-1}(A)\in\mathcal{I}$. We recall the
Katětov pre-order on ideals. If $\mathcal{I}$ and $\mathcal{J}$ are ideals on
$X$ and $Y$, respectively, we write $\mathcal{I}\leq_{K}\mathcal{J}$ if there
is a function $f\colon Y\to X$ (called a Katětov reduction) such that
$f^{-1}(A)\in\mathcal{J}$ for all $A\in\mathcal{I}$. In most of the cases, the
function $f$ can be assumed to be onto.

######  Proposition 2.2.

[3] Let $\mathcal{I}$ and $\mathcal{J}$ be ideals on $X$ and $Y$,
respectively. Suppose $\mathcal{J}$ contains an infinite set and
$\mathcal{I}\leq_{K}\mathcal{J}$, then there is an onto map $f\colon Y\to X$
such that $f^{-1}(A)\in\mathcal{J}$ for all $A\in\mathcal{I}$.

In general, Katětov reductions are not bijective but this naturally suggests a
variant of $\leq_{K}$ which was studied in [3]. They defined
$\mathcal{I}\sqsubseteq\mathcal{J}$ if there is a bijective Katětov reduction
from $\mathcal{I}$ to $\mathcal{J}$. In other words,
$\mathcal{I}\sqsubseteq\mathcal{J}$ if there is an ideal
$\mathcal{J}^{\prime}\subseteq\mathcal{J}$ such that $\mathcal{I}$ is
isomorphic to $\mathcal{J}^{\prime}$. We will use this variant in the sequel.

The Fubini product is defined as follows. For two ideals $\mathcal{I}$ and
$\mathcal{J}$ on $\mathbb{N}$, $\mathcal{I}\times\mathcal{J}$ is the ideal on
$\mathbb{N}\times\mathbb{N}$ given by:

$\displaystyle
A\in\mathcal{I}\times\mathcal{J}\Longleftrightarrow\\{n\in\mathbb{N}\,\colon\,\\{m\in\mathbb{N}\,\colon\,(n,m)\in
A\\}\not\in\mathcal{J}\\}\in\mathcal{I}.$

Another concept that will be used in the sequel is the following. Given a
collection $\mathcal{A}$ of subsets of $\mathbb{N}$, the orthogonal of
$\mathcal{A}$ (see [12, 23]) is the following family of sets

${\mathcal{A}}^{\perp}=\\{B\subseteq\mathbb{N}:(\forall
A\in{\mathcal{A}})(A\cap B\textrm{ is finite})\\}.$

An ideal $\mathcal{I}$ is called Fréchet if
$\mathcal{I}=\mathcal{I}^{\perp\perp}$. We include some examples in section
5.1.

Let $\\{K_{n}:\;n\in F\\}$ be a partition of a countable set $X$, where
$F\subseteq\mathbb{N}$. For $n\in F$, let $\mathcal{I}_{n}$ be an ideal on
$K_{n}$. The direct sum, denoted by $\bigoplus\limits_{n\in
F}\mathcal{I}_{n}$, is defined by

$A\in\bigoplus_{n\in F}\mathcal{I}_{n}\Leftrightarrow(\forall n\in F)(A\cap
K_{n}\in\mathcal{I}_{n}).$

In general, given a sequence of ideals $\mathcal{I}_{n}$ over a countable set
$X_{n}$, we define $\bigoplus_{n}\mathcal{I}_{n}$ by taking a partition
$\\{K_{n}:\;n\in\mathbb{N}\\}$ of $\mathbb{N}$ and an isomorphic copy
$\mathcal{I}_{n}^{\prime}$ of $\mathcal{I}_{n}$ on $K_{n}$ and let
$\bigoplus_{n}\mathcal{I}_{n}$ be $\bigoplus_{n}\mathcal{I}_{n}^{\prime}$. It
should be clear that $\bigoplus_{n}\mathcal{I}_{n}$ is, up to isomorphism,
independent of the partition and the copies used. If all $\mathcal{I}_{n}$ are
equal (isomorphic) to $\mathcal{I}$ we will write $\mathcal{I}^{\omega}$
instead of $\bigoplus_{n}\mathcal{I}_{n}$.

###  2.3. The space $c_{0,\mathcal{I}}(X)$

We begin by introducing the notion of $\mathcal{I}$-convergence for an ideal
$\mathcal{I}$ on $\mathbb{N}$. This concept is due to Kostyrko, Šalát and
Wilczyński [17].

######  Definition 2.3.

Let $\mathcal{I}$ be an ideal on $\mathbb{N}$ and $X$ be a Banach space. A
sequence $(x_{n})$ in $X$ is $\mathcal{I}$-convergent to $x\in X$, and we
write $\mathcal{I}-\lim x_{n}=x$, if for each $\varepsilon>0$,
$\\{n\in\mathbb{N}\,\colon\,\|x_{n}-x\|\geq\varepsilon\\}\in\mathcal{I}$.

The main objetive of this paper is to study the following subspace of
$\ell_{\infty}$.

$\displaystyle
c_{0,\mathcal{I}}(X)=\\{(x_{n})\in\ell_{\infty}(X)\,\colon\,\mathcal{I}-\lim
x_{n}={\bf 0}\\}.$

The upcomming result gives basic properties of $\mathcal{I}$-convergence (see
also [18, Theorem 2.3]). If ${\bf x}\in\ell_{\infty}(X)$ and $\varepsilon>0$,
we will use through out the whole paper the following notation:

$\displaystyle A(\varepsilon,{\bf
x})=\\{n\in\mathbb{N}\,\colon\,\|x_{n}\|\geq\varepsilon\\}.$

######  Proposition 2.4.

Let $X$ be a Banach space and $\mathcal{I}$ be an ideal on $\mathbb{N}$.

  1. (1)

If $c\in\mathbb{R}$ and $\mathcal{I}-\lim x_{n}=x$, then $\mathcal{I}-\lim
cx_{n}=cx$.

  2. (2)

If $\mathcal{I}-\lim x_{n}=x$ and $\mathcal{I}-\lim y_{n}=y$, then
$\mathcal{I}-\lim x_{n}+y_{n}=x+y$.

  3. (3)

If $\mathcal{I}-\lim x_{n}={\bf 0}$ and $\|y_{n}\|\leq\|x_{n}\|$ for all
$n\in\mathbb{N}$, then $\mathcal{I}-\lim y_{n}={\bf 0}$.

  4. (4)

$c_{0,\mathcal{I}}(X)$ is closed subspace of $\ell_{\infty}(X)$.

  5. (5)

If $X$ is a Banach lattice, then $c_{0,\mathcal{I}}(X)$ is a closed ideal of
$\ell_{\infty}(X)$.

###### Proof.

(1), (2) and (3) follow immediately from definition. (4): If ${\bf
y}\in\overline{c_{0,\mathcal{I}}(X)}^{\|\cdot\|_{\infty}}$ and $\varepsilon>0$
is given, there is ${\bf x}\in c_{0,\mathcal{I}}(X)$ such that $\|{\bf
y-x}\|<\varepsilon/2$. Since $A(\varepsilon,{\bf y})\subset
A(\varepsilon/2,{\bf x})$, we conclude that ${\bf y}\in c_{0,\mathcal{I}}(X)$.
Finally for (5), if ${\bf y}\in c_{0,\mathcal{I}}(X)$ and $|{\bf x}|\leq|{\bf
y}|$, then $\|x_{n}\|\leq\|y_{n}\|$ for each $n\in\mathbb{N}$. By (3), ${\bf
x}\in c_{0,\mathcal{I}}(X)$. ∎

When $X=\mathbb{R}$ we write $c_{0,\mathcal{I}}$ instead of
$c_{0,\mathcal{I}}(X)$. The next result provides another description of
$c_{0,\mathcal{I}}$. Recall that if ${\bf y}=(y_{n})$ is a sequence,
$\mathrm{supp}({\bf y})$ denotes the support of ${\bf y}$, that is, the set
$\\{n\in\mathbb{N}\,\colon\,y_{n}\neq 0\\}$.

######  Proposition 2.5.

Let $X$ be a Banach space and $\mathcal{I}$ a proper ideal on $\mathbb{N}$.

  1. (1)

$c_{0,\mathcal{I}}(X)=\overline{\\{{\bf
y}\in\ell_{\infty}(X)\,\colon\,\mathrm{supp}({\bf
y})\in\mathcal{I}\\}}^{\|\cdot\|_{\infty}}$.

  2. (2)

$c_{0,\mathcal{I}}=\overline{\mathrm{span}\\{\chi_{A}\,\colon\,A\in\mathcal{I}\\}}^{\|\cdot\|_{\infty}}$.

  3. (3)

$A\in\mathcal{I}$ iff $\chi_{A}\in c_{0,\mathcal{I}}$.

###### Proof.

(1) We first show that if ${\bf y}=(y_{n})\in\ell_{\infty}(X)$ and
$\mathrm{supp}({\bf y})\in\mathcal{I}$, then ${\bf y}\in
c_{0,\mathcal{I}}(X)$. In fact, let $\varepsilon>0$ be given. Clearly
$A(\varepsilon,{\bf y})\subset\mathrm{supp}({\bf y})$, and therefore
$A(\varepsilon,{\bf}y)\in\mathcal{I}$. Since $c_{0,\mathcal{I}}(X)$ is closed
in $\ell_{\infty}(X)$, then $\supseteq$ holds in the equation above.
Conversely, let ${\bf x}=(x_{n})\in c_{0,\mathcal{I}}(X)$. Fix $\varepsilon>0$
and let $A=A(\varepsilon,{\bf x})\in\mathcal{I}$. Pick a finite sequence of
real numbers
$\varepsilon=\lambda_{1}<\ldots<\lambda_{k}=\|(x_{n})\|_{\infty}+1$ such that
$\lambda_{i+1}-\lambda_{i}\leq\varepsilon$. Let $A_{i}=\\{n\in
A\,\colon\,\lambda_{i}\leq\|x_{n}\|<\lambda_{i+1}\\}$ and ${\bf
y}^{i}=(y_{n}^{i})$, where

$\displaystyle y_{n}^{i}=\begin{cases}\lambda_{i}x_{n}/\|x_{n}\|,&n\in
A_{i};\\\ 0,&\mbox{otherwise,}\end{cases}$

for $1\leq i<k$. Notice that $A_{i}=\mathrm{supp}({\bf y}^{i})\in\mathcal{I}$
for each $1\leq i<k$ and $\|{\bf x}-({\bf y}^{1}+\ldots+{\bf
y}^{k-1})\|_{\infty}\leq\varepsilon$. This shows $\subseteq$.

(2) Suppose that $X=\mathbb{R}$. Now we let

$\displaystyle A_{i}^{+}=\\{n\in A_{i}\,\colon\,x_{n}>0\\},\quad
A_{i}^{-}=\\{n\in A_{i}\,\colon\,x_{n}<0\\},$

${\bf y}_{1}^{i}=\lambda_{i}\chi_{A_{i}^{+}}$ and ${\bf
y}_{2}^{i}=-\lambda_{i}\chi_{A_{i}^{-}}$ for $1\leq i<k$. We have $\|{{\bf
x}-({\bf y}_{1}^{1}+\cdots+{\bf y}_{1}^{k-1}+{\bf y}_{2}^{1}+\cdots+{\bf
y}_{2}^{k-1})}\|\leq\varepsilon$.

(3) The only if part is obvious. Suppose $\chi_{A}\in c_{0,\mathcal{I}}$, then
$A=A(1/2,{\bf x})\in\mathcal{I}$ and we are done. ∎

We recall that an ideal $\mathcal{I}$ over $\mathbb{N}$ can be identified (via
characteristic functions) with a subset of the Cantor space
$\\{0,1\\}^{\mathbb{N}}$, so that we can talk about Borel, meager ideals, etc.
It is worth keeping in mind the following result when dealing with meager
ideals.

######  Theorem 2.6.

[19] Let $\mathcal{I}$ be an ideal on $\mathbb{N}$. If $\mathcal{I}$ is
meager, then $c_{0,\mathcal{I}}$ is not complemented in $\ell_{\infty}$, and,
thus, not isomorphic to $\ell_{\infty}$.

##  3\. Isomorphisms between $c_{0,\mathcal{I}}$ spaces and the Katetov order
on ideals

The classical Banach-Stone theorem states that given two locally compact
Hausdorff spaces $K$ and $S$, $C_{0}(K)$ and $C_{0}(S)$ are isometric iff $K$
and $S$ are homeomorphic [1, Theorem 4.1.5]. Holsztyński [13] extends this
result by showing that if $C_{0}(K)$ embeds isometrically in $C_{0}(S)$, then
$K$ is a continuous image of a subspace of $S$. Our goal in this section is to
prove versions of these results in the setting of $c_{0,\mathcal{I}}$ spaces.
We first look at isometry and later to Banach lattice isometries. We first
recall some classical notions.

Recall that $\beta\mathbb{N}$ is the Stone-Cech compactification of
$\mathbb{N}$ which is usually identified with the collection of all
ultrafilters on $\mathbb{N}$. For a set $A\subset\mathbb{N}$, we let
$A^{*}=\\{p\in\beta\mathbb{N}\,\colon\,A\in p\\}$. The family
$\\{A^{*}\,\colon\,A\subset\mathbb{N}\\}$ defines a basis for the topology of
$\beta\mathbb{N}$. As usual, we identify each $n\in\mathbb{N}$ with the
principal ultrafilter $\\{A\subseteq\mathbb{N}:n\in A\\}$. Every principal
ultrafilter is an isolated point of $\beta\mathbb{N}$. We set
$U_{\mathcal{I}}=\\{p\in\beta\mathbb{N}\,\colon\,\mathcal{I}\cap
p\neq\emptyset\\}$. Since $\displaystyle
U_{\mathcal{I}}=\bigcup\\{A^{*}\,\colon\,A\in\mathcal{I}\\}$,
$U_{\mathcal{I}}$ is open in $\beta\mathbb{N}$. Under the previous
identification, $\mathbb{N}\subseteq U_{\mathcal{I}}$ for every ideal
$\mathcal{I}$.

Given a bounded sequence $(x_{n})$ of reals numbers and an ultrafilter $p$ on
$\mathbb{N}$, we denote by $p-\lim x_{n}$ the only $x\in\mathbb{R}$ such that
$\\{n\in\mathbb{N}:|x_{n}-x|<\varepsilon\\}\in p$ for all $\varepsilon>0$.
Notice that if $p$ is the principal ultrafilter $m$, then
$p-\lim_{n}x_{n}=x_{m}$.

It is well known that $c_{0,\mathcal{I}}$ and $C_{0}(U_{\mathcal{I}})$ are
isometric (see for instance [11, p. 255] and [16, p. 12]), nevertheless we
include here a proof written in terms of $p$-limits which will be needed in
the sequel.

######  Proposition 3.1.

Let $\mathcal{I}$ be an ideal on $\mathbb{N}$. The map
$\Phi_{\mathcal{I}}\colon c_{0,\mathcal{I}}\to C_{0}(U_{\mathcal{I}})$ given
by

$\displaystyle\Phi_{\mathcal{I}}({\bf x})(p)=p-\lim x_{n},$

where ${\bf x}=(x_{n})$ and $p\in U_{\mathcal{I}}$, is a Banach lattice
isometry. Moreover, $A\in\mathcal{I}$ iff $\chi_{A^{*}}\in
C_{0}(U_{\mathcal{I}})$ and $\Phi_{\mathcal{I}}(\chi_{A})=\chi_{A^{*}}$ for
each $A\in\mathcal{I}$.

###### Proof.

Let ${\bf x}=(x_{n})\in c_{0,\mathcal{I}}$ be given. We prove first that
$\Phi_{\mathcal{I}}({\bf x})\in C_{0}(U_{\mathcal{I}})$. Fix $p\in
U_{\mathcal{I}}$ and let $r>0$. To show that $\Phi_{\mathcal{I}}({\bf x})$ is
continuous at $p$, suppose that $|\Phi_{\mathcal{I}}({\bf x})(p)-a|=|(p-\lim
x_{n})-a|<r$. Let $\delta>0$ be such that $|(p-\lim x_{n})-a|<r-\delta$ and
$A=\\{n\in\mathbb{N}\,\colon\,|x_{n}-a|<r-\delta\\}$. Note that $p\in A^{*}$
and for $q\in A^{*}$ we have $|(q-\lim x_{n})-a|\leq r-\delta<r$. Hence, $p\in
A^{*}\subset\Phi_{\mathcal{I}}^{-1}({\bf x})((a-r,a+r))$. Thus
$\Phi_{\mathcal{I}}({\bf x})$ is continuous at $p$.

To see that $\Phi_{\mathcal{I}}({\bf x})$ vanishes at infinity, let
$\varepsilon>0$ and consider
$A_{\varepsilon}=\\{n\in\mathbb{N}\,\colon\,|x_{n}|>\varepsilon/2\\}$. If
$p\in U_{\mathcal{I}}$ satisfies $|\Phi_{\mathcal{I}}({\bf x})(p)|=|p-\lim
x_{n}|\geq\varepsilon$, then $A_{\varepsilon}\in p$, that is $p\in
A_{\varepsilon}^{*}$. Hence $\\{p\in
U_{\mathcal{I}}\,\colon\,|\Phi_{\mathcal{I}}({\bf
x})(p)|\geq\varepsilon\\}\subset A_{\varepsilon}^{*}$. Thus $\\{p\in
U_{\mathcal{I}}\,\colon\,|\Phi_{\mathcal{I}}({\bf x})(p)|\geq\varepsilon\\}$
is compact for each $\varepsilon>0$.

Clearly $\Phi_{\mathcal{I}}$ is linear. Now we prove that $\Phi_{\mathcal{I}}$
is an isometry. Let ${\bf x}\in c_{0,\mathcal{I}}$ be given and $a=\|{\bf
x}\|_{\infty}$. Note that $|\Phi_{\mathcal{I}}({\bf x})(p)|=|p-\lim
x_{n}|\leq\|{\bf x}\|_{\infty}$. On the other hand, for $0<\delta<a$, we have
that
$A_{\delta}=\\{n\in\mathbb{N}\,\colon\,|x_{n}|>a-\delta\\}\in\mathcal{I}$.
Take $p\in A_{\delta}^{*}$. Whence, $p\in U_{\mathcal{I}}$ and
$\Phi_{\mathcal{I}}({\bf x})(p)=p-\lim x_{n}>a-\delta/2$. Since $\delta>0$ was
arbitrary, $\|\Phi_{\mathcal{I}}({\bf x})\|\geq a$.

To see that $\Phi_{\mathcal{I}}$ is onto, fix $f\in C_{0}(U_{\mathcal{I}})$.
Consider the sequence ${\bf x}=(x_{n})$ given by $x_{n}=f(n)$ for
$n\in\mathbb{N}$. For every $\varepsilon>0$ we have

$\displaystyle A(\varepsilon,{\bf x})\subset\\{p\in
U_{\mathcal{I}}\,\colon\,|f(p)|\geq\varepsilon\\}\subset
U_{\mathcal{I}}=\bigcup\\{A^{*}\,\colon\,A\in\mathcal{I}\\}.$

By compactness, there are $A_{1},\ldots,A_{n}\in\mathcal{I}$ such that
$A(\varepsilon,{\bf x})\subset\bigcup_{j=1}^{n}A_{j}$. Whence
$A(\varepsilon,{\bf x})\in\mathcal{I}$ and thus ${\bf x}\in
c_{0,\mathcal{I}}$. Since $f$ is continuous, we have

$\displaystyle\Phi_{\mathcal{I}}({\bf x})(p)=p-\lim x_{n}=p-\lim
f(n)=f(p),\quad\mbox{for all $p\in U_{\mathcal{I}}$.}$

Hence, $\Phi_{\mathcal{I}}({\bf x})=f$.

The identity $\Phi_{\mathcal{I}}(\chi_{A})=\chi_{A^{*}}$ for each
$A\in\mathcal{I}$ follows from the fact that
$\chi_{A^{*}}(p)=p-\lim\chi_{A}(n)$ for every $p\in\beta\mathbb{N}$ and all
$A\subset\mathbb{N}$. Finally, it is easy to see that $\Phi_{\mathcal{I}}$ is
a Banach lattice isomorphism. ∎

The proof of the following result is straightforward.

######  Proposition 3.2.

Let $\mathcal{I}$ and $\mathcal{J}$ be ideals on $\mathbb{N}$ and
$h:\mathbb{N}\to\mathbb{N}$ be an isomorphism between $\mathcal{I}$ and
$\mathcal{J}$. Let $T_{h}\colon c_{0,\mathcal{I}}\to c_{0,\mathcal{J}}$ be
given by $T_{h}({\bf x})={\bf x}\circ h$. Then $T_{h}$ is a Banach lattice
isometry.

######  Theorem 3.3.

Let $\mathcal{I}$ and $\mathcal{J}$ be ideals on $\mathbb{N}$. Then
$c_{0,\mathcal{I}}$ and $c_{0,\mathcal{J}}$ are isometric if, and only if,
$\mathcal{I}$ and $\mathcal{J}$ are isomorphic.

###### Proof.

One direction is Proposition 3.2. For the other, let $T\colon
c_{0,\mathcal{I}}\to c_{0,\mathcal{J}}$ be an isometric isomorphism. Then
$\Phi_{\mathcal{J}}\circ T\circ\Phi_{\mathcal{I}}^{-1}\colon
C_{0}(U_{\mathcal{I}})\to C_{0}(U_{\mathcal{J}})$ is an isometry. By the
classical Banach-Stone theorem, there are a homeomorphism $f\colon
U_{\mathcal{J}}\to U_{\mathcal{I}}$ and a continuous map $\sigma\colon
U_{\mathcal{J}}\to\\{-1,1\\}$ such that $\Phi_{\mathcal{J}}\circ
T\circ\Phi_{\mathcal{I}}^{-1}(F)=\sigma F\circ f$ for all $F\in
C_{0}(U_{\mathcal{I}})$. Thus $(\Phi_{\mathcal{J}}\circ T)({\bf
x})=\sigma\Phi_{\mathcal{I}}({\bf x})\circ f$ for all ${\bf x}\in
c_{0,\mathcal{I}}$. It follows from Proposition 3.1 that if $A\in\mathcal{I}$
is given, then

$\displaystyle(\Phi_{\mathcal{J}}\circ
T)(\chi_{A})(p)=\sigma(p)\Phi_{\mathcal{I}}(\chi_{A})(f(p))=\sigma(p)\chi_{A^{*}}(f(p))=\sigma(p)\chi_{f^{-1}(A^{*})}(p),\quad\mbox{for
all $p\in U_{\mathcal{J}}$.}$ (1)

On the other hand, if $p\in U_{\mathcal{J}}$, we have

$\displaystyle\chi_{f^{-1}(A^{*})}(p)=p-\lim\chi_{A^{*}}(f(n))=p-\lim\chi_{A}(f(n))=p-\lim\chi_{f^{-1}(A)}(n)=\chi_{(f^{-1}(A))^{*}}(p).$
(2)

By combining (1) and (2) we obtain that $(\Phi_{\mathcal{J}}\circ
T)(\chi_{A})=\sigma\chi_{(f^{-1}(A))^{*}}$ for all $A\in\mathcal{I}$. Hence,

$\displaystyle T\chi_{A}=s\chi_{f^{-1}(A)}\quad\mbox{for all
$A\in\mathcal{I}$,}$

where $s=\sigma\restriction\mathbb{N}$. So $h=f\restriction\mathbb{N}$ is a
bijection from $\mathbb{N}$ to $\mathbb{N}$ such that $A\in\mathcal{I}$ iff
$h^{-1}(A)\in\mathcal{J}$. Thus $\mathcal{I}$ and $\mathcal{J}$ are
isomorphic. ∎

######  Theorem 3.4.

Let $\mathcal{I}$ and $\mathcal{J}$ be ideals on $\mathbb{N}$. Then,
$\mathcal{I}\sqsubseteq\mathcal{J}$ if, and only if, there is an into Banach
isometry $T\colon c_{0,\mathcal{I}}\to c_{0,\mathcal{J}}$ such that
$T[c_{0,\mathcal{I}}]$ is an ideal.

###### Proof.

Suppose $\mathcal{I}\sqsubseteq\mathcal{J}$ and let
$h:\mathbb{N}\to\mathbb{N}$ be a bijective Katětov reduction. Let $T_{h}\colon
c_{0,\mathcal{I}}\to c_{0,\mathcal{J}}$ given by $T_{h}({\bf x})={\bf x}\circ
h$. By Proposition 3.2, $T_{h}$ is the required into Banach isometry (which is
moreover a lattice isometry).

Conversely, suppose such $T$ exists. By Proposition 4.1, there is an ideal
$\mathcal{J}^{\prime}$ such that
$T[c_{0,\mathcal{I}}]=c_{0,\mathcal{J}^{\prime}}$. By Theorem 3.3,
$\mathcal{I}$ is isomorphic to $\mathcal{J}^{\prime}$. Clearly
$\mathcal{J}^{\prime}\subseteq\mathcal{J}$, thus
$\mathcal{I}\sqsubseteq\mathcal{J}$. ∎

Next we will look at a condition stronger than isometries, namely, we work
with Banach lattice isometries. This will be related to the Katětov pre-order
on ideals.

Let $\mathcal{I}$ and $\mathcal{J}$ be ideals on $\mathbb{N}$ such that
$\mathcal{I}\leq_{K}\mathcal{J}$, that is, there is
$h\colon\mathbb{N}\to\mathbb{N}$ such that $h^{-1}(A)\in\mathcal{J}$ whenever
$A\in\mathcal{I}$. We can always assume $h$ to be onto (see Proposition 2.2).
Recall the map $T_{h}$ as in Proposition 3.2.

######  Proposition 3.5.

Let $\mathcal{I}$ and $\mathcal{J}$ be ideals on $\mathbb{N}$ with
$\mathcal{I}\leq_{K}\mathcal{J}$, $h:\mathbb{N}\to\mathbb{N}$ be an onto
Katětov reduction from $\mathcal{I}$ to $\mathcal{J}$ and $T_{h}$ as above.
Then,

  1. (1)

$T_{h}$ is an into Banach lattice isometry. Moreover, it satisfies the
following:

    1. (a)

For each $n\in\mathbb{N}$, there is $A\in\mathcal{I}$ satisfying
$T\chi_{A}(n)=1$.

    2. (b)

For all $\emptyset\neq\mathcal{F}\subseteq\mathcal{I}$ such that
$\bigcap_{A\in\mathcal{F}}A=\emptyset$, we have
$\bigwedge_{A\in\mathcal{F}}T(\chi_{A})=0$.

  2. (2)

$h\colon\mathbb{N}\to\mathbb{N}$ is bijective iff $T_{h}[c_{0,\mathcal{I}}]$
is an ideal of $c_{0,\mathcal{J}}$.

###### Proof.

(1) is straightforward.

(2) Suppose $h$ is bijective and let ${\bf x}=(x_{n})\in c_{0,\mathcal{I}}$
and ${\bf y}=(y_{n})\in c_{0,\mathcal{J}}$ be such that $|{\bf
y}|\leq|T_{h}({\bf x})|=T_{h}(|{\bf x}|)$. For each $n\in\mathbb{N}$, let
$z_{n}=y_{h^{-1}(n)}$ and ${\bf z}=(z_{n})$. Then $|z_{n}|\leq|x_{n}|$ for all
$n\in\mathbb{N}$, thus ${\bf z}\in c_{0,\mathcal{I}}$ and clearly $T_{h}({\bf
z})={\bf y}$. Conversely, suppose that $T_{h}[c_{0,\mathcal{I}}]$ is an ideal
and that there are $n\neq m$ such that $h(n)=h(m)=k$. Then
$T_{h}(\chi_{\\{k\\}})=\chi_{h^{-1}\\{k\\}}\geq\chi_{\\{n\\}}$. Thus, there is
${\bf x}\in c_{0,\mathcal{I}}$ such that $T_{h}({\bf x})=\chi_{\\{n\\}}$. In
particular $0=T_{h}({\bf x})(m)=x_{h(m)}=x_{h(n)}=T_{h}({\bf x})(n)=1$, a
contradiction. ∎

Now we proceed to prove the converse of Proposition 3.5.

######  Theorem 3.6.

Let $\mathcal{I}$ and $\mathcal{J}$ be ideals on $\mathbb{N}$. Let $T\colon
c_{0,\mathcal{I}}\to c_{0,\mathcal{J}}$ be an into Banach lattice isometry
such that $T[c_{0,\mathcal{I}}]$ is an ideal containing $c_{0}$. Then
$T=T_{h}$ for some bijection $h:\mathbb{N}\to\mathbb{N}$.

###### Proof.

First, we claim that for every $A\in\mathcal{I}$ there is $B\in\mathcal{J}$
such that $T(\chi_{A})=\chi_{B}$. In fact, we show that for all
$m\in\mathbb{N}$ there is $k\in\mathbb{N}$ such that
$T(\chi_{\\{m\\}})=\chi_{\\{k\\}}$. Let $k\in\mathbb{N}$ and $a>0$ be such
that $a\chi_{\\{k\\}}\leq T(\chi_{\\{m\\}})$. Since $T[c_{0,\mathcal{I}}]$ is
an ideal, there is ${\bf y}\in c_{0,\mathcal{I}}$ such that $T({\bf
y})=a\chi_{\\{k\\}}$. Clearly $0<{\bf y}\leq\chi_{\\{m\\}}$ and $\|{\bf
y}\|=a$. Therefore ${\bf y}=a\chi_{\\{m\\}}$. Thus
$T(\chi_{\\{m\\}})=\chi_{\\{k\\}}$.

Now let $A\in\mathcal{I}$ and put $B=\\{n\in\mathbb{N}:T(\chi_{A})(n)=1\\}$.
We claim that $\chi_{B}=T(\chi_{A})$. Clearly $\chi_{B}\leq T(\chi_{A})$.
Since $T[c_{0,\mathcal{I}}]$ is an ideal, there is ${\bf x}\in
c_{0,\mathcal{I}}$ such that $T({\bf x})=\chi_{B}$. For each $k\in A$, let
$m_{k}\in\mathbb{N}$ be such that $T(\chi_{\\{k\\}})=\chi_{\\{m_{k}\\}}$.
Notice that $m_{k}\in B$ since $T(\chi_{\\{k\\}})\leq T(\chi_{A})$ for all
$k\in A$. Thus $T(\chi_{\\{k\\}})=\chi_{\\{m_{k}\\}}\leq\chi_{B}=T({\bf x})$.
So, $\chi_{\\{k\\}}\leq{\bf x}$ for all $k\in A$. Therefore $\chi_{A}\leq{\bf
x}$ and hence $T(\chi_{A})\leq T({\bf x})=\chi_{B}$ and we are done.

Next we show show that for all $n\in\mathbb{N}$, there is $m\in\mathbb{N}$
such that $T(\chi_{\\{m\\}})=\chi_{\\{n\\}}$. In fact, let $n\in\mathbb{N}$.
Since $\chi_{\\{n\\}}\in c_{0}$, there is ${\bf x}\in c_{0,\mathcal{I}}$ with
${\bf x}>{\bf 0}$ and $T({\bf x})=\chi_{\\{n\\}}$. Let $m\in\mathbb{N}$ be
such that $x_{m}>0$. Since $x_{m}\chi_{\\{m\\}}\leq{\bf x}$, we have
$x_{m}T(\chi_{\\{m\\}})\leq\chi_{\\{n\\}}$. So, $m=n$ and ${\bf
x}=\chi_{\\{m\\}}$.

From the first claim, there is $f\colon\mathbb{N}\to\mathbb{N}$ such that
$T(\chi_{\\{m\\}})=\chi_{\\{f(m)\\}}$. Since $T$ is an isometry, $f$ is
injective. Since $c_{0}\subseteq T[c_{0,\mathcal{I}}]$, by (2) $f$ is onto.
Let $h=f^{-1}$. From the proof of (1) it follows that
$T(\chi_{A})=\chi_{f(A)}$ for every $A\in\mathcal{I}$. This shows that
$T=T_{h}$. ∎

######  Theorem 3.7.

Let $\mathcal{I}$ and $\mathcal{J}$ be ideals on $\mathbb{N}$ and $T\colon
c_{0,\mathcal{I}}\to c_{0,\mathcal{J}}$ be an into Banach lattice isometry.
Suppose $T$ has the following properties:

  1. (1)

for each $n\in\mathbb{N}$, there is $A\in\mathcal{I}$ satisfying
$T\chi_{A}(n)=1$.

  2. (2)

For all $\emptyset\neq\mathcal{F}\subseteq\mathcal{I}$ such that
$\bigcap_{A\in\mathcal{F}}A=\emptyset$, we have
$\bigwedge_{A\in\mathcal{F}}T(\chi_{A})=0$.

Then $\mathcal{I}\leq_{K}\mathcal{J}$.

###### Proof.

Let $T\colon c_{0,\mathcal{I}}\to c_{0,\mathcal{J}}$ be an into Banach lattice
isometry and define $\widetilde{T}\colon C_{0}(U_{\mathcal{I}})\to
C_{0}(U_{\mathcal{J}})$ by $\widetilde{T}=\Phi_{\mathcal{J}}\circ
T\circ\Phi_{\mathcal{I}}^{-1}$. Notice that $\widetilde{T}$ is a Banach
lattice isometry and for each $n\in\mathbb{N}$, there is $A\in\mathcal{I}$
such that $\widetilde{T}\chi_{A^{*}}(n)=1$. Following a proof of Holsztyński
theorem (as presented in [22]) we define, for $p\in U_{\mathcal{I}}$,

$\begin{array}[]{rcl}\Delta(p)&=&\\{q\in
U_{\mathcal{J}}\,\colon\,\widetilde{T}^{*}\delta_{q}(\\{p\\})=1\\},\quad\mbox{and}\\\
\\\ \Delta&=&\displaystyle\bigcup_{p\in U_{\mathcal{I}}}\Delta(p).\end{array}$

The following is a crucial ingredient, its proof can be found in [22].

######  Claim 3.8.

The following statements are valid:

  1. (1)

$\Delta(p)\neq\emptyset$ for all $p\in U_{\mathcal{I}}$.

  2. (2)

$\Delta(p)\cap\Delta(p^{\prime})=\emptyset$ if $p\neq p^{\prime}$.

  3. (3)

The map $\phi\colon\Delta\to U_{\mathcal{I}}$ defined by $\phi(q)=p$ iff
$q\in\Delta(p)$ is continuous and onto.

  4. (4)

$\widetilde{T}F(q)=\sigma(q)F(\phi(q))$ for all $F\in C_{0}(U_{\mathcal{I}})$
and $q\in\Delta$, where $\sigma\colon\Delta\to\mathbb{R}$ is continuous and
$|\sigma(q)|=1$ for all $q\in\Delta$.

Since $\widetilde{T}$ is a Banach lattice isometry, we have $\sigma(q)=1$ for
each $q\in\Delta$.

######  Claim 3.9.

For each $p\in U_{\mathcal{I}}$ we have

$\displaystyle\Delta(p)=\\{q\in
U_{\mathcal{J}}\,\colon\,\widetilde{T}F(q)=1,\mbox{whenever $F\geq{\bf 0}$ and
$F(p)=1=\|F\|$}\\}.$

Proof of Claim 3.9: Suppose that $\widetilde{T}\delta_{q}(\\{p\\})=1$. Let
$F\in C_{0}(U_{\mathcal{I}})$ be such that $F\geq{\bf 0}$ and $F(p)=1=\|F\|$.
Since $\chi_{\\{p\\}}\leq F$, we have
$\widetilde{T}F(q)=\widetilde{T}^{*}\delta_{q}(F)\geq\widetilde{T}^{*}\delta_{q}(\chi_{\\{p\\}})=1$.

For the other inclusion, let $q\in U_{\mathcal{J}}$ be such that
$\widetilde{T}F(q)=1$ whenever $F\geq{\bf 0}$ and $F(p)=1=\|F\|$. Also let
$\mathcal{V}_{p}$ be a fundamental system of open neighborhoods of $p$. If
$\varepsilon>0$ is given, there is an open $V\subset U_{\mathcal{I}}$ with
$p\in V$ and $\widetilde{T}^{*}\delta_{q}(V\setminus\\{p\\})<\varepsilon$. For
each $W\in\mathcal{V}_{p}$, let $F_{W}\in C_{0}(U_{\mathcal{I}})$ be such that
$F_{W}(p)=1=\|F_{W}\|$, $F_{W}\geq{\bf 0}$ and $F_{W}(U_{\mathcal{I}}\setminus
W)=\\{0\\}$. Take $W_{0}\in\mathcal{V}_{p}$ with $W_{0}\subset V$. So,

$\displaystyle
1=\widetilde{T}F_{W_{0}}(q)=\int_{U_{\mathcal{I}}}F_{W_{0}}\,d\widetilde{T}^{*}\delta_{q}=\int_{W_{0}}F_{W_{0}}\,d\widetilde{T}^{*}\delta_{q}\leq\widetilde{T}^{*}\delta_{q}(\\{p\\})+\widetilde{T}^{*}\delta_{q}(V\setminus\\{p\\})\leq
1+\varepsilon.$

Since $\varepsilon>0$ is arbitrary, we conclude that
$\widetilde{T}^{*}\delta_{q}(\\{p\\})=1$, that is, $q\in\Delta(p)$. Thus we
have proved Claim 3.9.

######  Claim 3.10.

For each $p\in U_{\mathcal{I}}$ we have

$\displaystyle\Delta(p)=\\{q\in
U_{\mathcal{J}}\,\colon\,\mbox{$\widetilde{T}\chi_{A^{*}}(q)=1$ for all
$A\in\mathcal{I}$ with $p\in A^{*}$}\\}.$

Proof of Claim 3.10: The inclusion $\subseteq$ is clear from Claim 3.9. For
the other direction, let $\widetilde{\Delta}(p)$ be the the right hand side of
the identity above and $q\in\widetilde{\Delta}(p)$. Let $\varepsilon>0$ be
given. If $F\in C_{0}(U_{\mathcal{I}})$ satisfies $F\geq{\bf 0}$ and
$F(p)=1=\|F\|$, then $B=\\{n\in\mathbb{N}\,\colon\,|F(n)-1|<\varepsilon\\}\in
p$. Let $A\in\mathcal{I}\cap p$. Since $q\in\widetilde{\Delta}(p)$,
$\widetilde{T}\chi_{(A\cap B)^{*}}(q)=1$. As $(1-\varepsilon)\chi_{(A\cap
B)^{*}}\leq F$ and $T$ is lattice order preserving, we have that
$1-\varepsilon\leq TF(q)$. Since $\varepsilon$ was arbitrary, $TF(q)=1$ and
$q\in\Delta(p)$. Thus we have proved Claim 3.10.

######  Claim 3.11.

$\mathbb{N}\subset\Delta$.

Proof of Claim 3.11: Fix $n\in\mathbb{N}$ and let
$\mathcal{F}_{n}=\\{A\in\mathcal{I}\,\colon\,T\chi_{A}(n)=1\\}$. Thus, from
the hypothesis (1), $\mathcal{F}_{n}$ is non-empty. Now if
$A,B\in\mathcal{F}_{n}$, then $T\chi_{A}(n)=1$ and $T\chi_{B}(n)=1$. Since
$T\chi_{(A\cap B)}=T(\chi_{A}\wedge\chi_{B})=T(\chi_{A})\wedge T(\chi_{B})$,
we conclude that $T\chi_{(A\cap B)}(n)=1$. Hence $\mathcal{F}_{n}$ has the
finite intersection property. Let $p\in\beta\mathbb{N}$ be such that
$\mathcal{F}_{n}\subset p$. We will show that $n\in\Delta(p)$.

Notice that $\widetilde{T}\chi_{A^{*}}(q)=q-\lim_{m}T\chi_{A}(m)$, for every
$q\in U_{\mathcal{I}}$ and $A\in\mathcal{I}$. In particular, when $q$ is the
principal ultrafilter $n$, we get $\widetilde{T}\chi_{A^{*}}(n)=T\chi_{A}(n)$.
Thus
$\mathcal{F}_{n}=\\{A\in\mathcal{I}\,\colon\,\widetilde{T}\chi_{A^{*}}(n)=1\\}$.

As $\mathcal{F}_{n}\subseteq p$, we have that $p\in\mathcal{U}_{\mathcal{I}}$.
Let $A\in\mathcal{F}_{n}$ be fixed. To show that $n\in\Delta(p)$ it suffices
to have, by Claim 3.10, that $\widetilde{T}\chi_{B^{*}}(n)=1$ for all
$B\in\mathcal{I}$ with $p\in B^{*}$. Fix such $B$ and suppose, towards a
contradiction, that $\widetilde{T}\chi_{B^{*}}(n)\neq 1$. From Claim 3.8 (4),
we have $\widetilde{T}\chi_{B^{*}}(n)=\chi_{B^{*}}(\phi(n))$. Thus
$\widetilde{T}\chi_{B^{*}}(n)=0$. On the other hand,

$\displaystyle 1=T\chi_{A}(n)=T\chi_{(A\cap B)}(n)+T\chi_{(A\setminus B)}(n).$

As $0\leq T\chi_{A\cap B}(n)\leq T\chi_{B}(n)$, we conclude that
$T\chi_{(A\setminus B)}(n)=1$ and $A\setminus B\in p$, which is a
contradiction. Thus we have proved Claim 3.11.

Now we show that $\mathbb{N}\subseteq\phi^{-1}(\mathbb{N})$ where $\phi$ is as
in Claim 3.8. Let $n\in\mathbb{N}$. Then
$\mathcal{F}_{n}=\\{A\in\mathcal{I}\,\colon\,T\chi_{A}(n)=1\\}$. We claim that
$\bigcap\\{A:\;A\in\mathcal{F}_{n}\\}\neq\emptyset$. In fact, we have
$\chi_{\\{n\\}}\leq T\chi_{A}$ for all $A\in\mathcal{F}_{n}$. Thus

$0<\bigwedge_{A\in\mathcal{F}_{n}}T\chi_{A}.$

Thus, by the hypothesis (2), there is $m$ such that $m\in A$ for all
$A\in\mathcal{F}_{n}$. Therefore $n\in\Delta(m)$, that is, $\phi(n)=m$.

To end the proof, we let $h:=\phi\restriction\mathbb{N}$. By Claim 3.8(3), $h$
is onto. We show that $h:\mathbb{N}\to\mathbb{N}$ is a Katětov reduction. We
have

$\displaystyle
T\chi_{A}(n)=\widetilde{T}\chi_{A^{*}}(n)=\chi_{A^{*}}(\phi(n))=\chi_{A}(\phi(n))=\chi_{\phi^{-1}(A)}(n)=\chi_{h^{-1}(A)}(n)$
(3)

for all $A\in\mathcal{I}$ and $n\in\mathbb{N}$. Thus, if $A\in\mathcal{I}$,
then $\chi_{h^{-1}(A)}\in c_{0,\mathcal{J}}$ which means that
$h^{-1}(A)\in\mathcal{J}$. This shows that $\mathcal{I}\leq_{K}\mathcal{J}$
and we are done. ∎

We end this section commenting about the role of the condition (1) in Theorem
3.7. We show first a simple way to construct subspaces of $c_{0,\mathcal{I}}$,
a particular case was already observed in [19].

######  Proposition 3.12.

Let $\mathcal{I}$ be an ideal on $\mathbb{N}$ and $A\subseteq\mathbb{N}$. Then
$c_{0,\mathcal{I}\restriction A}$ is Banach lattice isometric to a closed
ideal of $c_{0,\mathcal{I}}$.

###### Proof.

Recall that $\mathcal{I}\restriction A$ is an ideal on $A$ and thus
$c_{0,\mathcal{I}\restriction A}$ consists of sequences on $A$. For each ${\bf
x}=(x_{n})_{n\in A}\in c_{0,\mathcal{I}\restriction A}$, let $T({\bf
x}):=(\chi_{A}(n)x_{n})_{n\in\mathbb{N}}$. Clearly $T({\bf x})\in
c_{0,\mathcal{I}}$. The map $\Psi$ is the required Banach lattice isometry. It
is not difficult to check that $T(c_{0,\mathcal{I}\restriction A})$ is an
ideal of $c_{0,\mathcal{I}}$. ∎

In the previous result suppose $\mathcal{I}$ is not $\mathcal{P}(\mathbb{N})$
and $A\in\mathcal{I}$ is infinite, then $\mathcal{I}\restriction
A=\mathcal{P}(A)$ and thus $\mathcal{I}\restriction A\not\leq_{K}\mathcal{I}$.
In this case condition (1) in Theorem 3.7 fails because $T(\chi_{B})(n)=0$ for
all $B$ and any $n\not\in A$.

##  4\. Some lattices properties of $c_{0,\mathcal{I}}$

In this section we present some properties of $c_{0,\mathcal{I}}$ as an ideal
of the Banach lattice $\ell_{\infty}$. First, we characterize the closed
ideals of $\ell_{\infty}$. Part (iii) below says that it is somewhat harmless
to assumed that every closed ideal of $\ell_{\infty}$ contains $c_{0}$. The
particular case of (i) below for a maximal ideal was shown in [18, Theorem
4.1].

######  Proposition 4.1.

Let $Y$ be a closed sublattice of $\ell_{\infty}$. Then

  * (i)

$Y$ is an ideal iff there is an ideal $\mathcal{I}$ (not containing necesarely
${\sf Fin}$) on $\mathbb{N}$ such that $Y=c_{0,\mathcal{I}}$.

  * (ii)

$c_{0}\subseteq c_{0,\mathcal{I}}$ iff ${\sf Fin}\subseteq\mathcal{I}$.

  * (iii)

For each $A\subseteq\mathbb{N}$, let $Z_{A}=\\{{\bf
x}\in\ell_{\infty}:x_{n}=0\;\mbox{for all $n\in A$}\\}$. For each closed ideal
$Y$ of $\ell_{\infty}$, let $A=\\{n\in\mathbb{N}:\chi_{\\{n\\}}\not\in Y\\}$,
then $Y=\langle Y\cup c_{0}\rangle\cap Z_{A}$.

###### Proof.

(i) We have already seen that every $c_{0,\mathcal{I}}$ is an ideal of
$\ell_{\infty}$. Conversely, if $Y$ is an ideal of $\ell_{\infty}$, let
$\mathcal{I}=\\{A\subseteq\mathbb{N}:\;\chi_{A}\in Y\\}$. It is easy to verify
that $\mathcal{I}$ is an ideal. From Proposition 2.5, it follows that
$c_{0,\mathcal{I}}\subseteq Y$. For the other inclusion, fix ${\bf x}\in Y$
and $\varepsilon>0$. By an argument analogous to that used in the proof of
Proposition 2.5(2), there is ${\bf
y}\in\mathrm{span}\\{\chi_{A}\,\colon\,A\in\mathcal{I}\\}$ such that $\|{\bf
x}-{\bf y}\|<\varepsilon$.

(ii) is straightforward. (iii) Firstly, we prove that $Y\subseteq Z_{A}$.
Assume that $Y\setminus Z_{A}\neq\emptyset$ and let ${\bf z}=(z_{n})\in
Y\setminus Z_{A}$. So, there is $m\in A$ such that $z_{m}\neq 0$. Since
$|z_{m}|\chi_{\\{m\\}}\leq|{\bf z}|$ and $Y$ is an ideal,
$z_{m}\chi_{\\{m\\}}\in Y$. Hence, $\chi_{\\{m\\}}\in Y$ which is impossible.
On the other hand, it is clear that $Y\subseteq\langle Y\cup c_{0}\rangle$.
For the other inclusion, let $\mathcal{I}$ be the ideal given by (i), i.e,
such that $c_{0,\mathcal{I}}=Y$. By Theorem 5.1 below we have

$\displaystyle\langle Y\cup c_{0}\rangle=\langle Y+c_{0}\rangle=\langle
c_{0,\mathcal{I}}+c_{0,{\sf Fin}}\rangle=\langle c_{0,\mathcal{I}\sqcup{\sf
Fin}}\rangle=c_{0,\mathcal{I}\sqcup{\sf Fin}}.$

Now, if ${\bf x}\in\langle Y\cup c_{0}\rangle\cap Z_{A}$, then ${\bf x}\in
c_{0,\mathcal{I}\sqcup{\sf Fin}}$. Let $\varepsilon>0$ be given. Observe that
$A({\bf x},\varepsilon)\cap A=\emptyset$. If $C\in{\sf Fin}$ and
$B\in\mathcal{I}$ satisfy $A({\bf x},\varepsilon)=B\cup C$, we have
$C\in\mathcal{I}$ since $C\cap A=\emptyset$. So, $A({\bf
x},\varepsilon)\in\mathcal{I}$. Whence, ${\bf x}\in c_{0,\mathcal{I}}=Y$. ∎

######  Definition 4.2.

  1. (1)

Two elements $x,y\in\ell_{\infty}$ are called $c_{0}$-disjoint, and we write
$x\perp_{c_{0}}y$, if $|x|\wedge|y|\in c_{0}$.

  2. (2)

Two subspaces $S$ and $T$ of $\ell_{\infty}$ are called $c_{0}$-disjoint if
$s\perp_{c_{0}}t$ for each $s\in S$ and $t\in T$.

  3. (3)

The $c_{0}$-disjoint complement of a subspace $Y$ of $\ell_{\infty}$, denoted
by $c_{0}[Y]$, is

$c_{0}[Y]=\\{x\in\ell_{\infty}\,\colon\,\mbox{$x\perp_{c_{0}}y$ for all $y\in
Y$}\\}=\\{x\in\ell_{\infty}\colon\;\mbox{$|x|\wedge|y|\in c_{0}$ for all $y\in
Y$}\\}.$

######  Theorem 4.3.

Let $\mathcal{I}$ and $\mathcal{J}$ be two ideals over $\mathbb{N}$. Then

  * (i)

$c_{0,\mathcal{I}}$ and $c_{0,\mathcal{J}}$ are $c_{0}$-disjoint iff
$\mathcal{I}\subseteq\mathcal{J}^{\perp}$. In particular, $c_{0,\mathcal{I}}$
and $c_{0,\mathcal{I}^{\perp}}$ are $c_{0}$-disjoint for every ideal
$\mathcal{I}$.

  * (ii)

$c_{0,\mathcal{I}^{\perp}}=c_{0}[c_{0,\mathcal{I}}]$.

  * (iii)

$\mathcal{I}$ is Fréchet iff
$c_{0}[c_{0}[c_{0,\mathcal{I}}]]=c_{0,\mathcal{I}}$.

###### Proof.

  * (i)

Suppose $\mathcal{I}\subseteq\mathcal{J}^{\perp}$ and let ${\bf x}=(x_{n})\in
c_{0,\mathcal{I}}$ and ${\bf y}=(y_{n})\in c_{0,\mathcal{J}}$. Let
$\varepsilon>0$. Then $A(\varepsilon,{\bf
x})=\\{n\in\mathbb{N}:|x_{n}|\geq\varepsilon\\}\in\mathcal{I}$. Thus
$A(\varepsilon,{\bf x})\in\mathcal{J}^{\perp}$. On the other hand,
$A(\varepsilon,{\bf
y})=\\{n\in\mathbb{N}:|y_{n}|\geq\varepsilon\\}\in\mathcal{J}$. Thus
$A(\varepsilon,{\bf x})\cap A(\varepsilon,{\bf y})$ is finite. Thus there is
$n_{0}$ such that for all $n>n_{0}$, $|x_{n}|<\varepsilon$ or
$|y_{n}|<\varepsilon$. Thus $\min\\{|x_{n}|,|y_{n}|\\}<\varepsilon$ for all
$n>n_{0}$. Hence $|{\bf x}|\wedge|{\bf y}|\in c_{0}$.

Conversely, suppose $\mathcal{I}\not\subseteq\mathcal{J}^{\perp}$. Let
$A\in\mathcal{I}\setminus\mathcal{J}^{\perp}$. Then there is $B\subseteq A$
infinite such that $B\in\mathcal{J}\cap\mathcal{I}$. Hence $\chi_{B}\in
c_{0,\mathcal{I}}\cap c_{0,\mathcal{J}}$ and clearly $\chi_{B}\not\in c_{0}$
as $B$ is infinite.

For the second claim, we recall that
$\mathcal{I}\subseteq\mathcal{I}^{\perp\perp}$ for every ideal $\mathcal{I}$.

  * (ii)

The inclusion $\subseteq$ follows from (i). Conversely, suppose ${\bf
x}=(x_{n})\not\in c_{0,\mathcal{I}^{\perp}}$ and let $1>\varepsilon>0$ be such
that $A(\varepsilon,{\bf x})\not\in\mathcal{I}^{\perp}$. Thus there is
$B\in\mathcal{I}$ such that $C:=A(\varepsilon,{\bf x})\cap B$ is infinite.
Then $\chi_{C}\in c_{0,\mathcal{I}}$ and $\varepsilon\chi_{C}\leq|{\bf
x}|\wedge\chi_{C}$. Thus $|{\bf x}|\wedge\chi_{C}\not\in c_{0}$.

  * (iii)

Suppose $\mathcal{I}$ is Fréchet. By (ii), we have

$c_{0}[c_{0}[c_{0,\mathcal{I}}]]=c_{0}[c_{0,\mathcal{I}^{\perp}}]=c_{0,\mathcal{I}^{\perp\perp}}=c_{0,\mathcal{I}}.$

Conversely, to see that $\mathcal{I}$ is Fréchet it suffices to show that
$\mathcal{I}^{\perp\perp}\subseteq\mathcal{I}$. Let
$B\in\mathcal{I}^{\perp\perp}$. By (ii) and our assumption, $\chi_{B}\in
c_{0,\mathcal{I}^{\perp\perp}}=c_{0}[c_{0,\mathcal{I}^{\perp}}]=c_{0}[c_{0}[c_{0,\mathcal{I}}]]=c_{0,\mathcal{I}}$.
From Proposition 2.5, we conclude $B\in\mathcal{I}$.∎

Recall that an ideal $\mathcal{I}$ on $\mathbb{N}$ is tall if every infinite
subset of $\mathbb{N}$ contains a infinite member of $\mathcal{I}$. A subset
$A$ of a Banach lattice $E$ is called order dense in $B$ if for each ${\bf
0}\neq x\in B$ there exists $a\in A$ such that ${\bf 0}<|a|\leq|x|$. Observe
that $c_{0,\mathcal{I}}$ is order dense in $\ell_{\infty}$ for any ideal
$\mathcal{I}$ (containing ${\sf Fin}$).

######  Theorem 4.4.

Let $\mathcal{I}$ be an ideal on $\mathbb{N}$. The following statements are
equivalent:

  1. (1)

$\mathcal{I}$ is tall;

  2. (2)

$c_{0,\mathcal{I}}\setminus c_{0}$ is order dense in $\ell_{\infty}\setminus
c_{0}$;

  3. (3)

$c_{0,\mathcal{I}^{\perp}}=c_{0}$.

###### Proof.

(1) $\Rightarrow$ (2): Suppose that $\mathcal{I}$ is tall and let ${\bf
0}\neq{\bf x}=(x_{n})\in\ell_{\infty}\setminus c_{0}$ be given. For some
$\varepsilon_{0}>0$ we have $A(\varepsilon_{0},{\bf x})$ is infinite. The
tallness of $\mathcal{I}$ implies that there exists $B\in\mathcal{I}$ infinite
satisfying $B\subset A(\varepsilon_{0},{\bf x})$. So, ${\bf
y}=\chi_{B}\cdot{\bf x}\in c_{0,\mathcal{I}}\setminus c_{0}$ and ${\bf
0}<|{\bf y}|\leq|{\bf x}|$.

(2) $\Rightarrow$ (3): Suppose there is ${\bf x}\in c_{0,\mathcal{I}^{\perp}}$
such that ${\bf x}\not\in c_{0}$. Since $c_{0,\mathcal{I}}\setminus c_{0}$ is
order dense in $\ell_{\infty}\setminus c_{0}$, there exists ${\bf y}\in
c_{0,\mathcal{I}}\setminus c_{0}$ satisfying $0<|{\bf y}|\leq|{\bf x}|$. Thus,
$A(\varepsilon,{\bf y})\subset A(\varepsilon,{\bf x})$ for each
$\varepsilon>0$. There is $\varepsilon_{0}>0$ with $A(\varepsilon_{0},{\bf
x})$ infinite. Since $A(\varepsilon_{0},{\bf x})\in\mathcal{I}^{\perp}$ and
$A(\varepsilon,{\bf y})\in\mathcal{I}$, it follows that $A(\varepsilon,{\bf
y})$ is finite for each $0<\varepsilon<\varepsilon_{0}$ which contradicts that
${\bf y}\not\in c_{0}$. So, $c_{0,\mathcal{I}^{\perp}}=c_{0}$.

(3) $\Rightarrow$ (1): The equality $c_{0,\mathcal{I}^{\perp}}=c_{0}$ means
that $\mathcal{I}^{\perp}=\sf{Fin}$. Thus, $\mathcal{I}$ is tall. ∎

##  5\. Banach spaces isomorphic to $c_{0,\mathcal{I}}(X)$

In this section we show that, for some ideals, $c_{0,\mathcal{I}}$ is
isomorphic to a known classical Banach space. We do it for ideals of the form
$\mathcal{I}\sqcup\mathcal{J}$,
$\bigoplus\limits_{n\in\mathbb{N}}\mathcal{I}_{n}$,
$\mathcal{I}^{\omega\perp}$ and the Fubini product
$\mathcal{I}\times\mathcal{J}$.

######  Theorem 5.1.

Let $X$ be a Banach space. If $\mathcal{I}$ and $\mathcal{J}$ are ideals on
$\mathbb{N}$, then
$c_{0,\mathcal{I}\sqcup\mathcal{J}}(X)=c_{0,\mathcal{I}}(X)+c_{0,\mathcal{J}}(X)$.

###### Proof.

If ${\bf x}=(x_{n})\in c_{0,\mathcal{I}}(X)$ and ${\bf y}=(y_{n})\in
c_{0,\mathcal{J}}(X)$, then $A(\varepsilon,{\bf x+y})\subset
A(\varepsilon/2,{\bf x})\cup A(\varepsilon/2,{\bf y})$ for all
$\varepsilon>0$. So, $c_{0,\mathcal{I}}(X)+c_{0,\mathcal{J}}(X)\subset
c_{0,\mathcal{I}\sqcup\mathcal{J}}(X)$. On the other hand, if ${\bf
z}=(z_{n})\in\ell_{\infty}(X)$ and $\mathrm{supp}({\bf
z})\in\mathcal{I}\sqcup\mathcal{J}$, take $A\in\mathcal{I}$ and
$B\in\mathcal{J}$ with $\mathrm{supp}({\bf z})=A\cup B$ and $A\cap
B=\emptyset$. By setting ${\bf x}=(\chi_{A}(n)z_{n})$ and ${\bf
y}=(\chi_{B}(n)z_{n})$, we have ${\bf z=x+y}$, ${\bf x}\in c_{0,\mathcal{I}}$
and ${\bf y}\in c_{0,\mathcal{J}}$. By Proposition 2.5(1) we conclude that
$c_{0,\mathcal{I}\sqcup\mathcal{J}}(X)\subset
c_{0,\mathcal{I}}(X)+c_{0,\mathcal{J}}(X)$. ∎

######  Theorem 5.2.

Let $X$ be a Banach space, $\\{K_{n}:\;n\in\mathbb{N}\\}$ be a partition of
$\mathbb{N}$ and $\mathcal{I}_{n}$ be an ideal on $K_{n}$ for each
$n\in\mathbb{N}$. If
$\mathcal{I}=\bigoplus\limits_{n\in\mathbb{N}}\mathcal{I}_{n}$, then
$c_{0,\mathcal{I}}(X)$ is isometric to
$\ell_{\infty}((c_{0,\mathcal{I}_{n}}(X))_{n\in\mathbb{N}})$. In particular,
$c_{0,\mathcal{J}^{\omega}}(X)$ is isometric to
$\ell_{\infty}(c_{0,\mathcal{J}}(X))$ for any ideal $\mathcal{J}$.

###### Proof.

Let

$\displaystyle\Psi\colon c_{0,\mathcal{I}}(X)$
$\displaystyle\longrightarrow\ell_{\infty}((c_{0,\mathcal{I}_{n}}(X))_{n\in\mathbb{N}})$
$\displaystyle(x_{n})$ $\displaystyle\longmapsto((x_{n})_{n\in
K_{m}})_{m\in\mathbb{N}}.$

  1. (1)

To see that $\Psi$ is a well defined linear isometry, let ${\bf x}=(x_{n})\in
c_{0,\mathcal{I}}(X)$ and $\varepsilon>0$. Then

$\displaystyle A(\varepsilon,{\bf
x})\in\mathcal{I}\quad\Longleftrightarrow\mbox{$\\{n\in
K_{m}\,\colon\,\|x_{n}\|\geq\varepsilon\\}\in\mathcal{I}_{m}$ for each
$m\in\mathbb{N}$.}$

Thus, $(x_{n})_{n\in K_{m}}\in c_{0,\mathcal{I}_{m}}(X)$ for all
$m\in\mathbb{N}$. On the other hand,

$\displaystyle\|\Psi({\bf x})\|=\sup_{m\in\mathbb{N}}\|(x_{n})_{n\in
K_{m}}\|_{\infty}=\sup_{m\in\mathbb{N}}\sup_{n\in
K_{m}}\|x_{n}\|=\sup_{n\in\mathbb{N}}\|x_{n}\|=\|{\bf x}\|.$

  2. (2)

To see that $\Psi$ is onto, let $\displaystyle{\bf
y}=((y_{n}^{m}))_{m\in\mathbb{N}}\in\ell_{\infty}((c_{0,\mathcal{I}_{m}}(X))_{m\in\mathbb{N}})$
be given. Define ${\bf x}=(x_{n})\in\ell_{\infty}(X)$ by $x_{n}=y_{n}^{m}$ iff
$n\in K_{m}$. If $\varepsilon>0$ and $m\in\mathbb{N}$ are given, then

$\displaystyle\\{n\in\mathbb{N}\,\colon\,\|x_{n}\|\geq\varepsilon\\}\cap
K_{m}=\\{n\in
K_{m}\,\colon\,\|y_{n}^{m}\|\geq\varepsilon\\}\in\mathcal{I}_{m}.$

So $A(\varepsilon,{\bf x})\in\mathcal{I}$ for each $\varepsilon>0$ i.e. ${\bf
x}\in c_{0,\mathcal{I}}(X)$. Clearly $\Psi({\bf x})={\bf y}$.∎

######  Corollary 5.3.

Let $X$ be a Banach space. Let $\\{A,B\\}$ be a partition of $\mathbb{N}$ and
$\mathcal{I}$ and $\mathcal{J}$ be ideals on $A$ and $B$, respectively. Then
$c_{0,\mathcal{I}\oplus\mathcal{J}}(X)$ is isometric to
$c_{0,\mathcal{I}}(X)\oplus_{\infty}c_{0,\mathcal{J}}(X)$.

######  Theorem 5.4.

Let $X$ be a Banach space, $\\{K_{n}:\;n\in\mathbb{N}\\}$ be a partition of
$\mathbb{N}$, $\mathcal{I}_{n}$ be an ideal on $K_{n}$ for each
$n\in\mathbb{N}$ and
$\mathcal{I}=\bigoplus\limits_{n\in\mathbb{N}}\mathcal{I}_{n}$. Then
$c_{0,\mathcal{I}^{\perp}}(X)$ is isometric to
$c_{0}((c_{0,\mathcal{I}_{j}^{\perp}}(X))_{j\in\mathbb{N}})$. In particular,
$c_{0,\mathcal{J}^{\omega\perp}}(X)$ is isometric to
$c_{0}(c_{0,\mathcal{J}^{\perp}}(X))$ for any ideal $\mathcal{J}$.

###### Proof.

Recall that

$\displaystyle A\in\mathcal{I}^{\perp}\Longleftrightarrow\mbox{$(\exists
N\in\mathbb{N})(A\subseteq\bigcup_{i\leq N}K_{i}$ and $(\forall i\leq
N)\;(A\cap K_{i}\in\mathcal{I}_{i}^{\perp}))$.}$

For any ${\bf x}\in c_{0,\mathcal{I}^{\perp}}(X)$, we set ${\bf
y}^{j}=(x_{n})_{n\in K_{j}}$ for each $j\in\mathbb{N}$. We have the following:

  1. (1)

${\bf y}^{j}\in c_{0,\mathcal{I}_{j}^{\perp}}(X)$ for all $j\in\mathbb{N}$. In
fact, let $j\in\mathbb{N}$ and $\delta>0$ be given. There is $N\in\mathbb{N}$
such that $A(\delta,{\bf x})\subset K_{1}\cup\cdots\cup K_{N}$ and
$A(\delta,{\bf x})\cap K_{i}\in\mathcal{I}_{i}^{\perp}$ for all
$i\in\\{1,\ldots,N\\}$. Observe that $A(\delta,{\bf x})\cap
K_{j}=A(\delta,{\bf y}^{j})$. Thus, if $j\leq N$, we have that $A(\delta,{\bf
y}^{j})\in\mathcal{I}_{i}^{\perp}$. Otherwise, $A(\delta,{\bf
y}^{j})=A(\delta,{\bf x})\cap K_{j}=\emptyset$. In either case, we conclude
that ${\bf y}^{j}\in c_{0,\mathcal{I}_{j}^{\perp}}(X)$.

  2. (2)

The sequence $({\bf y}^{j})_{j\in\mathbb{N}}$ converges to 0. If
$\varepsilon>0$ is given, there is $N\in\mathbb{N}$ such that
$A(\varepsilon,{\bf x})\subset K_{1}\cup\cdots\cup K_{N}$. If $j>N$ and $n\in
K_{j}$, we have $\|x_{n}\|<\varepsilon$. So $\displaystyle\|{\bf
y}^{j}\|=\sup_{n\in K_{j}}\|x_{n}\|\leq\varepsilon$ if $j>N$.

The two previous claims show that

$\displaystyle\Psi\colon c_{0,\mathcal{I}^{\perp}}(X)$
$\displaystyle\longrightarrow
c_{0}((c_{0,\mathcal{I}_{j}^{\perp}}(X))_{j\in\mathbb{N}})$ $\displaystyle{\bf
x}=(x_{n})$ $\displaystyle\longmapsto({\bf y}^{j})_{j\in\mathbb{N}}$

is well defined and is clearly a linear isometry.

To see that $\Psi$ is onto, let $({\bf y}^{j})\in
c_{0}((c_{0,\mathcal{I}_{j}^{\perp}}(X))_{j\in\mathbb{N}})$ be given and
consider the sequence ${\bf x}=(x_{n})_{n}$ defined by $x_{n}=y^{j}_{n}$ if
$n\in K_{j}$. Let $\varepsilon>0$, there is $N\in\mathbb{N}$ such that $\|{\bf
y}^{j}\|=\displaystyle\sup_{n\in K_{j}}\|y_{n}^{j}\|<\varepsilon$ for all
$j>N$. Thus, $A(\varepsilon,{\bf x})\subset K_{1}\cup\cdots\cup K_{N}$. Also
note that $A(\varepsilon,{\bf x})\cap K_{j}=\\{n\in
K_{j}\,\colon\,\|y_{n}^{j}\|\geq\varepsilon\\}\in\mathcal{I}_{j}^{\perp}$ for
all $j\leq N$, which yields $(x_{n})\in c_{0,\mathcal{I}^{\perp}}(X)$.
Clearly, $\Psi({\bf x})=({\bf y}^{j})$. ∎

To study the Banach space $c_{0,\mathcal{I}\times\mathcal{J}}(X)$ we need to
compute the norm of the quotient $\ell_{\infty}/c_{0,\mathcal{I}}$. For that
end we recall the definition of $\mathcal{I}-\limsup$ of a sequence.

######  Definition 5.5.

[6] Let $\mathcal{I}$ be an ideal on $\mathbb{N}$. For each sequence ${\bf
x}=(x_{n})$ in $\mathbb{R}$, we set

$\displaystyle B_{\bf
x}=\\{b\in\mathbb{R}\,\colon\,\\{k\in\mathbb{N}\,\colon\,x_{k}>b\\}\not\in\mathcal{I}\\}$

and

$\displaystyle\mathcal{I}-\limsup x_{n}=\begin{cases}\sup B_{\bf x},&B_{\bf
x}\neq\emptyset,\\\ -\infty,&B_{\bf x}=\emptyset.\end{cases}$

Note that a sequence $(x_{n})$ is $\mathcal{I}$-convergent to 0 iff
$\mathcal{I}-\limsup|x_{n}|=0$ (see [6, Theorem 4]).

######  Lemma 5.6.

Let $\mathcal{I}$ be an ideal on $\mathbb{N}$.

  1. (1)

If $x_{n}\leq y_{n}$ for each $n\in\mathbb{N}$, then $\mathcal{I}-\limsup
x_{n}\leq\mathcal{I}-\limsup y_{n}$.

  2. (2)

If $c\in\mathbb{R}$, then $\mathcal{I}-\limsup c+x_{n}=c+\mathcal{I}-\limsup
x_{n}$.

###### Proof.

(1) If $\mathcal{I}-\limsup x_{n}=-\infty$, there is nothing to prove. Suppose
that $B_{\bf x}\neq\emptyset$ and let $\varepsilon>0$ be given. If $b\in
B_{\bf x}$ satisfies $\mathcal{I}-\limsup x_{n}-\varepsilon<b$, then
$\\{k\in\mathbb{N}\,\colon\,y_{k}>b\\}\not\in\mathcal{I}$. Hence $b\in B_{\bf
y}$ and $\mathcal{I}-\limsup x_{n}-\varepsilon\leq\mathcal{I}-\limsup y_{n}$.
Since $\varepsilon>0$ was arbitrary, the result follows.

(2) Just observe that

$\displaystyle\\{b\in\mathbb{R}\,\colon\,\\{k\in\mathbb{N}\,\colon\,x_{k}+c>b\\}\not\in\mathcal{I}\\}=c+\\{b^{\prime}\in\mathbb{R}\,\colon\,\\{k\in\mathbb{N}\,\colon\,x_{k}>b^{\prime}\\}\not\in\mathcal{I}\\}.\qed$

######  Lemma 5.7.

Let $X$ be a Banach space and ${\bf x}=(x_{n})\in\ell_{\infty}(X)$. Then

$\displaystyle\mathcal{I}-\limsup\|x_{n}\|=\|(x_{n})+c_{0,\mathcal{I}}(X)\|.$

###### Proof.

Let ${\bf z}\in c_{0,\mathcal{I}}(X)$. Then $\|x_{n}\|\leq\|{\bf
x-z}\|+\|z_{n}\|$ for all $n\in\mathbb{N}$. By Lemma 5.6 we have
$\mathcal{I}-\limsup\|x_{n}\|\leq\|{\bf x-z}\|$. Thus,
$\mathcal{I}-\limsup\|x_{n}\|\leq\inf\\{\|{\bf x-z}\|\,\colon\,{\bf z}\in
c_{0,\mathcal{I}}(X)\\}$. Now let $r\in\mathbb{R}$ be such that

$\displaystyle\mathcal{I}-\limsup\|x_{n}\|<r<\inf\\{\|{\bf
x-z}\|\,\colon\,{\bf z}\in c_{0,\mathcal{I}}(X)\\}.$

Note that $A:=A(r,{\bf x})\in\mathcal{I}$. If $A=\emptyset$, we are done. If
not, for $n\in A$ we have ${\bf z}=x_{n}\chi_{\\{n\\}}\in
c_{0,\mathcal{I}}(X)$ and $r<\|{\bf x-z}\|=0$, which is impossible. Whence,
$\mathcal{I}-\limsup\|x_{n}\|=\inf\\{\|{\bf x-z}\|\,\colon\,{\bf z}\in
c_{0,\mathcal{I}}(X)\\}=\|(x_{n})+c_{0,\mathcal{I}}(X)\|$. ∎

######  Theorem 5.8.

Let $X$ be a Banach space and $\mathcal{I}$ and $\mathcal{J}$ be ideals on
$\mathbb{N}$. Then
$c_{0,\mathcal{I}\times\mathcal{J}}(X)/c_{0,{\mathcal{J}}^{\omega}}(X)$ is
isomorphic to $c_{0,\mathcal{I}}(\ell_{\infty}(X)/c_{0,\mathcal{J}}(X))$.

###### Proof.

We start by proving:

######  Claim 5.9.

Let ${\bf x}=(x_{n})\in\ell_{\infty}(X)$ and $\delta>0$ be given.

  1. (1)

If $A(\delta,{\bf x})\in\mathcal{I}$, then
$\mathcal{I}-\limsup_{j}\|x_{j}\|\leq\delta$.

  2. (2)

If $\mathcal{I}-\limsup_{j}\|x_{j}\|\leq\delta$, then $A(2\delta,{\bf
x})\in\mathcal{I}$.

###### Proof of claim..

(1): Suppose that $\mathcal{I}-\limsup_{j}\|x_{j}\|>\delta$. There is
$b\in\mathbb{R}$ such that $\delta<b$ and
$\\{j\in\mathbb{N}\,\colon\,\|x_{j}\|>b\\}\not\in\mathcal{I}$. But we have
$\\{j\in\mathbb{N}\,\colon\,\|x_{j}\|>b\\}\subset A(\delta,{\bf
x})\in\mathcal{I}$, which is absurd. So,
$\mathcal{I}-\limsup_{j}\|x_{j}\|\leq\delta$.

(2): If $A(2\delta,{\bf x})=\\{j\in\mathbb{N}\,\colon\,\|x_{j}\|\geq
2\delta\\}\not\in\mathcal{I}$, we have
$\\{j\in\mathbb{N}\,\colon\,\|x_{j}\|>\delta^{\prime}\\}\not\in\mathcal{I}$
for all $\delta^{\prime}<2\delta$. Hence,
$\delta^{\prime}\leq\mathcal{I}-\limsup_{j}\|x_{j}\|$. Since
$\delta^{\prime}<2\delta$ were arbitrary, we conclude that
$2\delta\leq\mathcal{I}-\limsup_{j}\|x_{j}\|$, a contradiction. ∎

Now we continue with the proof of the theorem. If ${\bf
x}=(x_{(n,m)})\in\ell_{\infty}(\mathbb{N}\times\mathbb{N})(X)$, we set ${\bf
x}_{(n)}=(x_{(n,m)})_{m\in\mathbb{N}}$. Let

$\displaystyle\Psi\colon c_{0,\mathcal{I}\times\mathcal{J}}(X)$
$\displaystyle\to c_{0,\mathcal{I}}(\ell_{\infty}(X)/c_{0,\mathcal{J}}(X))$
$\displaystyle{\bf x}=(x_{(n,m)})$ $\displaystyle\mapsto({\bf
x}_{(n)}+c_{0,\mathcal{J}}(X))_{n\in\mathbb{N}}.$

From Claim 5.9 and Proposition 5.7 we have

$\displaystyle{\bf x}=(x_{(n,m)})\in c_{0,\mathcal{I}\times\mathcal{J}}(X)$
$\displaystyle\Longleftrightarrow(\forall\,\varepsilon>0)\,(\\{n\in\mathbb{N}\,\colon\,\\{m\in\mathbb{N}\,\colon\,(n,m)\in
A(\varepsilon,{\bf x})\\}\not\in\mathcal{J}\\}\in\mathcal{I})$
$\displaystyle\Longleftrightarrow(\forall\,\varepsilon>0)\,(\\{n\in\mathbb{N}\,\colon\,\\{m\in\mathbb{N}\,\colon\,\|x_{(n,m)}\|\geq\varepsilon\\}\not\in\mathcal{J}\\}\in\mathcal{I})$
$\displaystyle\Longleftrightarrow(\forall\,\varepsilon>0)\,(\\{n\in\mathbb{N}\,\colon\,\mathcal{J}-\limsup\|{\bf
x}_{(n)}\|\geq\varepsilon\\}\in\mathcal{I})$
$\displaystyle\Longleftrightarrow(\forall\,\varepsilon>0)\,(\\{n\in\mathbb{N}\,\colon\,\|{\bf
x}_{(n)}+c_{0,\mathcal{J}}(X)\|\geq\varepsilon\\}\in\mathcal{I})$
$\displaystyle\Longleftrightarrow({\bf
x}_{(n)}+c_{0,\mathcal{J}}(X))_{n\in\mathbb{N}}\in
c_{0,\mathcal{I}}(\ell_{\infty}(X)/c_{0,\mathcal{J}}(X)).$

So, $\Psi$ is well defined. Now, if $({\bf
y}^{n}+c_{0,\mathcal{J}}(X))_{n\in\mathbb{N}}\in
c_{0,\mathcal{I}}(\ell_{\infty}(X)/c_{0,\mathcal{J}}(X))$ is given, we set
$x_{(n,m)}=y_{m}^{n}$ for $n,m\in\mathbb{N}$. It is not difficult to check
that ${\bf x}=(x_{(n,m)})\in c_{0,\mathcal{I}\times\mathcal{J}}(X)$ and
$\Psi({\bf x})=({\bf y}^{n}+c_{0,\mathcal{J}}(X))_{n\in\mathbb{N}}$. Now,

$\displaystyle\|\Psi({\bf
x})\|=\sup_{n\in\mathbb{N}}\left(\mathcal{J}-\limsup\|{\bf
x}_{(n)}\|\right)\leq\sup_{n\in\mathbb{N}}\sup_{m\in\mathbb{N}}\|x_{(n,m)}\|=\|{\bf
x}\|_{\infty}.$

Also we have

$\displaystyle\Psi({\bf x})={\bf 0}$
$\displaystyle\Longleftrightarrow(\forall\,n\in\mathbb{N})\,(\,{\bf
x}_{(n)}\in c_{0,\mathcal{J}}(X)\,)$
$\displaystyle\Longleftrightarrow(\forall\,\varepsilon>0)\,(\forall\,n\in\mathbb{N})\,(\,A(\varepsilon,{\bf
x}_{(n)})\in{\mathcal{J}}\,)$
$\displaystyle\Longleftrightarrow(\forall\,\varepsilon>0)\,(\,A(\varepsilon,{\bf
x})\in{\mathcal{J}}^{\omega}\,)$ $\displaystyle\Longleftrightarrow{\bf x}\in
c_{0,{\mathcal{J}}^{\omega}}(X).\qed$

To state the next result, we recall that if $X$ and $Y$ are Banach spaces,
$X\stackrel{{\scriptstyle\vee}}{{\otimes}}Y$ denotes the injective tensor
product of $X$ and $Y$ (see [7, Chapter 1]).

######  Theorem 5.10.

Let $X$ be a Banach space and $\mathcal{I}$ be an ideal on $\mathbb{N}$. Then
$c_{0,\mathcal{I}}\stackrel{{\scriptstyle\vee}}{{\otimes}}X$ is isometric to a
subspace of $c_{0,\mathcal{I}}(X)$.

###### Proof.

If $u=\sum_{j=1}^{m}(x_{n}^{j})_{n}\otimes y_{j}\in c_{0,\mathcal{I}}\otimes
X$, then

$\displaystyle|u|_{\vee}$ $\displaystyle=\sup_{x^{*}\in
B_{c_{0,\mathcal{I}}^{*}},y^{*}\in
B_{X^{*}}}\left|\sum_{j=1}^{m}x^{*}((x_{n}^{j})_{n})y^{*}(y_{j})\right|$
$\displaystyle=\sup_{x^{*}\in B_{c_{0,\mathcal{I}}^{*}},y^{*}\in
B_{X^{*}}}\left|x^{*}\left(\sum_{j=1}^{m}(x_{n}^{j}y^{*}(y_{j}))_{n}\right)\right|$
$\displaystyle=\sup_{y^{*}\in
B_{X^{*}}}\left\|\sum_{j=1}^{m}(x_{n}^{j}y^{*}(y_{j}))_{n}\right\|$
$\displaystyle=\sup_{y^{*}\in
B_{X^{*}}}\left\|\sum_{j=1}^{m}(y^{*}(x_{n}^{j}y_{j}))_{n}\right\|$
$\displaystyle=\sup_{y^{*}\in
B_{X^{*}}}\left\|\left(y^{*}\left(\sum_{j=1}^{m}x_{n}^{j}y_{j}\right)\right)_{n}\right\|$
$\displaystyle=\left\|\left(\sum_{j=1}^{m}x_{n}^{j}y_{j}\right)_{n}\right\|.$

So the map

$\displaystyle\sum_{j=1}^{m}(x_{n}^{j})_{n}\otimes y_{j}\in
c_{0,\mathcal{I}}\otimes
X\mapsto\left(\sum_{j=1}^{m}x_{n}^{j}y_{j}\right)_{n}\in c_{0,\mathcal{I}}(X)$

extends to an isometry from
$c_{0,\mathcal{I}}\stackrel{{\scriptstyle\vee}}{{\otimes}}X$ to
$c_{0,\mathcal{I}}(X)$. ∎

######  Remark 5.11.

The above isometry is not onto in general. Indeed, if
$\mathcal{I}=\mathcal{P}(\mathbb{N})$, then $c_{0,\mathcal{I}}$ is
$\ell_{\infty}$ and it is known that the existence of an isometry from
$\ell_{\infty}\stackrel{{\scriptstyle\vee}}{{\otimes}}X$ onto
$\ell_{\infty}(X)$ implies that $X$ contains a complemented copy of $c_{0}$
(see [20]). However, when $\mathcal{I}={\sf Fin}$, the above isometry is onto
as it is established in [7, Theorem 1.1.11].

######  Question 5.12.

For which ideals $\mathcal{I}$, is
$c_{0,\mathcal{I}}\stackrel{{\scriptstyle\vee}}{{\otimes}}X$ isometric to
$c_{0,\mathcal{I}}(X)$?

###  5.1. Some examples

In [12] was studied the smallest collection $\mathcal{B}$ of ideals on
$\mathbb{N}$ containing the ideal of finite sets and closed under countable
direct sums and the operation of taking orthogonal. They defined by recursion
a sequence of ideals $P_{\alpha}$ and $Q_{\alpha}$ for $\alpha<\omega_{1}$.
For every limit ordinal $\alpha<\omega_{1}$ we fix an increasing sequence
$(\upsilon_{n}^{\alpha})_{n}$ of ordinals such that
$\sup_{n}(\upsilon_{n}^{\alpha})=\alpha$.

  * (i)

$P_{0}=\mathcal{P}(\mathbb{N})$ and $Q_{0}=(P_{0})^{\perp}={\sf Fin}$.

  * (ii)

$P_{\alpha+1}=(P_{\alpha}^{\perp})^{\omega}$.

  * (iii)

$P_{\alpha}=\oplus_{n}(P_{\upsilon_{n}^{\alpha}})^{\perp}$, for
$\alpha<\omega_{1}$ a limit ordinal.

  * (iv)

$Q_{\alpha}=(P_{\alpha})^{\perp}$ for every $\alpha<\omega_{1}$.

Then $\mathcal{B}=\\{P_{\alpha},Q_{\alpha}:\alpha<\omega_{1}\\}$ has exactly
$\aleph_{1}$ non isomorphic ideals.

For example, $P_{1}={\sf Fin}^{\omega}$ is the sum countably many times the
ideal ${\sf Fin}$ of all finite subsets of $\mathbb{N}$, it is a well known
ideal sometimes denoted by $\\{\emptyset\\}\times{\sf Fin}$. Its orthogonal
$Q_{1}={\sf Fin}^{\omega\perp}$ is also denoted ${\sf
Fin}\times\\{\emptyset\\}$. All these ideals are Fréchet (that is, they
satisfy that $\mathcal{I}^{\perp\perp}=\mathcal{I}$). In particular,
$(Q_{\alpha})^{\perp}=P_{\alpha}$.

######  Theorem 5.13.

[12]

(i) Every ideal in $\mathcal{B}$ is isomorphic to either $P_{\alpha}$,
$Q_{\alpha}$ or $P_{\alpha}\oplus Q_{\alpha}$ for some $\alpha<\omega_{1}$.

(ii) For every $\alpha<\beta<\omega_{1}$, there are subsets $A,B$ of
$\mathbb{N}$ such that $P_{\alpha}\approx Q_{\beta}\restriction A$ and
$Q_{\alpha}\approx P_{\beta}\restriction B$.

(iii) All ideals in $\mathcal{B}$ are Borel subsets of $2^{\mathbb{N}}$, and
thus they are meager ideals.

Let ${\sf WO}(\mathbb{Q})$ denote the ideal of all well founded subsets of
${\sf WO}(\mathbb{Q})$. For simplicity, we will write ${\sf WO}$ instead of
${\sf WO}(\mathbb{Q})$. Observe that ${\sf WO}^{\perp}$ is the ideal of well
founded subsets of $(\mathbb{Q},<^{*})$ where $<^{*}$ is the reversed order of
$\mathbb{Q}$. In fact, the map $x\mapsto-x$ from $\mathbb{Q}$ onto
$\mathbb{Q}$ is an isomorphism between ${\sf WO}$ and $WO^{\perp}$. In
particular, ${\sf WO}$ is a Fréchet ideal. We recall also that ${\sf WO}$ is a
meager ideal since it is a co-analytic subset of the Cantor cube
$\\{0,1\\}^{\mathbb{Q}}$. From the results in [12] we have

######  Theorem 5.14.

Every ideal $\mathcal{I}$ in $\mathcal{B}$ is isomorphic to a restriction
${\sf WO}\restriction A$ for some $A\subseteq\mathbb{Q}$.

######  Theorem 5.15.

The following hold for every countable ordinal $\alpha$.

  * (i)

$c_{0,P_{\alpha+1}}$ is isometric to $\ell_{\infty}(c_{0,Q_{\alpha}})$.

  * (ii)

$c_{0,Q_{\alpha+1}}$ is isometric to $c_{0}(c_{0,P_{\alpha}})$.

  * (iii)

$c_{0,P_{\alpha}}$ is isometric to
$\ell_{\infty}((c_{0,Q_{\upsilon_{n}^{\alpha}}})_{n})$ for $\alpha$ a limit
ordinal.

  * (iv)

$c_{0,Q_{\alpha}}$ is isometric to
$c_{0}((c_{0,P_{\upsilon_{n}^{\alpha}}})_{n})$ for $\alpha$ a limit ordinal.

###### Proof.

(i) Since $P_{\alpha+1}=(Q_{\alpha})^{\omega}$, the claim follows from Theorem
5.2. (ii) As $Q_{\alpha+1}=(Q_{\alpha})^{\omega\perp}$ and
$(Q_{\alpha})^{\perp}=P_{\alpha}$, the claim follows from Theorem 5.4. (iii)
Let $\alpha$ be a limit ordinal. The result follows from Theorem 5.2 as
$P_{\alpha}=\oplus_{n}(P_{\upsilon_{n}^{\alpha}})^{\perp}=\oplus_{n}Q_{\upsilon_{n}^{\alpha}}$.
(iv) Since
$Q_{\alpha}=(P_{\alpha})^{\perp}=(\oplus_{n}(P_{\upsilon_{n}^{\alpha}})^{\perp})^{\perp}$,
the result follows from Theorem 5.4. ∎

From Proposition 3.12 and Theorem 5.14 we immediately get the following.
Observe that $c_{0,{\sf WO}}$ is not isomorphic to $\ell_{\infty}$ by Theorem
2.6.

######  Theorem 5.16.

For every $\alpha<\omega_{1}$, $c_{0,P_{\alpha}}$ and $c_{0,Q_{\alpha}}$ are
isometric to a closed subspace of $c_{0,{\sf WO}}$.

Some particular instances of Theorem 5.15 are the following:

  * $c_{0,{\sf Fin}}=c_{0}$.

  * $c_{0,{\sf Fin}^{\perp}}=\ell_{\infty}$.

  * $c_{0,{\sf Fin}^{\perp\omega}}$ is isometric to $\ell_{\infty}(\ell_{\infty})=\ell_{\infty}$.

  * $c_{0,{\sf Fin}^{\omega}}$ is isometric to $\ell_{\infty}(c_{0})$.

  * $c_{0,{\sf Fin}^{\omega\perp}}$ is isometric to $c_{0}(\ell_{\infty})$.

  * $c_{0,{\sf Fin}^{\omega\perp\omega}}$ is isometric to $\ell_{\infty}(c_{0}(\ell_{\infty}))$.

  * $c_{0,{\sf Fin}^{\omega\perp\omega\perp}}$ is isometric to $c_{0}(\ell_{\infty}(c_{0}))$.

Cembranos and Mendoza [5] showed that $\ell_{\infty}(c_{0})$ and
$c_{0}(\ell_{\infty})$ are not isomorphic. Since ${\sf Fin}^{\omega}$ and
${\sf Fin}^{\omega\perp}$ are not isomorphic 111A quick way to see this is
noticing that ${\sf Fin}^{\omega\perp}$ is $F_{\sigma}$ and ${\sf
Fin}^{\omega}$ is not., from Theorem 3.3, we obtain the following weak version
of their result.

######  Theorem 5.17.

[5] $\ell_{\infty}(c_{0})$ is not isometric to $c_{0}(\ell_{\infty})$.

Concerning to the Banach space $c_{0,{\sf Fin\times Fin}}$, Theorem 5.8
implies that $c_{0}(\ell_{\infty}/c_{0})$ is isomorphic to
$c_{0,\sf{FIN}\times\sf{FIN}}/c_{0,\sf{FIN}^{\omega}}$. On the other hand, the
following result gives a quotient representation of $c_{0,{\sf Fin\times
FIN}}$.

######  Theorem 5.18.

$c_{0,{\sf Fin}\times{\sf Fin}}$ is isomorphic to $\ell_{\infty}(c_{0})\times
c_{0}(\ell_{\infty})/K$, where $K$ is isomorphic to $c_{0}$.

###### Proof.

Consider the maps $\Psi_{1}\colon c_{0,{\sf
Fin^{\omega}}}\to\ell_{\infty}((c_{0}(K_{n}))_{n\in\mathbb{N}})$ and
$\Psi_{2}\colon c_{0,{\sf Fin^{\omega\perp}}}\to
c_{0}((\ell_{\infty}(K_{n}))_{n\in\mathbb{N}})$ defined in Theorems 5.2 and
5.4, respectively. Let $S\colon c_{0,{\sf Fin^{\omega}}}\times c_{0,{\sf
Fin^{\omega\perp}}}\to c_{0,{\sf Fin^{\omega}}}+c_{0,{\sf Fin^{\omega\perp}}}$
be defined by $S(a,b)=a+b$, for $a\in c_{0,{\sf Fin^{\omega}}}$ and $b\in
c_{0,{\sf Fin^{\omega\perp}}}$. If $\Psi_{1}^{-1}\times\Psi_{2}^{-1}$ is the
natural map from $\ell_{\infty}((c_{0}(K_{n}))_{n\in\mathbb{N}})\times
c_{0}((\ell_{\infty}(K_{n}))_{n\in\mathbb{N}})$ onto $c_{0,{\sf
Fin^{\omega}}}\times c_{0,{\sf Fin^{\omega\perp}}}$, then

$\displaystyle
S\circ(\Psi_{1}^{-1}\times\Psi_{2}^{-1})\colon(x,y)\in\ell_{\infty}((c_{0}(K_{n}))_{n\in\mathbb{N}})\times
c_{0}((\ell_{\infty}(K_{n}))_{n\in\mathbb{N}})\to\Phi_{1}^{-1}(x)+\Phi_{2}^{-1}(y)\in
c_{0,{\sf Fin^{\omega}}}+c_{0,{\sf Fin^{\omega\perp}}}$

is linear, continuous and onto. By definition of ${\sf Fin}\times{\sf Fin}$ we
have

${\sf Fin}\times{\sf Fin}=\\{A\cup B:\;A\in{\sf Fin}^{\omega},\;B\in{\sf
Fin}^{\omega\perp}\\}.$

So, $c_{0,{\sf Fin\times FIN}}=c_{0,{\sf Fin}^{\omega}}+c_{0,{\sf
Fin}^{\omega\perp}}$ by Theorem 5.1. From the fact
$\ell_{\infty}((c_{0}(K_{n}))_{n\in\mathbb{N}})$ and
$c_{0}((\ell_{\infty}(K_{n}))_{n\in\mathbb{N}})$ are isometric to
$\ell_{\infty}(c_{0})$ and $c_{0}(\ell_{\infty})$, respectively, we get the
first part of the result. Finally, note that $K=\ker
S\circ(\Psi_{1}^{-1}\times\Psi_{2}^{-1})$ is isomorphic to $c_{0}(c_{0})$.
Indeed, we have

$\displaystyle K$ $\displaystyle=(\Psi_{1}^{-1}\times\Psi_{2}^{-1})^{-1}(\ker
S)$
$\displaystyle=(\Psi_{1}^{-1}\times\Psi_{2}^{-1})^{-1}(\\{(a,-a)\,\colon\,a\in
c_{0,{\sf Fin^{\omega}}}\cap c_{0,{\sf Fin^{\omega\perp}}}\\})$
$\displaystyle=(\Psi_{1}^{-1}\times\Psi_{2}^{-1})^{-1}(\\{(a,-a)\,\colon\,a\in
c_{0,{\sf Fin}}\\})$ $\displaystyle=\\{(x,-x)\,\colon\,x\in
c_{0}((c_{0}(K_{n}))_{n\in\mathbb{N}})\\}.$

So, the map $x\in c_{0}((c_{0}(K_{n}))_{n\in\mathbb{N}})\mapsto(x,-x)\in K$ is
an isomorphism. ∎

###  5.2. Grothendieck property

Recall that a Banach space is Grothendieck (or has the Grothendieck property)
if every operator from $X$ to $c_{0}$ is weakly compact. It is known that the
Grothendieck property pass to quotients and complemented subspaces [11,
Proposition 3.1.4].

Now we are interested in the Grothendieck property for the Banach spaces
$c_{0,\mathcal{I}}$. In [11, Problem 9] it is asked about the ideals
$\mathcal{I}$ such that $c_{0,\mathcal{I}}$ is Grothendieck. Concerning to
this question, we have the following results.

######  Theorem 5.19.

For each ordinal $1\leq\alpha<\omega_{1}$, $c_{0,P_{\alpha}}$ and
$c_{0,Q_{\alpha}}$ are not Grothendieck spaces.

###### Proof.

We proceed by induction on $\alpha$. If $\alpha=1$, $c_{0,P_{1}}$ and
$c_{0,Q_{1}}$ are isometric to $\ell_{\infty}(c_{0})$ and
$c_{0}(\ell_{\infty})$, respectively. Note that $c_{0,P_{1}}$ is not
Grothendieck because $c_{0}$ is a quotient of it. On the other hand,
$c_{0,Q_{1}}$ is not Grothendieck by [11, Proposition 5.3.8] since
$c_{0}(\ell_{\infty})$ is isometric to
$c_{0}\stackrel{{\scriptstyle\vee}}{{\otimes}}\ell_{\infty}$ [7, Theorem
1.1.11].

Now, let $\beta<\omega_{1}$ and suppose that $c_{0,P_{\beta}}$ and
$c_{0,Q_{\beta}}$ are not Grothendieck. By Theorem 5.15, $c_{0,P_{\beta+1}}$
and $c_{0,Q_{\beta+1}}$ are isometric to $\ell_{\infty}(c_{0,Q_{\beta}})$ and
$c_{0}(c_{0,P_{\beta}})$, respectively. Since $c_{0,Q_{\beta}}$ and
$c_{0,P_{\beta}}$ are quotient of $\ell_{\infty}(c_{0,Q_{\beta}})$ and
$c_{0}(c_{0,P_{\beta}})$, respectively, we conclude that $c_{0,P_{\beta+1}}$
and $c_{0,Q_{\beta+1}}$ do not have the Grothendieck property. Now, let
$\alpha$ be a limit ordinal and assume that $c_{0,P_{\beta}}$ and
$c_{0,Q_{\beta}}$ do not have the Grothendieck property for all
$\beta<\alpha$. Fix an increasing sequence $(\upsilon_{n}^{\alpha})_{n}$ of
ordinals such that $\sup_{n}(\upsilon_{n}^{\alpha})=\alpha$. By Theorem 5.15,
$c_{0,P_{\alpha}}$ and $c_{0,Q_{\alpha}}$ are isometric to
$\ell_{\infty}((c_{0,Q_{\upsilon_{n}^{\alpha}}})_{n})$ and
$c_{0}((c_{0,P_{\upsilon_{n}^{\alpha}}})_{n})$, respectively. Since
$c_{0,Q_{\upsilon_{n}^{\alpha}}}$ and $c_{0,P_{\upsilon_{n}^{\alpha}}}$ do not
have the Grothendieck property for any $n\in\mathbb{N}$, we infer that
$\ell_{\infty}((c_{0,Q_{\upsilon_{n}^{\alpha}}})_{n})$ and
$c_{0}((c_{0,P_{\upsilon_{n}^{\alpha}}})_{n})$ are not Grothendieck spaces. ∎

######  Theorem 5.20.

Let $\mathcal{I}$ be an ideal on $\mathbb{N}$. If $\mathcal{I}$ is a meager,
then $c_{0,{\sf Fin}\times\mathcal{I}}$ is not a Grothendieck space.

###### Proof.

Suppose that $c_{0,{\sf Fin}\times\mathcal{I}}$ is Grothendieck. By Theorem
5.8, $c_{0}(\ell_{\infty}/c_{0,\mathcal{I}})$ is Grothendieck. By [11,
Proposition 5.3.8], $\ell_{\infty}/c_{0,\mathcal{I}}$ must by finite-
dimensional. Consequently, $c_{0,\mathcal{I}}$ is complemented in
$\ell_{\infty}$. By Theorem 2.6, $\mathcal{I}$ is a non-meager ideal. ∎

It is known that $c_{0,\mathcal{I}}$ is complemented if $\mathcal{I}$ is a
maximal ideal ([19]). A more general fact is the following result for which we
include a proof for the sake of completeness.

######  Proposition 5.21.

[16, p. 2] If $\mathcal{I}$ is an intersection of finitely many maximal ideals
on $\mathbb{N}$, then $c_{0,\mathcal{I}}$ is complemented in $\ell_{\infty}$
and, in particular, it is Grothendieck.

###### Proof.

Let $\mathcal{J}_{1},\ldots,\mathcal{J}_{n}$ be maximal ideals on $\mathbb{N}$
such that $\mathcal{I}=\bigcap_{k=1}^{n}\mathcal{J}_{k}$. Consider the map
$\Phi\colon\ell_{\infty}\to\mathbb{R}^{n}$ defined by

$\displaystyle\Phi({\bf x})=(\mathcal{J}_{1}^{*}-\lim
x_{n},\ldots,\mathcal{J}_{n}^{*}-\lim x_{n}),\quad{\bf x}=(x_{n}).$

It is not difficult to check that $\Phi$ is an onto bounded linear operator.
Note that $\ker\Phi=c_{0,\mathcal{I}}$. So, $\ell_{\infty}/c_{0,\mathcal{I}}$
is finite dimensional. Hence, $c_{0,\mathcal{I}}$ is complemented in
$\ell_{\infty}$. ∎

######  Theorem 5.22.

If $\mathcal{I}$ is an intersection of finitely many maximal ideals on
$\mathbb{N}$, then $c_{0,\mathcal{I}^{\omega}}$ is complemented in
$\ell_{\infty}$ and, in particular, is Grothendieck.

###### Proof.

From the hypothesis, we know that $c_{0,\mathcal{I}}$ is complemented in
$\ell_{\infty}$ (Proposition 5.21). Let $Y\subset\ell_{\infty}$ be a subspace
such that $\ell_{\infty}=c_{0,\mathcal{I}}\oplus Y$. So,

$\displaystyle\ell_{\infty}\sim\ell_{\infty}(\ell_{\infty})\sim\ell_{\infty}(c_{0,\mathcal{I}})\oplus_{\infty}\ell_{\infty}(Y)\sim
c_{0,\mathcal{I}^{\omega}}\oplus_{\infty}\ell_{\infty}(Y),$

by Theorem 5.2. Whence $c_{0,\mathcal{I}^{\omega}}$ is isomorphic to a
complemented subspace of $\ell_{\infty}$. ∎

######  Question 5.23.

Is $c_{0,{\sf WO}}$ a Grothendieck space?

## Acknowledgments

Part of the research of this paper was developed during a postdoctoral stay of
the first author supported by Fundação de Apoio à Pesquisa do Estado de São
Paulo, FAPESP, Processo 2021/01144-2, and UIS.

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