You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: |
    It is shown that a set $\mathcal{J}\left(  \mu\right)  $ of Banach lattices of real-valued measurable functions, defined on a measure space $\left(
    \Omega,\Sigma,\mu\right)  $, may be equipped with a some natural ordering under which it becomes a distributive lattice, which is Dedekind complete provided $\mu$ is a probability measure. Moreover, some natural operations on considered spaces are in Galois connexion. These results are of most interest for symmetric Banach spaces.
address: 'B.E. Ukrecolan, 33-81, Iskrinskaya str., 61005, Kharkiv-5, Ukraine'
author:
- Eugene Tokarev
title: On Algebraic Properties of Sets of Banach Ideal Function Spaces
---

[^1]

Introduction
============

This paper is devoted to study algebraic properties of a set $\mathcal{J}\left(  \mu\right)  $ of Banach ideal spaces of real valued $\mu$-measurable functions. Namely, it will be shown that a quite natural ordering ”$\subset^{1}$” on $\mathcal{J}\left(  \mu\right)  $ makes this set to be a lattice; some restrictions on spaces from $\mathcal{J}\left(  \mu\right)  $ mark out sublattices of $\left\langle \mathcal{J}\left(  \mu\right)
,\subset^{1}\right\rangle $ having nice algebraic properties. Compositions of some natural operations on Banach ideal spaces (such as either the operation of conversion of a given space $E$ into its dual $E^{\prime}$ or the operation to pick out all elements of $E$ having an absolutely continuous norm to generate a new space $E_{0}$) may be chosen in a such way that they will be in the Galois connexion.

Section 2 is devoted to recall some definitions and notations that touch on Banach ideal spaces. The commonly used terminology is widely changed from one paper to another. Below mainly will be used the terminology of reviews \[1\] and \[2\]. For all results that are mentioned below without proofs the reader refers to these reviews.

Original results are contained in sections 3, 4 and 5..

Definitions
===========

Let $\left(  \Omega,\Sigma,\mu\right)  $ be a measure space, i.e., an abstract set $\Omega$ with a $\sigma$-algebra $\Sigma$ of its subsets and a countably additive function (measure) $\mu$, defined on $\Sigma$ with the range in $\mathbb{R}_{+}$.

Let $L_{0}\left(  \mu\right)  =L_{0}\left(  \Omega,\Sigma,\mu\right)  $ be the set of (classes of) $\mu$-measurable real valued functions, defined on $\Omega$. Certainly, $L_{0}\left(  \mu\right)  $ is a vector space under usual operations of addition of functions and multiplication by a scalar. $L_{0}\left(  \mu\right)  $ is also a lattice under a natural partial order ($x\left(  \omega\right)  \leq y\left(  \omega\right)  $ means that $x\left(
\omega\right)  \leq y\left(  \omega\right)  $ a.e.).

An ideal Banach function space (shortly: BIS) $E\left(  \mu\right)  $ is a vector subspace of $L_{0}\left(  \mu\right)  $, which is equipped with a *Banach norm* $\left\|  \cdot\right\|  _{E}$ (i.e. $E\left(
\mu\right)  $ is complete in the norm topology), which is* monotone*, i.e., such that

$y\left(  \omega\right)  \in E\left(  \mu\right)  ,$ $x\left(  \omega\right)
\in L_{0}\left(  \mu\right)  $ *and* $\left|  x\left(  \omega\right)
\right|  \leq\left|  y\left(  \omega\right)  \right|  $* implies that* $\left\|  x\right\|  _{E}\leq\left\|  y\right\|  _{E}$.

Classical examples of BIS are Lebesgue-Riesz spaces $L_{p}(\mu)$ ($1\leq
p\leq\infty$). However there are examples of such measure space $\left(
\Omega,\Sigma,\mu\right)  $ (which does not have the direct sum property; definitions see below) that the space $L_{p}(\mu)$ is not complete (i.e., is not a Banach space). Certainly, it may be completed (by the usual procedure of completition) and the resulting Banach space will be of kind $L_{p}(\mu^{\prime})$ as well. However the measure space $\left(  \Omega^{\prime
},\Sigma^{\prime},\mu^{\prime}\right)  $, where the complete space $\overline{L_{p}(\mu)}=L_{p}(\mu^{\prime})$ will be defined differs from the initial one.

So, it is necessary to put some restrictions on the measure space $\left(
\Omega,\Sigma,\mu\right)  $.

A measure space $\left(  \Omega,\Sigma,\mu\right)  $ is said to be admissible if is satisfies the following conditions:

-   If $A\in\Sigma$; $\mu\left(  A\right)  =0$ and $B\subset A$ then $B\in\Sigma$ and $\mu\left(  B\right)  =0$ (the measure $\mu$ is complete).

-   If $A\subset\Omega$ and every $B\in\Sigma$ with $\mu(B)<\infty$ is so that $A\cap B\in\Sigma$, then $A\in\Sigma$.

-   If $A\in\Sigma$ and $\mu\left(  A\right)  =\infty$ then there exists $B\subset A$, $B\in\Sigma$ such that $\mu(B)<\infty$ (the measure $\mu$ is semifinite).

-   There exists a set $\{A_{i}:i\in I\}$ of pairwice disjoint subsets of $\Omega$ with $\mu\left(  A_{i}\right)  <\infty$ for every $i\in I$ so that

    -   Every $B\in\Sigma$ of finite measure $\mu\left(  B\right)  <\infty$ may be represented as $$B=\cup\{B\cap A_{i}:i\in I_{0}\}\cup A_{0},$$ where $I_{0}$ is a countable subset of $I$ and $\mu\left(  A_{0}\right)  =0$ (the measure $\mu$ has the direct sum property).

It is known (see e.g. \[1\] and \[2\]) that for every admissible measure space $\left(  \Omega,\Sigma,\mu\right)  $ each BIS $E\left(  \mu\right)  $ is conditionally Dedekind complete, and a set of all integral functionals on $E\left(  \mu\right)  $ is total over it.

For any $A\in\Sigma$ the triple $\left(  A,\Sigma_{A},\mu_{A}\right)  $, where $\Sigma_{A}=\{B\cap A:B\in\Sigma\}$ and $\mu_{A}\left(  B\right)  =\mu\left(
B\cap A\right)  $ is a restriction of $\mu$ is an admissible set provided $\left(  \Omega,\Sigma,\mu\right)  $ is admissible.

It will be said that BIS $E\left(  \mu\right)  $ is of *maximal width in* $\left(  \Omega,\Sigma,\mu\right)  $ if $$\{z\in S\left(  \mu\right)  :zy=0\text{ \ for all }y\in E\left(  \mu\right)
\}=0.$$

Let $\left(  \Omega,\Sigma,\mu\right)  $ be an admissible measure space. A set $\mathcal{J}\left(  \mu\right)  $ is the set of all BIS $E\left(  \mu\right)
$ that are of maximal width in $\left(  \Omega,\Sigma,\mu\right)  $.

So, $E\left(  \mu_{A}\right)  \in\mathcal{J}\left(  \mu\right)  $ if and only if $\mu\left(  \Omega\backslash A\right)  =0$.

Let $I=\left\langle I,\ll\right\rangle $ be a partially ordered set. It will be said that $I$ is a *directed set* if for any $i_{1}$, $i_{2}\in I$ there exists $i\in I$ such that $i_{1}\ll i$ and $i_{2}\ll i$.

A sequence $\{x_{i}:i\in I\}$, which is indexed by elements of the directed set $I=\left\langle I,\ll\right\rangle $ will be called a* net*. It will be written $x_{i}\downarrow$ if $i\ll j$ implies $x_{i}\geq x_{j}$. If $x_{i}\downarrow$ and $\inf_{i\in I}\left(  x_{i}\right)  =x_{0}$, we shall write $x_{i}\downarrow x_{0}$.

Let $E=E\left(  \mu\right)  $ be a BIS; $x\in E$. It will be said that the norm of $x$ is *order continuous* (shortly: $\left(  o\right)
$-*continuous*) provided the condition $\left|  x\right|  \geq
x_{i}\downarrow0$ implies that $\lim_{I}\left\|  x_{i}\right\|  _{E}=0$.

The set of all elements of $E$ having the $\left(  o\right)  $-continuous norm is denoted by $E_{0}$.

Certainly, $E_{0}$ is a closed Banach subspace of $E$ and is an *ideal* in $E$:

*if* $x\in E_{0}$*;* $y\in E$* and* $\left|  y\right|
\leq\left|  x\right|  $* then* $y\in E_{0}$.

Notice that $E_{0}$ need not to be of maximal width in $\left(  \Omega
,\Sigma,\mu\right)  $; moreover it may be trivial. E.g., $\left(  L_{\infty
}\left[  0,1\right]  \right)  _{0}=\{0\}$.

Recall that a subset $F$ of a BIS $E$ is said to be a *foundation in* $E$ if it is an ideal in $E$ and is of maximal width in $\left(
\Omega,\Sigma,\mu\right)  $).

Let $E\left(  \mu\right)  \in\mathcal{J}\left(  \mu\right)  $. A *dual* space $E^{\prime}$ is the space of all elements $f(t)\in L_{0}(\mu)$ such that $$\left\|  f\right\|  _{E^{\prime}}=\sup\{\int\nolimits_{\Omega}f\left(
t\right)  x\left(  t\right)  d\mu:\left\|  x\right\|  _{E}=1\}<\infty.$$

$E^{\prime}$ may be identified with a subset of the conjugate space $E^{\ast}
$: every element $f\in E^{\prime}$ generates the (*integral*) functional $f\in E^{\ast}$ by the rule: $$\left\langle f,x\right\rangle =\int\nolimits_{\Omega}f\left(  t\right)
x\left(  t\right)  d\mu.$$

Certainly, $E^{\prime}$ is a Banach space under the norm $\left\|
\cdot\right\|  _{E^{\prime}}$ and is a BIS (of maximal width), which belongs to $\mathcal{J}\left(  \mu\right)  $.

There may be situations when $E_{0}$ is a foundation in $E$ but $\left(
E^{\prime}\right)  _{0}$ is not a foundation in $E^{\prime}$. E.g., $E=E_{0}=L_{1}\left[  0,1\right]  $; $E^{\prime}=L_{\infty}\left[  0,1\right]
$ and $\left(  E^{\prime}\right)  _{0}=\{0\}$. The paper $[3]$ contains an example of such BIS $E$ that $E_{0}=\left(  E^{\prime}\right)  _{0}=\{0\}$.

Let $\left(  \Omega,\Sigma,\mu\right)  $ be an admissible measure space. A set $\mathcal{J}_{0}\left(  \mu\right)  $ is the set of all Banach ideal spaces $E=E\left(  \mu\right)  $ such that $E_{0}=E_{0}\left(  \mu\right)  $ is a foundation in $E\left(  \mu\right)  $ and $\left(  E^{\prime}\right)  _{0} $ is a foundation in $E^{\prime}$.

Let $\left(  \Omega,\Sigma,\mu\right)  $ be an admissible measure space.

Let $E\subset F$ be BIS. Define an operator $i\left(  E,F\right)
:E\rightarrow F$, which asserts to any $x\in E$ the same function $x\in F$. This operator is called *the inclusion operator*. Its norm (that is the infimum of all possible constants $c(E,F)$) is called *the inclusion constant*.

The relation $E\subset^{1}F$ means that the inclusion constant $c(E,F)=1$.

The class $\mathcal{J}\left(  \mu\right)  $ is partially ordered by the relation $E\subset^{1}F$

$\left\langle \mathcal{J}\left(  \mu\right)  ,\subset^{1}\right\rangle $ is a partially ordered set, in which two BIS $E$ and $F$ are identified if and only if $E\subset^{1}F$ and $F\subset^{1}E$. This means that $E$ and $F$ are identical as sets, as vector lattices, as topological vector spaces and as Banach spaces.

On $\mathcal{J}\left(  \mu\right)  $ the more general relation $E\subset^{c}F
$ may be defined. It means that the inclusion constant $c(E,F)$ is bounded. This relation partially orders the quotient set $\mathcal{J}^{\approx}\left(
\mu\right)  =\mathcal{J}\left(  \mu\right)  /\approx$, where the equivalence relation $E\approx F$ means that $E\subset^{c}F$ and $F\subset^{c}E$.

Below it will be regarded only the partially ordered set $\left\langle
\mathcal{J}\left(  \mu\right)  ,\subset^{1}\right\rangle $. The study of the set $\left\langle \mathcal{J}^{\approx}\left(  \mu\right)  ,\subset
^{c}\right\rangle $ is reserved to readers.

It is worthwhile to note that the definition of relations $\subset^{1}$ and $\subset^{c}$ in a general case has some defects (which are eliminated while symmetric Banach spaces will be considered).

E.g., spaces $E=L_{p}\left[  0,1/2\right]  \oplus_{1}L_{q}\left[
1/2,1\right]  $ and $F=L_{q}\left[  0,1/2\right]  \oplus_{1}L_{p}\left[
1/2,1\right]  $  for $p\neq q$ must be considered as different. Indeed, they are not compatible neither in the sense of $\subset^{1}$ nor in the sense of $\subset^{c}$.

The following summary assembles all results about the order $\subset^{1}$ and operations $E\rightsquigarrow E_{0}$; $E\rightsquigarrow E^{\prime}$ that will be needed later.

Let $\left(  \Omega,\Sigma,\mu\right)  $ be an admissible set; $\mathcal{J}\left(  \mu\right)  $ - the corresponding set of Banach ideal spaces of $\mu
$-measurable functions defined on $\Omega$ that are of maximal width. Let $E$, $F\in\mathcal{J}\left(  \mu\right)  $. Then

1.  $E_{0}\subset^{1}E$;

2.  $E\subset^{1}F$ implies that $E_{0}\subset^{1}F_{0}$;

3.  $\left(  E_{0}\right)  _{0}=E_{0}$;

4.  $E\subset^{1}E^{\prime\prime}$;

5.  $E^{\prime}=E^{\prime\prime\prime}$;

6.  If $E\subset^{1}F$ then $F^{\prime}\subset^{1}E^{\prime}$.

$\left\langle \mathcal{J}\left(  \mu\right)  ,\subset^{1}\right\rangle $ as a lattice
=====================================================================================

Consider an admissible measure space $\left(  \Omega,\Sigma,\mu\right)  $ and the corresponding set $\mathcal{J}\left(  \mu\right)  $ equipped with the partial order $\subset^{1}$.

Let $E$, $F\in\mathcal{J}\left(  \mu\right)  $. According to \[4\] define a pair of BIS: $E\cap F$ and $E+F$.

$E\cap F$ consists of all functions $x$, that are common to $E$ and $F$: $f\in
E\cap F$ and is equipped with the norm $$\left\|  f\right\|  _{E\cap F}=\max\{\left\|  f\right\|  _{E},\left\|
f\right\|  _{F}\}.$$

$E+F$ is formed by functions of kind $f=u+v$, $u\in E$; $v\in F$, such that $$\left\|  f\right\|  _{E+F}=\inf\{\left\|  u\right\|  _{E}+\left\|  v\right\|
_{F}:u+v=f\}<\infty;$$

Define on $\left\langle \mathcal{J}\left(  \mu\right)  ,\subset^{1}\right\rangle $ binary operations, $\vee$ and $\wedge$. Namely, put $$E\vee F:=E+F;\text{ \ \ }E\wedge F:=E\cap F,$$

$\left\langle \mathcal{J}\left(  \mu\right)  ,\vee,\wedge\right\rangle $ is a lattice.

Recall that an algebraic structure $A$, endowed with a pair $\vee,\wedge$ of binary mappings $$\vee:A^{2}\rightarrow A;\text{ \ }\wedge:A^{2}\rightarrow A$$ is a lattice, if for any $a$, $b$, $c\in A$ $$\begin{aligned}
a\vee b &  =b\vee a;\text{ \ }a\wedge b=b\wedge a\\
a\vee\left(  b\vee c\right)   &  =\left(  a\vee b\right)  \vee c;\text{
\ }a\wedge\left(  b\wedge c\right)  =\left(  a\wedge b\right)  \wedge c\\
a\vee\left(  a\wedge b\right)   &  =a\wedge\left(  a\vee b\right)  =a.\end{aligned}$$ It is an easy exercise to verify lattice axioms for $\left\langle
\mathcal{J}\left(  \mu\right)  ,\vee,\wedge\right\rangle $.

In a general case $\left\langle \mathcal{J}\left(  \mu\right)  ,\subset
^{1}\right\rangle $ is not Dedekind complete.

Indeed, consider a BIS $E$ and a sequence $\left\langle E,\left\|
\cdot\right\|  _{n}\right\rangle _{n\in\mathbb{N}}$, where $\left\|
x\right\|  _{n}=n\left\|  x\right\|  $ for all $n\in\mathbb{N}$. Certainly, such a sequence does not have any upper bound.

This difficulty may be overcame if $\mu$ is a probability measure, say, $\mathbb{P}$, which does not contain atoms of positive measure.

Let $\mathcal{J}\left(  \mathbb{P}\right)  $ be a set of all BIS that satisfies the norming condition $\left\|  \chi_{\Omega}\left(  t\right)
\right\|  =1$, where $\chi_{A}\left(  t\right)  $ is the indicator function of $A\in\Sigma$:$\ $$$\chi_{A}\left(  t\right)  =1\text{ \ for \ }t\in A;\text{ \ }\chi_{A}\left(
t\right)  =0\text{ \ for \ }t\notin A$$ and are of maximal width on the probability space. The following theorem is valid.

The lattice $\left\langle \mathcal{J}\left(  \mathbb{P}\right)  ,\vee
,\wedge\right\rangle $ is Dedekind complete.

Indeed, if $E\left(  \mathbb{P}\right)  $ satisfies the norming conditions then $$L_{\infty}\left(  \mathbb{P}\right)  \subset^{1}E\left(  \mathbb{P}\right)
\subset^{1}L_{1}\left(  \mathbb{P}\right)  .$$ The greatest lower bound of a family $\left(  E_{i}\right)  _{i\in I}$ of BIS that satisfy norming conditions is the space $\Cap E_{i}$, which consists of all elements that are common to all $E_{i}$’s. Its norm is given by $$\left\|  x\right\|  _{\Cap E_{i}}=\sup\{\left\|  x\right\|  _{E_{i}}:i\in I\}.$$ Since every set of BIS that satisfy the norming condition is bounded (by $L_{1}\left(  \mathbb{P}\right)  $), the least upper bound of $\left(
E_{i}\right)  _{i\in I}$ may be obtained as the greatest lower bound of a family of all upper bounds of $\left(  E_{i}\right)  _{i\in I}$.

Another way is to define (according to $[4]$) a space $\Cup E_{i}$ as an ideal in $L_{0}\left(  \mathbb{P}\right)  $ that consists of all $x\in L_{0}\left(
\mathbb{P}\right)  $, which has the representation of kind $$x=\sum\nolimits_{i\in I}u_{i}\text{ \ where \ }u_{i}\in E_{i};\text{ \ }\sum\nolimits_{i\in I}\left\|  u_{i}\right\|  _{E_{i}}<\infty.$$ It is equipped with the norm $$\left\|  x\right\|  _{\Cup E_{i}}=\inf\{\sum\nolimits_{i\in I}\left\|
u_{i}\right\|  _{E_{i}}:x=\sum\nolimits_{i\in I}u_{i};\text{ \ }u_{i}\in
E_{i}\}.$$ From results of $[4]$ it follows that $\sup\nolimits_{i\in I}\left(
E_{i}\right)  =\Cup E_{i}$.

Now return to an arbitrary admissible measure space $\left(  \Omega,\Sigma
,\mu\right)  $ and consider the lattice $\left\langle \mathcal{J}\left(
\mu\right)  ,\vee,\wedge\right\rangle $.

Let $E$, $F\in\mathcal{J}\left(  \mu\right)  $; $E\subset^{1}F$. An order interval $[E,F]_{\subset^{1}}$ is given by $$\lbrack E,F]_{\subset^{1}}:=\{H\in\mathcal{J}\left(  \mu\right)  :E\subset
^{1}H\subset^{1}F\}.$$ It will be denoted by $[E,F]$ for simplicity. For any order interval $[E,F]$ there may be defined mappings $\lambda=\lambda_{E,F}$ and $\rho=\rho_{E,F}$ as follows. Put for $H\in\mathcal{J}\left(  \mu\right)  $$$\lambda_{E,F}(H)=(H\cap E)+F;\text{ \ }\rho_{E,F}(H)=(H+F)\cap E.$$

The following result is obvious.

*For any* $W\in\mathcal{J}\left(  \mu\right)  $ $$\lambda_{E,F}(W)\in\lbrack E,F];\text{ \ \ }\rho_{E,F}(W)\in\lbrack E,F];$$

*For any* $H\in\lbrack E,F]$ $$\lambda(H)=H;\text{ \ \ }\rho(H)=H.$$

So $\lambda_{E,F}$ and $\rho_{E,F}$ may be regarded as projections of $\mathcal{J}\left(  \mu\right)  $ on the interval $[E,F]$. It is clear that $\lambda^{2}=\rho\lambda=\lambda$; $\rho^{2}=\lambda\rho=\rho$.

Recall that lattices, which have the property $\lambda=\rho$, are said to be *modular.*

The lattice $\left\langle \mathcal{J}\left(  \mu\right)  ,\vee,\wedge
\right\rangle $ is modular.

From ($H\cap E)\subset^{1}\rho(H)$ and $F\subset^{1}\rho(H)$ it follows that for all $H\in\mathcal{J}\left(  \mu\right)  $ $$(H\cap E)+F\subset^{1}(H+F)\cap E,$$ i.e., $\lambda(H)\subset^{1}\rho(H)$ for all $H\in\mathcal{J}\left(
\mu\right)  $.

Let $x\in(H+F)\cap E$. Recall that $[E,F]$ is an order segment, i.e. $E\subset^{1}F$. Then $x\in H+F$ and, hence, $x=u+v$, where $v\in H$ and $u\in
F$. Its norm $\left\|  x\right\|  _{H+F}=\inf\{\left\|  v\right\|
_{H}+\left\|  u\right\|  _{F}\}$. Hence, the norm $\left\|  x\right\|  ^{1}$ of $x$, which is regarded as an element of $(H+F)\cap E$, is equal to $$\left\|  x\right\|  ^{1}=\max\{\inf\{\left\|  v\right\|  _{H}+\left\|
u\right\|  _{F}:v+u=x\};\text{ \ \ }\left\|  v+u\right\|  _{E}\}.$$ The norm $\left\|  x\right\|  ^{2}$ of $x$, when $x$ is regarded as element of $(H\cap E)+F$, is equal to $$\left\|  x\right\|  ^{2}=\inf\{\left\|  u\right\|  _{F},\max\{\left\|
v\right\|  _{H},\left\|  v\right\|  _{E}\}:u+v=x\}.$$

Clearly, $$\begin{aligned}
\left\|  x\right\|  ^{1} &  =\max\{\inf\{\left\|  v\right\|  _{H}+\left\|
u\right\|  _{F};\text{ \ }\left\|  x\right\|  _{E}\}:v+u=x\}\\
&  \geq\max\{\inf\{\left\|  v\right\|  _{H}+\left\|  u\right\|  _{F};\text{
\ }\left\|  v\right\|  _{E}\}:v+u=x\}\\
&  \geq\inf\{\left\|  u\right\|  _{F},\max\{\left\|  v\right\|  _{H},\text{
\ }\left\|  v\right\|  _{E}\}:u+v=x\}=\left\|  x\right\|  ^{2}.\end{aligned}$$

Hence, $(H+F)\cap E\subset^{1}(H\cap E)+F$ and, consequently, $\lambda
_{E,F}=\rho_{E,F}$ for any interval $[E,F]$.

From the property of $\left\langle \mathcal{J}\left(  \mu\right)  ,\vee
,\wedge\right\rangle $ to be modular follows **the first theorem of uniqueness.**

Let $E$, $F\in\mathcal{J}\left(  \mu\right)  $; $E\subset^{1}F$. If there exists such $G\in\mathcal{J}\left(  \mu\right)  $ that $G\cap E=G\cap F$ and $G+E=G+F$ then $E=F$.

Let $G\cap E=G\cap F=X$ and $G+E=G+F=Y$.

Then $X\subset^{1}E\subset^{1}F\subset^{1}Y$ and $\left(  G\cap F\right)
+E=E\subset^{1}\left(  G+E\right)  \cap F=F$. If $E\neq F$ then $\mathcal{J}\left(  \mu\right)  $ is not modular.

In fact, the more powerful result is true.

Recall that a lattice $A$ is said to be *distributive* if for any $a$, $b$, $c\in A$ the following equalities hold: $$a\vee\left(  b\wedge c\right)  =\left(  a\vee b\right)  \wedge\left(  a\vee
c\right)  ;\text{ \ }a\wedge\left(  b\vee c\right)  =\left(  a\wedge b\right)
\vee\left(  a\wedge c\right)  .$$ It is well known that these equalities are not independent; any of them is a consequence of other one. Both of them are equivalent to the inequality: for any $a$, $b$, $c\in A$ $$\left(  a\vee b\right)  \wedge c\leq a\vee\left(  b\wedge c\right)$$ ($a\leq b$ means that $a\wedge b=a$). From the last inequality it follows that every distributive lattice is modular.

The lattice $\left\langle \mathcal{J}\left(  \mu\right)  ,\vee,\wedge
\right\rangle $ is distributive.

It is sufficient to show that for any $E$, $F$, $G\in\mathcal{J}\left(
\mu\right)  $ the following inequality holds: $$\left(  E+F\right)  \cap G\subset^{1}E+\left(  F\cap G\right)  .$$

Let $w\in\left(  E+F\right)  \cap G$. Its norm is $$\begin{aligned}
\left\|  w\right\|  ^{1} &  =\max\{\inf\{\left\|  u\right\|  _{E}+\left\|
v\right\|  _{F}:u+v=w\},\left\|  w\right\|  _{G}\}\\
&  =\max\{\left\|  w\right\|  _{E+F},\left\|  w\right\|  _{G}\}.\end{aligned}$$

Assume that $\left\|  w\right\|  _{E+F}\leq\left\|  w\right\|  _{G}$. Since $\left\|  w\right\|  _{E+G}\leq\left\|  w\right\|  _{G}$, the norm $\left\|
w\right\|  ^{2}$ of the same element $w\in E+\left(  F\cap G\right)  $ may be estimated as follows: $$\begin{aligned}
\left\|  w\right\|  ^{2} &  =\inf\{\left\|  u\right\|  _{E}+\max\{\left\|
v\right\|  _{F},\left\|  v\right\|  _{G}\}:u+v=w\}\\
&  \leq\inf\{\max\{\left\|  u\right\|  _{E}+\left\|  v\right\|  _{F},\left\|
u\right\|  _{E}+\left\|  v\right\|  _{G}\}:u+v=w\}\\
&  \leq\max\{\left\|  w\right\|  _{E+F},\left\|  w\right\|  _{E+G}\}\leq\left\|  w\right\|  _{G}=\left\|  w\right\|  ^{1}.\end{aligned}$$

If we assume that $\left\|  w\right\|  _{E+F}\geq\left\|  w\right\|  _{G}$, then $\left\|  w\right\|  _{E+F}\geq\left\|  w\right\|  _{E+G}$ as well. Hence, $$\max\{\left\|  w\right\|  _{E+F},\left\|  w\right\|  _{E+G}\}=\left\|
w\right\|  _{E+F}=\left\|  w\right\|  ^{1}.$$ Certainly, this implies the desired inequality.

As a corollary we have **the second theorem of uniqueness**.

Let $E$, $F$, $G\in\mathcal{J}\left(  \mu\right)  $ be such that $E\cap
G=F\cap G$; $E+G=F+G$. Then either $G=E$ or $F=E$ or $G=F$.

Let $E\neq G$; $F\neq G$ and $E+G=F+G$; $E\cap G=F\cap G$. Then from distributivity it follows that $$E=E\cap\left(  F+G\right)  =\left(  E\cap F\right)  +\left(  E\cap G\right)
=E\cap F.$$

Hence $E\subset^{1}F$ and either $E=F$ or $\mathcal{J}\left(  \mu\right)  $ is not modular (cf. theorem 4). Since every distributive lattice is modular, $E=F$.

A pair of BIS $E$ and $F$ is uniquely determined by their sum $E+F$ and intersection $E\cap F$.

According to the M. Stone’s theorem $\left[  5\right]  $ every distributive lattice $A$ is isomorphic (as lattice) to a some ring of sets. Moreover, as this ring of sets it may be chosen the ring of compact open sets of the so-called Stonian space $\frak{S}\left(  A\right)  $ of the lattice $A$ – the topological $T_{0}$-space, which is uniquely (up to a homeomorphism) determined by $A$ and has the following properties:

-   The base of open sets of $\frak{S}\left(  A\right)  $ forms compact open sets;

-   Intersection of two compact open sets is compact;

-   If $K\subset\frak{S}\left(  A\right)  $ is closed then $\cap
    \{U_{i}:i\in I\}\cap K\neq\varnothing$ for an arbitrary set of compact open sets $\{U_{i}:i\in I\}$ ($I\neq\varnothing$) of $\frak{S}\left(  A\right)  $ so that

    -   For any $i$, $j\in I$ there is $l\in I$ such that $U_{l}\subset
        U_{i}\cap U_{j}$;

    -   $U_{i}\cap K\neq\varnothing$ for all $i\in I$.

If, in addition, $A$ has the maximal element, then $\frak{S}\left(  A\right)
$ is compact itself.

E.g., in the aforementioned case $\left\langle \mathcal{J}\left(
\mathbb{P}\right)  ,\vee,\wedge\right\rangle $, the Stonian space $\frak{S}\left(  \mathcal{J}\left(  \mathbb{P}\right)  \right)  $ is compact. Other examples may be given by using Banach symmetric spaces (see the concluding section).

Closure operators and Galois connexions on $\left\langle
\mathcal{J}\left(  \mu\right)  ,\subset^{1}\right\rangle $
==========================================================

Recall some algebraic definitions.

Let $\left\langle L,<\right\rangle $ be a lattice. A mapping $\pi:L\rightarrow
L$ is said to be a closure operator, if for all $a$, $b\in L$

-   $a<b$ implies that $\pi\left(  a\right)  <\pi\left(  b\right)  $;

-   $a<\pi\left(  a\right)  $;

-   $\pi\circ\pi\left(  a\right)  =\pi\left(  a\right)  $.

(Cf. $[6]$). Let $\left\langle L,<\right\rangle $ and $\left\langle L^{\prime
},<^{\prime}\right\rangle $ be lattices; $k:L\rightarrow L^{\prime}$ and $k^{\prime}:L^{\prime}\rightarrow L$ be mappings. The pair $\left(
k,k^{\prime}\right)  $ is said to be the Galois connexion between $L$ and $L^{\prime}$ if

-   $a<b\Rightarrow k(a)<^{\prime}k(b)$ for $a$, $b\in L$;

-   $a^{\prime}<^{\prime}b^{\prime}\Rightarrow k^{\prime}(a^{\prime
    })<k^{\prime}(b^{\prime})$ for $a^{\prime}$, $b^{\prime}\in L^{\prime}$;

-   $k^{\prime}\circ k\left(  a\right)  <a$ for $a\in L$;

-   $k\circ k^{\prime}(a^{\prime})<^{\prime}a^{\prime}$ for $a^{\prime}\in
    L^{\prime}$.

Below it will be needed the following simple result.

Let $E\in\mathcal{J}_{0}\left(  \mu\right)  $. Then

1.  $\left(  E_{0}\right)  ^{\prime\prime}\subset^{1}E^{\prime\prime}$;

2.  $E^{\prime\prime}\subset^{1}\left(  \left(  E^{\prime}\right)
    _{0}\right)  ^{\prime}$;

3.  $\left(  \left(  \left(  E_{0}\right)  ^{\prime}\right)  _{0}\right)
    ^{\prime}\subset^{1}\left(  \left(  E^{\prime}\right)  _{0}\right)  ^{\prime}
    $;

4.  $\left(  E_{0}\right)  ^{\prime\prime}\subset^{1}\left(  \left(  \left(
    E_{0}\right)  ^{\prime}\right)  _{0}\right)  ^{\prime}$.

Since we assume that $E\in\mathcal{J}_{0}\left(  \mu\right)  $, both $E_{0}$ and $\left(  E^{\prime}\right)  _{0}$ are nontrivial. So,

1\. $E_{0}\subset^{1}E\Rightarrow E^{\prime}\subset^{1}\left(  E_{0}\right)
^{\prime}\Rightarrow\left(  E_{0}\right)  ^{\prime\prime}\subset^{1}E^{\prime\prime}$.

2\. $\left(  E^{\prime}\right)  _{0}\subset^{1}E^{\prime}\Rightarrow
E^{\prime\prime}\subset^{1}\left(  \left(  E^{\prime}\right)  _{0}\right)
^{\prime}$.

3\. $E_{0}\subset^{1}E\Rightarrow E^{\prime}\subset^{1}\left(  E_{0}\right)
^{\prime}\Rightarrow\left(  E^{\prime}\right)  _{0}\subset^{1}\left(  \left(
E_{0}\right)  ^{\prime}\right)  _{0}\Rightarrow\left(  \left(  \left(
E_{0}\right)  ^{\prime}\right)  _{0}\right)  ^{\prime}\subset^{1}\left(
\left(  E^{\prime}\right)  _{0}\right)  ^{\prime}$.

4\. $\left(  \left(  E_{0}\right)  ^{\prime}\right)  _{0}\subset^{1}\left(
E_{0}\right)  ^{\prime}\Rightarrow\left(  E_{0}\right)  ^{\prime\prime}\subset^{1}\left(  \left(  \left(  E_{0}\right)  ^{\prime}\right)
_{0}\right)  ^{\prime}$.

Consider the lattice $\mathcal{J}_{0}^{\ast}\left(  \mu\right)  $, equipped with the inverse order $\vartriangleleft$: $E\vartriangleleft F$ is the same as $F\subset^{1}E$ for $E$, $F\in\mathcal{J}_{0}\left(  \mu\right)  $. Put $\mathcal{J}_{0}:=\left\langle \mathcal{J}_{0}\left(  \mu\right)  ,\subset
^{1}\right\rangle $; $\mathcal{J}_{0}^{\ast}:=\left\langle \mathcal{J}_{0}\left(  \mu\right)  ,\vartriangleleft\right\rangle $; $\mathcal{J}:=\left\langle \mathcal{J}\left(  \mu\right)  ,\subset^{1}\right\rangle $

The mapping $\left(  _{0}\right)  :\mathcal{J}_{0}^{\ast}\left(  \mu\right)
\rightarrow\mathcal{J}_{0}^{\ast}\left(  \mu\right)  $ is a closure operator on $\mathcal{J}_{0}^{\ast}$.

The mapping $\left(  ^{\prime\prime}\right)  :\mathcal{J}_{0}\left(
\mu\right)  \rightarrow\mathcal{J}_{0}\left(  \mu\right)  $ is a closure operator on $\mathcal{J}$.

The proof is an obvious consequence of definitions and the summary.

It may be defined the most important sublattices of the $\mathcal{J}\left(
\mu\right)  $

The lattice $\mathcal{J}_{00}\left(  \mu\right)  $ consists of all BIS $E$ with the absolute continuous norm (i.e., such that $E=E_{0}$).

The lattice $\mathcal{J}^{\prime}\left(  \mu\right)  $ contains all BIS $E$ of kind $E=F^{\prime}$ for some BIS $F$.

Lattices $\mathcal{J}_{00}\left(  \mu\right)  $ and $\mathcal{J}^{\prime
}\left(  \mu\right)  $ both are distributive (and, hence, modular). If the measure $\mu$ is a probability measure $\mathbb{P}$ and all BIS $E$ from $\mathcal{J}\left(  \mathbb{P}\right)  $ satisfy the norming condition (i.e. if $\mathcal{J}\left(  \mathbb{P}\right)  $ is Dedekind complete) then sublattices $\mathcal{J}_{00}\left(  \mathbb{P}\right)  $ and $\mathcal{J}^{\prime}\left(  \mathbb{P}\right)  $ both are Dedekind complete too.

As it is well known, the set of fixed points of a closure operator that acting on the distributive Dedekind complete lattice $A$ is a sublattice of $A$ that holds these properties. Clearly, the set of fixed points of $\left(
_{0}\right)  $ is exactly $\mathcal{J}_{00}\left(  \mu\right)  $. To show that $\mathcal{J}^{\prime}\left(  \mu\right)  $ coincides with the set of fixed points of the closure operator $\left(  ^{\prime\prime}\right)  $ it is enough to notice that $E=F^{\prime}$ for some BIS $F$ if and only if $E=E^{\prime
\prime}$. Certainly, if $E=E^{\prime\prime}$ then $E=F^{\prime}$ for $F=E^{\prime}$. Conversely, if $E=F^{\prime}$ then $E^{\prime\prime}=F^{\prime\prime\prime}=\left(  F^{\prime}\right)  ^{\prime\prime}=F^{\prime}$.

Correspondences $E\rightsquigarrow E_{0}$ and $E\rightsquigarrow E^{\prime}$ may be regarded as mappings. It will be written $\left(  _{0}\right)  :$ $E\rightarrow E_{0}$; $\left(  ^{\prime}\right)  :E\rightarrow E^{\prime}$.

Using the mappings $\left(  _{0}\right)  $ and $\left(  ^{\prime}\right)  $ it may be constructed a pair of mappings, say $k$ and $k^{\prime}$ that are in the Galois connexion.

Let $k:\mathcal{J}_{0}\left(  \mu\right)  \rightarrow\mathcal{J}_{0}^{\ast
}\left(  \mu\right)  $ and $k^{\prime}:\mathcal{J}_{0}^{\ast}\left(
\mu\right)  \rightarrow\mathcal{J}_{0}\left(  \mu\right)  $ are given by $$kE=\left(  E_{0}\right)  ^{\prime};\text{ \ }k^{\prime}E=\left(  E^{\prime
}\right)  _{0}.$$

The pair $\left(  k,k^{\prime}\right)  $ is the Galois connexion between $\mathcal{J}_{0}^{\ast}$ and $\mathcal{J}_{0}$.

Let us check up properties from the definition 5.

-   $E\subset^{1}F\Rightarrow E_{0}\subset^{1}F_{0}\Rightarrow\left(
    F_{0}\right)  ^{\prime}\subset^{1}\left(  E_{0}\right)  ^{\prime}\Rightarrow\left(  E_{0}\right)  ^{\prime}\vartriangleleft\left(
    F_{0}\right)  ^{\prime} $.

-   $E\vartriangleleft F\Rightarrow F\subset^{1}E\Rightarrow\left(
    E^{\prime}\right)  _{0}\subset^{1}\left(  F^{\prime}\right)  _{0}$.

-   $((\left(  F^{\prime}\right)  _{0})_{0})^{\prime}=(\left(  F^{\prime
    }\right)  _{0})^{\prime}$. By the theorem 4, $F^{\prime\prime}\subset
    ^{1}(\left(  F^{\prime}\right)  _{0})^{\prime}$ and, hence, $$F\subset^{1}F^{\prime\prime}\subset^{1}(\left(  F^{\prime}\right)
    _{0})^{\prime}.$$

-   $(\left(  \left(  E_{0}\right)  ^{\prime}\right)  ^{\prime})_{0}=\left(
    \left(  E_{0}\right)  ^{\prime\prime}\right)  _{0}\subset^{1}E$. Hence, $E\vartriangleleft\left(  \left(  E_{0}\right)  ^{\prime\prime}\right)  _{0}$.

Compositions $k\circ k^{\prime}$and $k^{\prime}\circ k$ are closure operators on $\mathcal{J}_{0}^{\ast}$ and $\mathcal{J}_{0}$ respectively.

This is an obvious consequence of definitions. 

So, we obtain some more closure operators on $\mathcal{J}_{0}\left(
\mu\right)  $. Notice that $k\circ k^{\prime}:E\rightarrow(\left(  E^{\prime
}\right)  _{0})^{\prime}$ and $k^{\prime}\circ k:E\rightarrow\left(  \left(
E_{0}\right)  ^{\prime\prime}\right)  _{0}$. The second mapping is coincide with the usual $\left(  _{0}\right)  :E\rightarrow E_{0}$. However the first one pick out from $\mathcal{J}^{\prime}\left(  \mu\right)  $ those BIS that are dual to BIS having the absolutely continuous norm.

Symmetric Banach spaces
=======================

Results from previous sections are of the most interest when a special class of BIS - the class of *symmetric Banach spaces* is considered.

Recall the definition.

A Banach ideal space $E$ of (classes of) measurable real functions, which are defined on the admissible measure space $\left(  \Omega,\Sigma,\mu\right)  $ is said to be *symmetric* if for any functions $x=x\left(  t\right)  $ and $y=y\left(  t\right)  $ of $E$ the following condition holds:

-   If $x\in E$ and functions $\left|  y(t)\right|  $ and $\left|
    x(t)\right|  $ are equimeasurable then $y\in E$ and $\left\|  y\right\|
    _{E}=\left\|  x\right\|  _{E}$.

Let $\mathcal{S}\left(  \Omega,\Sigma,\mu\right)  $ be a class of all symmetric Banach spaces (they in the future will be referred to as symmetric spaces). This class contains Lebesgue-Riesz spaces $L_{p}\left(  \mu\right)
$; Orlicz spaces $L_{M}\left(  \mu\right)  $ and so on. Usually properties 1 and 2 are supplemented with the following *norming condition*: $$\left\|  \chi_{e}\right\|  _{E}=1\text{ \ for any set }e\in\Sigma\text{ \ of
the measure }\mu\left(  e\right)  =1.$$

This condition implies that $\mathcal{S}\left(  \Omega,\Sigma,\mu\right)  $ is a Dedekind complete distributive lattice because of the known theorem of inclusion: $$L_{1}\left(  \mu\right)  \cap L_{\infty}\left(  \mu\right)  \subset
^{1}E\left(  \mu\right)  \subset^{1}L_{1}\left(  \mu\right)  +L_{\infty
}\left(  \mu\right)  .$$ For a probability measure $\mathbb{P}$ these inclusions looks like $$L_{\infty}\left(  \mathbb{P}\right)  \subset^{1}E\left(  \mathbb{P}\right)
\subset^{1}L_{1}\left(  \mathbb{P}\right)  .$$

For a purely atomic measure (with mass of every point is equal to 1) we obtain so called *symmetric discrete (or sequence) spaces* defined on an arbitrary set. The only characteristic that distinguishes corresponding classes of discrete spaces is the cardinality of $\Omega$. For $\operatorname{card}\Omega=\varkappa$  the class $\mathcal{S}\left(
\Omega,\Sigma,\mu\right)  $ will be denoted by $\mathcal{S}\left(
\varkappa\right)  $. Inclusions in this case looks like: $$l_{1}\left(  \varkappa\right)  \subset^{1}E\left(  \varkappa\right)
\subset^{1}l_{\infty}\left(  \varkappa\right)  .$$

Notice that in the case of symmetric spaces the finiteness of $\mu$ implies that it is non-atomic.

As it follows from the preceding consideration, all lattices of symmetric Banach spaces may be participate into three parts:

-   $\mathcal{S}^{\left(  1\right)  }$ - lattices of symmetric spaces, defined on a probability (non atomic) space;

-   $\mathcal{S}^{\left(  \infty\right)  }$ - lattices of symmetric spaces, defined on a non atomic space of infinite measure;

-   $\mathcal{S}^{\left(  \mathbf{D}\right)  }$ - lattices of symmetric sequence spaces: $$\mathcal{S}=\mathcal{S}^{\left(  1\right)  }\cup\mathcal{S}^{\left(
    \infty\right)  }\cup\mathcal{S}^{\left(  \mathbf{D}\right)  }.$$

Our nearest aim is to show that all lattices $\left\langle \mathcal{S}\left(
\mu\right)  ,\subset^{1}\right\rangle $ from a given class $\mathcal{S}^{\left(  \mathbf{?}\right)  }$, where $\mathbf{?}\in\{1,\infty,\mathbf{D}\}$ are pairwice lattice isomorphic, i.e. that there are at most three different lattices amongst all of kind $\mathcal{S}\left(  \mu\right)  .$

Lattices $\mathcal{S}\left(  \mu\right)  $ and $\mathcal{S}\left(  \nu\right)
$ that belong to the same class $\mathcal{S}^{\left(  \mathbf{?}\right)  }$, where $\mathbf{?}\in\{1,\infty,\mathbf{D}\}$ are lattice-isomorphic.

The one-to-one correspondence between members of these lattices may be shown by using the operation of replanting of symmetric spaces from one measure to another. Such operation was suggested by A.A. Mekler \[7\].

Namely, let $E\left(  \mu\right)  \in$ $\mathcal{S}\left(  \mu\right)  $. To $x\left(  t\right)  \in E\left(  \mu\right)  $ corresponds its *distribution function* $$n_{x}\left(  s\right)  =\operatorname{mes}\left(  \{t\in\lbrack0,1]:x\left(
t\right)  >s\}\right)$$ and its *non-increasing rearrangement* $$x^{\ast}\left(  t\right)  =\inf\{s\in\lbrack0,\infty):n_{\left|  x\right|
}\left(  s\right)  <t\}.$$

Obviously, $x^{\ast}\left(  t\right)  $ is defined either on $\left[
0,1\right]  $ or on $[0,\infty)$ (both with the Lebesgue measure) or at $\mathbb{N}$ (with mass 1 in every point) and, hence is an element of the corresponding vector lattice $L_{0}\left[  0,1\right]  $ (resp., $L_{0}\left[
0,\infty\right]  $) or $L_{0}\left[  \mathbb{N}\right]  $. Notice that usually $L_{0}\left[  \mathbb{N}\right]  $ is denoted by $s$).

Let, for distinctness, $x^{\ast}\left(  t\right)  \in L_{0}\left[  0,1\right]
$.

It is clear that the set $$\widetilde{E}\left[  0,1\right]  =\{x^{\ast}\left(  \sigma t\right)  :x\left(
t\right)  \in E\left(  \mu\right)  ;\text{ }\sigma\in\operatorname{Aut}\}$$ (where $\operatorname{Aut}$ denotes the set of all preserving measure automorphisms of $\left[  0,1\right]  $) is an ideal in $L_{0}\left[
0,1\right]  $, which, being equipped with the norm $$\left\|  x^{\ast}\left(  t\right)  \right\|  _{\widetilde{E}}:=\left\|
x\right\|  _{E\left(  \mu\right)  }\text{; \ }\left\|  y\left(  t\right)
\right\|  _{\widetilde{E}}:=\left\|  y^{\ast}\left(  t\right)  \right\|
_{\widetilde{E}},$$  becomes a Banach symmetric function space on $\left[  0,1\right]  $. 

Moreover, this space is uniquely determined by $E\left(  \mu\right)  $.

Notice that for any probability measure $\nu$ the space $\widetilde{E}\left[
0,1\right]  $ in the same way uniquely defines the corresponding space $\widetilde{E}\left(  \nu\right)  $.

It will be said that the space $\widetilde{E}\left(  \nu\right)  $ is obtained from $E\left(  \mu\right)  $ by the* replanting Mekler’s procedure:*$\ $$$\operatorname{Mekl}_{\mu,\nu}:\mathcal{S}\left(  \mu\right)  \rightarrow
\mathcal{S}\left(  \nu\right)  ;$$$$E\left(  \mu\right)  \rightarrowtail\widetilde{E}\left(  \nu\right)$$

So, every symmetric space $E=E\left(  \mu\right)  $ generates a tower $$\left\lfloor E\right\rfloor =\{\widetilde{E}\left(  \Omega,\Sigma,\nu\right)
=\operatorname{Mekl}_{\mu,\nu}(E(\mu))\}$$ where $\left(  \Omega,\Sigma,\nu\right)  $ runs all probability spaces.

Obviously, the procedure $\operatorname{Mekl}_{\mu,\nu}$ holds the relation $\subset^{1}$and lattice operations: $$E\subset^{1}F\Rightarrow\operatorname{Mekl}_{\mu,\nu}(E)\subset^{1}\operatorname{Mekl}_{\mu,\nu}(F);$$$$\begin{aligned}
\operatorname{Mekl}_{\mu,\nu}(E\cap F) &  =\operatorname{Mekl}_{\mu,\nu
}(E)\cap\operatorname{Mekl}_{\mu,\nu}(F);\\
\operatorname{Mekl}_{\mu,\nu}(E+F) &  =\operatorname{Mekl}_{\mu,\nu
}(E)+\operatorname{Mekl}_{\mu,\nu}(F)\text{;}$$$$\operatorname{Mekl}_{\mu,\nu}(E_{0})=\left(  \operatorname{Mekl}_{\nu
}(E)\right)  _{0};\text{ \ \ }\operatorname{Mekl}_{\mu,\nu}(E^{\prime
})=\left(  \operatorname{Mekl}_{\mu,\nu}(E)\right)  ^{\prime}$$

and, hence, generates the lattice isomorphism between lattices $\mathcal{S}\left(  \mu\right)  $ and $\mathcal{S}\left(  \nu\right)  $.

Similarly for lattices that are belong to $\mathcal{S}^{\left(  \mathbf{?}\right)  }$, where either $\mathbf{?}=\infty$, or $\mathbf{?}=\mathbf{D}$.

It is not clear: whether are lattice-isomorphic lattices of different types (e.g., from $\mathcal{S}^{\left(  1\right)  }$and $\mathcal{S}^{\left(
\infty\right)  }$). It may be shown that lattices $\mathcal{S}^{\left(
1\right)  }$ and $\mathcal{S}^{\left(  \mathbf{D}\right)  }$ are isomorphic to quotient lattices of $\mathcal{S}^{\left(  \infty\right)  }$

There exists (up to lattice-isomorphism) at most three lattices $\mathcal{S}=\mathcal{S}^{\left(  \mathbf{?}\right)  }$ ($?\in\{1,\infty,\mathbf{D}\}$) of symmetric spaces.

All of them are Dedekind complete, distributive (and, hence, modular).

Operations $kE=\left(  E_{0}\right)  ^{\prime}\ $and  $k^{\prime}E=\left(
E^{\prime}\right)  _{0}$ interrelate the ”fundamental part” $\mathcal{S}_{0}$ (see the definition 3) of each of these lattices with itself in the reverse order (i.e., with $\mathcal{S}_{0}^{\ast}$ by the given notation):

The pair $\left(  k.k^{\prime}\right)  $ is the Galois connexion between $\mathcal{S}_{0}^{\ast}$ and $\mathcal{S}_{0}$

For the class $\mathcal{S}$ of all symmetric spaces, classes $\mathcal{S}_{00}$ and $\mathcal{S}^{\prime}$ (see the definition 6) are just those, whose elements are usually called rearrangement invariant Banach spaces

References
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[^1]: This paper is in final form and no version of it will be submitted for publication elsewhere.