\documentclass{article}

\usepackage{amsmath,amsthm,amscd,amsxtra,amsfonts,amssymb,latexsym}
%\usepackage{amstex,amssymb}

\newcommand{\TeXButton}[2]{#2}
%\newcommand{\cal}[1]{\mathcal #1}
\newcommand{\stackunder}[2]{\underset{#1}{#2}}
\newcommand{\NEG}{\not}
%\newcommand{\QATOP}[2]{#1 \atop #2}
\setcounter{secnumdepth}{3}
\newcommand{\mint}{\hbox{\small\rm min}}
\newcommand{\maxt}{\hbox{\small\rm max}}


\begin{document}

\title{Tree structures, Entropy, and Action Principles for Neighbourhood Topologies}
\author{Hanno Hammer\\
Department of Applied Mathematics and Theoretical Physics, \\
%EndAName
Silver Street, Cambridge, UK}
%\keywords{Tree structures. Entropy. Neighbourhood Topology. Action Principles}
\maketitle

\begin{abstract}
We recast the idea of decision trees as they emerge in Information theory
and Complexity theory into a set theoretical language; the result we call
tree structures over a given set. We identify all main structural elements
of tree structures, the most important of which is the tree function,
defined as a sum over certain quantities at every nod in the tree. We show
in detail that the minimization of the tree function on, possibly
constrained, sets of tree structures over a given set renders the functional
form of entropy, or of Wiener-Shannon information, depending on the context.
 We suggest three natural axioms defining tree
structures, which are valid also when the underlying set is infinite; in
this case the resulting trees are fractal-like objects. These axioms turn out to be related to the neighbourhood axioms
describing neighbourhoods on a topological space. In fact we will show that
the paths in a tree structure, which are totally ordered subsets of a tree,
can be regarded as a countable neighbourhood basis, which in turn defines a
topology on the underlying set.  The tree function then assigns nonnegative numbers to every neighbourhood topology on the
underlying set that arises from a tree structure. On the preferred topology,
this number becomes minimal, in which case it represents an entropy-like
quantity. This has the distinct flavour of an action principle,
distinguishing certain topologies by means of minimization of the tree
function. This result hints at a deep relation between entropy-like
quantities and preferred neighbourhood topologies on sets.
\end{abstract}

\tableofcontents

\section{Introduction}

Decision trees are known in Information Theory \cite{Baierlein} and
Complexity Theory as tools for determining important problems in these
fields. They can be used to approach computational aspects of entropy and
Wiener-Shannon information. The initial motivation for this work was
somewhat related: Consider the functional form $-\sum p_i\ln \left(
p_i\right) $ of the Wiener-Shannon information of a probability distribution 
$\left( p_i\right) $; this functional form is intuitively convincing, but it
is not clear from the outset how this intuition comes about. We ask, is
there an underlying {\bf operational} approach to this form, involving only
natural numbers and a finite number of Yes-No-questions about the
probability distribution? The answer to this question is known in the
literature; however, in tackling the problem on a level somewhat more formal
than usual, we arrive at the concept of ''tree structures'' as developed
here. This work has been inspired by ideas we have found in the book \cite
{Eppler}. After completion, ideas concerning the splitting of sets by
partitions and assignment of information to tree-like objects were found in
the book \cite{Baierlein}, and decision trees together with some of the
features discussed in our work were found in \cite{Farhi}. Also, from the
introduction to the book \cite{Falconer} we have learned that there might be
a connection to the theory of fractal geometry. On the other hand, one of
the main results of this work, namely that the entropy is a quantity related
to preferred neighbourhood topologies on a set, seems to have not yet been
observed in the literature. We now briefly describe our approach to tree
structures:

{\bf What are tree structures:\quad }A tree structure ${\cal B}\left(
X\right) $ over a given set $X$ is a subset of the power set ${\cal P}X$ of $%
X$ that is obtained by a continuous splitting of its elements $b\in {\cal B}%
\left( X\right) $ into smaller and ever smaller subsets; this splitting will
be described in terms of partitions of sets. Tree structures can be defined
over sets of arbitrary cardinality, countable or non-countable. In case that
the underlying set $X$ is infinite, the tree structures over $X$ are
fractal-like objects. There are three natural axioms governing tree
structures, which are independent of whether the set $X$ is countable or
non-countable; in the infinite case, a tree structure build over such a set
will be a fractal in general. The axioms describing tree structures will be
shown to give rise to preferred topologies on a set $X$. For a given set $X$%
, there exist many tree structures over $X$. Some of them have distinct
features, as we will see later.

{\bf How do tree structures arise:\quad }Tree structures first arise in
modelling processes of information gaining. We will see that a tree
structure encodes the operational aspect of the problem of information
gaining, which yields the concept of entropy/information. In such a process
we assign a nonnegative (in this work, a natural) number to the outcome of
an interaction between a unit that seeks to find a distinct but unknown
element $x_0$ of a set $X$, and a unit that possesses this information, but
renders only information about ''neighbourhoods'' of the distinct element,
as these neighbourhoods zoom more and more into $x_0$. These
''neighbourhoods'' will be given a topological meaning at the end.

{\bf What are the typical structural elements:\quad }The elements of ${\cal B%
}\left( X\right) $ will be called the ''nods'' in the tree, and provide one
of the main structural elements; the second main structural element will
turn out to be ''paths'' of nods, which are subsets of the tree structure $%
{\cal B}\left( X\right) $ on which a natural total order is defined. Every
such path describes the ''zooming-in'' or the encircling of one of the
elements $x\in X$, as we move one level further in the tree structure. To
every path in a finite tree structure, a natural number, called the
''amount'' of the path, can be assigned, which expresses how many
Yes-No-questions are necessary to single out the element $x$ in the given
tree ${\cal B}\left( X\right) $. Of central importance will be the sum over
the amounts of all complete paths in the tree (the term complete will be
introduced below); this sum will be called the ''tree function''. Theorem 
\ref{TrF} shows that the tree function is indeed a sum of products of the
so-called ''characters'' on all nods in the tree. Thus we can assign a value
of the tree function to every tree structure ${\cal B}\left( X\right) $ over 
$X$.

{\bf The natural question concerning tree structures:\quad }Assigning a
value of the tree function to every tree over $X$, we will ask, on which
trees the tree function takes its minimum; these trees will be called
''minimal''. It will turn out that the result depends only on the number $%
n=\#X$ of elements of $X$, but not on the set $X$ itself. This question can
be generalized, as constraints on the admissible trees can be imposed. The
admissible trees then preserve a prescribed initial partition of $X$, which
reflects a choice of ''weights'' $\left( w_i\right) $ for the path amounts
in such a tree. This is analogous to choosing a probability distribution $%
\left( p_i\right) $ for the paths in the admissible trees.

{\bf The first main result concerning tree structures:\quad }We will show
that, if there are no constraints, then the minimal value of the tree
function is close to $n\cdot \lg \left( n\right) $, where $\lg \left(
n\right) $ is the integer that comes closest to the logarithm of $n$ with
respect to the basis $2$. Thus, the mean value of the amounts of $n$ paths
in a complete tree over $X$ comes close to $\lg \left( n\right) $, which is
the information gained in finding a distinct element among $n$ ''equally
weighted'' elements; or the entropy of $n$ distinct states, depending on the
context. One of the central results of this work is, that the functional
form $\lg \left( n\right) $ of the entropy so defined is itself the result
of a process of minimization, i.e. there is a more general functional form
underlying, namely the expression $\sum\limits_{b\in {\cal B}\left( X\right)
}n\left( b\right) \cdot \left[ m\left( b\right) -1\right] $ of the tree
function on the trees, where $b$ denotes the nodes in the tree, and $n,m$
are the characters of the nod, see section \ref{TreeFunc}. After solving the
unconstrained problem in detail, we describe the essential steps to extend
the results to the constrained case; here the elements $x\in X$ are endowed
with weights $w_i$ (in this work $w_i$ are natural numbers), so that the
value of the tree function on the minimal tree belonging to such a
distribution of weights comes close to $n\cdot \lg \left( n\right) -\sum
w_i\cdot \lg \left( w_i\right) $. Here, in the second term we recognize the
Wiener-Shannon information of a series $\left( w_i\right) $ of weights, or
the statistical entropy%
$$
-\sum \frac{w_i}n\cdot \lg \left( \frac{w_i}n\right) =\frac 1n\cdot \left[
n\cdot \lg \left( n\right) -\sum w_i\cdot \lg \left( w_i\right) \right] 
$$
of the probability distribution $\left( \frac{w_i}n\right) $, depending on
the context. Again, we have the striking result that the functional form $%
-\sum w_i\cdot \lg \left( w_i\right) $ of the entropy is itself the result
of a process of optimization of a more general expression, namely the tree
function, and the entropy, as usually known, is only the minimal value of
this more general function.

{\bf The second main result concerning tree structures:\ }In the last
section we will see how every tree structure over a set $X$ defines a
neighbourhood topology on $X$. As we vary the tree structures, so vary the
topologies on $X$. If there is a tree function on the set of all tree
structures, it will single out preferred neighbourhood topologies, namely
those, for which the tree function becomes minimal. This defines an action
principle for neighbourhood topologies on the set $X$, where the value of
the action=tree function on the minimal trees is an entropy-like quantity.

--- The plan of the paper is as follows: In section \ref{Part} we recall the
definition of partitions of sets. In section \ref{motiv} we outline how a
tree structure encodes the operational aspect of the problem of information
gaining, which yields the concept of entropy/information. Basic properties
of notions pertaining to tree structures are given in section \ref{BasProp},
where we also introduce three axioms describing tree structures. In section 
\ref{OrdSet} we recall some elementary facts on ordered sets; in section \ref
{ZvonX} we show how the set of all partitions of a given set $X$ is a
partially ordered set. After this preparation we define paths in a tree
structure in section \ref{Paths}. Ideas concerning subtrees, sum, reduction,
extension of trees etc. are introduced in sections \ref{Subtree} and \ref
{SumUnEtc}. In section \ref{PartComp} we show how a tree over $X$ selects a
distinct subset of partitions of the underlying set $X$; here we introduce
the important concepts of minimal and maximal partitions of the underlying
set $X$ in the tree ${\cal B}$. After defining the characters of a nod in
section \ref{CharNod}, we come to the central notions in our theory: In
section \ref{AmFunc} we introduce amount functions on sets of tree
structures; after the technical section \ref{IndPart}, which contains
several splitting lemmata for amount functions, this is extended in section 
\ref{TreeFunc} to the definition of the tree function on the set of all tree
structures over $X$. The problem of minimizing trees is taken up in section 
\ref{MinClass}. We introduce the concept of divisions in section \ref
{divisions}, and explain its relation to partitions in section \ref
{PartAndDiv}. In sections \ref{OptDiv} and \ref{OptimalTrees} we introduce
optimal divisions of sets, and the concept of optimal trees based on optimal
divisions. After technical issues in section \ref{MinClassInT} and \ref
{BasIntLog}, the optimal amount is defined in section \ref{OptAm}. Section 
\ref{PreOptTrees} contains tools that are central to the proof of the main
theorem about the minimality of optimal trees. This theorem is approached in
a series of propositions given in section \ref{SectMinimalOptDiv}. Section 
\ref{MPAmQuDev} reflects the same statements from the point of view of the
mean path amount in a tree over $X$. In section \ref{ExtMinProb} we outline
how to find constrained minimal trees on which the functional form of the
tree function contains the entropy $-\sum w_i\lg \left( w_i\right) $. In the
last section, \ref{TopolAndTree}, we show how tree structures define
neighbourhood topologies on $X$, and how the tree function selects distinct
topologies according to a minimal principle.


\section{Partitions \label{Part}}

Let $X$ be a non-empty set. A {\it partition }$z$ of $X$ is a system of
mutually disjoint non-empty subsets $\mu \subset X$ whose union is $X$, i.e.

\begin{description}
\item[(P1)]  $\bigcup\limits_{\mu {\cal \;}\in \;z}\mu =X$ ,

\item[(P2)]  $\mu \neq \mu ^{\prime }\;\Rightarrow \;\mu \bigcap \mu
^{\prime }=\emptyset $ .
\end{description}

The {\it power set }${\cal P}X$ of $X$ is the set of all subsets of $X$,
including the empty set; that is to say, ${\cal P}X$ contains the elements 
\begin{equation}
\label{pp4t1fo1}\emptyset \;;\;\left\{ x\right\} \text{ for }x\in
X\;;\;\left\{ x,y\right\} \text{ for }x,y\in X\text{, }x\neq y\;;\;\ldots
\;;\; X \quad . 
\end{equation}
We see that every partition is a subset $z\subset {\cal P}X$ of the power
set of $X$. Hence it constitutes an element of the power set of the power
set, $z\in {\cal PP}X$.

The partition $z_0\equiv \left\{ X\right\} $ will be called the {\it trivial
partition}. If necessary, a nontrivial partition will be called {\it proper}%
. If $X$ is countable, we will say that the partition $z$ is {\it complete}
in case that every element $\mu $ of $z$ contains precisely one element of $%
X $, i.e. $\#\mu =1$ for all $\mu \in z$.

The set of all partitions $z$ of $X$ will be denoted by ${\cal Z}\left(
X\right) $. The set of all nontrivial partitions will be denoted by ${\cal Z}%
^{*}\left( X\right) $, i.e. ${\cal Z}^{*}\left( X\right) ={\cal Z}\left(
X\right) -\left\{ z_0\right\} $.

\section{Movitation for tree structures \label{motiv}}

We want to show how tree structures arise in the course of modelling
processes of information gaining. We now describe such a model: Let $X$ be a
non-empty finite set, $0<n\equiv \#X<\infty $. Let $x_0\in X$ be arbitrary.
We want to find a numerical measure for the information that is gained when $%
x_0$ has been identified as a distinct object amongst $n$ objects. Consider
the interaction of two (information processing) units, the first one
(storage unit) of which has stored the knowledge about $x_0$, and the second
unit tries to identify $x_0$ amongst all $n$ elements of $X$. The only
knowledge permitted to the second unit (search unit) is, that all $n$
choices are equally likely. The search unit suggests a partition of $X$; if
the number of elements in the partition is $m_1$, then the search unit has
to pose at most $\left( m_1-1\right) $ yes-no-questions to the storage unit
in order to identify the element of the partition that contains $x_0$. Now
the search unit suggests a partition of the subset that contains $x_0$, and
so on. This gives the following scheme: On level $1$, we have a partition $%
z\left( X\right) \in {\cal Z}^{*}\left( X\right) $ with $\#z\left( X\right)
=m_1>1$ elements, i.e. 
\begin{equation}
\label{pp3t1fo2}z\left( X\right) =\left\{ X_1,\ldots ,X_{m_1}\right\} \quad
. 
\end{equation}
On level $2$ we partition all the subsets $X_i$ in (\ref{pp3t1fo2}):
Decompose $X_1$ into $m_2\left( 1\right) $ non-empty subsets, $X_2$ into $%
m_2\left( 2\right) $ non-empty subsets, $\ldots \ $, $X_{m_1}$ into $%
m_2\left( m_1\right) $ subsets; here the subscripts $1,2$ in $m_1,m_2$ refer
to the levels $1$ and $2$, respectively. Hence for $i_1=1,\ldots ,m_1$ we
have partitions $z\left( X_{i_1}\right) \in {\cal Z}^{*}\left(
X_{i_1}\right) $ with cardinal number $\#z\left( X_{i_1}\right) \equiv
m_2\left( i_1\right) $, 
\begin{equation}
\label{pp3t1fo3}z\left( X_{i_1}\right) =\left\{ X_{i_1,1},\ldots
,X_{i_1,m_2\left( i_1\right) }\right\} \quad . 
\end{equation}
Now we continue along these lines: $X_{1,1}$ is decomposed into $m_3\left(
1,1\right) $ subsets; $X_{1,m_2\left( 1\right) }$ is decomposed into $%
m_3\left( 1,m_2\left( 1\right) \right) $ subsets; $\ldots $ ; $%
X_{m_1,m_2\left( m_1\right) }$ is decomposed into $m_3\left( m_1,m_2\left(
m_1\right) \right) $ subsets, i.e. for $i_1=1,\ldots ,m_1$, $i_2=1,\ldots
,m_2\left( i_1\right) $ we introduce a partition $z\left( X_{i_1,i_2}\right)
\in {\cal Z}^{*}\left( X_{i_1,i_2}\right) $ with cardinal number $\#z\left(
X_{i_1,i_2}\right) =m_3\left( i_1,i_2\right) $ such that 
\begin{equation}
\label{pp3t1fo4}z\left( X_{i_1,i_2}\right) =\left\{ X_{i_1,i_2,1},\ldots
,X_{i_1,i_2,m_3\left( i_1,i_2\right) }\right\} \quad , 
\end{equation}
etc. Any of the subsets $X_{i_1,i_2\cdots }$ emerging in this process is an
element of the power set ${\cal P}X$ of $X$. The totality of all these
subsets is a certain subset of the power set of $X$; we term it a {\it tree
structure }or simply a {\it tree}${\cal \ B}\left( X\right) $ {\it over }$X$%
. Hence,%
$$
{\cal B}\left( X\right) =\left\{ X,X_1,\ldots ,X_{m_1},X_{1,1},\ldots
,X_{1,m_2\left( 1\right) },\ldots ,X_{m_1,1},\ldots ,X_{m_1,m_2\left(
m_1\right) },\right. 
$$
$$
\left. X_{1,1,1},\ldots ,X_{1,1,m_3\left( 1,1\right) },\ldots
,X_{1,m_2\left( 1\right) ,1},\ldots ,X_{1,m_2\left( 1\right) ,m_3\left(
1,m_2\left( 1\right) \right) },\ldots \right. 
$$
\begin{equation}
\label{pp3t1fo5}\left. \ldots ,X_{m_1,m_2\left( m_1\right) ,m_3\left(
m_1,m_2\left( m_1\right) \right) },\ldots \right\} \quad . 
\end{equation}
For a given set $X$ (which later on need not necessarily be finite or even
countable), let ${\cal MB}\left( X\right) $ denote the set of all tree
structures over $X$. The elements of a given tree structure ${\cal B}\left(
X\right) $ can obviously be labelled by finite series%
$$
\emptyset ,\left( 1\right) ,\ldots ,\left( m_0\right) ,\left( 1,1\right)
,\ldots ,\left( 1,m_1\left( 1\right) \right) ,\ldots ,\left( m_0,1\right)
,\ldots ,\left( m_0,m_1\left( m_0\right) \right) , 
$$
\begin{equation}
\label{pp3t1fo6}\left( 1,1,1\right) ,\ldots ,\left( 1,1,m_2\left( 1\right)
\right) ,\ldots ,\left( m_0,m_1\left( m_0\right) ,m_2\left( m_0,m_1\left(
m_0\right) \right) \right) ,\ldots \quad . 
\end{equation}
Every such series will be called a {\it path} in the tree structure ${\cal B}%
\left( X\right) $. Let a general path in ${\cal B}\left( X\right) $ be
denoted by $\left( i_1,\cdots ,i_\kappa \right) $. If $X$ is finite, then
the quantities 
\begin{equation}
\label{pp3t1fo7}n\left( i_1,\cdots ,i_\kappa \right) \equiv \#X_{i_1,\cdots
,i_\kappa } 
\end{equation}
are natural numbers. In this case we say that the tree structure ${\cal B}%
\left( X\right) $ is {\it finite}. For a given tree structure ${\cal B}%
\left( X\right) $ the path $\left( i_1,\cdots ,i_\kappa \right) $ is said to
be {\it complete} if $n\left( i_1,\cdots ,i_\kappa \right) =1$, otherwise it
is called {\it incomplete}. Hence, in a finite tree structure with $n=\#X$
there are precisely $n$ distinct, complete paths.

\TeXButton{Abst}{\vspace{0.8ex}}

--- We have seen how tree structures emerge naturally in processes modelling
information gaining. The basic properties of tree structures, as they
present themselves from the above analysis, will be compiled in the next
section.

\section{Basic properties of tree structures \label{BasProp}}

We now suggest three natural axioms describing a set of subsets of $X$ as
given in (\ref{pp3t1fo5}). A {\it tree structure }${\cal B}\left( X\right) $%
{\it \ over }$X$ is defined to be a system of nonempty subsets $b\subset X$
of $X$ (hence a subset of the power set ${\cal P}X$ of $X$), with the
properties:

\begin{description}
\item[(A1)]  $X\subset {\cal B}\left( X\right) $ .

\item[(A2)]  If $b,b^{\prime }\in {\cal B}\left( X\right) $, then $b%
\subsetneqq b^{\prime }$\ or\ $b^{\prime } \subsetneqq 
b$\ or$\;b=b^{\prime }\;$or$\;b\cap b^{\prime }=\emptyset \;$[This
is ''exclusive or''].

\item[(A3)]  For all$\;b,b^{\prime }\in {\cal B}\left( X\right) $ there
exists $\tilde b\in {\cal B}\left( X\right) $ such that $b,b^{\prime
}\subset \tilde b\;$.
\end{description}

\TeXButton{Abst}{\vspace{0.8ex}}Elements $b$ of ${\cal B}\left( X\right) $
will sometimes be referred to as the {\it nods in the tree }${\cal B}\left(
X\right) $. An element $b$ of ${\cal B}\left( X\right) $ will be called {\it %
primitive}, if $b$ contains only one element, i.e. $b=\left\{ y\right\} $
for some $y\in X$. The tree structure ${\cal B}\left( X\right) $ will be
called {\it complete} if it contains all primitive elements, i.e. if $%
\left\{ y\right\} \in {\cal B}\left( X\right) $ for all $y\in X$.

An element $b$ of ${\cal B}\left( X\right) $ will be called {\it resolvable
in }${\cal B}\left( X\right) $ if and only if $b$ is {\bf not} primitive 
\underline{and} there exists $b^{\prime }\in {\cal B}\left( X\right) $ with $%
b^{\prime } \subsetneqq b$. Consequently, all primitive
elements are non-resolvable.

\TeXButton{Abst}{\vspace{0.8ex}}Although most of the definitions we will
introduce in this work will be stated as general as possible, and in
particular will include the case that $X$ is infinite, we will mostly deal
with {\bf finite} sets $X$ in this paper. The peculiarities that can arise
when $X$ is infinite will be discussed in a future publication.

\section{Ordered sets \label{OrdSet}}

For the following developments we need some definitions:

A non-empty set $X$ is called {\it ordered}, if a relation $"<"$ is defined
on $X$, satisfying:

\begin{description}
\item[(O1)]  For any two elements $a,b$ of $X$ either $a<b$ or $b<a$ or $a=b$
is true.

\item[(O2)]  If $a<b$ and $b<c$ then $a<c$.
\end{description}

If the nonempty set $X$ contains a nonempty ordered subset $T$, then $X$ is
said to be {\it partially ordered}. Hence every ordered set is partially
ordered. To distinguish this from a partial ordering we sometimes say that
an ordered set $X$ is {\it totally ordered}.

If $X$ contains an element $x_0$ for which $x_0<x$ for all $x\in X$ is true,
we call $x_0$ the principal{\it \ element} in $X$ [or in the pair $\left(
X,<\right) $, to be precise].

An ordered set is called {\it well-ordered}, if every nonempty subset $T$ of 
$X$ contains a principal element in $T$.

\section{${\cal Z}\left( X\right) $ as a partially ordered set \label{ZvonX}}

On the set ${\cal Z}\left( X\right) $ of all partitions of $X$, a natural
partial ordering can be introduced as follows: Let $z,z^{\prime }\in {\cal Z}%
\left( X\right) $. The relation $z<z^{\prime }$ is defined to be true if and
only if every $b^{\prime }\in z^{\prime }$ is contained in some $b\in z$
according to $b^{\prime }\subset b$, \underline{and} there exists a pair $%
\left( b,b^{\prime }\right) \in z\times z^{\prime }$ such that this
inclusion is proper, $b \supsetneqq b^{\prime }$. In this
case we say that the partition $z^{\prime }$ is a {\it refinement} of the
partition $z$. If both $z$ and $z^{\prime }$ are finite this implies in
particular that $\#z<\#z^{\prime }$.

Given two partitions $z,z^{\prime }$, clearly none of the relations $%
z<z^{\prime }$ or $z^{\prime }<z$ need be true; that is why the set ${\cal Z}%
\left( X\right) $ is only partially ordered. If $\#X>1$ then there exists a
nonempty subset of ${\cal Z}\left( X\right) \times {\cal Z}\left( X\right) $
which is totally ordered under $"<"$. If $\#X>2$, there exists a nonempty
subset of ${\cal Z}\left( X\right) ^{*}\times {\cal Z}\left( X\right) ^{*}$
on which $"<"$ is defined.

If $z=\left\{ b_1,\ldots ,b_k\right\} $ is a refinement of $z^{\prime }$,
then $z^{\prime }$ {\it preserves} the partition $z$ in the sense that no
element $b_i$ of $z$ is partitioned in the process $z^{\prime }\rightarrow z$%
. Thus, if $z$ is ''kept fixed'', we can think of the set of all partitions $%
z^{\prime }$ of $X$ which preserve $z$; obviously, these are precisely the
elements $z^{\prime }$ for which $z$ is a refinement; they comprise the set 
\begin{equation}
\label{pp4t1zvonx1}{\cal Z}\left( X,z\right) \equiv \left\{ z^{\prime }\in 
{\cal Z}\left( X\right) \mid z^{\prime }<z\right\} \quad . 
\end{equation}

From now on we assume that $X$ is finite.

\section{Paths in a tree structure \label{Paths}}

We now show that tree structures have a natural partial ordering. To this
end we observe that there exist distinct subsets in a tree structure, which
can be totally ordered: Let ${\cal B}\left( X\right) $ be a given tree
structure over $X$. Let $b\in {\cal B}\left( X\right) $. Then we call the
set 
\begin{equation}
\label{pp4t1fo8}q\left( b\right) \equiv \left\{ b^{\prime }\in {\cal B}%
\left( X\right) \mid b^{\prime }\supset b\right\} 
\end{equation}
the {\it path of }$b${\it \ in }${\cal B}\left( X\right) $. $q\left(
b\right) $ is certainly non-empty, since it always contains $X$ and $b$
itself. From the definition of $q\left( b\right) $ we see that a total
ordering $"<"$ on $q\left( b\right) $ for all pairs of elements $\left(
b^{\prime },b^{\prime \prime }\right) $ of $q\left( b\right) $ can be
defined by setting $b^{\prime }<b^{\prime \prime }$ if and only if $%
b^{\prime } \supsetneqq b$. This makes $q\left( b\right) $
totally ordered for all $b\in {\cal B}\left( X\right) $, and hence ${\cal B}%
\left( X\right) $ partially ordered. Since $q\left( b\right) $ is finite (as 
$X$ is finite), the cardinality $\#q\left( b\right) $ is a well-defined
natural number, denoted by $o\left( b\right) \equiv \#q\left( b\right) $,
which we will call the {\it length of the path }$q\left( b\right) $ in $%
{\cal B}\left( X\right) $.

Since $X$ is finite, all paths $q\left( b\right) $ are not only ordered, but
also well-ordered; a feature, that ceases to be true if $X$ is infinite.
This will be proven in a later paper.

\section{Subtrees \label{Subtree}}

Let $b\in {\cal B}\left( X\right) $. The set 
\begin{equation}
\label{pp4t1fo11}{\cal B}\left( X,b\right) \equiv \left\{ b^{\prime }\in 
{\cal B}\left( X\right) \mid b^{\prime }\subset b\right\} 
\end{equation}
will be called a {\it subtree of }${\cal B}\left( X\right) $ {\it over }$b$.
By definition, ${\cal B}\left( X,b\right) $ is a tree structure over $b$,
and hence ${\cal B}\left( X,b\right) \in {\cal MB}\left( b\right) $.

If $b$ is primitive, then ${\cal B}\left( X,b\right) =\left\{ b\right\} $.

\section{Sum, union, extension, reduction and completion of trees \label
{SumUnEtc}}

Let ${\cal B}\left( X\right) $ be a tree structure over $X$. Consider the
elements $X_1,\ldots ,X_{m\left( 1\right) }$ of level 1 in the partition $%
z\left( X\right) $ of $X$, cf. section \ref{motiv}. For every $X_i$, we can
think of the subtree ${\cal B}\left( X,X_i\right) $ over $X_i$. The relation
of the subtrees ${\cal B}\left( X,X_i\right) $, $i=1,\ldots ,m\left(
1\right) $, to the ''superior'' tree ${\cal B}\left( X\right) $ will be
described by saying that ${\cal B}\left( X\right) $ is the {\it sum} of the
trees ${\cal B}\left( X,X_i\right) $. Now we see how to extend this
definition to tree structures over sets which are not a priori subsets of a
given set: Let $m\in \TeXButton{N}{\mathbb{N}}$, let $X_1,\ldots ,X_m\neq
\emptyset $ be non-empty pairwise disjoint sets, i.e. $X_i\cap X_j=\emptyset 
$ for $i\neq j$. Let ${\cal B}\left( X_1\right) ,\ldots {\cal B}\left(
X_m\right) $ be tree structures over $X_1,\ldots ,X_m$. Then the set $%
\sum_{j=1}^m{\cal B}\left( X_j\right) $, defined by 
\begin{equation}
\label{pp4t1fo16}\sum_{j=1}^m{\cal B}\left( X_j\right) \equiv \left[
\bigcup\limits_{j=1}^m{\cal B}\left( X_j\right) \right] \cup \left[
\bigcup\limits_{j=1}^mX_j\right] \quad , 
\end{equation}
will be called the {\it sum of }${\cal B}\left( X_1\right) ,\ldots {\cal B}%
\left( X_m\right) .$ By construction, this is a tree structure over $%
\bigcup\limits_{j=1}^mX_j$ with subtrees ${\cal B}\left( X_1\right) ,\ldots 
{\cal B}\left( X_m\right) $.

Another construction is the {\it union of trees}. This is defined as
follows: Let ${\cal B}\left( X\right) $ be a tree structure, and let $b\in 
{\cal B}\left( X\right) $ be a {\bf non-resolvable}, {\bf non-primitive}
element. Then $\#b>1$. Although $b$ is not further partitioned in the tree $%
{\cal B}\left( X\right) $, we can nevertheless consider tree structures over 
$b$ without reference to ${\cal B}\left( X\right) $. Let ${\cal B}\left(
b\right) $ be such a tree over $b$. Then we can attach ${\cal B}\left(
b\right) $ to ${\cal B}\left( X\right) $ by identifying $b\in {\cal B}\left(
b\right) $ with $b\in {\cal B}\left( X\right) $; the resulting set is the
union ${\cal B}\left( b\right) \cup {\cal B}\left( X\right) $, and will be
called the {\it union of the trees }${\cal B}\left( b\right) $ and ${\cal B}%
\left( X\right) $. Conversely, we could remove the tree ${\cal B}\left(
X,b\right) $ from ${\cal B}\left( X\right) $ and adding $\left\{ b\right\} $
to the reduced set, ${\cal B}\left( X\right) -{\cal B}\left( X,b\right)
\mapsto \left[ {\cal B}\left( X\right) -{\cal B}\left( X,b\right) \right]
\cup \left\{ b\right\} ={\cal B}^{\prime }\left( X\right) $. Then ${\cal B}%
^{\prime }\left( X\right) $ is a tree structure by definition, which in this
context will be called the {\it tree }${\cal B}\left( X\right) ${\it \
reduced by }${\cal B}\left( X,b\right) $ (or ${\cal B}\left( b\right) $).

A somewhat related, but more general, concept is the {\it extension of trees}%
. Let ${\cal B}$ and ${\cal B}^{\prime }$ be two tree structures over the
same set $X$. We will say that ${\cal B}$ is an {\it extension of }${\cal B}%
^{\prime }$ if ${\cal B}^{\prime }\subset {\cal B}$. In this case we can
also refer to ${\cal B}^{\prime }$ as a {\it reduction of} ${\cal B}$. A
special case of extension is the {\it completion }$\left[ {\cal B}\right] $ 
{\it of a tree }${\cal B}$: This is defined to be a tree structure $\left[ 
{\cal B}\right] $ over the same set $X$ that extends ${\cal B}$ \underline{%
and} which is complete, i.e. the maximal partition in $\left[ {\cal B}%
\right] $ takes the form $\left\{ \left\{ x_1\right\} ,\left\{ x_2\right\}
,\ldots \right\} $, where $x_i$ runs through all elements of $X$. If $X$ is
finite, every tree structure ${\cal B}$ admits such a completion; but,
clearly, there are many completions $\left[ {\cal B}\right] $ for a given
tree structure ${\cal B}$ in general.

The relation of these ideas to the subtrees discussed in the last paragraph is
as follows: Assume that ${\cal B}\left( X\right) ={\cal B}\left( b\right)
\cup {\cal B}^{\prime }\left( X\right) $ is the union of a tree ${\cal B}%
\left( b\right) $ over $b$, and another tree ${\cal B}^{\prime }\left(
X\right) $ over $X$. Then certainly ${\cal B}\left( b\right) ={\cal B}\left(
X,b\right) $ as defined in (\ref{pp4t1fo11}), and ${\cal B}^{\prime }\left(
X\right) =\left[ {\cal B}\left( X\right) -{\cal B}\left( X,b\right) \right]
\cup \left\{ b\right\} $. The tree ${\cal B}^{\prime }\left( X\right) $
therefore is obtained from ${\cal B}\left( X\right) $ be simply cutting off
the ''branch'' containing the evolution, i.e. the partitions, of $b$, but
reattaching $b$ as a non-resolvable element. This contains an important

\subsection{Splitting principle}

Every tree ${\cal B}\left( X\right) $ can be expressed as the union of any
of its subtrees ${\cal B}\left( X,b\right) $ with a cutoff tree ${\cal B}%
^{\prime }\left( X\right) $, ${\cal B}\left( X\right) ={\cal B}\left(
X,b\right) \cup {\cal B}^{\prime }\left( X\right) $, where both trees are
subsets of the original tree, i.e. ${\cal B}\left( X,b\right) ,{\cal B}%
^{\prime }\left( X\right) \subset {\cal B}\left( X\right) $.

\section{Partitions compatible with the tree \label{PartComp}}

Consider a given tree element $b\in {\cal B}\left( X\right) $. The subtree $%
{\cal B}\left( X,b\right) $ defines a collection of partitions of $b$ which
forms a proper subset of the set of all possible partitions ${\cal Z}\left(
b\right) $ of $b$. This motivates the definitions 
\begin{equation}
\label{pp4t1fo17}\zeta \left( b\right) \equiv \left\{ z\in {\cal Z}\left(
b\right) \mid z\subset {\cal B}\left( X\right) \right\} ={\cal Z}\left(
b\right) \cap {\cal B}\left( X\right) \quad , 
\end{equation}
\begin{equation}
\label{pp4t1fo18}\zeta \left( b\right) ^{*}\equiv {\cal Z}\left( b\right)
^{*}\cap {\cal B}\left( X\right) \;=\;\zeta \left( b\right) -\left\{ \left\{
b\right\} \right\} \quad , 
\end{equation}
where $\left\{ b\right\} $ is the trivial partition of $b$. The elements of $%
\zeta \left( b\right) $ will be called the partitions {\it compatible with
the tree }${\cal B}\left( X\right) $.

The set $\zeta \left( X\right) $ obviously contains two distinct partitions;
firstly, the trivial partition $z_0\left( X\right) =\left\{ X\right\} $; and
secondly, the partition of $X$ which is constituted by the set of all
elements which are not resolvable in ${\cal B}\left( X\right) $. Similarly,
if $b\in {\cal B}\left( X\right) $ is arbitrary, then $\zeta \left( b\right) $
contains the trivial partition $z_0\left( b\right) =\left\{ b\right\} $ as
well as the partition of $b$ which is constituted by those elements $%
b^{\prime }\in {\cal B}\left( X,b\right) $ which are not resolvable in $%
{\cal B}\left( X,b\right) $, hence not resolvable in ${\cal B}\left(
X\right) $. This distinct partition will be denoted by $z_{\max }\left(
b\right) $, and will be called the {\it maximal partition} of $b${\it \ in
the tree }${\cal B}\left( X\right) $. If $b$ is non-resolvable in ${\cal B}%
\left( X\right) $ then $z_{\max }\left( b\right) =z_0\left( b\right) $,
whereas, if $b$ is resolvable, then $z_{\max }\left( b\right) \in \zeta
^{*}\left( b\right) $. If $b=X$, then the union of all paths $q\left(
b^{\prime }\right) $ with $b^{\prime }\in z_{\max }\left( X\right) $ renders
the whole tree structure ${\cal B}\left( X\right) $.

Given an element $b\in {\cal B}\left( X\right) $, then obviously $b$ is the
last element in the path $q\left( b\right) $ in the sense that $b^{\prime
}<b $ for all $b^{\prime }\in q\left( b\right) $. Now, if for $b\neq X$, $%
q\left( b\right) $ is a path in ${\cal B}\left( X\right) $, then so is $%
q\left( b\right) -\left\{ b\right\} $; the last element in ${\cal B}\left(
X\right) $ with respect to the path $q\left( b\right) -\left\{ b\right\} $
will be called the {\it predecessor }$b^{-}$ {\it of }$b${\it \ in the tree }%
${\cal B}\left( X\right) $. Clearly, distinct elements $b,b^{\prime }$ have
the same predecessor only if $b\cap b^{\prime }=\emptyset $.

The partition of $b\in {\cal B}\left( X\right) $ that is obtained when
stepping to the next level in the tree we call the {\it minimal partition of 
}$b${\it \ in }${\cal B}\left( X\right) $. If $b$ is non-resolvable, there
are no proper partitions of $b$, and hence we define $z_{\min }\left(
b\right) =\left\{ b\right\} $ to be the trivial partition in this case. The
number of elements, $m\left( b\right) $, in the minimal partition of $b\in 
{\cal B}\left( X\right) $ will play a crucial role, 
\begin{equation}
\label{pp4t1fo19}m\left( b\right) \equiv \#z_{\min }\left( b\right) \quad . 
\end{equation}
For non-resolvable elements we have $m\left( b\right) =1$.

\subsection{Ordering property of $z_{\mint}$ and $z_{\maxt}$}

Let $b\in {\cal B}\left( X\right) $. Then the minimal and maximal partitions 
$z_{\min }$ and $z_{\max }$ of a given element $b\in {\cal B}\left( X\right) 
$ have the following obvious but important ordering properties: For all
non-resolvable $b$ in ${\cal B}\left( X\right) $ we have 
\begin{equation}
\label{pp4t1fo21aa}z_0\left( b\right) =z_{\min }\left( b\right) =z_{\max
}\left( b\right) \quad . 
\end{equation}
If $b$ is resolvable in ${\cal B}\left( X\right) $, then for all $z\in \zeta
\left( b\right) $ with $z\neq z_0\left( b\right) ,z_{\min }\left( b\right) $
we have 
\begin{equation}
\label{pp4t1fo21ab}z_{\min }\left( b\right) <z\quad , 
\end{equation}
and for all $z\in \zeta \left( b\right) $ with $z\neq z_0\left( b\right)
,z_{\max }\left( b\right) $ we have 
\begin{equation}
\label{pp4t1fo21ac}z<z_{\max }\left( b\right) \quad . 
\end{equation}

\subsection{Split of trees according to minimal partition}

Given the minimal partition $z_{\min }\left( b\right) $ of any element $b\in 
{\cal B}\left( X\right) $, we can split the subtree ${\cal B}\left(
X,b\right) $ accordingly into a sum of subtrees. Let $z_{\min }\left(
b\right) =\left\{ b_1,\ldots ,b_k\right\} $, then ${\cal B}\left( X,b\right) 
$ is a sum of subtrees 
\begin{equation}
\label{pp4t1fo21a}{\cal B}\left( X,b\right) =\sum_{i=1}^k{\cal B}\left(
X,b_i\right) \quad . 
\end{equation}
We will make use of this formula in section \ref{AmFunc}.

\subsection{Reduction of trees by partitions}

Another reduction of trees ${\cal B}\left( X\right) $ can be obtained as
follows: Let $z\in \zeta \left( X\right) $ be compatible with ${\cal B}%
\left( X\right) $. $z$ has the form $z=\left\{ b_1,\ldots ,b_k\right\} $ for
some $k$, and all $b_i$ are elements of ${\cal B}\left( X\right) $. Now we
think of a new tree ${\cal B}\left( X,z\right) \equiv {\cal B}\left(
z\right) $ obtained from the original one by regarding the elements $b_i$ as 
{\bf non-resolvable} in ${\cal B}\left( X,z\right) $; in other words, all
paths in ${\cal B}\left( X,z\right) $ terminate at the elements $b_i$,
whereas in the original tree ${\cal B}\left( X\right) $ they could have
continued to be partitioned. Effectively, this means we cut off all subtrees
based on $b_i$, and stipulate that now $b_i$ be non-resolvable.

\section{Characters of a nod \label{CharNod}}

Let $z_{\min }\left( b\right) $ be the unique minimal partition of $b$ in $%
{\cal B}\left( X\right) $. Accordingly we define 
\begin{equation}
\label{pp4t1fo22}m\left( b\right) \equiv \#z_{\min }\left( b\right) \quad
;\quad n\left( b\right) \equiv \#b 
\end{equation}
for all $b\in {\cal B}\left( X\right) $. We have 
\begin{equation}
\label{pp4t1fo23}n\left( b\right) =\sum_{a\,\in z_{\min }\left( b\right)
}n\left( a\right) \quad . 
\end{equation}
For every $b\in {\cal B}\left( X\right) $ the following inequality holds: 
\begin{equation}
\label{pp4t1fo24}1\le m\left( b\right) \le n\left( b\right) \quad . 
\end{equation}
Furthermore, if $b\neq X$, then 
\begin{equation}
\label{pp4t1fo25}n\left( b\right) \leq n\left( b^{-}\right) -m\left(
b^{-}\right) +1\quad . 
\end{equation}
Statements (\ref{pp4t1fo23}, \ref{pp4t1fo24}) are obvious consequences of
the basic definitions. (\ref{pp4t1fo25}) follows, since all $m\left(
b^{-}\right) $ subsets of $b^{-}$ that constitute the minimal partition $%
z_{\min }\left( b^{-}\right) $ contain at least one element. Apart from $b$,
there exist $m\left( b^{-}\right) -1$ such subsets. Hence $\#b^{-}\ge
\#b+m\left( b^{-}\right) -1$, from which (\ref{pp4t1fo25}) follows on using $%
\#b^{-}=n\left( b^{-}\right) $ and $\#b=n\left( b\right) $.

The pair $\left[ n\left( b\right) ,m\left( b\right) \right] $ of numbers of
a given nod $b$ will play an important role in what follows; we shall refer
to $\left[ n\left( b\right) ,m\left( b\right) \right] $ as the {\it %
characters of the nod }$b$.

\section{Amount functions \label{AmFunc}}

Let $b\in {\cal B}\left( X\right) $ with $b\neq X$ . Since $X$ is finite, $b^{-}$ exists, %
and the number of elements in $z_{\min }\left( b^{-}\right) $ is $m\left(
b^{-}\right) $. Now we think of $b$ as being distinct in the set of elements 
$b^{\prime }$ comprising $z_{\min }\left( b^{-}\right) $. If we are
presented this set in order to find out which of the $b^{\prime }\in z_{\min
}\left( b^{-}\right) $ is the distinct one, we have to expend at most $%
m\left( b^{-}\right) -1$ questions. We can extend this reasoning to the
whole path $q\left( b\right) $: Since $b^{-}$ contains $b$, it is distinct
in the set of elements $b^{\prime \prime }$ comprising $z_{\min }\left(
b^{2-}\right) $, where $b^{2-}$ denotes the predecessor of $b^{-}$ in ${\cal %
B}\left( X\right) $. In order to determine
that just $b^{-}$ is distinct amongst all $b^{\prime \prime }$, we have to
expend at most $m\left( b^{2-}\right) -1$ questions. We can continue in this
way along the whole path $q\left( b\right) $ until no predecessor $b^{\left(
k+1\right) -}$ exists any longer, in other words, until $b^{k-}=X$. The
maximum number of questions necessary to find out that $b\in {\cal B}\left(
X\right) $ is the distinct object we were seeking out is therefore the sum
of all the expressions above,%
$$
\sum_{a\,\in q\left( b\right) -\left\{ b\right\} }\left[ m\left( a\right)
-1\right] \;=\;\left[ \sum_{a\,\in q\left( b\right) -\left\{ b\right\}
}m\left( a\right) \right] -o\left( b\right) +1\quad . 
$$
This leads us to the following

\subsection{Definition}

Let $b\in {\cal B}\left( X\right) $ and $b\neq X$ .\ Then 
\begin{equation}
\label{pp4t1fo26}e\left( b\right) \equiv \sum_{a\,\in q\left( b\right)
-\left\{ b\right\} }\left[ m\left( a\right) -1\right] \;=\;\left[
\sum_{a\,\in q\left( b\right) -\left\{ b\right\} }m\left( a\right) \right]
-o\left( b\right) +1 
\end{equation}
will be called the {\it amount of }$b$ {\it in the tree }${\cal B}\left(
X\right) $ . When emphasizing the fact that the amount is dependent on the
underlying tree structure we will also write $e\left( b\right) \equiv e_{%
{\cal B}\left( X\right) }\left( b\right) $.

\subsection{Definition}

Now let\ $z\in \zeta \left( X\right) ={\cal Z}\left( X\right) \cap {\cal B}%
\left( X\right) $ be an arbitrary partition of $X$ compatible with the tree $%
{\cal B}\left( X\right) $. Then every element $b\in z$ gives rise to the
uniquely defined path $q\left( b\right) \subset {\cal B}\left( X\right) $.
Hence it makes sense to speak of the {\it total amount }$G\left( z\right) $ 
{\it of }$z\in \zeta \left( X\right) $ {\it with respect to the tree }${\cal %
B}\left( X\right) $, defined by 
\begin{equation}
\label{pp4t1fo27}G\left( z\right) \equiv \sum_{b\,\in z}e_{{\cal B}\left(
X\right) }\left( b\right) \;=\;\sum_{b\,\in z}\sum_{a\,\in q\left( b\right)
-\left\{ b\right\} }\left[ m\left( a\right) -1\right] \quad . 
\end{equation}
By construction this is just the total amount $G_{{\cal B}\left( z\right) }$
of the reduced tree ${\cal B}\left( z\right) $. When emphasizing the fact
that the total amount is depending on the underlying tree structure we will
also write $G\left( z\right) \equiv G_{{\cal B}\left( X\right) }\left(
z\right) $. Now consider the total amount $G\left( z_{\max }\left( X\right)
\right) $. From 
\begin{equation}
\label{pp4t1fo28}G\left( z_{\max }\left( X\right) \right) =\sum_{a\,\in 
{\cal B}\left( X\right) -z_{\max }\left( X\right) }\left[ m\left( a\right)
-1\right] 
\end{equation}
we see that in this case we sum over all resolvable elements $b\in {\cal B}%
\left( X\right) $; hence the total amount for the maximal partition of $X$
in ${\cal B}\left( X\right) $ is depending on ${\cal B}\left( X\right) $
only; it therefore defines a map from the set of all tree structures over $X$
into the natural numbers, 
\begin{equation}
\label{pp4t1fo29}G:\left\{ 
\begin{array}{c}
{\cal MB}\left( X\right) \rightarrow \TeXButton{N}{\mathbb{N}} \\ z_{\max
}\left( X\right) \mapsto G\left( z_{\max }\left( X\right) \right) 
\end{array}
\right. \quad . 
\end{equation}
Accordingly, we write $G\left( z_{\max }\left( X\right) \right) \equiv G_{%
{\cal B}\left( X\right) }$ and call $G_{{\cal B}\left( X\right) }$ the {\it %
total amount of the tree structure} ${\cal B}\left( X\right) $. Definition (%
\ref{pp4t1fo29}) now suggests that we ask ourselves, which trees ${\cal B}%
\left( X\right) $ in ${\cal MB}\left( X\right) $ would actually minimize the
amount $G_{{\cal B}\left( X\right) }$.

\subsection{Notation conventions}

We introduce some notation conventions that will prove convenient in the
sequel.

If $b\in {\cal B}\left( X\right) $ and $q\left( b\right) $ is the associated
path in ${\cal B}\left( X\right) $, we denote $\dot q\left( b\right) \equiv
q\left( b\right) -\left\{ b\right\} $.

If ${\cal B}\left( X,b^{\prime }\right) $ is a subtree of ${\cal B}\left(
X\right) $, and if \ $b\in {\cal B}\left( X,b^{\prime }\right) $, then the
path of $b$ in ${\cal B}\left( X,b^{\prime }\right) $ will be denoted by $q_{%
{\cal B}\left( X,b^{\prime }\right) }\left( b\right) \equiv \left\{ a\in 
{\cal B}\left( X,b^{\prime }\right) \mid a\supset b\right\} $.

\subsection{Proposition}

For every\ $b\in {\cal B}\left( X\right) $ we have 
\begin{equation}
\label{pp4t1fo30}o\left( b\right) -1\le e\left( b\right) \le n\left(
X\right) -n\left( b\right) \le n\left( X\right) -1\quad . 
\end{equation}

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

Set\ $o\left( b\right) =:\kappa $\ and\ $q\left( b\right) =\left\{ \beta
_1,\ldots ,\beta _\kappa \right\} $, with $\beta _1=X$, $\beta _\kappa =b$,
then\ $\dot q\left( b\right) =\left\{ \beta _1,\ldots ,\beta _{\kappa
-1}\right\} $, and we must have 
\begin{equation}
\label{pp4t1fo31}e\left( b\right) =\sum_{j=1}^{\kappa -1}\left[ m\left(
\beta _j\right) -1\right] \;=\;\sum_{j=2}^\kappa \left[ m\left( \beta
_{j-1}\right) -1\right] \quad . 
\end{equation}

For all $j\in \left\{ 1,\ldots ,\kappa -1\right\} $ we must have $m\left(
\beta _j\right) \ge 2$. If this is inserted into (\ref{pp4t1fo31}) we obtain
the first inequality in (\ref{pp4t1fo30}).

If (\ref{pp4t1fo25}) is inserted into (\ref{pp4t1fo31}) we find%
$$
e\left( b\right) \le \sum_{j=2}^\kappa \left[ n\left( \beta _{j-1}\right)
-n\left( \beta _j\right) \right] =\sum_{j=1}^{\kappa -1}n\left( \beta
_j\right) -\sum_{j=2}^\kappa n\left( \beta _j\right) =n\left( \beta
_1\right) -n\left( \beta _\kappa \right) \;=\;n\left( X\right) -n\left(
b\right) \quad . 
$$
This yields the second inequality in (\ref{pp4t1fo30}).

The last inequality follows trivially from the second one for the special
case that $b$ is primitive, i.e. $\#b=1$. \TeXButton{BWE}
{\hfill
\vspace{2ex}
$\blacksquare$}

\section{Induced partitions \label{IndPart}}

Let\ $b\in {\cal B}\left( X\right) $. For every\ $z\in \zeta \left( X\right) 
$\ we can introduce the set\ $\sigma \left( z,b\right) \equiv z\cap {\cal B}%
\left( X,b\right) $. $\sigma \left( z,b\right) $\ can be empty, if all
elements $b^{\prime }\in z$ are ''coarser'' than $b$, i.e. $b$ is contained
in precisely one of the $b^{\prime }$, and has vanishing intersection with
the rest. If $\sigma \left( z,b\right) $ is non-empty, then $\sigma \left(
z,b\right) $ is a partition of $b$ compatible with ${\cal B}\left(
X,b\right) $, and hence with ${\cal B}\left( X\right) $; the details are
proved below. In this case, $\sigma \left( z,b\right) $ will be called a 
{\it partition of }$b${\it \ induced by }$z$. $\sigma $ defines a map 
\begin{equation}
\label{pp4t1fo32}\sigma :\left\{ 
\begin{array}{c}
\zeta \left( X\right) \times 
{\cal B}\left( X\right) \rightarrow {\cal PP}\left( X\right) \\ \left(
z,b\right) \mapsto \sigma \left( z,b\right) \equiv z\cap {\cal B}\left(
X,b\right) 
\end{array}
\right. \quad . 
\end{equation}

\subsection{Theorem}

Let $b\in {\cal B}\left( X\right) $ with $b \subsetneqq X$.
Then 
\begin{equation}
\label{pp4t1fo33}\sigma \left( \zeta \left( X\right) ,b\right) \equiv
\left\{ \sigma \left( z,b\right) \mid z\in \zeta \left( X\right) \right\}
=\zeta \left( b\right) \cup \left\{ \emptyset \right\} \quad , 
\end{equation}
where $\emptyset \in {\cal PP}b$ is the zero element in the Boolean algebra $%
{\cal PP}b$.

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

Let $\sigma \left( z,b\right) \in \sigma \left( \zeta \left( X\right)
,b\right) $. Then $\sigma \left( z,b\right) $ can be either $\left( 1\right) 
$ empty, $\left( 2\right) $ equal to $\left\{ b\right\} $, or $\left(
3\right) $ contain more than one element. Case $\left( 1\right) $, $\sigma
\left( z,b\right) =\emptyset $, occurs if and only if $z=\left\{
u_1,u_2,\ldots \right\} $, where $u_1\supsetneqq b$, and all
other elements $u_i$ with $i\ge 2$ have vanishing intersection with $b$.
This includes the possibility that there are no elements $u_i$ with $i\ge 2$
at all, in which case $z=\left\{ X\right\} $ is the trivial partition of $X$%
. Case $\left( 2\right) $, $\sigma \left( z,b\right) =\left\{ b\right\} $,
happens if and only if $z=\left\{ b,u_2,\ldots \right\} $. Case $\left(
3\right) $, $\#\sigma \left( z,b\right) \ge 2$, is true if and only if $%
z=\left\{ v_1,v_2,\ldots ,u_1,u_2,\ldots \right\} $, where all $v_i$ are
contained in $b$, and all $u_j$ have vanishing intersection with $b$. In
this case $\sigma \left( z,b\right) =\left\{ v_1,v_2,\ldots \right\} $,
where the right hand side must be a partition of $b$. Thus in the last two
cases we have $\sigma \left( z,b\right) \in \zeta \left( b\right) $. This
proves, that the LHS of (\ref{pp4t1fo33}) is included in the RHS.

Now prove the reverse: Since $b\subsetneqq X$, we certainly
have $o\left( b\right) \ge 2$. In this case the trivial partition $%
z_0=\left\{ X\right\} $ of $X$ satisfies the condition of case $\left(
1\right) $ above, which implies that the LHS of (\ref{pp4t1fo33}) contains $%
\emptyset $. Now let $z^{\prime }\in \zeta \left( b\right) $ arbitrary; let $%
z^{\prime \prime }$ be the maximal partition in the cutoff tree ${\cal B}%
^{\prime }\left( X\right) \equiv {\cal B}\left( X\right) -{\cal B}\left(
X,b\right) $. $z^{\prime \prime }$ preserves $b$ (i.e. does not partition
it) since $b$ is non-resolvable in ${\cal B}^{\prime }\left( X\right) $.
This implies that $z\equiv \left[ z^{\prime \prime }-\left\{ b\right\}
\right] \cup z^{\prime }$ is a partition of $X$ preserving the tree
structure ${\cal B}\left( X\right) $ and satisfying $\sigma \left(
z,b\right) =z\cap {\cal B}\left( X,b\right) =z^{\prime }$. But this means
that $z^{\prime }$ must be contained in the LHS of (\ref{pp4t1fo33}). 
\TeXButton{BWE}{\hfill
\vspace{2ex}
$\blacksquare$}

\subsection{Theorem: Refinement of partitions \label{lem1}}

Let\ ${\cal B}\in {\cal MB}\left( X\right) $.\ Let\ $z,z^{\prime }\in \zeta
\left( X\right) $\ with\ $z<z^{\prime }$. Then

\begin{description}
\item[(A)]  $\ z-\left( z\cap z^{\prime }\right) \neq \emptyset $.

\item[(B)]  $\sigma \left( z^{\prime },b\right) \in \zeta ^{*}\left( b\right) $
for all $b\in z-\left( z\cap z^{\prime }\right) $, and $\sigma \left(
z^{\prime },b\right) =z_0\left( b\right) =\left\{ b\right\} $ is the trivial
partition for all $b\in z\cap z^{\prime }$. Hence, $\sigma \left( z^{\prime
},b\right) \in \zeta \left( b\right) $ for all $b\in z$.

\item[(C)]
\begin{equation}
\label{pp4t1fo34}z^{\prime }-\left( z^{\prime }\cap z\right)
=\bigcup\limits_{b\,\in z-\left( z\cap z^{\prime }\right) }\sigma \left(
z^{\prime },b\right) \quad . 
\end{equation}
\end{description}

\TeXButton{Beweis}{\raisebox{0ex}{\it Proof :}
\vspace{1ex}}

Since a relation of the form $z<z^{\prime }$ exists, $z^{\prime }$ is a
refinement of $z$, and any element of $z^{\prime }$ is contained in some
element of $z$. $z\cap z^{\prime }$ contains all elements of $z$ that are
not partitioned under the refinement $z\mapsto z^{\prime }$. This means that
for all $b\in z\cap z^{\prime }$, $z^{\prime }\cap {\cal B}\left( X,b\right)
=\left\{ b\right\} $, hence $\sigma \left( z^{\prime },b\right) $ is the
trivial partition of $b$. This proves the second statement in $\left(
B\right) $. On the other hand, if $b\in z-\left( z\cap z^{\prime }\right) $,
then $b$ is undergoing a proper partition under the refinement $z\mapsto
z^{\prime }$. This says that $\sigma \left( z^{\prime },b\right) \in \zeta
^{*}\left( b\right) $, thus proving the first statement in $\left( B\right) $%
. Since $z^{\prime }$ is refined, there must exist at least one element of $%
z $ that undergoes a proper partition, which says that $z-\left( z\cap
z^{\prime }\right) $ cannot be empty, hence $\left( A\right) $. Statement $%
\left( C\right) $ just summarizes the reasoning of $\left( B\right) $; it is
valid even if $z-\left( z\cap z^{\prime }\right) $ were empty. 
\TeXButton{BWE}{\hfill
\vspace{2ex}
$\blacksquare$}

\TeXButton{Abst}{\vspace{0.8ex}}We now apply the propositions deduced above
to derive an important intermediary result about the behaviour of total
amount functions under refinements and split of trees:

\subsection{Splitting lemma 1 for total amount}

\begin{description}
\item[A )]  Let\ $z,z^{\prime }\in \zeta \left( X\right) $\ with\ $%
z<z^{\prime }$. Then we have 
\begin{equation}
\label{pp4t1fo37}G\left( z^{\prime }\right) =G\left( z\right)
+\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right) }\left[ \#\sigma
\left( z^{\prime },b\right) -1\right] \cdot e_{{\cal B}\left( z\right)
}\left( b\right) +\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right)
}G_{{\cal B}\left( z^{\prime },b\right) }\quad . 
\end{equation}
This says that the total amount of the reduced tree ${\cal B}\left(
z^{\prime }\right) $ after the refinement $z^{\prime }$ is composed of three
contributions: The total amount of the original reduced tree ${\cal B}\left(
z\right) $ with respect to the partition $z$, a contribution that links the
amounts $e_{{\cal B}\left( z\right) }\left( b\right) $ of the paths $q\left(
b\right) $ of $b$ in ${\cal B}\left( z\right) $ with the ''degree of
splitting'' $\#\sigma \left( z^{\prime },b\right) $ of the set $b$ under the
refinement $z\mapsto z^{\prime }$, and the sum of all amounts of the
subtrees ${\cal B}\left( z^{\prime },b\right) $ of the larger tree ${\cal B}%
\left( z^{\prime }\right) $.

\item[B)]  For the special case\ $z=z_{\min }\left( X\right) $,\ $z^{\prime
}=z_{\max }\left( X\right) $\ we obtain from (\ref{pp4t1fo37}): 
\begin{equation}
\label{pp4t1fo38}G_{{\cal B}\left( X\right) }=\left[ m\left( X\right)
-1\right] \cdot \sum\limits_{b\,\in \,z_{\min }\left( X\right) }\#z_{\max
}\left( b\right) \;+\;\sum\limits_{b\,\in z_{\min }\left( X\right) }G_{{\cal %
B}\left( X,b\right) }\quad . 
\end{equation}
\end{description}

If the maximal partition $z_{\max }\left( X\right) $ is complete, i.e. $%
z_{\max }\left( X\right) =\left\{ \left\{ x_1\right\} ,\left\{ x_2\right\}
,\ldots \right\} $, where $x_i$ are the elements of $X$, then $\#z_{\max
}\left( b\right) =\#b=n\left( b\right) $, hence 
\begin{equation}
\label{pp4t1fo38a}\sum_{b\in z_{\min }\left( X\right) }1=m\left( X\right)
\quad \text{and\quad }\sum\limits_{b\,\in \,z_{\min }\left( X\right)
}n\left( b\right) =n\left( X\right) \quad . 
\end{equation}
This gives 
\begin{equation}
\label{pp4t1fo38b}G_{{\cal B}\left( X\right) }=\left[ m\left( X\right)
-1\right] \cdot n\left( X\right) +\sum\limits_{b\,\in z_{\min }\left(
X\right) }G_{{\cal B}\left( X,b\right) } 
\end{equation}
in this case.

This says that $G_{{\cal B}\left( X\right) }$ splits into a ''level 1''
contribution depending solely on the numbers $n\left( X\right) $ and $%
m\left( X\right) $, and the total amounts of the subtrees ${\cal B}\left(
X,b\right) $. The latter contributions clearly are independent of the first
one, since different tree structures over $X$ may have concident pairs of
numbers $n\left( X\right) $, $m\left( X\right) $ for their minimal partition 
$z_{\min }\left( X\right) $ of $X$.

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

\underline{Ad A :}\quad Use (\ref{pp4t1fo27}) to find%
$$
G\left( z^{\prime }\right) =\sum\limits_{b^{\prime }\,\in z^{\prime }}e_{%
{\cal B}\left( z^{\prime }\right) }\left( b^{\prime }\right)
=\sum\limits_{b^{\prime }\,\in z^{\prime }\cap z}e_{{\cal B}\left( z^{\prime
}\right) }\left( b^{\prime }\right) \,+\sum\limits_{b^{\prime }\,\in
\,z^{\prime }-\left( z^{\prime }\cap z\right) }e_{{\cal B}\left( z^{\prime
}\right) }\left( b^{\prime }\right) = 
$$
$$
=\sum\limits_{b^{\prime }\,\in z^{\prime }\cap z}e_{{\cal B}\left( z^{\prime
}\right) }\left( b^{\prime }\right) +\sum\limits_{b^{\prime }\,\in
\,z^{\prime }-\left( z^{\prime }\cap z\right) }\sum\limits_{a^{\prime }\,\in
\,\dot q_{{\cal B}\left( z^{\prime }\right) }\left( b^{\prime }\right)
}\left[ m\left( a^{\prime }\right) -1\right] = 
$$
$$
=\sum\limits_{b^{\prime }\,\in z^{\prime }\cap z}e_{{\cal B}\left( z^{\prime
}\right) }\left( b^{\prime }\right) +ZS\quad , 
$$

where%
$$
ZS\;=\;\sum\limits_{b^{\prime }\,\in \,z^{\prime }-\left( z^{\prime }\cap
z\right) }\sum\limits_{a^{\prime }\,\in \,\dot q_{{\cal B}\left( z^{\prime
}\right) }\left( b^{\prime }\right) }\left[ m\left( a^{\prime }\right)
-1\right] \quad . 
$$
With the help of (\ref{pp4t1fo34}) we can split the sums in $ZS$ further:%
$$
ZS=\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right)
}\sum\limits_{b^{\prime }\,\in \sigma \left( z^{\prime },b\right)
}\sum\limits_{a^{\prime }\,\in \,\dot q_{{\cal B}\left( z^{\prime }\right)
}\left( b^{\prime }\right) }\left[ m\left( a^{\prime }\right) -1\right]
\quad . 
$$
Since\ $b\in z-\left( z\cap z^{\prime }\right) $\ and\ $b^{\prime }\in
\sigma \left( z^{\prime },b\right) $,\ we have\ $b\in \dot q_{{\cal B}\left(
z^{\prime }\right) }\left( b^{\prime }\right) $. But%
$$
\left\{ a^{\prime }\in \dot q_{{\cal B}\left( z^{\prime }\right) }\left(
b^{\prime }\right) \mid a^{\prime }\subset b\right\} =\dot q_{{\cal B}\left(
z^{\prime },b\right) }\left( b^{\prime }\right) \quad , 
$$
and%
$$
\left\{ a^{\prime }\in \dot q_{{\cal B}\left( z^{\prime }\right) }\left(
b^{\prime }\right) \mid a^{\prime }\supsetneqq b\right\}
=\dot q_{{\cal B}\left( z\right) }\left( b\right) 
$$
for all\ $b^{\prime }\in \sigma \left( z^{\prime },b\right) $ and $b\in
z-\left( z\cap z^{\prime }\right) $. Hence we can write%
$$
\dot q_{{\cal B}\left( z^{\prime }\right) }\left( b^{\prime }\right)
=\left\{ a^{\prime }\in \dot q_{{\cal B}\left( z^{\prime }\right) }\left(
b^{\prime }\right) \mid a^{\prime }\supsetneqq b\right\}
\cup \left\{ a^{\prime }\in \dot q_{{\cal B}\left( z^{\prime }\right)
}\left( b^{\prime }\right) \mid a^{\prime }\subset b\right\} = 
$$
$$
=\;\dot q_{{\cal B}\left( z\right) }\left( b\right) \;\cup \;\dot q_{{\cal B}%
\left( z^{\prime },b\right) }\left( b^{\prime }\right) \quad . 
$$
This yields%
$$
ZS=\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right)
}\sum\limits_{b^{\prime }\,\in \sigma \left( z^{\prime },b\right) }\left\{
\sum\limits_{a^{\prime }\,\in \dot q_{{\cal B}\left( z\right) }\left(
b\right) }\left[ m\left( a^{\prime }\right) -1\right] \quad
+\sum\limits_{a^{\prime }\,\in \dot q_{{\cal B}\left( z^{\prime },b\right)
}\left( b^{\prime }\right) }\left[ m\left( a^{\prime }\right) -1\right]
\right\} = 
$$
$$
=\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right)
}\sum\limits_{b^{\prime }\,\in \sigma \left( z^{\prime },b\right) }e_{{\cal B%
}\left( z\right) }\left( b\right) \;+\sum\limits_{b\,\in \,z-\left( z\cap
z^{\prime }\right) }\sum\limits_{b^{\prime }\,\in \sigma \left( z^{\prime
},b\right) }e_{{\cal B}\left( z^{\prime },b\right) }\left( b^{\prime
}\right) = 
$$
$$
=\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right) }\#\sigma \left(
z^{\prime },b\right) \cdot e_{{\cal B}\left( z\right) }\left( b\right)
\;+\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right)
}\sum\limits_{b^{\prime }\,\in \sigma \left( z^{\prime },b\right) }e_{{\cal B%
}\left( z^{\prime },b\right) }\left( b^{\prime }\right) \quad , 
$$
where we have used definition (\ref{pp4t1fo26}) for $e_{{\cal B}\left(
z\right) }\left( b\right) $. We now insert $ZS$ into the expression for $%
G\left( z^{\prime }\right) $:%
$$
G\left( z^{\prime }\right) =\sum\limits_{b^{\prime }\,\in z^{\prime }\cap
z}e_{{\cal B}\left( z^{\prime }\right) }\left( b^{\prime }\right)
\;+\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right) }\#\sigma \left(
z^{\prime },b\right) \cdot e_{{\cal B}\left( z\right) }\left( b\right) \;+ 
$$
$$
+\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right)
}\sum\limits_{b^{\prime }\,\in \sigma \left( z^{\prime },b\right) }e_{{\cal B%
}\left( z^{\prime },b\right) }\left( b^{\prime }\right) = 
$$
$$
=\sum\limits_{b\in z\cap z^{\prime }}e_{{\cal B}\left( z^{\prime }\right)
}\left( b\right) \;+\sum\limits_{b\in z-\left( z\cap z^{\prime }\right) }e_{%
{\cal B}\left( z\right) }\left( b\right) \;+ 
$$
\begin{equation}
\label{pp4t1fo39}+\;\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right)
}\left[ \#\sigma \left( z^{\prime },b\right) -1\right] \cdot e_{{\cal B}%
\left( z\right) }\left( b\right) \;+\sum\limits_{b\,\in \,z-\left( z\cap
z^{\prime }\right) }G_{{\cal B}\left( z^{\prime },b\right) }\quad . 
\end{equation}
Now observe that $e_{{\cal B}\left( z^{\prime }\right) }\left( b\right) =e_{%
{\cal B}\left( z\right) }\left( b\right) $, since in both cases the counting
stops at $b$. Thus,%
$$
\sum\limits_{b\in z\cap z^{\prime }}e_{{\cal B}\left( z^{\prime }\right)
}\left( b\right) +\sum\limits_{b\in z-\left( z\cap z^{\prime }\right) }e_{%
{\cal B}\left( z\right) }\left( b\right) = 
$$
$$
=\sum\limits_{b\in z}e_{{\cal B}\left( z\right) }\left( b\right) =G\left(
z\right) \quad . 
$$
Now (\ref{pp4t1fo37}) follows from (\ref{pp4t1fo39}).

\underline{Ad B :}\quad For $z=z_{\min }\left( X\right) $, $z^{\prime
}=z_{\max }\left( X\right) $ we have $G\left( z\right) =m\left( X\right)
\cdot \left[ m\left( X\right) -1\right] $, as follows from definition (\ref
{pp4t1fo27}), and $G\left( z^{\prime }\right) =G_{{\cal B}\left( X\right) }$%
. Furthermore,\ for $b\in z-\left( z\cap z^{\prime }\right) $ we have \ $%
\sigma \left( z^{\prime },b\right) =z_{\max }\left( b\right) \in \zeta
\left( b\right) $, and\ $G_{{\cal B}\left( z^{\prime },b\right) }=G_{{\cal B}%
\left( X,b\right) }$. Inserting these expressions into (\ref{pp4t1fo37}) we
obtain%
$$
G_{{\cal B}\left( X\right) }=m\left( X\right) \cdot \left[ m\left( X\right)
-1\right] + 
$$
$$
+\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right) }\left[ \#z_{\max
}\left( b\right) -1\right] \,\cdot \,\stackunder{=\;m\left( X\right) -1}{%
\underbrace{e_{{\cal B}\left( X\right) }\left( b\right) }}%
+\sum\limits_{b\,\in \,z-\left( z\cap z^{\prime }\right) }G_{{\cal B}\left(
X,b\right) }= 
$$
$$
=\left[ m\left( X\right) -1\right] \cdot \left\{ m\left( X\right)
+\sum\limits_{b\,\in \,z_{\min }\left( X\right) }\left[ \#z_{\max }\left(
b\right) -1\right] \right\} + 
$$
$$
+\sum\limits_{b\,\in z_{\min }\left( X\right) }G_{{\cal B}\left( X,b\right)
}\quad . 
$$
Here we have allowed $b$ running over elements in $z\cap z^{\prime }$ as
well; this is admissible, since for these elements $\#z_{\max }\left(
b\right) =1$, which makes their contribution to the first sum vanish; and
furthermore, $G_{{\cal B}\left( X,b\right) }=0$, since the associated tree $%
{\cal B}\left( X,b\right) $ is trivial, ${\cal B}\left( X,b\right) =\left\{
b\right\} $. Now the first and third term in the $\left\{ \cdot \right\} $%
-brackets cancel, since $\#z_{\min }\left( X\right) =m\left( X\right) $.
This proves (\ref{pp4t1fo38}). \TeXButton{BWE}
{\hfill
\vspace{2ex}
$\blacksquare$}

\subsection{Splitting lemma 2}

Let ${\cal B}^{\prime }\left( X\right) $ be the reduced tree ${\cal B}%
^{\prime }\left( X\right) =\left[ {\cal B}\left( X\right) -{\cal B}\left(
X,b\right) \right] \cup \left\{ b\right\} $ such that ${\cal B}\left(
X\right) $ is the union of the trees ${\cal B}^{\prime }\left( X\right) $
and ${\cal B}\left( X,b\right) $. Then 
\begin{equation}
\label{pp4t1fo40}G_{{\cal B}\left( X\right) }=G_{{\cal B}^{\prime }\left(
X\right) }+\left[ \#z_{\max }\left( b\right) -1\right] \cdot e_{{\cal B}%
^{\prime }\left( X\right) }\left( b\right) +G_{{\cal B}\left( X,b\right)
}\quad , 
\end{equation}
where $z_{\max }\left( b\right) $ is the maximal partition of $b$ in the
full tree ${\cal B}\left( X\right) $.

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

Let $z_{\max }\left( X\right) =\left\{ u_1,\ldots ,u_k,v_2,v_3,\ldots
\right\} $ be the maximal partition of $X$ in ${\cal B}\left( X\right) $.
Without loss of generality we may assume that $\left\{ u_1,\ldots
,u_k\right\} $ is a partition of $b$, so that $\left\{ b,v_2,v_3,\ldots
\right\} \equiv z_{\max }^{\prime }$ is the maximal partition in the reduced
tree ${\cal B}^{\prime }\left( X\right) $. Now apply (\ref{pp4t1fo37}) with $%
z^{\prime }=z_{\max }$, $z=z_{\max }^{\prime }$. Inserting $G\left(
z^{\prime }\right) =G_{{\cal B}\left( X\right) }$, $G\left( z\right) =G_{%
{\cal B}^{\prime }\left( X\right) }$, $\#\sigma \left( z^{\prime },b\right)
=k=\#z_{\max }\left( b\right) $ and $G_{{\cal B}\left( z^{\prime },b\right)
}=G_{{\cal B}\left( X,b\right) }$ yields the result. \TeXButton{BWE}
{\hfill
\vspace{2ex}
$\blacksquare$}

The next theorem is the first main statement about the properties of amount
functions, in that it expresses the total amount of a tree as a function of
the pairs of numbers $\left[ n\left( b\right) ,m\left( b\right) \right] $ at
every {\it nod }$b\in {\cal B}\left( X\right) $ in the tree. To this end we
first define

\section{The tree function $E_{{\cal B}\left( X\right) }$ \label{TreeFunc}}

Let ${\cal B}\left( X\right) $ be a tree structure over $X$. The {\it tree
function }$E_{{\cal B}\left( X\right) }$ of the tree ${\cal B}$ is defined
to be the sum of the expressions $n\left( b\right) \cdot \left[ m\left(
b\right) -1\right] $ associated with the characters $\left( n,m\right) $ of
a nod $b$ {\bf over all nods} in the tree, 
\begin{equation}
\label{pp4t1int10}E_{{\cal B}\left( X\right) }\equiv \sum_{b\in {\cal B}%
\left( X\right) }n\left( b\right) \cdot \left[ m\left( b\right) -1\right]
\quad . 
\end{equation}

\subsection{Theorem: Tree function and total amount \label{TrF}}

Let ${\cal B}\in {\cal MB}\left( X\right) $.

\begin{description}
\item[A)]  Let $z_{\max }\left( X\right) $ denote the maximal partition of $%
X $ in ${\cal B}\left( X\right) $, and let $e_{{\cal B}\left( X\right)
}\left( b\right) $ denote the amount of the path $q\left( b\right) $ of $%
b\in {\cal B}\left( X\right) $ as defined in (\ref{pp4t1fo26}). Then 
\begin{equation}
\label{pp4t1fo41}E_{{\cal B}\left( X\right) }=\sum\limits_{b\in z_{\max
}\left( X\right) }n\left( b\right) \cdot e_{{\cal B}\left( X\right) }\left(
b\right) \quad . 
\end{equation}

\item[B)]  If the tree ${\cal B}\left( X\right) $ is complete, then $n\left(
b\right) =1$ for all $b\in z_{\max }\left( X\right) $, and hence 
\begin{equation}
\label{pp4t1fo42}E_{{\cal B}\left( X\right) }=\sum\limits_{b\in z_{\max
}\left( X\right) }e_{{\cal B}\left( X\right) }\left( b\right) =G_{{\cal B}%
\left( X\right) }\quad . 
\end{equation}
\end{description}

These results say that for a complete tree, the tree function coincides with
the total amount in the tree, whereas if the tree is incomplete, then the
tree function renders a {\bf weighted sum} of the path amounts $e_{{\cal B}%
\left( X\right) }\left( b\right) $, the weights being the cardinality of the
non-resolvable elements $b\in z_{\max }\left( B\right) $ in the incomplete
tree.

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

We first prove $\left( B\right) $ by induction with respect to $n\equiv \#X$%
. The case $n=1$ is trivially satisfied.

$\underline{n=2:}\quad $If ${\cal B}\left( X\right) $ is the trivial tree $%
\left\{ X\right\} $, then $G_{{\cal B}}=0$, and (\ref{pp4t1fo42}) is
satisfied. The only other possible tree is ${\cal B}\left( X\right) =\left\{
X,\left\{ x_1\right\} ,\left\{ x_2\right\} \right\} $, with\ $m\left(
X\right) =2$. The amount of $G_{{\cal B}}$ is $2$, which coincides with the
RHS of (\ref{pp4t1fo42}), as $2\left( 2-1\right) +1\left( 1-0\right)
+1\left( 1-0\right) =2$ .

$\underline{2\le n-1,\;n-1\rightarrow n:}\quad $We assume that (\ref
{pp4t1fo42}) is valid for all possible sets\ $X$\ with \ $\#X=k\in \left\{
2,\ldots \,,n-1\right\} $. We prove that (\ref{pp4t1fo42}) is valid for
sets\ $X$\ with\ $\#X=n$. To this end we decompose\ ${\cal B}\left( X\right) 
$\ into three disjoint subsets, defined by%
$$
{\cal K}_1\equiv \left\{ X\right\} \quad ,\quad {\cal K}_2\equiv \left\{
b\in z_{\min }\left( X\right) \mid \#b=1\right\} \quad ,\quad {\cal K}%
_3\equiv \bigcup\limits_{a\in z_{\min }\left( X\right) :\#a>1}{\cal B}\left(
X,a\right) \quad . 
$$
$\left\{ {\cal K}_1,{\cal K}_2,{\cal K}_3\right\} $ clearly is a partition
of ${\cal B}\left( X\right) $. Now consider the sum on the RHS of (\ref
{pp4t1fo42}). We split it according to $\sum\limits_{b\in {\cal B}\left(
X\right) }=\sum\limits_{b\in {\cal K}_1}+\sum\limits_{b\in {\cal K}%
_2}+\sum\limits_{b\in {\cal K}_3}$. This gives contributions%
$$
\sum\limits_{b\in {\cal K}_2}n\left( b\right) \cdot \left[ m\left( b\right)
-1\right] =0\quad , 
$$
and 
$$
\sum\limits_{b\in {\cal K}_1}n\left( b\right) \cdot \left[ m\left( b\right)
-1\right] =n\left( X\right) \cdot \left[ m\left( X\right) -1\right] \quad . 
$$
Since all subtrees\ ${\cal B}\left( X,a\right) $\ are disjoint, we have%
$$
\sum\limits_{b\in {\cal K}_3}n\left( b\right) \cdot \left[ m\left( b\right)
-1\right] =\sum\limits_{a\in {\cal z}_{\min }\left( X\right)
\,:\,\#a>1}\sum\limits_{b\in {\cal B}\left( X,a\right) }n\left( b\right)
\cdot \left[ m\left( b\right) -1\right] \quad . 
$$
According to assumption, however, we have%
$$
\sum\limits_{b\in {\cal B}\left( X,a\right) }n\left( b\right) \cdot \left[
m\left( b\right) -1\right] =G_{{\cal B}\left( X,a\right) } 
$$
for all subtrees ${\cal B}\left( X,a\right) $. Putting the last four
formulae together we deduce%
$$
\sum\limits_{b\in {\cal B}\left( X\right) }n\left( b\right) \cdot \left[
m\left( b\right) -1\right] =n\left( X\right) \cdot \left[ m\left( X\right)
-1\right] +\sum\limits_{a\in z_{\min }\left( X\right) }G_{{\cal B}\left(
X,a\right) }\quad , 
$$
where $a$ can range over the whole of $z_{\min }\left( X\right) $, since $G_{%
{\cal B}\left( X,a\right) }=0$, if $\#a=1$. But (\ref{pp4t1fo38b}) in
splitting lemma 1 says that the RHS of the last equation is $G_{{\cal B}%
\left( X\right) }$.

Now prove part $\left( A\right) $ of the theorem. Let ${\cal B}$ be an
arbitrary tree over $X$, let ${\cal B}_c$ be an arbitrary, but fixed,
completion of ${\cal B}$. The total amount of ${\cal B}$ is $G_{{\cal B}}$,
the total amount of ${\cal B}_c$ will be written as $G_c$. We now observe
that the completion ${\cal B}_c$ can be regarded as the union of ${\cal B}=%
{\cal B}\left( X\right) $ with all subtrees ${\cal B}_c\left( b\right) $,
where $b\in z_{\max }\left( X\right) $ ranges in the maximal partition $%
z_{\max }\left( X\right) $ of $X$ {\bf in }${\cal B}$ (whereas the maximal
partition $\widetilde{z_{\max }\left( X\right) }$ of $X$ in ${\cal B}_c$ is
complete). Here ${\cal B}_c\left( b\right) $ is defined to be the subtree of 
$b$ in the completion ${\cal B}_c$ of ${\cal B}$. Hence we can apply
splitting lemma 2, (\ref{pp4t1fo40}), repeatedly; this gives 
\begin{equation}
\label{pp4t1fo43}G_c=G_{{\cal B}}+\sum_{b\in z_{\max }\left( X\right)
}\left[ \#\widetilde{z_{\max }\left( b\right) }-1\right] \cdot e_{{\cal B}%
\left( X\right) }\left( b\right) +\sum_{b\in z_{\max }\left( X\right) }G_{%
{\cal B}_c\left( b\right) }\quad , 
\end{equation}
where $\widetilde{z_{\max }\left( b\right) }$ is the maximal partition of $b$
in the complete tree ${\cal B}_c$. Now we denote the characters of a nod $%
b\in {\cal B}$ by $n\left( b\right) ,m\left( b\right) $, and the characters
of a nod $b\in {\cal B}_c$ by $\tilde n\left( b\right) ,\tilde m\left(
b\right) $. Then we have $\#\widetilde{z_{\max }\left( b\right) }=\#b=\tilde
n\left( b\right) =n\left( b\right) $. In equation (\ref{pp4t1fo43}), the LHS
and the last sum on the RHS belong to complete trees; hence we can apply the
result from part $\left( B\right) $ immediately, and (\ref{pp4t1fo43})
becomes%
$$
\sum_{b\in {\cal B}_c}\tilde n\left( b\right) \cdot \left[ \tilde m\left(
b\right) -1\right] =G_{{\cal B}}+\sum_{b\in z_{\max }\left( X\right) }\left[
n\left( b\right) -1\right] \cdot e_{{\cal B}\left( X\right) }\left( b\right)
+ 
$$
$$
+\sum_{b\in z_{\max }\left( X\right) }\sum_{a\in {\cal B}_c\left( b\right)
}\tilde n\left( a\right) \cdot \left[ \tilde m\left( a\right) -1\right]
\quad . 
$$
Now ${\cal B}_c$ can be partitioned as ${\cal B}_c=\left[ {\cal B}-z_{\max
}\left( X\right) \right] \cup \left[ \bigcup_{b\in z_{\max }\left( X\right) }%
{\cal B}_c\left( b\right) \right] $; if the sum on the LHS of the last
equation is decomposed accordingly, we see that the contribution from the
second part in the partition of ${\cal B}_c$ cancels the last expression on
the RHS, so that we obtain 
\begin{equation}
\label{pp4t1fo44}\sum_{b\in {\cal B}-z_{\max }\left( X\right) }\tilde
n\left( b\right) \cdot \left[ \tilde m\left( b\right) -1\right] =G_{{\cal B}%
}+\sum_{b\in z_{\max }\left( X\right) }\left[ n\left( b\right) -1\right]
\cdot e_{{\cal B}\left( X\right) }\left( b\right) \quad . 
\end{equation}
Now observe that elements $b\in z_{\max }\left( X\right) $ are
non-resolvable in the tree ${\cal B}$, and therefore have $\tilde m\left(
b\right) =1$, as opposed to $m\left( b\right) \ge 1$. Thus we can extend the
sum on the LHS of the last formula to range in all of ${\cal B}$; but then
this sum obviously represents the tree function of the tree ${\cal B}$,%
$$
\sum_{b\in {\cal B}}\tilde n\left( b\right) \cdot \left[ \tilde m\left(
b\right) -1\right] =\sum_{b\in {\cal B}}n\left( b\right) \cdot \left[
m\left( b\right) -1\right] =E_{{\cal B}}\quad . 
$$
Furthermore, recall from (\ref{pp4t1fo27}) that the total amount $G_{{\cal B}%
}$ of ${\cal B}$ is given as%
$$
G_{{\cal B}}\equiv \sum_{b\,\in z_{\max }\left( X\right) }e_{{\cal B}\left(
X\right) }\left( b\right) \quad . 
$$
Thus, formula (\ref{pp4t1fo44}) becomes%
$$
E_{{\cal B}}=G_{{\cal B}}+\sum_{b\in z_{\max }\left( X\right) }n\left(
b\right) \cdot e_{{\cal B}\left( X\right) }\left( b\right) \;-G_{{\cal B}%
}\quad , 
$$
which yields (\ref{pp4t1fo41}). This proves theorem \ref{TrF}. 
\TeXButton{BWE}{\hfill
\vspace{2ex}
$\blacksquare$}

%\chapter{Minimal Problems in Tree Structures}

\section{Minimal classes \label{MinClass}}

We now come to discuss the problem of minimizing the tree function on
certain sets of tree structures. We will need a couple of new notions which
we introduce in the sequel.

Let ${\cal MB}\left( X\right) $\ be the set of all tree structures over\ $X$%
. To every\ ${\cal B}\in {\cal MB}\left( X\right) $\ we can uniquely assign
the minimal partition $z_{\min }\left( X\right) $ induced by ${\cal B}$ on $%
X $; this assignment will be denoted by $z_{\min }:{\cal MB}\left( X\right)
\rightarrow \zeta \left( X\right) $, ${\cal B}\mapsto z_{\min }\left( {\cal B%
}\right) \equiv z_{\min }\left( X\right) $ in ${\cal B}$. Given $z\in \zeta
\left( X\right) $, the inverse image\ $z_{\min }^{-1}\left( z\right) $\ is
the set of all tree structures ${\cal B}$ over $X$ with the same minimal
partition $z_{\min }\left( X\right) $ of $X$.

Let $n=\#X$. For\ $1\le m\le n$,\ let ${\cal M}\left( X,m\right) $ denote
the set of all tree structures over $X$ whose minimal partition $z_{\min
}\left( X\right) $ contains $m$ elements. Since all ${\cal M}\left(
X,m\right) $ are disjoint, this defines a partition of ${\cal MB}\left(
X\right) $, 
\begin{equation}
\label{pp4t2fo3}{\cal MB}\left( X\right) =\bigcup\limits_{1\le m\le n}{\cal M%
}\left( X,m\right) \quad . 
\end{equation}

Recall that the tree function $E:{\cal MB}\left( X\right) \rightarrow 
\TeXButton{N}{\mathbb{N}}_{+}$ sends every tree over $X$ to the sum over all 
$n\left( b\right) \left[ m\left( b\right) -1\right] $, as $b$ ranges through
all nodes in the tree. We are interested in the minima of this map, as $E$
is restricted to certain subsets of ${\cal MB}\left( X\right) $. We observe
that it makes no sense to ask for the global minimum of $E$ on ${\cal MB}%
\left( X\right) $, as the answer is trivial: In this case the minimum
clearly is taken on the trivial tree ${\cal B}=\left\{ X\right\} $, since $%
E_{{\cal B}}=G_{{\cal B}}=0$. Meaningful results are obtained, however, if
we first focus on the subset of all {\bf complete} trees ${\cal CP}\left(
X\right) \subset {\cal MB}\left( X\right) $; this inclusion is proper for $%
\#X\ge 2$. We write ${\cal CP}\left( X,m\right) $ for the set of all
complete trees with $m$ elements in the minimal partition of $X$. On the
complete trees, the tree function $E$ coincides with the global amount $G$,
as follows from theorem \ref{TrF}. Now we define 
\begin{equation}
\label{pp4t2fo5}\min \left( n\right) \equiv \min _{{\cal B}\in {\cal CP}%
\left( X\right) }E\left( {\cal B}\right) \quad , 
\end{equation}
and 
\begin{equation}
\label{pp4t2fo6}\min \left( n,m\right) \equiv \min _{{\cal B}\in {\cal CP}%
\left( X,m\right) }E\left( {\cal B}\right) \quad . 
\end{equation}
Clearly, as indicated in the notation, $\min \left( n\right) $ is a function
of $n$ only, and $\min \left( n,m\right) $ is a function of $n$ and $m$
only. These minima exist, since all tree functions take their values in the
non-negative natural numbers. Thus it makes sense to speak of the set of all
trees ${\cal MIN}\left( X\right) \equiv E^{-1}\left( \min \left( X\right)
\right) \cap {\cal CP}\left( X\right) $, on which the tree function $E$
actually attains its minimum. Similarly, introduce 
\begin{equation}
\label{pp4t2fo8}{\cal MIN}\left( X,m\right) \equiv E^{-1}\left( \min \left(
X,m\right) \right) \cap {\cal CP}\left( X,m\right) \quad . 
\end{equation}
We term ${\cal MIN}\left( X\right) \;$the {\it global minimal class in \ }$%
{\cal CP}\left( X\right) $. ${\cal MIN}\left( X,m\right) $\ will be called 
{\it minimal class in\ }${\cal CP}\left( X,m\right) $.

We now can prove:

\subsection{Proposition \label{min1}}

Let\ ${\cal B}\in {\cal CP}\left( X,m\right) $. Then ${\cal B}\in {\cal MIN}%
\left( X,m\right) $ if and only if ${\cal B}\left( X,b\right) \in {\cal MIN}%
\left( b\right) $ for all $b\in z_{\min }\left( X\right) $.

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

From definition (\ref{pp4t1int10}) we deduce 
\begin{equation}
\label{pp4t2fo9a}E_{{\cal B}}=n\left( X\right) \cdot \left[ m\left( X\right)
-1\right] +\sum_{b\in z_{\min }\left( X\right) }E_{{\cal B}\left( X,b\right)
}\quad , 
\end{equation}
from which the statement follows immediately. \TeXButton{BWE}
{\hfill
\vspace{2ex}
$\blacksquare$}

--- We now subsequently will approach the problem of finding the minima of $%
E $ on the sets (\ref{pp4t2fo5}, \ref{pp4t2fo6}) introduced above.

\section{Divisions \label{divisions}}

Given a natural number $n$, we can decompose $n$ into $m$ terms according to 
$n=n_1+\cdots +n_m$ with $1\le m\le n$ in many different ways, and for
values of $m$ ranging from $1$ to $n$. We observe that, for a given $m$, the
numbers $n_i$ can range between $0$ and $n$, and that the $n_i$ need not be
mutually different. A decomposition of $n$ in this form will be called a 
{\it division of }$n${\it \ into }$m$ {\it terms}. We can regard it as an $m$%
-tupel $u=\left( n_1,\ldots ,n_m\right) $ with nonnegative integer
components, $n_i\ge 0$, such that $\sum n_i=n$. The set of all divisions of $%
n$ into $m$ terms will be denoted by $U\left( n,m\right) $. If $n$ is fixed
and $m$ varies from $1$ to $n$, the collection of all $U\left( n,m\right) $
defines a partition of the {\it set of all divisions} $U\left( n\right) $%
{\it \ of }$n$, 
\begin{equation}
\label{pp4t2fo11}U\left( n\right) \equiv \bigcup\limits_{1\le m\le n}U\left(
n,m\right) \quad . 
\end{equation}
We introduce the {\it trivial division}\ $u_0=\left( n\right) $, and denote
the set of all nontrivial divisions of $n$ by $U^{*}\left( n\right) $, $%
U^{*}\left( n\right) =U\left( n\right) -\left\{ u_{0}\right\} $.

$U\left( n,m\right) $ is a proper subset of 
\begin{equation}
\label{pp4t2div1}H\left( m,n\right) \equiv \left\{ u\in \TeXButton{R}
{\mathbb{R}}^m\mid \sum_{i=1}^mu_i=n\right\} \subset \TeXButton{R}
{\mathbb{R}}^m\quad , 
\end{equation}
which is a hyperplane in $\TeXButton{R}{\mathbb{R}}^m$ whose least
(Euclidean) distance to the origin is $\frac n{\sqrt{m}}$. The element of $%
H\left( m,n\right) $ associated with the least distance will be denoted by $%
\bar u$; it has components $\bar u=\left( \frac nm,\ldots ,\frac nm\right) $%
. Usually, $n/m$ is not integer, so that $\bar u\not \in U\left( n,m\right) $%
. However, there are always elements $\bar n$ of $U\left( n,m\right) $ that
come closest to $\bar u$. The minimal distance between these elements $\bar
n $ and $\bar u$ ranges between $0$ and $\frac{\sqrt{m}}2$. If $\bar u$
coincides with a point in $U\left( n,m\right) $, then $\bar n=\bar u$ is
uniquely defined. The bigger the distance between $\bar u$ and lattice
points, the more elements $\bar n$ there are. If $\bar u$ lies in the center
of a cube formed by elements of $U\left( n,m\right) $, then there are $2^m$
candidates for $\bar n$, their distance from $\bar u$ being $\frac{\sqrt{m}}%
2 $ precisely. In this case $m$ must be even, as follows from $\frac
nm+\frac 12\in \TeXButton{Z}{\mathbb{Z}}$. Whenever there is more than one $%
\bar n$, they must be related by permutation of components.

There is another way to describe a division $n=n_1+\cdots +n_m$; this is in
terms of {\it occupation numbers }$t_k$ for all natural numbers $k$ between $%
0$ and $n$ (and in turn, even beyond), which express how often $k$ appears
as one of the terms $n_i$ in a given decomposition of $n$. Obviously, the
description of a decomposition of $n$ into $m$ terms is determined by the
set of occupation numbers $\left( t_1,t_2,\ldots \right) $ uniquely up to
permutation of the terms $n_i$ in the sum. Here comes the detailed
definition:

Let $n\in \TeXButton{N}{\mathbb{N}}$, let $1\le m\le n$. The $n$-tupel\ $%
t\equiv \left( t_1,t_2,\ldots \,\right) \in \TeXButton{N}{\mathbb{N}}%
_0\times \TeXButton{N}{\mathbb{N}}_0\times \cdots $ will be called {\it %
occupation numbers of the} {\it division of }$n$ {\it into }$m${\it \ terms}%
, if $t$ satisfies 
\begin{equation}
\label{pp4t2fo10}\sum\limits_{k=1}^nt_k=m\quad \text{\underline{and}\quad }%
\sum\limits_{k=1}^nk\cdot t_k=n\quad . 
\end{equation}
The first sum says that the number of terms in the decomposition of $n$ is $%
m $; the second sum is just the decomposition of $n$. Clearly, for $k>n$ all
occupation numbers $t_k$ must vanish. For this reason we will now focus on
the finite sequences $t=\left( t_1,t_2,\ldots ,t_n\right) $ of occupation
numbers rather than the infinite ones, so that $t$ ranges in $\TeXButton{N}
{\mathbb{N}}_0^n$.

The trivial division as expressed by occupation numbers is $t_0\equiv \left(
0,\ldots \,,0,1\right) $, i.e. $t_n=1$, and all other components vanishing.
The set of all occupation numbers of divisions of $n$ into $m$ terms will be
denoted by $T\left( n,m\right) $; the set of all occupation numbers of
divisions of $n$ will be written as $T\left( n\right) $. The occupation
numbers of nontrivial divisions comprise the set $T^{*}\left( n\right) $.
Clearly, $t_n=0$ for every nontrivial $t\in T^{*}\left( n\right) $.

The relation between divisions $u$ and their associated occupation numbers $%
t $ is as follows: Every division $u=\left( n_1,\ldots ,n_m\right) $ defines
a unique $n$-tupel of occupation numbers $\kappa \left( u\right) =\left(
t_1,\ldots ,t_n\right) $ by 
\begin{equation}
\label{pp4t2div2}t_a\equiv \sum_{i=1}^m\delta _{a,n_i}\quad ; 
\end{equation}
it follows readily that this indeed satisfies (\ref{pp4t2fo10}).
Furthermore, every $n$-tupel $t$ of occupation numbers defines a division $u$
of $n$ by $m$ according to the following scheme: First identify $m\equiv
\sum_{a=1}^nt_a$, with\ $1\le m\le n$; next, for every\ $j\in \left\{
1,\ldots \,,m\right\} $, define 
\begin{equation}
\label{pp4t2div3}u_j\equiv \min \left\{ a\in \left\{ 1,\ldots \,,n\right\}
\mid \sum\limits_{i=1}^at_i\ge j\right\} \quad ; 
\end{equation}
it is easy to see that $\sum u_j=m$, where $u_j\le n$ by construction. Also
by construction we see that the $u_j$ are naturally ordered, $u_1\le u_2\le
\cdots \le u_m$. Now the inverse image $\kappa ^{-1}\left( t\right) $ of an
occupation number $t$ is just the set of all divisions $u^{\prime }$ that
differ from the naturally ordered division $u$ constructed above by
permutation of components. Thus, every such inverse image has a naturally
ordered representative. We conclude that there is a 1--1 relation between
naturally ordered divisions of $n$ and occupation numbers.

\section{Partitions and divisions \label{PartAndDiv}}

Let $n=\#X$, let $z$ be an arbitrary partition of $X$, not necessarily
related to a tree structure over $X$. Assume that the partition $z$ contains 
$m$ elements, $m\equiv \#z$, where $z=\left\{ b_1,\ldots \,,b_m\right\} $. $%
z $ defines a division $u\left( z\right) $ of $n$ into $m$ terms by $%
u=\left( \#b_1,\ldots ,\#b_m\right) $. This defines the $u${\it -map }$u:%
{\cal Z}\left( X\right) \rightarrow U\left( n\right) $, $z\mapsto u\left(
z\right) $. The associated occupation number will be written as $\tau \left(
z\right) $ and has components 
\begin{equation}
\label{pp4t2fo12}t_a\equiv \sum\limits_{b\,\in z}\delta _{a\,,\,\#b} 
\end{equation}
for $a=1,\ldots ,n$. $t_a$\ will be called the $a$-{\it th occupation number
of the partition }$z$. This defines the $\tau $-{\it map }$\tau :{\cal Z}%
\left( X\right) \rightarrow T\left( n\right) $, $z\mapsto \tau \left(
z\right) $; it sends every partition of $X$ to the associated $n$-tupel of
occupation numbers. The $u$-, $\tau $-maps are obviously surjective, since
for every division of $n$ into $m$ terms one can construct an associated 
partition of $X$.

From the surjectivity of $u$ and $\tau $ and the fact that the map $z_{\min
} $ sends ${\cal MB}\left( X\right) $ onto the set of all partitions ${\cal Z%
}\left( X\right) $ we find $U\left( n\right) =\left( u\circ z_{\min }\right)
\left( {\cal MB}\left( X\right) \right) $ and $T\left( n\right) =\left( \tau
\circ z_{\min }\right) \left( {\cal MB}\left( X\right) \right) $, and
furthermore, $U\left( n,m\right) =\left( u\circ z_{\min }\right) \left( 
{\cal M}\left( X,m\right) \right) $ and $T\left( n,m\right) =\left( \tau
\circ z_{\min }\right) \left( {\cal M}\left( X,m\right) \right) $.

The distinct occupation number $t_{\min }\left( X\right) \equiv \left( \tau
\circ z_{\min }\right) \left( X\right) $ will be called the {\it minimal
division of }$n=\#X$ in ${\cal B}$.

--- For\ $n\in \TeXButton{N}{\mathbb{N}}_0$ ,\ $m\in \TeXButton{N}
{\mathbb{N}}$\ let 
\begin{equation}
\label{pp4t2fo19}\left[ \frac nm\right] \equiv \max \left\{ n^{\prime }\in 
\TeXButton{N}{\mathbb{N}}_0\mid n^{\prime }\cdot m\le n\right\} 
\end{equation}
denote the {\it integer quotient of }$n$ by $m$.

\section{Optimal division \label{OptDiv}}

Let\ $1\le m\le n$. Let $\nu =\left[ \frac nm\right] $ be the integer
quotient of $n$ by $m$; then $n=\nu \cdot m+r$ for $r<m$. We construct a
division of $n$ into $m$ terms according to 
\begin{equation}
\label{pp4t2opt1}\left( \nu ,\ldots ,\nu ,\nu +1,\ldots ,\nu +1\right) \quad
, 
\end{equation}
with $\left( m-r\right) $ occurrences of $\nu $ and $r$ occurrences of $%
\left( \nu +1\right) $. The associated occupation number is denoted as $\bar
t\equiv \bar t\left( n,m\right) =\left( \bar t_1,\ldots \,,\bar t_n\right) $%
, with $\bar t_\nu =m-r$, $\bar t_{\nu +1}=r$, and $\bar t_\lambda =0$ for $%
\lambda \not \in \left\{ \nu ,\nu +1\right\} $. Consider the inverse image $%
\kappa ^{-1}\left( \bar t\right) $ of $\bar t$ under $\kappa $; every
representative of this set will be called {\it optimal division of }$n$ by $%
m $, and be denoted by $\bar n$. Obviously, the optimal divisions come
closest to the $m$-tupel $\bar u=\left( \frac nm,\ldots ,\frac nm\right) \in
H\left( m,n\right) \subset \TeXButton{R}{\mathbb{R}}^m$, where $\bar u$ is
the element in $H\left( m,n\right) $ with least distance to the origin;
thus, they coincide with the objects $\bar n$ introduced in section \ref
{divisions}. We observe that $\kappa ^{-1}\left( \bar t\right) $ is the set
of all elements $\bar n$ of $U\left( n,m\right) $ for which the Euclidean
norm 
\begin{equation}
\label{pp4t2opt1a}\left\| \bar n-\bar u\right\| \le \frac{\sqrt{m}}2\quad . 
\end{equation}

We now prove an important lemma about optimal divisions:

\subsection{Lemma on optimal divisions \label{LmOptDiv}}

Let $\left\| u\right\| =\sqrt{\sum_{i=1}^m\left( u_i\right) ^2}$ denote the
Euclidean norm of an element $u\in \TeXButton{R}{\mathbb{R}}^m$. Let $%
u=\left( n_1,\ldots ,n_m\right) $ be an element of $U\left( n,m\right) $.
Then there exists a {\bf finite} sequence $u^0,u^1,\ldots ,u^f$ of elements
in $U\left( n,m\right) $ with $u^0\equiv u$, $u^f=\bar n$ for some $\bar
n\in \kappa ^{-1}\left( \bar t\right) $, such that 
\begin{equation}
\label{pp4t2opt2}\left\| u^0\right\| \ge \left\| u^1\right\| \ge \cdots \ge
\left\| u^f\right\| \quad , 
\end{equation}
and the step $u^\alpha \mapsto u^{\alpha +1}$ involves alteration of {\bf two%
} components of $u^\alpha $ only.

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

Denote $M\equiv \left\{ 1,\ldots ,m\right\} $ for short. For $\left(
i,j\right) \in M^2$, $i\neq j$, we define an operation $S_{ij}$ on elements $%
u\in \TeXButton{R}{\mathbb{R}}^m$ by 
\begin{equation}
\label{pp4t2opt3}S_{ij}\left( u_1,\ldots ,u_m\right) \equiv \left(
u_1,\ldots ,u_i+1,\ldots ,u_j-1,\ldots ,u_n\right) \quad , 
\end{equation}
i.e. all components except $u_i$ and $u_j$ remain the same. By construction, 
$S_{ij}$ preserves $H\left( m,n\right) $, for if $u\in H\left( m,n\right) $,
then so is $S_{ij}u$.

--- We prove the statement: Let $u\in U\left( n,m\right) $, let $\Delta
\equiv u-\bar u$. If $\Delta _j-\Delta _i\le 1$ for all $i,j\in M^2$, then $%
u=\bar n\in \kappa ^{-1}\left( \bar t\right) $. {\it Proof:} If $\Delta =0$
then the statement is trivial; hence assume $\Delta \neq 0$. Let $\Delta
_{\max }$ denote the maximal element in $\left\{ \Delta _1,\ldots ,\Delta
_m\right\} $. $\Delta _{\max }$ is certainly $>0$; for $\sum \Delta _i=\sum
u_i-\sum \bar u_i=0$, and there must be nonzero components $\Delta _i$. Our
starting assumption says that $\Delta _{\max }-\Delta _i\in \left\{
0,1\right\} $. Thus, we have $\left( m-r\right) $ components of the form $%
\Delta _i=\Delta _{\max }$, and $r$ components of the form $\Delta _j=\Delta
_{\max }-1$, where $0\le r<m$. Their sum must vanish, hence $\Delta _{\max
}=\frac rm$. An easy computation now gives $\left\| \Delta \right\| ^2=\frac{%
r\left( m-r\right) }m$. If $m$ is fixed, the expression on the RHS is zero
for $r=0$, and becomes maximal for $r=\frac m2$, in which case it takes the
value $\frac{\sqrt{m}}2$. Hence $\left\| u-\bar u\right\| \le \frac{\sqrt{m}}%
2$, which implies that $u=\bar n$, by (\ref{pp4t2opt1a}). This proves the statement.

--- Now we prove our lemma. We describe step 1 in constructing the series (%
\ref{pp4t2opt2}): Let $\Delta ^0\equiv u^0-\bar u$. If $u^0=\bar n$, there
is nothing to prove. If $u^0\not \in \bar n$, we conclude from the above
statement that there exists a pair $\left( i,j\right) \in M^2$ with $i\neq j$
such that $\Delta _j^0-\Delta _i^0\ge 2$. Now define the new element $%
u^1\equiv S_{ij}u^0$ for this choice $\left( i,j\right) $. Let $\Delta
^1\equiv u^1-\bar u=S_{ij}\Delta ^0$. Since $\bar u$ has least distance to
the origin, it is perpendicular to the hyperplane $H\left( m,n\right) $,
whereas $\Delta ^0,\Delta ^1$ lie in this plane. Hence, by Pythagoras,%
$$
\quad \left\| u^0\right\| =\left\| \bar u\right\| +\left\| \Delta ^0\right\|
\quad ,\quad \left\| u^1\right\| =\left\| \bar u\right\| +\left\| \Delta
^1\right\| \quad , 
$$
or $\left\| u^1\right\| -\left\| u^0\right\| =\left\| S_{ij}\Delta
^0\right\| -\left\| \Delta ^0\right\| $. The last expression is just $%
2\left( \Delta _i^0-\Delta _j^0+1\right) $, which must be $\le -2$ owing to $%
\Delta _j^0-\Delta _i^0\ge 2$. Thus $\left\| u^1\right\| \le \left\|
u^0\right\| -2$, and only two components of $u^0$, namely $u_i^0$ and $u_j^0$
have been altered. This finishes step 1. In step 2 we check whether $%
u^1=\bar n$; if yes, the process terminates; if no, it continues in the same
manner. Since every step $\alpha $ involves a decrease of $\left\| \Delta
^\alpha \right\| $ by at least $-2$, the process must terminate after a
finite number of steps. \TeXButton{BWE}{\hfill
\vspace{2ex}
$\blacksquare$}

\section{Optimal trees \label{OptimalTrees}}

A tree structure ${\cal O}={\cal O}\left( X\right) $ over the set $X$ is
called {\it optimal over }$X$, if ${\cal O}$ is {\bf complete}, and%
$$
\tau \left( z_{\min }\left( b\right) \right) =\bar t\left( n\left( b\right)
,2\right) 
$$
for all {\bf resolvable} $b\in {\cal O}$. This means that every nod $b$ not
belonging to the maximal partition $z_{\max }\left( X\right) $ is
partitioned into two halves, when stepping to the next level in the tree;
and every non-resolvable nod contains only one element. The set of all
optimal trees over $X$ forms (for $\#X>2$) a proper subset of\ ${\cal CP}%
\left( X\right) $, which will be denoted by ${\cal OB}\left( X\right) $. In
general, ${\cal OB}\left( X\right) $ contains more than one element.

\section{Minimal classes in $T\left( n,m\right) $ \label{MinClassInT}}

Every minimal tree ${\cal B}\in {\cal MIN}\left( X\right) $ maps into a
certain partition $z$ under $z_{\min }$, and into a certain occupation
number $t$ under the $\tau $-map. This sequence of maps will be written as $%
dv$, for short: $dv\equiv \tau \circ z_{\min }$. We shall be interested in
the image of ${\cal MIN}\left( X\right) $ under this sequence, which will be
denoted by 
\begin{equation}
\label{pp4t2fo14}T_{\min }\left( n\right) \equiv dv\,\left( {\cal MIN}\left(
X\right) \right) \quad , 
\end{equation}
and will be called the {\it global minimal class in }$T\left( n\right) $.\
For\ $1\le m\le n$, the set 
\begin{equation}
\label{pp4t2fo15}T_{\min }\left( n,m\right) \equiv dv\,\left( {\cal MIN}%
\left( X,m\right) \right) 
\end{equation}
will be termed {\it minimal class in\ }$T\left( n,m\right) $. Note that we
now have several distinct classes of occupation numbers in $T\left( n\right) 
$; we have the class containing all optimal divisions of $n$ by $m$, $%
\left\{ \bar t\left( n,1\right) ,\bar t\left( n,2\right) ,\ldots ,\bar
t\left( n,m\right) \right\} $; and on the other hand the classes $T_{\min
}\left( n,m\right) $. The relation between these will be disclosed in the
following developments.

Now let $t\in T\left( n\right) $. We can study its inverse image $\left(
dv\right) ^{-1}\left( t\right) \cap {\cal CP}\left( X\right) $ in ${\cal CP}%
\left( X\right) $. To every tree in this set we can assign the associated
tree function $E$; thus it makes sense to ask, on which trees ${\cal B}\in
\left( dv\right) ^{-1}\left( t\right) \cap {\cal CP}\left( X\right) $ for a
given division $t$ of $X$ the tree function $E$ assumes its minimum. This
minimum will be denoted by $\min \left( t\right) $; hence 
\begin{equation}
\label{pp4t2fo16}\min \left( t\right) \equiv \min _{{\cal B}\in \left(
dv\right) ^{-1}\left( t\right) \cap {\cal CP}\left( X\right) }E\left(
t\right) \quad . 
\end{equation}
The associated subset of trees in $\left( dv\right) ^{-1}\left( t\right)
\cap {\cal CP}\left( X\right) $ that actually take this minimum will be
written as ${\cal MIN}\left( t\right) $, 
\begin{equation}
\label{pp4t2fo17}{\cal MIN}\left( t\right) \equiv E^{-1}\left( \min \left(
t\right) \right) \cap \left( dv\right) ^{-1}\left( t\right) \cap {\cal CP}%
\left( X\right) \quad . 
\end{equation}

\section{Bases and integer logarithm \label{BasIntLog}}

Let\ $L\in \TeXButton{N}{\mathbb{N}}$ with\ $L\ge 2$. Then the set 
\begin{equation}
\label{pp4t2fo21}\TeXButton{B}{\mathbb{B}}_L\equiv \left\{ L^k\mid k\in 
\TeXButton{N}{\mathbb{N}}_0\right\} 
\end{equation}
will be called {\it basis over\ }$L$. The set $\TeXButton{B}{\mathbb{B}}_2$\
is also called\ {\it binary basis}. If no confusion is likely, $\TeXButton{B}
{\mathbb{B}}_L$ will be simply denoted by $\TeXButton{B}{\mathbb{B}}$.

Furthermore, for a natural number $n\in \TeXButton{N}{\mathbb{N}}$, we
introduce the {\it integer logarithm }$\lg _L\left( n\right) $ {\it of }$n$ 
{\it with respect to }$L$, by 
\begin{equation}
\label{pp4t2fo22}\lg _L\left( n\right) \equiv \max \left\{ k^{\prime }\in 
\TeXButton{N}{\mathbb{N}}_0\mid L^{k^{\prime }}\le n\right\} \quad . 
\end{equation}
If no confusion is likely, the integer logarithm of $n$ with respect to $2$
will simply be written $\lg \left( n\right) \equiv \lg _2\left( n\right) $.
Clearly, $\lg _L$ is a monotonically increasing function on $\TeXButton{N}
{\mathbb{N}}$.

The integer logarithm obeys rules reminiscent from standard analysis; we
quote them without proof:

\subsection{Properties of integer logarithm}

\begin{enumerate}
\item  Let\ $n,n^{\prime }\in \TeXButton{N}{\mathbb{N}}$. Then 
\begin{equation}
\label{pp4t2fo26}\lg _L\left( n\cdot n^{\prime }\right) =\lg _L\left(
n\right) +\lg _L\left( n^{\prime }\right)  \quad .
\end{equation}

\item  Let\ $p\in \TeXButton{N}{\mathbb{N}}_0$. Then 
\begin{equation}
\label{pp4t2fo29}\lg _L\left( n^p\right) =p\cdot \lg _L\left( n\right) \quad
. 
\end{equation}
\end{enumerate}

\section{Optimal amount \label{OptAm}}

\subsection{Theorem: Amount of optimal trees}

Let $\#X=n$ and\ ${\cal O}\in {\cal OB}\left( X\right) $. Let $\lg \left(
n\right) $ denote the integer logarithm of $n$ with respect to $2$. Then 
\begin{equation}
\label{pp4t2fo30}E_{{\cal O}}=G_{{\cal O}}=n\cdot \lg \left( n\right)
+2\cdot \left[ n-2^{\lg \left( n\right) }\right] \quad . 
\end{equation}
This value is constant for all ${\cal O}\in {\cal OB}\left( X\right) $, and
depends only on $n$. Thus it will be denoted by $E\left( n\right) =G\left(
n\right) $.

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

By induction with respect to $n$. The statement is clear for $\underline{n=1}
$, since $\lg \left( 1\right) =0$.

$\underline{1\le n-1,\;n-1\rightarrow n:}$\quad We assume that (\ref
{pp4t2fo30}) holds for all $1\le n^{\prime }\le n-1$. Let $X$ be a set with\ 
$\#X=n$, let\ ${\cal O}\in {\cal OB}\left( X\right) $. Use formula (\ref
{pp4t1fo38b}) in splitting lemma 1, together with the fact, that $m\left(
X\right) =2$. This gives 
\begin{equation}
\label{pp4t2fo31}G_{{\cal O}}=n+\sum\limits_{b\,\in z_{\min }\left( X\right)
}G_{{\cal O}\left( X,b\right) }\quad . 
\end{equation}
We have\ $\#b\le n-1$ for all $b\in z_{\min }\left( X\right) $, hence $G_{%
{\cal O}\left( X,b\right) }=\#b\cdot \lg \left( \#b\right) +2\cdot \left[
\#b-2^{\lg \left( \#b\right) }\right] $ by assumption. Since\ $\tau \left(
z_{\min }\left( X\right) \right) =\bar t\left( n,2\right) $, we must
distinguish whether $n$ is even or odd. In both cases, the equation 
\begin{equation}
\label{pp4t2fo32}\lg \left( 2\nu +1\right) =\lg \left( 2\nu \right) =\lg
\left( \nu \right) +1 
\end{equation}
is crucial.

\underline{Case 1 :}\quad $n=2\nu +1$. Apply (\ref{pp4t2fo31}), then a short
computation yields%
$$
G_{{\cal O}}=3n+\nu \cdot \lg \left( \nu \right) +\left( \nu +1\right) \cdot
\lg \left( \nu +1\right) -2^{\lg \left( \nu \right) +1}-2^{\lg \left( \nu
+1\right) +1}\quad . 
$$
Two subcases must be distinguished: $\lg \left( \nu \right) =\lg \left( \nu
+1\right) $, or $\lg \left( \nu \right) +1=\lg \left( \nu +1\right) $.
Application of (\ref{pp4t2fo32}) then shows that for both subcases, (\ref
{pp4t2fo30}) is satisfied.

\underline{Case 2 :}\quad $n=2\nu $. This case is even more straightforward,
and proceeds along the same lines as above. \TeXButton{BWE}
{\hfill
\vspace{2ex}
$\blacksquare$}

The next lemma is a straightforward consequence of (\ref{pp4t2fo30}):

\subsection{Lemma}

The optimal amount is a monotonous function of $n$. In particular, 
\begin{equation}
\label{pp4t2fo33}G\left( n+1\right) -G\left( n\right) =\lg \left( n\right)
+2 
\end{equation}
for all $n\in \TeXButton{N}{\mathbb{N}}$.

Using this lemma we can prove

\subsection{Proposition \label{Kriterium}}

Let $n_1^{\prime }+n_2^{\prime }=n_1+n_2$. Then 
\begin{equation}
\label{pp4t2fo34}G\left( n_1^{\prime }\right) +G\left( n_2^{\prime }\right)
\ge G\left( n_1\right) +G\left( n_2\right) 
\end{equation}
if and only if 
\begin{equation}
\label{pp4t2fo35}\left( n_1^{\prime }\right) ^2+\left( n_2^{\prime }\right)
^2\ge n_1^2+n_2^2\quad . 
\end{equation}

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

We can assume $n_1\ge n_2$ without loss of generality. We must have $%
n_1^{\prime }=n_1+\Delta $, $n_2^{\prime }=n_2-\Delta $ for some $\Delta $.
First assume $\Delta \ge 0$. Then (\ref{pp4t2fo34}) is equivalent to%
$$
\sum_{j=0}^{\Delta -1}\lg \left( n_1+j\right) \ge \sum_{j=0}^{\Delta -1}\lg
\left( n_2-\Delta +j\right) \quad , 
$$
which is true iff $n_1\ge n_2-\Delta $, hence iff $\Delta +n_1-n_2\ge 0$,
hence iff  $2\Delta \left( \Delta +n_1-n_2\right) \ge 0$, on account of $%
\Delta \ge 0$. But the last expression is just $\left( n_1+\Delta \right)
^2+\left( n_2-\Delta \right) ^2-n_1^2-n_2^2$. This proves the proposition
for $\Delta \ge 0$.

Now assume $0>\Delta \equiv -\rho $. Then (\ref{pp4t2fo34}) is equivalent to%
$$
\sum_{j=0}^{\rho -1}\lg \left( n_2+j\right) \ge \sum_{j=0}^{\rho -1}\lg
\left( n_1-\rho +j\right) \quad , 
$$
which is true iff $n_2\ge n_1-\rho $, hence iff $-2\rho \left( -\rho
+n_1-n_2\right) \ge 0$, on account of $-\rho <0$. But the last expression is
just $\left( n_1-\rho \right) ^2+\left( n_2+\rho \right) ^2-n_1^2-n_2^2$,
which proves the statement for $\Delta <0$. \TeXButton{BWE}
{\hfill
\vspace{2ex}
$\blacksquare$}

This proposition will help proving the important

\subsection{Theorem \label{TheoOptDiv}}

Let $u=\left( n_1,\ldots ,n_m\right) \in U\left( n,m\right) $, let $\bar
n=\left( \bar n_1,\ldots ,\bar n_m\right) \in \kappa ^{-1}\left( \bar
t\right) $ be an optimal division of $n$ by $m$. Then 
\begin{equation}
\label{pp4t2fo36}\sum_{i=1}^mG\left( n_i\right) \ge \sum_{i=1}^mG\left( \bar
n_i\right) \quad . 
\end{equation}

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

Clearly, the RHS of (\ref{pp4t2fo36}) is independent of the representative $%
\bar n\in \kappa ^{-1}\left( \bar t\right) $, as the representatives differ
only by permutation of components. According to lemma \ref{LmOptDiv} there
exists a finite sequence $u^0,u^1,\ldots ,u^f$ of elements in $U\left(
n,m\right) $ with $u^0\equiv u$, and $u^f=\bar n$ for some $\bar n\in \kappa
^{-1}\left( \bar t\right) $, such that $\left\| u^0\right\| \ge \left\|
u^1\right\| \ge \cdots \ge \left\| u^f\right\| $, and the step $u^\alpha
\mapsto u^{\alpha +1}$ involves alteration of {\bf two} components of $%
u^\alpha $ only, namely $u_i^{\alpha +1}=u_i^\alpha +1$, and $u_j^{\alpha
+1}=u_j^\alpha -1$. Using the arguments in the proof of lemma \ref{LmOptDiv}
we see that%
$$
\left( u_i^\alpha +1\right) ^2+\left( u_j^\alpha -1\right) ^2-\left(
u_i^\alpha \right) ^2-\left( u_j^\alpha \right) ^2=2\left( \Delta _i^\alpha
-\Delta _j^\alpha +1\right) \le -2\quad , 
$$
and hence proposition \ref{Kriterium} says that $G\left( u_i^{\alpha
+1}\right) +G\left( u_j^{\alpha +1}\right) \le G\left( u_i^\alpha \right)
+G\left( u_j^\alpha \right) $. Since all other components $u_k^\alpha $ for $%
k\neq i,j$ remain the same, we have%
$$
\sum_{k=1}^mG\left( u_k^\alpha \right) \ge \sum_{k=1}^mG\left( u_k^{\alpha
+1}\right) \quad . 
$$
This inequality holds for every step $\alpha $ involved, hence (\ref
{pp4t2fo36}) follows. \TeXButton{BWE}{\hfill
\vspace{2ex}
$\blacksquare$}

\section{Preoptimized trees \label{PreOptTrees}}

The concept of preoptimization is required as a necessary intermediate step
in order to solve the problem of finding the global minimal class ${\cal MIN}%
\left( n\right) $. Let $n=\#X$. We define: A tree ${\cal B}\in {\cal MB}%
\left( X\right) $ is called {\it preoptimized} if every subtree ${\cal B}%
\left( X,b\right) $ of ${\cal B}$ based on elements $b\in z_{\min }\left(
X\right) $ in the minimal partition of $X$ in ${\cal B}$ is {\bf optimal}.
Thus, the only ''degrees of freedom'' of varying a preoptimized tree are the
different choices of minimal partitions $z_{\min }\left( X\right) $, where
these choices can be effectively described by the set of all divisions $%
U\left( n\right) $ of $n$ into $m$ terms, for $m=1,\ldots ,n$. Every
preoptimized tree is complete. The subset of all preoptimized trees over $X$
in ${\cal MB}\left( X\right) $ will be denoted by $p{\cal O}\left( X\right)
\subset {\cal CP}\left( X\right) $. This contains the disjoint subsets $p%
{\cal O}\left( X,m\right) $ of preoptimized trees with $m$ elements in the
minimal partition $z_{\min }\left( X\right) $. Hence we have a partition of $%
p{\cal O}\left( X\right) $ according to $p{\cal O}\left( X\right)
=\bigcup\limits_{1\le m\le n}p{\cal O}\left( X,m\right) $. Furthermore, we
define $p{\cal O}^{*}\left( X\right) =\bigcup\limits_{2\le m\le n}p{\cal O}%
\left( X,m\right) =p{\cal O}\left( X\right) -\left\{ \left\{ X\right\}
\right\} $.

On the subsets just described, the tree function $E$ coincides with the
total amount by theorem \ref{TrF}, since all trees are complete. It takes
the minima $p\min \left( X\right) \equiv \min \limits_{{\cal B}\in p{\cal O}%
^{*}\left( X\right) }E\left( {\cal B}\right) $, and $p\min \left( X,m\right)
\equiv \min \limits_{{\cal B}\in p{\cal O}\left( X,m\right) }E\left( {\cal B}%
\right) $. Accordingly, we can introduce the set of all preoptimized trees
for which $G$ actually takes the corresponding minimum: 
\begin{equation}
\label{pp4t3fo1}
\begin{array}{c}
{\cal MIN}_p\left( X\right) \equiv {\cal MIN}_p\left( n\right) \equiv
E^{-1}\left( p\min \left( X\right) \right) \cap p{\cal O}^{*}\left( X\right)
\quad , \\ {\cal MIN}_p\left( X,m\right) \equiv {\cal MIN}_p\left(
n,m\right) \equiv E^{-1}\left( p\min \left( X,m\right) \right) \cap p{\cal O}%
\left( X,m\right) \quad . 
\end{array}
\end{equation}
Obviously, $dv\left( p{\cal O}^{*}\left( X\right) \right) =T^{*}\left(
n\right) $, and $dv\left( p{\cal O}\left( X,m\right) \right) =T\left(
n,m\right) $. Hence\ $p{\cal O}\left( X,m\right) $\ can be partitioned
according to 
\begin{equation}
\label{pp4t3fo2}p{\cal O}\left( X,m\right) =\bigcup\limits_{t\,\in T\left(
n,m\right) }\left[ dv^{-1}\left( t\right) \cap p{\cal O}\left( X\right)
\right] \quad . 
\end{equation}

Now let $m=\#z_{\min }\left( X\right) $, let $t\in T\left( n\right) $, and $%
{\cal B}\in dv^{-1}\left( t\right) \cap p{\cal O}\left( X\right) $. From (%
\ref{pp4t2fo9a}) we have 
\begin{equation}
\label{pp4t3fo3}E_{{\cal B}}=n\left( m-1\right) +\sum_{b\in z_{\min }\left(
X\right) }E_{{\cal B}\left( X,b\right) }\quad ; 
\end{equation}
but since all subtrees ${\cal B}\left( X,b\right) $ are optimal, $E_{{\cal B}%
\left( X,b\right) }$ coincides with $G\left( \#b\right) $ according to (\ref
{pp4t2fo30}), and the first term $n\left( m-1\right) $ is constant for fixed 
$t$. Thus $E_{{\cal B}}$ is constant on $dv^{-1}\left( t\right) \cap p{\cal O%
}\left( X\right) $ and hence descends to a map, again denoted by $E:T\left(
n\right) \rightarrow \TeXButton{N}{\mathbb{N}}$, $E\left( t\right) =E\left( 
{\cal B}\right) $ for any choice of representative ${\cal B}\in
dv^{-1}\left( t\right) \cap p{\cal O}\left( X\right) $. Now (\ref{pp4t3fo3})
can be expressed as 
\begin{equation}
\label{pp4t3fo4}E\left( t\right) =n\left( m-1\right)
+\sum\limits_{k=1}^nt_k\cdot G\left( k\right) 
\end{equation}
for all $t\in T\left( n,m\right) $. Furthermore, we write $E\left( u\right)
\equiv E\left( t\right) $ for any division $u\in \kappa ^{-1}\left( t\right) 
$.

In the next section we will compare the values $E\left( t\right) $ with $%
E\left( \bar t\right) $ at the optimal division $\bar t\in T\left(
n,m\right) $.

\section{Minimality of the optimal division \label{SectMinimalOptDiv}}

In this section we show that the preoptimized trees for which the minimal
partition $z_{\min }\left( X\right) $ is optimal, or equivalently, for which 
$t=\bar t$, are actually the minimal ones, i.e. they lie in ${\cal %
MIN}\left( n,m\right) $. First we show that they are the minimal ones in the
set of all preoptimized trees $p{\cal O}\left( n,m\right) $:

\subsection{Theorem \label{MinOptDiv}}

Let $\bar t=\bar t\left( n,m\right) $ be the occupation number of the 
optimal division of $n$ by $m$, as defined in (\ref{pp4t2opt1}). Then 
\begin{equation}
\label{pp4t3fo5}E\left( t\right) \ge E\left( \bar t\right) 
\end{equation}
for all $t\in T\left( n,m\right) $. Hence 
\begin{equation}
\label{pp4t3fo6}p\min \left( n,m\right) =E\left( \bar t\right) \quad , 
\end{equation}
and the inverse image of $\bar t$ in $p{\cal O}\left( X\right) $ must
therefore lie in ${\cal MIN}_p\left( n,m\right) $, 
\begin{equation}
\label{pp4t3fo7}\left( dv\right) ^{-1}\left( \bar t\right) \cap p{\cal O}%
\left( X\right) \subset {\cal MIN}_p\left( n,m\right) \quad . 
\end{equation}

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

Let $t=\left( t_1,\ldots ,t_n\right) \in T\left( n,m\right) $. Let $u,\bar n$
be arbitrary representatives of $\kappa ^{-1}\left( t\right) $, $\kappa
^{-1}\left( \bar t\right) $, respectively; this means that $u$ and $\bar n$
are divisions of $n$ by $m$, $u=\left( n_1,\ldots ,n_m\right) $ and $\bar
n=\left( \bar n_1,\ldots ,\bar n_m\right) $, such that (\ref{pp4t3fo4}) can
be expressed as%
$$
E\left( t\right) =n\left( m-1\right) +\sum\limits_{j=1}^mG\left( n_j\right)
\quad , 
$$
with a similar expression for $E\left( \bar t\right) $. This implies that
the inequality in (\ref{pp4t3fo5}) will be satisfied if and only if 
$$
\sum\limits_{j=1}^mG\left( n_j\right) \ge \sum\limits_{j=1}^mG\left( \bar
n_j\right) \quad . 
$$
But the last statement is true due to (\ref{pp4t2fo36}) in theorem \ref
{TheoOptDiv}. \TeXButton{BWE}{\hfill
\vspace{2ex}
$\blacksquare$}

\subsection{Remark}

The inclusions in (\ref{pp4t3fo7}) are proper in general. This means that
there exist elements in ${\cal MIN}_p\left( n,m\right) $ which are {\bf not}
optimal. As an example, consider\ $n=6$, $X=\left\{ 1,\ldots ,6\right\} $,
with%
$$
G\left( 6\right) =6+G\left( 3\right) +G\left( 3\right) =6+5+5=16\quad ; 
$$
now compare with the complete tree ${\cal B}=p{\cal O}\left( 2\right) +p%
{\cal O}\left( 4\right) $, which is a sum of the preoptimized trees $p{\cal O%
}\left( 2\right) $ and $p{\cal O}\left( 4\right) $, respectively. ${\cal B}$
is non-optimal, since the minimal partition $z_{\min }\left( X\right) $ is
based on the non-optimal division\ $\left( 2,4\right) $ of $6$. We find $%
G\left( 2\right) +G\left( 4\right) =2+8=10$, and $G_{{\cal B}}=6+10=16=p\min
\left( 6,2\right) $, although ${\cal B}$ is not optimal.

--- The next theorem explains how $p\min \left( n,m\right) $ changes for
fixed $n$ as $m$ increases:

\subsection{Theorem \label{VaryM}}

Let\ $n\ge 2$ and $1\le m<n$. Then 
\begin{equation}
\label{pp4t3fo8}p\min \left( n,m+1\right) >p\min \left( n,m\right) \quad . 
\end{equation}
\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

The case $m=1$ yields $p\min \left( n,1\right) =0$, whereas $p\min \left(
n,2\right) =G\left( n\right) >0$ for $n\ge 2$. Thus we certainly have $p\min
\left( n,1\right) <p\min \left( n,2\right) $. Therefore assume now that $%
m\ge 2$. Let $n$ be optimally divided by $\left( m+1\right) $ according to $%
n=\nu \cdot \left( m+1\right) +r$, where $\nu =\left[ \frac n{m+1}\right] $.
The naturally ordered representative of $\kappa ^{-1}\left( \bar t\right) $
is%
$$
\bar n=\left( \stackunder{m+1-r}{\underbrace{\nu ,\ldots \,,\nu }}\,,\,%
\stackunder{r}{\underbrace{\nu +1,\ldots \,,\nu +1}}\right) \quad . 
$$
According to this decomposition we have%
$$
p\min \left( n,m\right) =E_{*}\left( \bar t\right) =n\left( m-1\right)
+\left( m-r\right) \cdot G\left( \nu \right) +r\cdot G\left( \nu +1\right)
\quad , 
$$
as follows from theorem \ref{MinOptDiv} and (\ref{pp4t3fo4}). The value of $%
E $ on $\bar n$ is 
\begin{equation}
\label{pp4t3fo9}E\left( \bar n\right) =p\min \left( n,m+1\right) =nm+\left(
m+1-r\right) \cdot G\left( \nu \right) +r\cdot G\left( \nu +1\right) \quad . 
\end{equation}
Now define a new division $u$ of $n$ into $m$ terms by $u=\left( u_1,\ldots
,u_m\right) \equiv \left( \bar n_2,\ldots ,\bar n_m,\bar n_1+\bar
n_{m+1}\right) $; the value of $E$ on $u$ is 
\begin{equation}
\label{pp4t3fo10a}E\left( u\right) =n\left( m-1\right) +\left( m-r\right)
\cdot G\left( \nu \right) +\left( r-1\right) \cdot G\left( \nu +1\right)
+G\left( 2\nu +1\right) \quad . 
\end{equation}
The formulas 
\begin{equation}
\label{pp4t3fo10}G\left( 2\nu +1\right) =G\left( \nu \right) +G\left( \nu
+1\right) +\left( 2\nu +1\right) \quad ,\quad G\left( 2\nu \right) =2G\left(
\nu \right) +2\nu 
\end{equation}
will be used. Two cases must be distinguished: $r=0$ or $r>0$. Assume $r>0$:
In this case we have $u_m=\bar n_1+\bar n_{m+1}=2\nu +1$. Use (\ref
{pp4t3fo10a}, \ref{pp4t2fo30}) and formula (\ref{pp4t3fo10}) to compute 
\begin{equation}
\label{pp4t3fo11}p\min \left( n,m+1\right) -E\left( u\right) =\nu \left(
m-1\right) +\left( r-1\right) \quad . 
\end{equation}
Since $r\ge 1$ and $m\ge 2$, the RHS is $>0$. For the case $r=0$ we obtain 
\begin{equation}
\label{pp4t3fo12}p\min \left( n,m+1\right) -E\left( u\right) =\nu \left(
m-1\right) \quad , 
\end{equation}
which is again greater than zero. Finally, take into account (\ref{pp4t3fo5}%
), which implies in this context that 
\begin{equation}
\label{pp4t3fo13}E\left( u\right) \ge p\min \left( n,m\right) \quad . 
\end{equation}
Now (\ref{pp4t3fo11}--\ref{pp4t3fo13}) imply the result in (\ref{pp4t3fo8}). 
\TeXButton{BWE}{\hfill
\vspace{2ex}
$\blacksquare$}

The next theorem explains the role of the optimal trees ${\cal OB}\left(
X\right) $ in this context:

\subsection{Theorem \label{OptTree}}

Let\ $\#X=n\ge 2$. Then the optimal trees minimize the tree function on the
set of all preoptimized trees with $2$ elements in $z_{\max }\left( X\right) 
$, and hence on all preoptimized trees. In symbols, 
\begin{equation}
\label{pp4t3fo14}{\cal OB}\left( X\right) \subset {\cal MIN}_p\left(
X,2\right) \subset {\cal MIN}_p\left( X\right) \quad , 
\end{equation}
and hence 
\begin{equation}
\label{pp4t3fo14a}G\left( n\right) =p\min \left( n,2\right) =p\min \left(
n\right) \quad . 
\end{equation}

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

Proof of the first inclusion (\ref{pp4t3fo14}): ${\cal B}\in {\cal OB}\left(
X\right) $ implies that ${\cal B}$ is preoptimized, and the minimal
partition $z_{\max }\left( X\right) $ is optimal, i.e. $dv\left( {\cal B}%
\right) =\bar t\left( n,2\right) $; therefore\ ${\cal B}\in \left( dv\right)
^{-1}\left( \bar t\left( n,2\right) \right) \cap p{\cal O}\left( X\right) $.
But due to (\ref{pp4t3fo7}) this set is included in ${\cal MIN}_p\left(
X,2\right) $. \TeXButton{BWE}{\hfill
\vspace{2ex}
$\blacksquare$}

Now we come to the main theorem of this work:

\subsection{Theorem \label{GlobMin}}

Let $\#X=n\ge 2$.$\ $Then the optimal trees over $X$ belong to the globally
minimal trees over $X$, i.e. 
\begin{equation}
\label{pp4t3fo15}{\cal OB}\left( X\right) \subset {\cal MIN}\left( X\right)
\quad , 
\end{equation}
and hence $G\left( X\right) =\min \left( X\right) =\min \left( n\right) $.

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

Since all trees involved in the present discussion are complete, the tree
function $E$ coincides with the total amount $G$ of the tree, as follows
from theorem \ref{TrF}. We proof (\ref{pp4t3fo15}) by induction with respect
to\ $n=\#X$.

$\underline{n=2:}$\quad This is clear, since\ ${\cal OB}\left( X\right) =%
{\cal MIN}\left( X\right) $ in this case.

$\underline{2\le n-1\;,\;n-1\rightarrow n:}$\quad Assume 
\begin{equation}
\label{pp4t3fo16}{\cal OB}\left( n^{\prime }\right) \subset {\cal MIN}\left(
n^{\prime }\right) 
\end{equation}
for all$\;2\le n^{\prime }\le n-1$.We prove (\ref{pp4t3fo15}) for\ $\#X=n$
by showing that $G_{{\cal B}}\ge G\left( n\right) $ for every complete tree $%
{\cal B}\in {\cal CP}\left( X\right) $ over $X$. Let $u=\left( u_1,\ldots
,u_m\right) =\left( \#b_1,\ldots ,\#b_m\right) $, where $z_{\min }\left(
X\right) =\left( b_1,\ldots ,b_m\right) $ is the minimal partition of $X$ in 
${\cal B}$. Now apply (\ref{pp4t1fo38b}) in splitting lemma 1:%
$$
G_{{\cal B}}=n\left( m-1\right) +\sum\limits_{j=1}^mG_{{\cal B}\left(
X,b_j\right) }\quad . 
$$
By (\ref{pp4t3fo16}) we have $G_{{\cal B}\left( X,b_j\right) }\ge G\left(
u_j\right) $ for all $j=1,\ldots ,m$, since the subtrees ${\cal B}\left(
X,b_j\right) $ need not be optimal. Thus%
$$
G_{{\cal B}}\ge n\left( m-1\right) +\sum\limits_{j=1}^mG\left( u_j\right)
\equiv G_{{\cal B}^{\prime }}\quad , 
$$
where the RHS of the last formula defines the total amount of the
preoptimized tree \ ${\cal B}^{\prime }=\sum_{b\in z_{\min }\left(
X\right) }{\cal O}\left( b\right) \in p{\cal O}\left( X;m\right) $. Now $G_{%
{\cal B}^{\prime }}=E\left( t\right) $, where $t$ is the occupation number
of $u$, $t=\kappa \left( u\right) $; hence, by (\ref{pp4t3fo5}, \ref
{pp4t3fo6}) in theorem \ref{MinOptDiv}, we have $G_{{\cal B}^{\prime }}\ge
E\left( \bar t\right) =p\min \left( n,m\right) $, where $\bar t$ now is the
optimal division of $n$ by $m$. Using (\ref{pp4t3fo8}) in theorem \ref{VaryM}
we have $p\min \left( n,m\right) \ge p\min \left( n,2\right) $. Using (\ref
{pp4t3fo14a}) in theorem \ref{OptTree} we have $p\min \left( n,2\right)
=p\min \left( n\right) =G\left( n\right) $. Thus, putting all inequalities
together,%
$$
G_{{\cal B}}\ge G\left( n\right) =G\left( X\right) \quad , 
$$
which proves the theorem. \TeXButton{BWE}
{\hfill
\vspace{2ex}
$\blacksquare$}

\section{Mean path amount and quadratic deviation \label{MPAmQuDev}}

From section \ref{OptAm}, formula (\ref{pp4t2fo30}), we immediately see that
the mean path amount $\frac 1n\cdot \sum e_i$ will be close to $\lg \left(
n\right) $. We can make this statement more precise by defining the quantity 
\begin{equation}
\label{pp4t3fo17}\overline{e}_{{\cal B}}\equiv \min \left\{ \eta \in 
\TeXButton{N}{\mathbb{N}}\mid \eta \cdot n\ge G_{{\cal B}}\right\} \quad , 
\end{equation}
which will be called {\it mean path amount in }${\cal B}\left( X\right) $
henceforth. Thus $G_{{\cal B}}=\left( {\overline{e}_{{\cal B}}\cdot n\;-\;r } \right)$, where\ 
$r<n$. In particular, in an optimal tree ${\cal B}={\cal O}$, 
\begin{equation}
\label{pp4t3fo18}\bar e_{{\cal O}}=\left\{ 
\begin{array}{ccc}
\lg \left( n\right) & , & n\in 
\TeXButton{B}{\mathbb{B}} \\ \lg \left( n\right) +1 & , & n\NEG \in 
\TeXButton{B}{\mathbb{B}} 
\end{array}
\right. \quad . 
\end{equation}
With $r$ as given above we define the $n$-tupel 
\begin{equation}
\label{pp4t3fo19}\left( \bar e_1,\ldots \,\bar e_n\right) \equiv \left( 
\stackunder{\left( n-r\right) \times }{\underbrace{\bar e_{{\cal B}},\ldots
\,,\,\bar e_{{\cal B}}}}\,,\,\stackunder{r\times }{\underbrace{\bar e_{{\cal %
B}}-1,\ldots \,,\,\bar e_{{\cal B}}-1}}\right) \quad ; 
\end{equation}
thus, for any tree ${\cal B}$ over $X$ (which need not be optimal), we have 
\begin{equation}
\label{pp4t3fo22}G_{{\cal B}}=\sum\limits_{i=1}^ne_i=\sum\limits_{i=1}^n\bar
e_i=\left( n-r\right) \cdot \bar e_{{\cal B}}+r\cdot \left( \bar e_{{\cal B}%
}-1\right) \quad . 
\end{equation}
Introducing the $n$-tupel of deviations 
\begin{equation}
\label{pp4t3fo20}\Delta e\equiv \left( \Delta e_1,\ldots ,\Delta e_n\right)
\equiv \left( e_1-\bar e_1,\ldots \,,e_n-\bar e_n\right) 
\end{equation}
and the {\it total quadratic deviation in\ }${\cal B}\left( X\right) $ by 
\begin{equation}
\label{pp4t3fo21}\sigma _{tot}\left( {\cal B}\right) \equiv
\sum\limits_{i=1}^n\left( \Delta e_i\right) ^2\quad , 
\end{equation}
we find that, on using (\ref{pp4t3fo22}), 
\begin{equation}
\label{pp4t3fo23}\sigma _{tot}\left( {\cal B}\right)
=\sum\limits_{i=1}^n\left( e_i^2-\bar e_i^2\right) +2\cdot
\sum\limits_{i=n-r+1}^n\Delta e_i\quad . 
\end{equation}

We now present some statements about the mean path amount in optimal trees.
In every tree ${\cal B}$, the elements $\left\{ c_1,\ldots ,c_K\right\} $ in
the maximal partition $z_{\max }\left( X\right) $ can be labelled so that
the associated path amounts are monotonically decreasing, $e_1\ge e_2\ge
\cdots \ge e_K$. In particular, if $\bar e_{{\cal O}}$ is the mean path
amount in the optimal tree ${\cal B=O}$ as defined in (\ref{pp4t3fo18}), and
if $r\equiv n\cdot \bar e_{{\cal O}}-G_{{\cal O}}$, then it is easy to prove
that for\ $n\in \TeXButton{B}{\mathbb{B}}$, 
\begin{equation}
\label{pp4t3fo24}e_i=\lg \left( n\right) 
\end{equation}
for all $i=1,\ldots ,n$, whereas for $n\not \in \TeXButton{B}{\mathbb{B}}$, 
\begin{equation}
\label{pp4t3fo25}e_i=\left\{ 
\begin{array}{ccc}
\lg \left( n\right) +1 & , & i=1,\ldots ,n-r \\ 
\lg \left( n\right) & , & i=n-r+1,\ldots ,n 
\end{array}
\right. \quad . 
\end{equation}
It then follows that for every optimal tree we have 
\begin{equation}
\label{pp4t3fo26}e_i=\bar e_i 
\end{equation}
for all $i=1,\ldots ,n$.

We can now express the minimality properties of optimal trees in terms of
mean path amounts and minimal quadratic deviations:

\subsection{Theorem}

Let ${\cal O}$ be an optimal tree over $X$,\ ${\cal O}\in {\cal OB}\left(
X\right) $. Then

\begin{enumerate}
\item  The mean path amount in an optimal tree is minimal compared with all
other trees over $X$, i.e. 
\begin{equation}
\label{pp4t3fo27}\overline{e}_{{\cal O}}\le \overline{e}_{{\cal B}} 
\end{equation}
for all ${\cal B}\in {\cal MB}\left( X\right) $.

\item  The total quadratic deviation of an optimal tree vanishes, and hence
is minimal compared with all other trees over $X$, i.e. 
\begin{equation}
\label{pp4t3fo28}\sigma _{tot}\left( {\cal O}\right) =0\quad . 
\end{equation}
\end{enumerate}

\TeXButton{Beweis}{\raisebox{-1ex}{\it Proof :}
\vspace{1ex}}

Statement (\ref{pp4t3fo27}) follows immediately from theorem \ref{GlobMin}
and the definition (\ref{pp4t3fo17}) of $\bar e_{{\cal B}}$. Statement (\ref
{pp4t3fo28}) follows immediately from formula (\ref{pp4t3fo23}) on using the
result (\ref{pp4t3fo26}) for optimal trees. \TeXButton{BWE}
{\hfill
\vspace{2ex}
$\blacksquare$}

\section{Extended minimal problem \label{ExtMinProb}}

In the previous section we have solved the problem of minimizing the tree
function $E$ on the set of all complete trees over the set $X$. Finally, we
now show how to extend the framework we have worked in so far in order to
obtain tree functions that contain expressions like $\sum w_i\lg \left(
w_i\right) $ in the functional form of their minimal value, when restricted
to a certain class of tree structures over $X$. We approach this problem as
follows:

\subsection{Trees preserving a partition}

A complete tree has a maximal partition of $X$ which is complete, i.e. the
elements of $z_{\max }\left( X\right) $ are comprised by the $1$%
-element-subsets $\left\{ x\right\} $ for $x\in X$. Trivially, every
partition $z$ of $X$ preserves $z_{\max }\left( X\right) $, as $z_{\max }$
is a refinement of every partition $z$ of $X$. We now generalise this
reasoning to the case where $z_{\max }\left( X\right) $ is no longer
complete: We want to prescribe a partition $z$ to $X$ such that the relation 
$z^{\prime }<z$ is true for all $z^{\prime }\in \zeta \left( X\right) $
compatible with ${\cal B}$. In particular, for the maximal partition of $X$
in ${\cal B}$ we must have $z_{\max }\left( X\right) \le z$. In general, the
prescribed element $z$ that is preserved by the partitions compatible with $%
{\cal B}$ need not be an element of $\zeta \left( X\right) $ itself; in this
case it induces a non-trivial partition on at least one of the elements $%
b\in z_{\max }\left( X\right) $ which are non-resolvable in ${\cal B}$.
Alternatively, we can have $z=z_{\max }\left( X\right) $; in this case,
elements $b\in z_{\max }\left( X\right) $ can be partitioned no further.
This leads us naturally to the definition: Let ${\cal B}\in {\cal MB}\left(
X\right) $, let $z\in {\cal Z}\left( X\right) $ be a partition of $X$. $%
{\cal B}$ is called $z${\it -preserving}, if $z^{\prime }<z$ for all $%
z^{\prime }\in \zeta \left( X\right) $. ${\cal B}$ is called $z${\it %
-complete}, if ${\cal B}$ is $z$-preserving \underline{and} $z_{\max }\left(
X\right) =z$ in ${\cal B}$. The set of all $z$-preserving trees over $X$
will be denoted as ${\cal MB}\left( X,z\right) $; the set of all $z$%
-complete trees will be written as ${\cal CP}\left( X,z\right) $. Clearly, $%
{\cal CP}\left( X,z\right) \subsetneqq {\cal MB}\left(
X,z\right) $ in general.

The question we alluded to at the beginning of this section can now be
posed: On which elements of ${\cal CP}\left( X,z\right) $ does the tree
function $E$ take its minimum in ${\cal CP}\left( X,z\right) $? This minimum
will be denoted by $\min \left( z\right) $. The set of all $z$-complete
trees for which $E$ actually takes this minimum is denoted as ${\cal MIN}%
\left( X,z\right) $, and coincides with the intersection $E^{-1}\left( \min
\left( z\right) \right) \cap {\cal CP}\left( X,z\right) $. Without proof, we
now summarize the necessary steps to find the solution; a detailed
derivation of all statements involved will be given elsewhere.

We have to start with prescribing a partition $z$; this will be the maximal
partition $z_{\max }\left( X\right) $ of $X$ in all trees ${\cal B}\in {\cal %
CP}\left( X,z\right) $ that are $z$-complete. We assume that $z$ contains $K$
elements, $z=\left\{ c_1,\ldots ,c_K\right\} $. Furthermore, we continue to
note $n=\#X$. Since $c_1\cup \cdots \cup c_K=X$, we obviously must have $%
K\le n$, where equality $K=n$ pertains to the special case of complete trees
that has been examined in the previous sections. In this case every
element $c_i$ of $z$ is primitive, and hence trivially non-resolvable. We
denote the $K$-tupel of cardinalities $\#c_i=n\left( c_i\right) $ by $w$, 
\begin{equation}
\label{pp4t3fo29}w\equiv \left( w_1,\ldots ,w_K\right) \equiv \left( n\left(
c_1\right) ,\ldots ,n\left( c_K\right) \right) \quad . 
\end{equation}
We now assume without loss of generality that the $c_i$ are labelled so that 
$w_1\le w_2\le \cdots \le w_K$. Clearly, $\sum_{i=1}^Kw_i=n$.

Assume that $w$ is fixed. From (\ref{pp4t1int10}) and (\ref{pp4t1fo41}) in
theorem \ref{TrF} we know that 
\begin{equation}
\label{pp4t3fo30}E_{{\cal B}}=\sum\limits_{c\in z}n\left( c\right) \cdot e_{%
{\cal B}}\left( c\right) =\sum_{i=1}^Kw_i\cdot e_{{\cal B}}\left( c_i\right) 
\end{equation}
for ${\cal B}\in {\cal CP}\left( X,z\right) $, where $e_{{\cal B}}\left(
c_i\right) $ is the path amount of $c_i$ in the tree ${\cal B}$. For the
sake of simplicity we stick to the notation $e_i\equiv e_{{\cal B}}\left(
c_i\right) $ in the following. Now define two trees to be equivalent if and
only if they differ by permutation of the elements of the maximal partition $%
z_{\max }\left( X\right) =z$ of $X$. In this case the $K$-tupels of path
amounts $e=\left( e_1,\ldots ,e_K\right) $ and $e^{\prime }\equiv \left(
e_1^{\prime },\ldots ,e_K^{\prime }\right) $ are related by $e^{\prime }=\pi
\left( e\right) $, where $\pi $ is a permutation of $K$ objects. Now from
the second sum in (\ref{pp4t3fo30}) it is clear that, for a given tree $%
{\cal B}$ with maximal partition $z$, the minimum of $E$ on the set of all
trees equivalent to ${\cal B}$ will be taken on the permutation of $e$ for
which 
\begin{equation}
\label{pp4t3fo31}e_1^{\prime }\ge e_2^{\prime }\ge \cdots \ge e_K^{\prime
}\quad . 
\end{equation}
For in any other case we had elements $w_i\le w_j$ with $i<j$ but $e_i<e_j$;
if we put $e_j^{\prime }\equiv e_i$, $e_i^{\prime }\equiv e_j$, the
difference $E^{\prime }-E$ of tree functions would be%
$$
w_ie_j+w_je_i-w_ie_i-w_je_j\quad , 
$$
which is certainly negative under the above assumptions. Thus, the optimal
tree within an equivalence class of trees associated with a fixed division $%
n=w_1+\cdots +w_K$ of $n$ into $K$ terms is always the one with property (%
\ref{pp4t3fo31}). Thus, we can assume $e_1\ge e_2\ge \cdots \ge e_K$ for all
path amounts from now on.

A detailed investigation now shows that the minimal class ${\cal MIN}\left(
X,z\right) $ contains trees with the distinct property that all $m$%
-characters are equal to $2$, as was already the case in the simpler
discussion above; i.e., $m\left( b\right) =2$ for all $b\in {\cal B}$. Thus,
to every resolvable nod $b\in {\cal B}$ there belongs a division $n\left(
b\right) =n\left( b_1\right) +n\left( b_2\right) $ of $n\left( b\right) $
into two terms, where $b_1\cup b_2=b$. However, in general this division
will no longer be optimal, in that $n\left( b_1\right) $, $n\left(
b_2\right) $ may deviate considerably from the mean value $\frac{n\left(
b\right) }2$. Now think of $w$ as being an element of the set $H\left(
K,n\right) $, as defined in (\ref{pp4t2div1}). Furthermore, for $%
n=\sum_{i=1}^Kw_i$, consider the set of integers 
\begin{equation}
\label{pp4t3fo32}\left\{ \left[ \frac n{2^1}\right] ,\left[ \frac
n{2^2}\right] ,\ldots ,\left[ \frac n{2^{\lg \left( n\right) }}\right]
=1\right\} \quad .
\end{equation}
Consider the set of all divisions $d=\left( d_1,\ldots ,d_K\right) $ of $n$
into $K$ terms, where all these terms lie in the set (\ref{pp4t3fo32}), i.e. 
$d_i=\left[ \frac n{2^{e_i}}\right] $ for all $i$, and furthermore, where $%
d_1\le d_2\le \cdots \le d_K$. This defines a series of quantities $\left(
e_i\right) $ with $e_1\ge e_2\ge \cdots \ge e_K$. By construction, all these 
$d$ are elements of $H\left( K,n\right) $; hence there exists (at least) one
division $d_{\min }$ which comes closest to $w$ (with respect to the $l_1$-
or $l_2$-norm) in $H\left( K,n\right) $. Then we have the result that the
minimal value of the tree function $E_{{\cal B}}=\sum_{i=1}^Kw_i\cdot e_i$
(for fixed $w$) is taken on the $K$-tupel $e=\left( e_1,\ldots ,e_K\right) $
associated with $d_{\min }$. In this case, $E_{{\cal B}}$ comes closest to%
$$
n\cdot \lg \left( n\right) -\sum w_i\cdot \lg \left( w_i\right) \quad , 
$$
up to a correction term [cf. formula (\ref{pp4t2fo30}) in section \ref{OptAm}%
], whose precise form will be given elsewhere. There it will be shown that
these data are sufficient to reconstruct a tree which has all the properties
mentioned above, and in particular for which the $K$-tupel of path amounts $e
$ is determined by $d_{\min }$.

\section{Tree structures and neighbourhood topology \label{TopolAndTree}}

Finally, we want to put forward arguments how tree structures define a
topology on the underlying set $X$. We now allow the set $X$ to have
arbitrary cardinality; in particular, $X$ can be non-countable. Furthermore,
we assume that the trees are without constraints, so that all paths have
infinite length, without terminating at a finite non-resolvable tree
element. Such a tree will be called {\it unconstrained} in the following. We
may want to build in weights, which we can describe by a function $%
w:X\rightarrow \TeXButton{N}{\mathbb{N}}$ or $\TeXButton{R}{\mathbb{R}}$.
Let $x\in X$; we recall that the path $q\left( \left\{ x\right\} \right) $
of $x$ in ${\cal B}\left( X\right) $ is defined to be the set of all
elements $b^{\prime }$ in the tree structure that contain $\left\{ x\right\} 
$, $q\left( \left\{ x\right\} \right) =\left\{ b^{\prime }\in {\cal B}\left(
X\right) \mid x\in b^{\prime }\right\} $. It must be emphasized that $%
\left\{ x\right\} $ need not be an element of the tree itself; although this
is true for finite $X$, in the infinite case we can have paths $q$ such that
every element $b^{\prime }$ of the path contains $x$, without $\left\{
x\right\} $ being an element of the tree. However, regardless of whether $%
\left\{ x\right\} \in q$ or not, it will always be true that for every $x\in
X$ there exists a unique path containing $x$. This path will be denoted by $%
q\left( x\right) $. Thus, the paths in such a tree will be uniquely labelled
by the elements $x\in X$. If the characters $m\left( b\right) $ at every nod 
$b$ remain finite, the path will always be a countable subset of the tree $%
{\cal B}$. We now show that, employing the tree structure ${\cal B}$ over $X$%
, we can define a neighbourhood topology on $X$. We recall \cite{HeuserII,Brown}%
 that this is defined to be a collection ${\cal N}\equiv
\bigcup_{x\in X}{\cal N}\left( x\right) $ of sets $N$, where the elements of 
$N\in {\cal N}\left( x\right) $ are distinct subsets of $X$ called
neighbourhoods of $x$ (in the given topology), satisfying the axioms \cite
{Brown}

\begin{description}
\item[(N1)]  \ If $N$ is a neighbourhood of $x$, then $x\in N$.

\item[(N2)]  \ If $N$ is a subset of $X$ containing a neighbourhood of $x$,
then $N$ is a neighbourhood of $x$.

\item[(N3)]  \ The intersection of two neighbourhoods of $x$ is again a
neighbourhood of $x$.

\item[(N4)]  \ Any neighbourhood $N$ of $x$ contains a neighbourhood $M$ of $%
x$ such that $N$ is a neighbourhood of each point of $M$.
\end{description}

$X$ together with ${\cal N}$ is called a topological space. Furthermore, a
base for the neighbourhoods at $x$ is a set $Bas\left( x\right) $ of
neighbourhoods of $x$ such that every neighbourhood $N$ of $x$ contains an
element $b\in Bas\left( x\right) $. Now we define the path $q\left( x\right) 
$ to be a neighbourhood base for $x$, and a subset $N\subset X$ to be a
neighbourhood of $x$ if and only if there exists a $b\in q\left( x\right) $
that is contained in $N$. We have to verify that this yields indeed a
neighbourhood topology on $X$. (N1) and (N2) are satisfied by construction.
Let $N,N^{\prime }$ be two neighbourhoods of $x$; then there exist $%
b,b^{\prime }\in q\left( x\right) $ such that $b\in N$, $b^{\prime }\in
N^{\prime }$. But since $b,b^{\prime }$ are members of the same path, we
have either $b\subset b^{\prime }$ or $b^{\prime }\subset b$, hence $N\cap
N^{\prime }$ contains either $b$ or $b^{\prime }$ and is therefore a
neighbourhood of $x$, thus (N3) follows. Now let $N$ be a neighbourhood of $%
x $; then $N$ contains some $b\in q\left( x\right) $; but for every element $%
y\in b$ we have that $b$ lies in the path $q\left( y\right) $ of $y$, hence $%
N$ is a neighbourhood for every $y\in b$, which gives (N4). Thus we have
proven:

\subsection{Theorem}

Every unconstrained tree structure ${\cal B}\left( X\right) $ over $X$
defines a neighbourhood topology on $X$.

\TeXButton{Abst}{\vspace{0.8ex}}The rest of our arguments will be somewhat
heuristic. Assume that we have defined weights $w:X\rightarrow \TeXButton{R}
{\mathbb{R}}_{+}$ on $X$; and assume that somehow we can define a tree
function for tree structures over sets $X$ of arbitrary cardinality, which
renders a real, nonnegative number as a value on the tree ${\cal B}$, say.
We then can pose the same questions as above, namely, which trees over $X$
minimize the tree function $E$ for fixed weights $w$. The value of $E$ on
such a minimal tree will be an entropy-like quantity, and it will single out
a preferred topology on $X$. We see that this looks distinctively like an
action principle for topologies on the set $X$, the role of the action being
played by the tree function, the degrees of freedom being expressed by the
different trees over $X$, and the minimal value of the action=tree function $%
E$ being associated with the entropy of the weights $w:X\rightarrow 
\TeXButton{R}{\mathbb{R}}_{+}$.

\begin{thebibliography}{9}
\bibitem{Baierlein}  Ralph Baierlein, ''Atoms and Information Theory''. W.H.
Freeman and Company, 1971.
\bibitem{Eppler}  W. Meyer-Eppler, ''Grundlagen und Anwendungen der
Informationstheorie''. Springer-Verlag, 1969.

\bibitem{Farhi}  Edward Farhi, Sam Gutmann. ''Quantum Computation and
Decision Trees''. {\sf quant-phys/9706062}.

\bibitem{Falconer}  K. J. Falconer, ''The Geometry of Fractal Sets''.
Cambridge University Press, 1985.

\bibitem{HeuserII}  H. Heuser, ''Lehrbuch der Analysis, Teil 2''. B.G.
Teubner Stuttgart, 1988.

\bibitem{Brown}  Ronald Brown, ''Topology''. Ellis Harwood Limited, 1988.
\end{thebibliography}

\end{document}
