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\begin{document}
\chapter{Rough Mereology and \v{C}ech Topologies}
\markboth{Rough Mereology and \v{C}ech Topologies}{Rough Mereology
and \v{C}ech Topologies}
\markright{Rough Mereology and \v{C}ech Topologies}
\thispagestyle{empty}
Rough mereology [21], [22] proved to be a very interesting means
of introducing similarity relations in rough set theory.
In this Chapter we study \v{C}ech topologies which arise in
rough mereological universe. We show that \v{C}ech topologies
extend the quasi-\v{C}ech topologies defined by Clarke [1,2] in
the framework of ``calculus of individuals'' -- a version of
mereology going back to A.N.Whitehead.
\section{Introduction}
We begin with a presentation of basic mereological ideas put
forth by Stanis"law Le"s\-niew\-ski. Then, we introduce rough
mereological spaces following [22]. For completeness sake and
the convenience of the reader, we include proofs.
We do not attempt originality here and we mostly paraphraze
slightly the original proofs given by Le"sniewski [4].
Mereology was proposed in Lesniewski [4] as an alternative
set theory conceived to avoid some paradoxes of naive set theory
disputed at the time. The primitive notion of mereology is the
notion of being a part.
An alternative approach, based on the dual notion of extension
proposed by Whitehead (see [1]) was developed later into so called
``calculus of individuals'' based on the pri\-mi\-tive binary predicate
of being in the relation of connection (cf.\ [1], [2]).
The principal features of mereology are that (i) avoids the null-object
(the empty set) (introducing a null-object
would trivialize the theory) and (ii) it collapses the hierarchy
of the naive set theory (an element, a set, a power set etc$\ldots$)
by treating a collection of objects as an object. This makes
mereology a useful tool for expressing e.g.\ spatial and
temporal relations in computer processing of natural language.
In many applications we only have a partial information about
object. This is e.g.\ a starting point for rough set theory [15]
where any object $x$ is represented by its information vector
$\Inf(x)$ which permits us to define only the indiscernibility
class $[x]$ of $x$.
The degree to which an object $x$ belongs to a set $X$ of
objects is measured by the rough membership function $\mu(x,X)$.
In [22] the rough membership function has been generalized to a
rough inclusion $\mu$, $\mu(X,Y)$ being a degree to which on
object $X$ is included into an object $Y$. A rough inclusion can
be regarded as a hierarchy of relations of partial inclusion. It
was shown in [22], [23] that under some natural assumptions on $\mu$,
one can obtain the Lesniewski mereology as the theory of the relation
of being a part in degree 1, therefore a rough inclusion
provides a model for mereology. It was shown that one can obtain
by means of $\mu$ a model of the calculus of individuals.
In the calculus of individuals one can introduce ([1]) a quasi-topological
structure which under some additional axioms turns out to be a topological
structure with out the empty set.
In this chapter we will show how to introduce a \v{C}ech topological
structure into a rough mereological space and we will prove that
the introduced topological structure contains as a subspace the
quasi-topological structure of [1]. The results are taken from our
work [33].
Now we introduce some of the work of Le"sniewski which is the primitive
notion of mereology (a relation part). We choose to follow the
original presentation by Le"sniewski ([4]); for the convenience
of the reader we include all proofs in their original, slightly
modernized as to terminology, versions.
\section{Mereology of Le"sniewski}
The basic notion here is that of being a part.
\subsection{Axiom}
If object $X$ is a part of object $Y$, then object $Y$ is not a
part of object $X$.
\subsection{Axiom}
If object $X$ is a part of object $Y$ and and object $Y$ is a
part of object $Z$, then object $X$ is a part of object $Z$.
\begin{theorem}
{\it
\vspace*{-4mm}
No object is a part of itself.
}
\end{theorem}
\hbox{}
\begin{proof}
Obvious.
\end{proof}
Now from the relation part, the relation $ing$ of being
ingredient is defined as follows:
\begin{deff}
For any pair $X,Y$ of objects, $X ing Y$ iff $X$ part $Y$ or $X
=Y$.
\end{deff}
\begin{theorem}\hbox{}\hfill
{\it
\vspace{-6mm}
\begin{enumerate}
\item Every object is an ingredient of itself.
\item If $Y$ is a part of object $X$, then $Y$ is an ingredient
of object $X$.
\end{enumerate}
}
\end{theorem}
\begin{proof}
obvious. The relation $ingr$ is obviously transitive.
\end{proof}
\begin{theorem}\hbox{}\hfill\break
\vspace*{-10mm}
\noindent {\em If $X$ is an ingredient of object $Y$ and $Y$ is an ingredient
of $Z$, then $X$ is an ingredient of object $Z$.}
\end{theorem}
Similarly, the following statement is obvious by Theorem 5.2.2.
\begin{theorem}\hbox{}\hfill
{\it
If $Y$ is an ingredient of object $X$, then some ingredient of
object $Y$ is an ingredient of object $X$.
}
\end{theorem}
Again, 5.2.2. implies
\begin{theorem}\hbox{}\hfill
{\it
If $Y$ is an ingredient of object $X$, then some ingredient of
object $Y$ is an ingredient of some ingredient of object $X$.
}
\end{theorem}
\begin{deff}
The expresssion `set of objects $m$' is to denote every object $X$,
which satisfies the following condition:
if $Y$ is an ingredient of object $X$, then some ingredient of
object $Y$ is an ingredient of
some $m$, which is an ingredient of object $X$.
\end{deff}
\begin{example}
Let
$$
X=\{x: x\;\;\mbox{is a chemical atom}\}
$$
Since if $Y$ is an ingredient of $X$, then some ingredient of
object $Y$ is an ingredient of some atom who is an ingredient of
$X$.
\end{example}
\begin{example}
Consider the segments $XY$ of figure 1. and use the expression `$m$'
to denote every segment $XX_{1}$, $X_{1}X_{2}$, $X_{2}X_{3}$
and $X_{3}Y$, which is a part of segment $XY$. Segment $XX_{5}$
is not a set of objects $m$, because segment $X_{4}X_{5}$, which
is an ingredient of segment $XX_{5}$, has not even one ingredient,
which is an ingredient of any $m$, which is an ingredient of segment
$XX_{5}$, therefore it is not true, that if $Z$ is an ingredient
of segment $XX_{5}$, then some ingredient of object $Z$ is an
ingredient of some $m$, which is an ingredient of segment $XX_{5}$.
\end{example}
\eject
\hbox{}
\medskip
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\vspace*{-10mm}
\begin{deff}
The expressions' set of all objects $m$' and `class of objects
$m$' denote every object $X$, which satisfies the following conditions:
\begin{enumerate}
\item Every $m$ is an ingredient of object $X$.
\item If $Y$ is an ingredient of object $X$, then some ingredient of object
$Y$ is an ingredient of some $m$.
\end{enumerate}
Now we introduce the following two axioms (see [4]):
\begin{description}
\item[Axiom I.] If some object is $m$, then some object is the class
of objects $m$.
\item[Axiom II.] If $X$ is the class of objects $m$, and $Y$ is
the class of objects $m$, then $X$ is~$Y$.
\end{description}
\end{deff}
\begin{example}
Let $X$ be the set of all chemical atoms, then $X$ is the class
of elements, because
\begin{enumerate}
\item Every element is an ingredient of $X$,
\item If $Y$ is an ingredient of $X$, then some ingredient of
object $Y$ is an ingredient of some element.
\end{enumerate}
\end{example}
\begin{example}
The segment $XX_{1}$ of figure 2 is not the class of parts of
segment $XY$, because not every part of segment $XY$ is an
ingredient of segment $XX_{1}$. Also the segment $XY$ is
not the class of segment $XX_{1}$, because in this case
condition 2 of Definition 5.2.3 does not hold.
\end{example}
\medskip
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\begin{theorem}\hbox{}\hfill\break
\vspace*{-11mm}
{\it
\begin{enumerate}
\item $X$ is the class of objects $m$ implies $X$ is a set of
objects $m$.
\item Object $X$ is the class of ingredients of $X$.
\item If some object is a part of object $X$, then $X$ is the
class of parts of object $X$.
\item Object $X$ is the class of objects $X$.
\end{enumerate}
}
\end{theorem}
\vfill\eject
\begin{proof}
\vspace*{-3mm}
\begin{enumerate}
\item Suppose that:
\begin{itemize}
\item[(a)] $X$ is the class of objects $m$ in
accordance with definition 5.2.3; we may say:
\item[(b)] Every $m$ is an ingredient of object $X$.
\item[(c)] If $Y$ is an ingredient of object $X$, then
some ingredient of object $Y$ is an ingredient
of some $m$.
From (c) we infer on the basis of (b), that, if
$Y$ is an ingredient of object $X$, then some
ingredient of object $Y$ is an ingredient of
some $m$, which is an ingredient of object $X$.
From this and from definition 5.2.2.:
\item [(d)] $X$ is a set of objects $m$.
Therefore (a) implies (d). It follows from
this that, if $X$ is the class of objects $m$,
then $X$ is a set of objects $m$, which was
to be proved.
\end{itemize}
\item By the law of identity we may write:
\begin{itemize}
\item [(a)] Every ingredient of object $X$ is an ingredient
of object $X$. From Theorem 5.2.5 we have:
\item [(b)] If $Y$ is an ingredient of object $X$, then
some ingredient of object $Y$ is an ingredient
of some ingredient of object $X$.
\end{itemize}
\item Assume that,
\begin{itemize}
\item [(a)] Some object is a part of object $X$.
It follows that some object is an object $Y$,
such that:
\item [(b)] $Y$ is a part of object $X$.
Now from (a) and from theorem 5.2.3 part(2) we get:
\item [(c)] Every part of object $X$ is an ingredient of object $X$.
From (b) and (c) we get:
\item [(d)] $Y$ is an ingredient of object $X$.
From theorem 5.2.5 part (1) we have that:
\item [(e)] $Y$ is an ingredient of object $Y$.
From (d) and (e) we get:
\item [(f)] Some ingredient of object $X$ is an ingredient
of object $Y$.
From (b) and (f) we get:
Some ingredient of object $X$ is an ingredient
of some part of object $X$.
\item [(g)] If $Z$ is a part of object $X$, then some ingredient
of object $Z$ is an ingredient of some part of
object $X$, since if we suppose differently it
will imply that some part $Z_{1}$ of object
$Z$ is such that no ingredient of object $Z_{1}$,
is an ingredient of any part of object $X$,
from this and theorem 5.2.3 part (1) we get
that this object $Z_{1}$ is not an ingredient
of any part of object $X$, from which we get
a contradiction; we write:
\item [(h)] Every ingredient of object $X$ is either object
$X$ or a part of object $X$.
From (f), (g) and (h) we get:
\item [(i)] If $Y$ is an ingredient of object $X$, then
some ingredient of object $Y$ is an ingredient
of some part of object $X$.
From (c) and (i) and Definition 5.2.3 we get:
\item [(j)] $X$ is the class of parts of object $X$.
\end{itemize}
\item From theorem 5.2.2 part(1): $X$ is an ingredient of $X$,
so we may write:
\begin{itemize}
\item Every $X$ is an ingredient of object $X$. From
theorem 5.2.4 we know that:
\item If $Y$ is an ingredient of object $X$, then some
ingredient of object $Y$ is an ingredient of object
$X$. Therefore from (a), (b) and Definition 5.2.3
we get~4.
\end{itemize}
\end{enumerate}
\end{proof}
\begin{deff}
The expression {\em element of object} $X$ denotes any object $Y$
when, with some meaning of the expression $m$ the following two
conditions are maintained:
\begin{enumerate}
\item $X$ is the class of objects $m$.
\item $Y$ is $m$.
\end{enumerate}
\end{deff}
\begin{example}
Segment $XY$ in figure 3 is an element of $XZ$, because, if the
expression `$m$' is used with the meaning of the expression segment
which is $XY$ or $XZ$, then
\begin{enumerate}
\item Segment $XZ$ is the class of objects $m$,
\item Segment $XY$ is $m$.
\end{enumerate}
\end{example}
\smallskip
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In the following theorem, we collect the basic properties of
mereology of Le"sniewski, let us observe that 1,2 below imply
that being an element is equivalent to being an ingredient, and
10, 12 imply that any class of sets of objects is identical
with itself as the class of objects i.e. the Cantorian hierarchy
of sets is collapsed to the level of objects.
\begin{theorem}
\vspace*{-6mm}
{\it
\begin{enumerate}
\item If $Y$ is an ingredient of object $X$, then $Y$ is an
element of object $X$.
\item If $Y$ is an element of object $X$, then $Y$ is an
ingredient of object $X$.
\item If $Y$ is a part of object $X$, then $Y$ is an
element of object $X$.
\item Every object is an element of itself.
\item If $X$ is an element of $Y$ and $Y$ is an element of $Z$,
then $X$ is an element of $Z$.
\item If $X$ is the class of objects $m$, then every $m$ is an element
of object $X$.
\item If $X$ is a set of objects $m$, then some $m$ is an
element of object $x$.
\item If $X$ is $m$, then $X$ is a set of objects $m$.
\item If $X$ is a set of objects $m$, and every $m$ is $n$
then $X$ is a set of objects $n$.
\item If $X$ is the class of sets of objects $m$, then $X$ is
the class of objects $m$.
\item If $X$ is a set of objects $m$, then $X$ is an ingredient
of the class of objects $m$.
\item If $X$ is the class of objects $m$, then $X$ is the
class of sets of objects $m$.
\item If it is true, that if $Y$ is an ingredient of object $X_{1}$
then some ingredient of object $Y$ is an ingredient of object
$X$, then $X_{1}$ is an ingredient of object $X$.
\item Every object $X$ is the class of elements of that object $X$.
\item Every set is its own element.
\item No object is the class of sets which are not their own elements.
\item The assertion: if $X$ is an element of a set of object $m$,
then $X$ is $m$ is false.
\end{enumerate}
}
\end{theorem}
The proofs are straight forward; we include the exemplary
proofs of statements 1,2, only.
\hbox{}
\begin{proof}
\begin{enumerate}
\item Assume that:
\begin{itemize}
\item[(a)] $Y$ is an ingredient of object $X$. By theorem
5.2.6 part (2) we may say:
\item[(b)] $X$ is the class of ingredients of objects $X$.
Using the expression $m$ with the meaning of
the expression ingredient of object $X$, we can get from (a)
and (b) that:
\item[(c)] $X$ is the class of objects $X$.
\vfil\eject
\item[(d)] $Y$ is $x$.
From (c), (d) and Definition 5.2.2 we get:
\item[(e)] $Y$ is an element of object $X$. Therefore
assuming (a) we get (e). Thus if $Y$ is an ingredient of $X$,
then $Y$ is an element of object $X$.
\end{itemize}
\item Suppose that:
\begin {itemize}
\item[(a)] $Y$ is an element of object $X$. From (a) and Definition 5.2.2
we see that there exists a meaning of the expression `$m$',
such that:
\item[(b)] $X$ is the class of objects $m$.
\item[(c)] $Y$ is $X$.
From (b) and Definition 5.2.3, we get:
\item[(d)] Every $m$ is an ingredient of object $X$.
From (c) and (d) we get:
\item[(e)] $Y$ is an ingredient of object $X$.
Therefore, assuming (a) we get (c) it follows that if
$Y$ is an element object $X$, then $Y$ is an
ingredient of object $X$.
\end{itemize}
\item This follows from Theorem 5.2.3.
\item Suppose it is false. It follows that some object is such
an object $X$, that:
\begin{itemize}
\item[(a)] $X$ is not an element of object $X$.\newline
From (a) we get:
\item[(b)] $X$ is not ingredient of object $X$. Since
$X$ is an ingredient of object $X$, then $X$ is an
element of $X$ (by part (1) of this theorem), which
is a contradiction to (a). (b) is a contradiction
to theorem 5.2.3. Therefore 4 is true.
\end{itemize}
\item Suppose that:
\begin{itemize}
\item[(a)] $X$ is an element of object $Y$ and $Y$ is an
element of object $Z$.\newline
Therefore by this theorem part 2 we get:
\item[(b)] $X$ is an ingredient of object $Y$.
\item[(c)] $Y$ is an ingredient of object $Z$.\newline
From (b), (c) and theorem 5.2.3 we get:
\item[(d)] $X$ is an ingredient of $Z$.\newline
From (d) and this theorem part 1 we get
\item[(e)] $X$ is an element of $Z$.\newline
Therefore from (a) we arrive at (e) and the
proof is finished.
\end{itemize}
\end{enumerate}
\end{proof}
We comment on the notion of a subset in mereology.
\begin{deff}
The expression 'subset
of object $X'$ denotes any object $X_{1}$ which satisfies
the following condition: every element of object $X_{1}$ is an
element of object $X$.
\end{deff}
\begin{example}
Segment $PQ$ in figure 4 is a subset of segment $PR$ because
segment $PQ$ satisfies the condition of Definition 5.2.5: every
element of segment $PQ$ is an element of segment $PR$.
\end{example}
\begin{example}
Segment $PR$ in figure 4 is not a subset of segment $PQ$ because
it is not the case that every element of segment $PR$ is an
element of segment $PQ$: here for instance the segment $PR$ is
an element of segment $PR$, is not an element of segment $PQ$
(look at theorem 5.2.7 part (4)).
\end{example}
\smallskip
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\begin{deff}
The expression `proper subset of object $X$' denotes any such
subset $X_{1}$ of object $X$, which is not $X$.
\end{deff}
\begin{example}
Segment $PQ$ in figure 4 is a proper subset of segment $PR$, because
segment $PQ$ is a such a subset of segment $PR$ which is not
segment $PR$.
\end{example}
\begin{theorem}
{\it
If $X_{1}$ is an ingredient of object $X$, then $X_{1}$ is a
subset of object~$X$.
}
\end{theorem}
\begin{proof}
Suppose that the theorem is false. It follows that some objects are
such objects $P$ and $Q$:
\begin{itemize}
\item[(a)] $P$ is an ingredient of object $Q$, but
\item[(b)] $P$ is not a subset of object $Q$.\newline
From (b) and Definition 5.2.5 we get:
\item[(c)] Some element of object $P$ is not an element
of object $Q$, from which we see that some object $R$
is such that.
\item[(d)] $R$ is an element of object $P$, but
\item[(e)] $R$ is not an element of object $Q$.\newline
From (e) we get:
\item[(f)] $R$ is not an ingredient of object $Q$, since if $R$
were an ingredient of object $Q$, then from this
and theorem 5.2.7 part (1), $R$ is an element of
object $Q$ which is a contradition to (e).
\vfil\eject
From Theorem 5.2.7 part (2) and (d) we get:
\item[(g)] $R$ is an ingredient of $P$.\newline
From (a), (g) and Theorem 5.2.3 we get:
\item[(h)] $R$ is an ingredient of object $Q$.\newline
Now (h) is a contradiction to (f).
\end{itemize}
\end{proof}
\begin{theorem}
{\it
If $X_{1}$ is a subset of object $X$, then $X_{1}$ is an
ingredient of object~$X$.
}
\end{theorem}
\begin{proof}
Suppose the theorem is false. We get from this that some objects
are such objects $P$ and $Q$:
\begin{itemize}
\item[(a)] $P$ is a subset of object $Q$, but
\item[(b)] $P$ is not an ingredient of object $Q$.\newline
From (b) we get:
\item[(c)] $P$ is not an element of object $Q$, if $P$ is an
element of $Q$, then from this and Theorem 5.2.7 part (2)
imply that $P$ is an ingredient of object $Q$,
which is a contradiction to (b). From Theorem 5.2.7
part (4) we know that:
\item[(d)] $P$ is an element of object $P$.\newline
From (d) and (c) we get:
\item[(e)] Some element of object $P$ is not an element of
object $Q$.\newline
From (e) and Definition 5.2.5 we get
\item[(f)] $P$ is not a subset of object $Q$.
Now (f) is a contradiction to (a).\newline
Theorem must be true.
\end{itemize}
\end{proof}
We prove similarly.
\begin{theorem}
{\it
If $X_{1}$ is a part of object $X$, then $X_{1}$ is a proper
subset of object $X$.
}
\end{theorem}
and
\begin{theorem}
{\it
If $X_{1}$ is a proper subset of object $X$, then $X_{1}$ is a
part of object $X$.
}
\end{theorem}
\begin{theorem}
{\it No object is a proper subset of itself.\/}
\end{theorem}
\begin{proof}
Suppose that some object $X$ is a proper subset of itself, then
this means that some object $X$ is a proper subset of object
$X$, then it follows from this and Theorem 5.2.11 that $X$ is a
part of object $X$, which contradicts Theorem 5.2.2.
\end{proof}
\begin{theorem}
{\it
Every object is a subset of itself.
}
\end{theorem}
\begin{proof}
Assume that some object $X$ is not a subset of itself. This means
that some object $X$ is not a subset of object $X$. Then it
follows from this and Theorem 5.2.8 that $X$ is not an ingredient
of object $X$, which contradicts Theorem 5.2.3 part (1).
\end{proof}
\begin{theorem}
{\it
If $X$ is a proper subset of object $X_{1}$, then $X_{1}$ is
not a proper subset of object $X$.
}
\end{theorem}
\begin{proof}
Suppose that:
\begin{itemize}
\item[(a)] $X$ is a proper subset of object $X_{1}$. From this
and Theorem 5.2.11, we get:
\item[(b)] $X$ is a part of object $X_{1}$. From this we get:
\item[(c)] $X_{1}$ is not a part of object $X$.\newline
From Theorem 5.2.11 and (c) we get:
\item[(d)] $X_{1}$ is not a proper subset of object $X$.
Thus (a) leads to (d) and theorem is proved.
\end{itemize}
\end{proof}
\begin{theorem}
{\it
If $X$ is a proper subset of object $X_{1}$, then $X_{1}$ is not a
subset of object $X$.
}
\end{theorem}
\begin{proof}
Suppose that the theorem is false. It follows from this, that
some objects $P$ and $Q$ are such that
\begin{itemize}
\item[(a)] $P$ is a proper subset of object $Q$, but
\item[(b)] $Q$ is a subset of object $P$.\newline
From (a) and Definition 5.2.6 we get
\item[(c)] $P$ is not $Q$. From (c) we get:
\item[(d)] $Q$ is not $P$.\newline
From (b), (d) and Definition 5.2.6 we get:
\item[(e)] $Q$ is a proper subset of object $P$. \newline
From Theorem 5.2.14 and (a) we get:
\item[(f)] $Q$ is not a proper subset of $P$. Now (f)
contradicts (e). Theorem is true.
\end{itemize}
\end{proof}
We prove in the same way.
\begin{theorem}
{\it
If $X$ is a subset of object $X_{1}$ and $X_{1}$ is a subset of
object $X_{2}$, then $X$ is a subset of object $X_{2}$.
}
\end{theorem}
\begin{theorem}
{\it
If $X$ is a proper subset of object $X_{1}$, and $X_{1}$ is a
subset of object $X_{2}$, then $X$ is a proper subset of object $X_{2}$.
}
\end{theorem}
\begin{theorem}
{\it
If $X$ is a subset of object $X_{1}$ and $X_{1}$ is a proper
subset of object $X_{2}$, then $X$ is a proper subset of
object $X_{2}$.
}
\end{theorem}
\begin{theorem}
{\it If $X_{1}$ is an element of $X$, then $X_{1}$ is a subset of $X$.\/}
\end{theorem}
\begin{proof}
Immediately from Theorem 5.2.7 part (2) and Theorem 5.2.8.
\end{proof}
\begin{theorem}
{\it If $X_{1}$ is a subset of $X$, then $X_{1}$ is an element of $X$.\/}
\end{theorem}
\begin{proof}
Immediately from Theorem 5.2.7 part (1) and Theorem 5.2.9.
\end{proof}
\begin{theorem}
{\it Every $X$ is the class of subsets of that $X$.\/}
\end{theorem}
\begin{proof}
From Theorem 5.2.9 we know that:
\begin{itemize}
\item[(a)] Every subset of $X$ is an ingredient of $X$. From
Theorem 5.2.5 we know that:
\item[(b)] If $Y$ is an ingredient of $X$, then some ingredient
of $Y$ is an ingredient of some ingredient of $X$.
From (b) and Theorem 5.2.8 we get:
\item[(c)] If $Y$ is an ingredient of $X$, then some ingredient
of $Y$ is an ingredient of some subset of $X$.
From (a), (c) and Definition 5.2.6 we obtain theorem.
\end{itemize}
\end{proof}
\begin{theorem}
{\it
If $X$ is a subset of $m$, every $m$ is $n$, then $X$ is a subset
of a class of objects $n$.
}
\end{theorem}
\begin{proof}
Suppose that:
\begin{itemize}
\item[(a)] $X$ is a subset of objects $m$, and every $m$ is $n$.\newline
From Theorem 5.2.7 part (9) and (a) we get:
\item[(b)] $X$ is a subset of objects $n$.\newline
From Theorem 5.2.7 part (11) and (b) we get:
\item[(c)] $X$ is an ingredient of the class of objects $n$.\newline
From Theorem 5.2.8 and (c) we get:
\item[(d)] $X$ is a subset of the class of objects $n$.
Thus assuming (a) we get (d) i.e. we get the theorem.
\end{itemize}
\end{proof}
\begin{deff}
The expression ``universe'' denotes the class of objects.
\end{deff}
\begin{theorem}
{\it Some object is the class of non-contradictory objects.\/}
\end{theorem}
\begin{proof}
We may state that every object is a non-contradictory object. We get from
this that some object is a non-contradictory object from this
and Axiom I the theorem is proved.
\end{proof}
\begin{theorem}
{\it The class of non-contradictory objects is the universe.\/}
\end{theorem}
\begin{proof}
By Definition 5.2.3 we may say:
\begin{itemize}
\item[(a)] Every non-contradictory object is an ingredient of the
class of non-contradictory objects.
\item[(b)] If $Y$ is an ingredient of the class of
non-contradictory objects, then some ingredient of
object $Y$ is an ingredient of some non-contradictory
object.
By the law of non-contradiction we get:
\item[(c)] Every object is non-contradictory object.\newline
From (a) and (c) we get:
\item[(d)] Every object is an ingredient of the class of
non-contradictory objects.\newline
From (b) we get:
\item[(e)] If $Y$ is an ingredient of the class of non-contradictory
objects, then some ingredient of $Y$ is an ingredient
of some object.\newline
From (d), (e) and Definition 5.2.3 we get:
\item[(f)] The class of non-contradictory objects is the class
of objects, from this and Definition 5.2.7 we obtain
the given theorem.
\end{itemize}
\end{proof}
\begin{theorem}
{\it If $X$ is the universe, and $X_{1}$ is the universe, then $X$ is
$X_{1}$.\/}
\end{theorem}
\begin{proof}
Suppose that:
\begin{itemize}
\item[(a)] $X$ is the universe
\item[(b)] $X_{1}$ is the universe.\newline
From (a) and Definition 5.2.7 we get:
\item[(c)] $X$ is the class of objects.\newline
From (b) and Definition 5.2.7 we get:
\item[(d)] $X_{1}$ is the class of objects.\newline
From (c), (d) and Axiom II we get:
\item[(e)] $X$ is $X_{1}$.
Thus (a) and (b) gets to (e) and we obtain the given theorem.
\end{itemize}
\end{proof}
A topological context may be now introduced into mereology.
\begin{deff}
The expression `object exterior to object $X$' denotes every
such object $X_{1}$, which satisfies the following condition:
No ingredient of $X$ is an ingredient of $X_{1}$.
\end{deff}
\begin{example}
Segment $PQ$ in figure 4 is an object exterior to segment $SR$,
because no ingredient of segment $SR$ is an ingredient of
segment $PQ$ - and segment $PS$ is not an object exterior to
segment $QR$ because segment $QS$ which is an ingredient of segment
$QR$ is also an ingredient of segment $PS$, therefore it is not
true, that no ingredient of segment $QR$ is an ingredient of
segment $PS$.
\end{example}
\begin{theorem}
{\it
If $X_{1}$ is an object exterior to object $X$, then $X$ is an
object exterior to object $X_{1}$.
}
\end{theorem}
\begin{proof}
Suppose that:
\begin{itemize}
\item[(a)] $X_{1}$ is an object exterior to object $X$.\newline
From (a) and Definition 5.2.8, we get
\item[(b)] No ingredient of $X_{1}$ is an ingredient of $X$.\newline
From (b) we get:
\item[(c)] No ingredient of $X_{1}$ is an ingredient of object $X$.\newline
From (c) and Definition 5.2.8 we get
\item[(d)] $X$ is an object exterior to $X_{1}$. Thus we
obtain the given theorem.
\end{itemize}
\end{proof}
\begin{theorem}
{\it No object is an object exterior to itself.\/}
\end{theorem}
\begin{proof}
Suppose that the Theorem is false: we get from this, that some
object is such an object $X$, that:
\begin{itemize}
\item[(a)] $X$ is an object exterior to object $X$. \newline
From (a) and Definition 5.2.8 we get:
\item[(b)] No ingredient of object $X$ is an ingredient of $X$.
Now (b) is a contradictory assertion. Thus the
theorem is true.
\end{itemize}
\end{proof}
\begin{deff}
The expression `complement of object $X_{1}$ with respect to
$X$' denotes any $X_{2}$, if the following conditions are satisfied:
\begin{enumerate}
\item $X_{1}$ is a subset of $X$,
\item $X_{2}$ is the class of elements of object $X$, exterior
to object $X_{1}$.
\end{enumerate}
\end{deff}
\begin{example}
Segment $PQ$ in figure 4 is not the complement of segment $SR$
with respect to segment $PR$, because condition (1) is indeed
not observed here (segment $SR$ is a subset of segment $PR$),
but condition (2) is not satisfied (segment $PQ$ is not the
class of elements of segment $PR$, exterior to segment $SR$).
\end{example}
\begin{theorem}
{\it
If $X_{1}$ is a part of object $X$, then some object is the
complement of object $X_{1}$ with respect to $X$.
}
\end{theorem}
\begin{proof}
Suppose that:
\begin{itemize}
\item[(a)] $X_{1}$ is a part of object $X$. From Theorem 5.2.10
and (a) we get:
\item[(b)] $X_{1}$ is a proper subset of object $X$. From (b) and
Definition 5.2.6, we get
\item[(c)] $X_{1}$ is a subset of object $X$, on the other hand
by Theorem 5.2.11 we get
\item[(d)] $X$ is not a subset of object $X_{1}$.\newline
From (d) and Theorem 5.2.8 we get:
\item[(e)] $X$ is not an ingredient of object $X_{1}$. We may
say that some object $X_{2}$ is such that,
\item[(f)] $X_{2}$ is an ingredient of object $X$,
\item[(g)] No ingredient of object $X_{2}$ is an ingredient of
object $X_{1}$, because if no object is object
$X_{2}$ agrees with (f) and (g), then it is true that if $Y$ is an
ingredient of object $X$, then same ingredient of
object $Y$ is an ingredient of object $X_{1}$, on the
other hand from this and from Theorem 5.2.7 part
(13), it follows that $X$ is an ingredient of object $X_{1}$,
which contradicts (e).\newline
From (g) and Definition 5.2.8 we get:
\item[(h)] $X_{1}$ is an object exterior to $X_{2}$. From (h)
and Theorem 5.2.26 we get:
\item[(i)] $X_{2}$ is an object exterior to $X_{1}$. From
Theorem 5.2.7 part (1) and (f) we get:
\item[(j)] $X_{2}$ is an element of $X$.\newline
From (j) and (i) we get:
\item[(k)] $X_{2}$ is an element of $X$ exterior to $X_{1}$.
From (k) and Axiom I we get:
\item[(l)] Some object $X_{3}$ is the class of elements of
object $X$, exterior to $X_{3}$.\newline
From (c), (l) and Definition 5.2.9 we get
\item[(m)] $X_{3}$ is the complement of object $X_{1}$ with
respect to $X$.\newline
Thus assuming (a) we arrive to (m) and so we obtain the
Theorem.
\end{itemize}
\end{proof}
\begin{theorem}
{\it
If $X_{2}$ is the complement of object $X_{1}$ with respect to
$X$, then $X_{2}$ is an object exterior to object $X_{1}$.
}
\end{theorem}
\begin{proof}
Suppose the theorem is false: from this we get objects $X$,
$X_{1}$ and $X_{2}$ such that:
\begin{itemize}
\item[(a)] $X_{2}$ is the complement of object $X_{1}$ with respect
to object $X$, but
\item[(b)] $X_{2}$ is not an object exterior to $X_{1}$.
From (a) and Definition 5.2.9 we get:
\item[(c)] $X_{2}$ is the class of elements of $X$, exterior to
object $X_{1}$.\newline
From (c) and Definition 5.2.7 we get:
\item[(d)] If $Y$ is an ingredient of $X_{2}$, then some
ingredient of object $Y$ is an ingredient of some
element of object $X$, which is an object exterior to
$X_{1}$.\newline
From (b) and Definition 5.2.8 we get:
\item[(e)] Some ingredient of $X_{1}$ is an ingredient of $X_{2}$.
We get from this that some $X_{3}$ is such that:
\item[(f)] $X_{2}$ is an ingredient of $X_{1}$,
\item[(g)] $X_{3}$ is an ingredient of $X_{2}$. \newline
From (d) and (g) we get:
\item[(h)] Some ingredient of $X_{3}$ is an ingredient
of some element of $X$, which is an object exterior
to $X_{1}$. \newline
We get from this that some object $X_{4}$, such that:
\item[(i)] $X_{4}$ is an ingredient of $X_{3}$.
\item[(j)] $X_{4}$ is an ingredient of some element of $X$,
which is an object exterior to $X_{1}$. From (j), we
get that some object $X_{5}$ such that:
\item[(k)] $X_{3}$ is an element of $X$,
\item[(l)] $X_{3}$ is an object exterior to $X_{1}$,
\item[(m)] $X_{4}$ is an ingredient of $X_{5}$.\newline
From (l) and Definition 5.2.8 we get
\item[(n)] No ingredient of $X_{1}$ is an ingredient of
$X_{5}$. \newline
From (i), (f), (g), and Theorem 5.2.3 we get
\item[(o)] $X_{4}$ is an ingredient of $X_{1}$.\newline
From (n) and (o) we get:
\item[(p)] $X_{4}$ is not an ingredient of $X_{5}$.
Now (p) is a contradiction to (m), and so we obtain
the theorem.
\end{itemize}
\end{proof}
\begin{theorem}
{\it
If $X_{2}$ is the complement of $X_{1}$ with respect to $X$,
then $X_{2}$ is a part of $X$.
}
\end{theorem}
\begin{proof}
Suppose that:
\begin{itemize}
\item[(a)] $X_{2}$ is the complement of $X_{1}$ with respect to $X$.\newline
From (a) and Definition 5.2.9 we get
\item[(b)] $X_{1}$ is a subset of $X$.
\item[(c)] $X_{2}$ is the class of elements of $X$, exterior to
$X_{1}$.\newline
From (c) and Definition 5.2.3 we get:
\item[(d)] If $Y$ is an ingredient of $X_{2}$, then some ingredient
of $Y$ is an ingredient of some element of $X$, which
is an object exterior to $X_{1}$.\newline
We get from this that:
\item[(e)] If $Y$ is an ingredient of $X_{2}$, then some ingredient
of $Y$ is an ingredient of some element of $X$.\newline
From (e) and Theorem 5.2.7 part (2) we get:
\item[(f)] If $Y$ is an ingredient of $X_{2}$, then some
ingredient of $Y$ is an ingredient of some ingredient
of $X$.\newline
\item[(g)] If $Y$ is an ingredient of $X_{2}$, then some ingredient
of $Y$ is an ingredient of $X$.\newline
From Theorem 5.2.7 part (13) we get:
\item[(h)] $X_{2}$ is an ingredient of $X$.\newline
From Theorem 5.2.29 and (a) we get:
\item[(i)] $X_{2}$ is an object exterior to $X_{1}$.\newline
From (i) and Definition 5.2.8 we get:
\item[(j)] No ingredient of $X_{1}$ is an ingredient of object
$X_{2}$.\newline
From Theorem 5.2.3 part (1) we get:
\item[(k)] $X_{1}$ is an ingredient of $X_{1}$. \newline
From (j) and (k) we get:
\item[(l)] $X_{1}$ is not an ingredient of $X_{2}$. \newline
From Theorem 5.2.9 and (b) we get:
\item[(m)] $X_{1}$ is an ingredient of $X$.\newline
From (l) and (m) we get:
\item[(n)] $X_{2}$ is not $X$.\newline
From (h) and (m) we get:
\item[(o)] $X_{2}$ is a part of $X$.
Thus assuming (a) we arrive to (o), so we obtain the
theorem.
\end{itemize}
\end{proof}
\begin{theorem}
{\it
If $X_{2}$ is the complement of $X_{1}$ with respect to $X$,
then $X_{1}$ is the complement of $X_{2}$ with respect to $X$.
}
\end{theorem}
\begin{proof}
Suppose that:
\begin{itemize}
\item[(a)] $X_{2}$ is the complement of $X_{1}$ with respect to
$X$.\newline
From this and Definition 5.2.9 we get:
\item[(b)] $X_{2}$ is the class of elements of $X$, exterior to $X$,\newline
On the other hand, by Theorem 5.2.30 we get:
\item[(c)] $X_{2}$ is a part of $X$.\newline
From (c) and Theorem 5.2.10 we get:
\item[(d)] $X_{2}$ is a subset of $X$.\newline
From (b) and Definition 5.2.3 we get:
\item[(e)] Every element of $X$, exterior to $X_{1}$, is an ingredient
of $X_{2}$, we may write:
\item[(f)] Every element of $X$, exterior to $X_{2}$, is an ingredient
of $X_{1}$.
By means of the following reasoning: we suppose, that
(f) is false, we get that some object $X_{3}$ is such
that:
\begin{enumerate}
\parsep=0pt \itemsep=0.5pt
\item $X_{3}$ is an element of $X$.
\item $X_{3}$ is an object exterior to $X_{2}$,
\item $X_{3}$ is not an ingredient of $X_{1}$, from (3) we get
that some object $X_{4}$ such that:
\item $X_{4}$ is an ingredient of $X_{3}$.
\item No ingredient of $X_{4}$ is an ingredient of $X_{1}$
because if no object is $X_{4}$ agrees with (4) and
(5) we get from this that it is true, that, if $Y$ is an
ingredient of $X_{3}$, then some ingredient of $Y$ is
an ingredient of $X_{1}$, on the other hand we get
from this and according to theorem 5.2.7 part (13)
that $X_{3}$ is an ingredient of $X_{1}$ which
contradicts (3) from (5) and Definition 5.2.8 we get:
\item $X_{1}$ is an object exterior to $X_{4}$.\newline
From Theorem 5.2.26 and (6) we get:
\item $X_{4}$ is an object exterior to $X_{1}$,\newline
from Theorem 5.2.7 part (1) and (4) we get:
\item $X_{4}$ is an element of $X_{3}$, \newline
from (8), (1) and 5.2.7 part (5) we get:
\item $X_{4}$ is an element of $X$.\newline
From (9) and (7) we get:
\item $X_{4}$ is an element of $X$ exterior to $X_{1}$,\newline
from (e) and (10) we get
\item $X_{4}$ is an ingredient of $X_{2}$.\newline
From (2) and Definition 5.2.8 we get:
\item No ingredient of $X_{2}$ is an ingredient of $X_{3}$.\newline
From (11) and (12) we get:
\item $X_{4}$ is not an ingredient of $X_{3}$.
Now (13) contradicts (4), so the supposition that (f)
is false, which led to this contradiction, must therefore
be false, therefore (f) is true.
\end{enumerate}
From (a) and Definition 5.2.9 we get:
\item[(g)] $X_{1}$ is a subset of $X$, on the other hand with
we get
\item[(h)] $X_{2}$ is an object exterior to $X_{1}$,\newline
from Theorem 5.2.9 and (g) we get:
\item[(i)] $X_{1}$ is an element of $X$.\newline
From Theorem 5.2.26 and (h) we get:
\item[(j)] $X_{1}$ is an object exterior to $X_{2}$. \newline
From (i) and (j) we get:
\item[(k)] $X_{1}$ is an element of $X$, exterior to $X_{2}$.\newline
From Theorem 5.2.4 we know:
\item[(l)] If $Y$ is an ingredient of $X_{1}$ then some ingredient of $Y$
is an ingredient of $X_{1}$.\newline
From (k) and (l) we get:
\item[(m)] If $Y$ is an ingredient of $X_{1}$, then some ingredient
of $Y$ is an ingredient of some element of $X$, which
is an object exterior to $X_{2}$.\newline
From (f), (j) and Definition 5.2.3 we get:
\item[(n)] $X_{1}$ is the class of elements of $X$, exterior to
$X_{2}$.\newline
From (d), (n) and Definition 5.2.9 we get
\item[(o)] $X_{1}$ is the complement of $X_{2}$ with respect to
object $X$.
Thus assuming (a) we arrive to (o) and we obtain the theorem.
\end{itemize}
\end{proof}
\begin{theorem}
{\it
If $X_{2}$ is the complement of $X_{1}$ with respect to $X$,
then $X_{1}$ is a part of $X$.
}
\end{theorem}
\begin{proof}
Follows from theorem 5.2.31 and 5.2.30.
\end{proof}
\begin{theorem}
{\it
If $X_{1}$ is a proper subset of $X$, then some object is the
complement of $X_{1}$ with respect to $X$.
}
\end{theorem}
\begin{proof}
Follows from Theorem 5.2.28 and 5.2.10.
\end{proof}
\begin{theorem}
{\it
If $X_{2}$ is the complement of $X_{1}$ with respect to $X$, and
$X_{3}$ is the complement of $X_{1}$, with respect to $X$, then
$X_{2}$ is $X_{3}$.
}
\end{theorem}
\begin{proof}
Suppose that
\begin{itemize}
\item[(a)] $X_{2}$ is the complement of $X_{1}$ with respect to $X$.
\item[(b)] $X_{3}$ is the complement of $X_{1}$ with respect to $X$.\newline
From (a) and Definition 5.2.9 we get
\item[(c)] $X_{2}$ is the class of elements of $X$, exterior to $X_{1}$.\newline
From (b) we get
\item[(d)] $X_{3}$ is the class of elements of $X$, exterior to $X_{1}$.\newline
From (c), (d) and Axiom II we get:
\item[(e)] $X_{2}$ is $X_{3}$.
Thus assuming (a) and (b) we arrive to (e). Therefore
the theorem is true.
\end{itemize}
\end{proof}
\begin{theorem}
{\it No object is the complement of itself with respect to some objects.\/}
\end{theorem}
\begin{proof}
Suppose that the theorem is false; we get from this that some
object $X$ and $X_{1}$ are such that:
\begin{itemize}
\item[(a)] $X_{1}$ is the complement of $X_{1}$ with respect to $X$.\newline
From 5.2.30 and (a) we get:
\item[(b)] $X_{1}$ is an object exterior to $X_{1}$.
Now (b) is a contradiction of Theorem 5.2.27.
Therefore the theorem must be true.
\end{itemize}
\end{proof}
\begin{theorem}
{\it
No object $X$ is the complement of a given object with respect
to $X$.
}
\end{theorem}
\begin{proof}
Suppose that the theorem is false, we get from this that some
object is an object $X_{1}$ such that:
\begin{itemize}
\item[(a)] $X$ is the complement of $X_{1}$ with respect to $X$.\newline
From Theorem 5.2.30 and (a) we get:
\item[(b)] $X$ is a part of $X$.
Now (b) contradicts (a). Therefore the theorem is true.
\end{itemize}
\end{proof}
\begin{theorem}
{\it
No object is the complement of a given $X$ with respect to the
same $X$.
}
\end{theorem}
\begin{proof}
Suppose that the theorem is false, we get from this that some
object is an object $X_{1}$ such that
\begin{itemize}
\item[(a)] $X_{1}$ is the complement of $X$ with respect to $X$.\newline
From Theorem 5.2.32 and (a) we get
\item[(b)] $X$ is a part of $X$
Now (b) contradicts Theorem 5.2.1. Therefore the
given theorem is true.
\end{itemize}
\end{proof}
\begin{theorem}
{\it
If $X_{2}$ is the complement of $X_{1}$ with respect to $X$, then
$X$ is the class of objects, which are $X_{1}$ and $X_{2}$.
}
\end{theorem}
\begin{proof}
Suppose that:
\begin{itemize}
\item[(a)] $X_{2}$ is the complement of $X_{1}$ with respect to
$X$.\newline
From (a) and Theorem 5.2.32 we get
\item[(b)] $X_{1}$ a part of $X$.\newline
On the other hand by Theorem 5.2.30 we get:
\item[(c)] $X_{2}$ is a part of $X$.\newline
From Theorem 5.2.3 part (2) and (b) we get
\item[(d)] $X_{1}$ is an ingredient of $X$.\newline
From Theorem 5.2.3 part (2) and (c) we get
\item[(e)] $X_{2}$ is an ingredient of $X$.\newline
From (d) and (e) we get:
\item[(f)] Every object which is $X_{1}$ or $X_{2}$ is an
ingredient of $X$.\newline
From (a) and Definition 5.2.7 we get:
\item[(g)] $X_{2}$ is the class of elements of $X$, exterior to
$X_{1}$\newline
From (g) and Definition 5.2.3 we get
\item[(h)] Every element of $X$, exterior to $X_{1}$, is an
ingredient of $X_{2}$.\newline
We may say that:
\item[(i)] If $Y$ is an ingredient of $X$, then some ingredient
of $Y$ is an ingredient of some object which is $X_{1}$ or $X_{2}$.
By means of the following reasoning: we suppose, that
(i) is false, we get from this that some object is
such an object $Y_{1}$ such that
\begin{enumerate}
\item $Y_{1}$ is an ingredient of $X$, but
\item No ingredient of $Y_{1}$ is an ingredient of the given
object, which is $X_{1}$ or $X_{2}$, from (2) we get
\item No ingredient of $Y_{1}$ is an ingredient of $X_{1}$,
\item No ingredient of $Y_{1}$ is an ingredient of $X_{2}$;\newline
from (3) and Definition 5.2.8 we get
\item $X_{1}$ is an object exterior to $Y_{1}$.\newline
From Theorem 5.2.26 and (5) we get
\item $Y_{1}$ is an object exterior to $X_{1}$.\newline
From Theorem 5.2.7 part (1) and (1) we get
\item $Y_{1}$ is an element of $X$.\newline
From (7) and (6) we get
\item $Y_{1}$ is an element of $X$, exterior to $X_{1}$.\newline
From (8) and (h) we get
\item $Y_{1}$ is an ingredient of $X_{2}$.\newline
From Theorem 5.2.3 part (1) we know:
\item $Y_{1}$ is an ingredient of $Y_{1}$.\newline
From (10) and (4) we get:
\item $Y_{1}$ is not an ingredient of $X_{2}$.
\item[] Now (11) contradicts (9). The supposition that (i) is
false which led to this contradiction must be false,
thus (i) is true.
\end{enumerate}
From (f), (i) and Definition 5.2.3 we get
\item[(j)] $X$ is the class of objects, which are $X_{1}$ of $X_{2}$.\newline
Therefore assuming (a) we get (j) and thus theorem is proved.
\end{itemize}
\end{proof}
This concludes the survey of mereology of Le"sniewski.
\section{Calculus of individuals}
This version of mereology is based on the predicate $C(X,Y)$
($X$ is connected to $Y$) satisfying the following conditions [1].
Introduced by A.N.Whitehead (who used the dual predicate
``extends over'', it was defined in the present form by Leonard
and Goodman (see [1] and references there).
\begin{deff}
\vspace*{-4.1mm}
\begin{itemize}
\parsep=0pt \itemsep=-0.4pt
\item [(i)] $C(X,X)$ for any object $X$,
\item [(ii)] $C(X,Y) \Rightarrow C(Y,X)$ for any pair $X,Y$
of objects,
((i), (ii) state that $C$ as a relation is a
tolerance relation ([17])).
\item [(iii)] $(\forall Z (C(X,Z) \Leftrightarrow C(Y,Z)) \Rightarrow
X = Y$ for any pair $X,Y$ of objects.
\end{itemize}
\end{deff}
Restoring from the predicate $C$ the mereological
relations {\it part\/}, $ingr$ is done by means of
the following definitions.
\begin{deff}
\vspace*{-4.1mm}
\begin{itemize}
\parsep=0pt \itemsep=-0.6pt
\item [(i)] $X {\it ingr\/} Y$ iff $\forall Z \; C(X,Z)\Rightarrow C(Y,Z)$,
\item [(ii)] $X$ {\it part\/} $Y$ iff $X {\it ingr\/} Y$ and it is not true
that $Y {\it ingr\/} X$.
\end{itemize}
\end{deff}
The predicate $C$ may be used to define other predicates.
\begin{deff}
\vspace*{-4.1mm}
\begin{itemize}
\parsep=0pt \itemsep=-0.2pt
\item [(i)] $O(X,Y)$ iff $\exists Z.\; Z {\it ingr\/} X$ and $Z {\it ingr\/} Y$
(the {\it overlapping\/} relation)
\item [(ii)] $EC(X,Y)$ iff $C(X,Y)$ and it is not true that
$O(X,Y)$ (the {\it external connectedness\/} relation).
\item [(iii)] $TP(X,Y)$ iff $X {\it ingr\/} Y$ and $\exists Z.\; E C(Z,X)$
and $EC(Z,Y)$ (the {\it being a tangential ingredient\/}
relation)
\item [(iv)] $NTP(X,Y)$ iff $X {\it ingr\/} Y$ and it is not true that
$\exists Z.\; EC(Z,X)$ and $EC (Z,Y)$.\newline
The relation $NTP$ permits us to define the interior
operator on objects [3].
\end{itemize}
\end{deff}
\begin{deff}
%\vspace*{-0.4mm}
For an object $X$ we define $iX$, the quasi-interior of $X$, as follows:
$$
iX = {\it class\/}\, (T: NTP (T,X)).
$$
\vfil\eject
As shown in [1] $i$ is a topological interior operator
under additional axioms ([1], Axiom 2.1); one may observe, that in general,
$i$ is a \v{C}ech interior operator ([3]).
\end{deff}
Let us quote the following formulas of calculus of
individuals:
$$
\forall X \forall Y \bigl[ \neg \exists Z E C(Z, X) \Rightarrow
P(X,Y) \equiv \forall Z \bigl(0(Z,X) \Rightarrow 0(Z,Y)\bigr)\bigr]\}
$$
and
$$
\forall X\forall Y\bigl[ \neg EC(X,Y) \equiv (0(X,Y) \equiv C(X,Y)
$$
which demonstrate the differences between calculus of
individuals and mereology of Le"sniewski.
\section{Rough mereology}
The rough mereology is an extension of Le"sniewski's
mereology through employing the relations of being a part in a degree.
The proposal to that is take as a primitive notion a function
(called a rough inclusion) returning the degree of
inclusion of one set into another.
Any rough inclusion can be regarded as the family of partial
inclusion relations (indexed by degree of inclusion being the
value of rough inclusion) the relation of partial inclusion in
degree $\alpha$ can also be interpreted as the relation of being
a part in degree $\alpha$. Notice here that the relation
of being a part in degree 1 satisfies the axioms of mereology,
which is the alternative set theory proposed by Stanislaw
Le"sniewski.
Now consider the function $\mu(X,Y)$ defined for $X,Y \subseteq
U$ as follows:
$$
\mu (X,Y) = \frac{|X\cap Y|}{|X|} \;\;\mbox{in case}\;\;
X \neq \phi\;\;\mbox{and}
$$
$$
\mu(\phi, Y) =1 \quad \mbox{for any $Y$}
$$
the function $\mu$ has the following properties:
\begin{itemize}
\item[(RI1)] $\mu(X,Y) \in [0,1]$
\item[(RI2)] $\mu(X,X)=1$
\item[(RI3)] $\mu(X,Y) = 1 \Rightarrow \forall Z.\;\,
\mu(Z,Y)\geq \mu (Z,X)$ (monotonicity)
\item[(RI3)$'$] $\mu(X,Y) = 1 = \mu(Y,X) \Rightarrow \forall Z.\;\,
\mu(X,Z) = \mu (Y,Z)$
\item[(RI4)] $\exists N \forall X \mu(N,X) = 1$
(The existence of null - object)
\item[(RI5)] For any pair $X,Y$ of objects
\begin{eqnarray*}
(\forall Z \neq N \cdot \mu (Z,X) = 1 & \Rightarrow &
\exists T \neq N \cdot \mu (T,Z) =1 \\
& = & (T,Y))) \Rightarrow \mu(X,Y) =1\;\,
\mbox{(inference rule)}.
\end{eqnarray*}
\item[(RI6)] For any collection $U$ there exists $X$ s.t.
\begin{itemize}
\item[(a)] $\forall Z \neq N\cdot \mu (Z,X) = 1
\Rightarrow \exists W\neq N \; \,
\exists T\in U\cdot \mu(W,Z) =1$,
$u(Z,T) =1$, $\mu(T,X) =1$.
\item[(b)] for any $Z\in U$, $\mu(Z,X) =1$
\item[(c)] for any $Y$ if $\mu(X,Y) <1$ then either
(a) or (b) is not true when $X$ is
replaced by $Y$.
\end{itemize}
\end{itemize}
Any function $\mu$ which satisfies (RI1)-(RI6) is called a
rough inclusion.
A rough mereological space will be any collection of objects
endowed with a rough inclusion. Given a rough inclusion $\mu$,
one introduces the mereological notions of a part and an
ingredient as follows [22].
\begin{deff}
\vspace*{-4.1mm}
\begin{itemize}
\parsep=0pt \itemsep=-0.6pt
\item [(i)] $X {\it part\/} Y$ iff $\mu(X,Y) = 1$ and $\mu(Y,X)<1$
\item [(ii)] $X {\it ingr\/} Y$ iff $\mu(X,Y) = 1$.
\end{itemize}
\end{deff}
\begin{deff}
%\vspace*{-4mm}
The relation $=_{\mu}$ defined as follows:
$$
X=_{\mu} Y\quad\mbox{iff}\quad \mu(X,Y)=1\quad\mbox{and}\quad
\mu(Y,X)=1
$$
\noindent is a congruence w.r.t. {\it part\/}. We will factor our collection
by $=_{\mu}$, denoting $=_{\mu}$ - classes of objects by object symbols.\newline
In particular, we denote by $N$ the class of null-objects.
\end{deff}
The notions of a set and a class are introduced in rough
mereology in a modified way, relative to the class of null-objects.
\begin{deff}
%\vspace*{-4mm}
For any collection ${\cal U}$ of objects, we say that an object $X$ is
a set of objects in ${\cal U}$, $X$ set ${\cal U}$ in short, if:
\begin{itemize}
\item [(i)] $\forall Z \neq_{\mu} N$. $\;[Z {\it ingr\/} X \Rightarrow
\exists W \neq_{\mu} N.\; \exists T.\; W {\it ingr\/} Z$,
$W {\it ingr\/} T,\; T {\it ingr\/} X\;\;\mbox{and}\newline
T\in {\cal U}]$;
similarly, $X$ is a class of objects in ${\cal U}, X$ class
${\cal U}$ in short, if
\item [(ii)] $\,\,\,$(I) \ $X$ set ${\cal U}$,\newline
\hbox{} (II) \ $Z {\it ingr\/} X$ for any $Z\in {\cal U}$,\newline
(III) \ if it is not true that $X {\it ingr\/} Y$, then
either\newline
\hbox{}\hspace{9.5mm} (I) or (II) is not true when $X$ is replaced by $Y$.
\end{itemize}
It follows then ([8]) that for any ${\cal U}$ there exists a unique
{\it class\/} ${\cal U}$.
\end{deff}
A rough inclusion $\mu$ on a universe $V$, generates a hierarchy
of similarity relations $\tau \mu,k$ defined for any given $k\in
(0,1)$ by means of the formula:
$$
x \tau_{\mu,k} y \quad \mbox{iff} \quad
\mu(x,y) \geq k \quad \mbox{and} \quad \mu(y,x) \geq k.
$$
Therefore, any rough inclusion can be regarded as a generalized
(parametrized) similarity relation on $V$.
The reader will find in [22] a discussion on how a rough
inclusion $\mu$ can be generated from an information/decision
system (a data table).
\section{Rough mereology $vs$. Connection (cf.\ [33])}
Consider a rough inclusion $\mu$ satisfying:
\begin{itemize}
\begin{itemize}
\item [(RI7)] for any pair $X,Y$ of objects,
$$ \mu(X,Y) <1 \Rightarrow \exists Z\neq_{\mu} N. \;
\mu(Z,X) = 1\; \mbox{and}\; \mu(Z,Y)= 0.
$$
\item [(RI8)] $\mu(X,Y) = 1$ and $\mu(X,Z) > 0 \Rightarrow \mu
(Y,Z) > 0$
\item [(RI9)] $X=_{\mu} Y \Leftrightarrow \forall Z. (\mu (Z,X) =\mu(Z,Y))$.
\item [(RI10)] $\forall X \neq_{\mu} N.\; \mu(X,N) = 0$.
\end{itemize}
\end{itemize}
Let us introduce the predicate $C$.
\begin{deff}
{\it
$C(X,Y)$ iff $\mu(X,Y) > 0$ and $\mu(Y,X) >0$
}
\end{deff}
\noindent Let us state the following properties of $C$.
\bigskip
\noindent {\bf Proposition 5.5.1}\hbox{}\hfill
\noindent $C$ satisfies the axioms 5.3.1 (i)-(iii) of the
predicate ``being connected''.
\bigskip
\noindent {\bf Proof} \
We check (i)-(iii) of 5.3.1. Concerning (i), $C(X,X)$ follows
from (RI2). Concerning (ii), if follows from
definition 5.5.1. Concerning (iii), assume that
$C(Z,X)$ iff $C(Z,Y)$ for any object $Z$. Assume that
$X\neq_{\mu} Y$, e.g. $\mu(X,Y) < 1$. By (RI7) there exists
$Z\neq_{\mu}N$ such that $\mu(Z,X)=1$ and $\mu(Z,Y)=0$.
By (RI8) we have $\mu(X,Z)>0$. i.e. $C(X,Z)$. As not
$C(Y,Z)$, we have a contradiction, hence $X=_{\mu} Y$.
Let us observe the following
\bigskip
\noindent {\bf Proposition 5.5.2}
{\it
\begin{itemize}
\parsep=0pt \itemsep=-0.5pt
\item [(a)] $X =_{\mu} Y \Rightarrow \forall Z.\; [C(Z,X)
\Leftrightarrow C(Z,Y)]$.
\item [(b)] for no object $X\neq_{\mu} N.\; C(X,N)$.
\end{itemize}
}
\medskip
\noindent {\bf Proof}\hbox{}\hfill
\begin{itemize}
\item [(a)] Let $X=_{\mu} Y$; consider any $Z$; assume $C(Z,X)$ i.e.
$\mu(Z,X) >0$ and $\mu(X,Z) >0$. By (RI3), from $\mu(X,
Y) = 1$ it follows that $\mu(Y,Z) \geq \mu(X,Z)>0$ hence
$\mu(Y,Z)>0$. By (RI9), $\mu(Z,Y) = \mu(Z,X) >0$
hence $\mu (Z, Y) >0$ and $C(Z,Y)$.
\item [(b)] follows from (RI10).
\end{itemize}
We now check that rough mereological definitions of {\it part\/} and
$ingr $, agree with definitions introduced by means of
$C(X,Y)$ in the calculus of individuals.
\bigskip
\noindent {\bf Proposition 5.5.3.}\hfill\break
{\it
$\mu(X,Y)=1$ \ iff \ $\forall Z.\; C(Z,X) \Rightarrow C(Z,Y)$.
}
\bigskip
\noindent {\bf Proof} \
Assume first that $\mu(X,Y) = 1$; let $C(Z,X)$ i.e.
$\mu(Z,X)>0$ and $\mu(X,Z)>0$. By (RI3) we have $\mu(Z,Y) >0$.
As $\mu(X,Y) = 1$ and $\mu(X,Z) > 0$, we have by (RI8) that
$\mu(Y,Z) >0$; thus $C(Z,Y)$. Thus being an ingredient in the
rough mereological sense implies being an ingredient in the calculus
of individuals sense. Assume now that $C(Z,X) \Rightarrow C(Z,Y)$
for all $Z$. Assume $\mu(X,Y) < 1$, then by (RI7), there exists
$Z\neq_{\mu} N$ such that $\mu(Z,X)=1$ and $\mu(Z,Y)=0$. Hence
not $C(Z,Y)$. By (RI8) we get $\mu(X,Z)>0$ and thus $C(Z,X)$
holds. It follows that $\mu(X,Y) = 1$.
\vspace*{-7mm}
\begin{flushright}
$\Box$
\end{flushright}
We have proved that
the relation $ingr $ defined by $X {\it ingr\/} Y$ iff $\mu(X,Y)=1$
agrees with the relation $ingr $ introduced by $C(X,Y)$.
\medskip
\noindent {\bf Corollary 5.5.4.}\hbox{}\hfill
\noindent The relation part defined by $X$ {\it part\/} $Y$ iff $\mu(X,Y) =
1$ and $\mu(Y,X) <1$ agrees with the relation {\it part\/} defined by
means of $C$ in the calculus of individuals.
\section{\v{C}ech topologies induced by a rough inclusion}
In this section, we present our result which shows that, under
natural assumption about a rough inclusion $\mu$, the \v{C}ech
topology generated from $\mu$. Contains as a subset, the quasi-\v{C}ech
topology generated from the connectivity operator $C$ in Clarke [1].
The existence of \v{C}ech topology in a rough mereological universe,
opens up -- as indicated in Marcus [7] -- the possibility of
stydying the topological learning theory in the framework of
rough set theory. This possibility will be explosed in further studies.
In 5.3.4, above we have recalled the notion of a quasi-interior
$iX$ of an object $X$ as introduced in the calculus of
individuals [1]. We now consider a rough inclusion $\mu$ which
satisfies (RI1)-(RI10). We will discuss the operator $iX$
introduced by the connective $C$ generated from $\mu$ as in
5.5.1 above. We show that $ix$ is a \v{C}ech interior
operator i.e. it satisfies the following conditions
\begin{deff}
{\it
An operator $i$ on objects, assigning for any object
$X$ a unique object $ix$, is a \v{C}ech interior operator if
the following hold ([3]):
\begin{itemize}
\parsep=0pt \itemsep=-0.5pt
\item [(i)] $iX\; {\it ingr\/} X$
\item [(ii)] if $X {\it ingr\/} Y$, then $iX\; {\it ingr\/}\; iY$,
\item [(iii)] $iN =_{\mu} N$.
\end{itemize}
}
\end{deff}
\medskip
\noindent {\bf Proposition 5.6.2. ([33])}\hfill\break
$iX=_{\mu}$ {\it class\/} $(T: NTP(T,X))$ is a \v{C}ech interior operator.
\bigskip
\noindent {\bf Proof}\hbox{}\hfill
\begin{itemize}
\item [(i)] We check that $iX\; {\it ingr\/} X$. Let $Z\neq_{\mu} N$ and
$Z {\it ingr\/}\, iX$. As $iX =_{\mu}$ {\it class\/}
$(T : NTP(T,X))$,
we have by (RI6) (a) that there exist objects
:$P\neq_{\mu} N$ and $T$ such that: $P {\it ingr\/} Z$ and
$P {\it ingr\/} T$, $T {\it ingr\/}\,iX$ and $T$ has the property
that $NTP (T,X)$. In particular, $T {\it ingr\/} X$ hence
$P {\it ingr\/} X$. By (RI5), we conclude that $iX\, {\it ingr\/}
X$.
\item [(ii)] Assume $X {\it ingr\/} Y$ and let $Z\neq_{\mu} N$ and $Z
{\it ingr\/}\,iX$. Let $P\neq_{\mu} N$ and $T$ be as in (i).
Then by (i) above, $P {\it ingr\/} X$ hence $P {\it ingr\/} Y$. Now
what does $NTP(T,X)$ means?
\begin{eqnarray*}
NTP(T,x) & \Leftrightarrow & T\, {\it ingr\/} X \; \mbox{and}\;
\sim \exists W \neq_{\mu} N.E C(W,T) \;\mbox{and}\; EC(W,X)\\
EC(W,T) &\Leftrightarrow & C(W,T)\; \mbox{and}\: \neg O(W,T)\\
EC(W,X) &\Leftrightarrow & C(W,X)\;\mbox{and}\; \neg O(W,X).
\end{eqnarray*}
\vspace*{-5mm}
We have:
\begin{eqnarray*}
C(W,T) & \Rightarrow & C(W,X)\\
O(W,T) & \Rightarrow & O(W,X) \; \mbox{so}\\
O(W,X) & \Rightarrow & O(W,T).
\end{eqnarray*}
\end{itemize}
\noindent {\bf Claim: }
\begin{eqnarray*}
NTP(T,X) & \Leftrightarrow & \forall W \neq_{\mu} N. \; EC(W,X)
\wedge \neg C(W,T)
\end{eqnarray*}
\smallskip
\noindent {\bf Proof of the claim.}
Figure 1.:
\begin{center}
\unitlength=0.55mm
\special{em:linewidth 0.4pt}
\linethickness{0.4pt}
\begin{picture}(56.00,90.00)
\put(46.00,81.00){\makebox(0,0)[cc]{$W$}}
\put(38.00,58.00){\makebox(0,0)[cc]{$T$}}
\put(28.00,39.00){\makebox(0,0)[cc]{$X$}}
\put(28.00,22.00){\makebox(0,0)[cc]{Figure 1.}}
\bezier{64}(51.00,89.00)(44.00,90.00)(41.00,82.00)
\bezier{68}(49.00,89.00)(56.00,90.00)(53.00,81.00)
\bezier{28}(49.00,76.00)(51.00,77.00)(53.00,81.00)
\bezier{84}(49.00,76.00)(39.00,72.00)(41.00,82.00)
\bezier{124}(43.00,75.00)(29.00,76.00)(28.00,59.00)
\bezier{136}(28.00,59.00)(28.00,41.00)(41.00,50.00)
\bezier{116}(43.00,75.00)(49.00,58.00)(41.00,50.00)
\bezier{212}(42.00,76.00)(20.00,78.00)(17.00,47.00)
\bezier{180}(17.00,47.00)(16.00,25.00)(37.00,35.00)
\bezier{96}(37.00,35.00)(47.00,41.00)(49.00,53.00)
\bezier{104}(43.00,75.00)(51.00,69.00)(49.00,53.00)
\end{picture}
\end{center}
\vfill\eject
$(\Rightarrow)$ Let $NTP (T,X)$ and $\exists W \neq_{\mu} N.\;
EC(W,X)\; \wedge \;C(W,T)$; then $O(W,T)$ because
$$
O(W,T) \Rightarrow O(X,W) \quad\mbox{then}\quad EC(W,T),
$$
so $\exists W \neq_{\mu} N.\; EC(W,X)\; \wedge \;EC(W,T)$,
a contradiction with $NTP(T,X)$.
Now, let $\forall W\neq_{\mu} N.\; EC(W,X)\Rightarrow \neg C(W,T)$.
Then clearly $NTP(T,X)$. This concludes the proof of the Claim.
\noindent We now prove from the claim that
$$
X {\it ingr\/}\, Y \Rightarrow iX\, {\it ingr\/}\, iY.
$$
Assume that we have $T {\it ingr\/} iX$; we need $T\, {\it ingr\/}\,iY$.
We check that $NTP(T,Y)$; indeed, assume that non $NTP(T,Y)$ i.e.
$$
\exists W \neq_{\mu} N. \; EC(W,Y) \; \wedge \; C(W,T)
$$
then:
$EC(W,X)$ (because: $C(W,T) \Rightarrow C(W,X); \; O(W,X) \Rightarrow
O(W,Y))$, a contradiction with $NTP(T,X)$. Hence $T {\it ingr\/} iY$. By (RI5),
$iX {\it ingr\/}\, iY$.\newline
For (iii) it follows from the definition.
\vspace*{-8mm}
\begin{flushright}
$\Box$
\end{flushright}
We have proved that in the rough mereological setting, $iX$ is a
\v{C}ech interior operator; clearly, the operator $i$ restricted to
non-null objects is the quasi - topological interior operator of
[2].
Now we recall [2] the notion of boolean intersection $X\wedge Y$
of objects $X,Y$.
\begin{deff}
{\it
$\; X\wedge Y =_{\mu}$ {\it class\/} $(T: T {\it ingr\/} X$ and
$T\,{\it ingr\/}\,Y)$ (cf.\ ([8])).
}
We would like to check under what conditions the \v{C}ech interior
operator $i$ is a topological interior operator. To this end, we
introduce a new condition. We recall that the relation {\it exterior\/} of
being exterior is the negation of the relation $O$ of overlapping.
\end{deff}
\begin{deff}
{\it
For a pair $X,Y$ of objects, we say $X$ is exterior to $Y$, $X$
{\it ext\/} $Y$ in short, if there is no $Z\neq_{\mu} N$ such
that $Z\,{\it ingr\/} X$ and $Z\,{\it ingr\/} Y$.
}
\end{deff}
Thus, $X ext Y \Leftrightarrow \; \mbox{non}\; O(X,Y)$.
We introduce a new condition on a rough in\-clu\-sion~$\mu$.
\begin{itemize}
\begin{itemize}
\item [(RI11)] if $Z ext (X\wedge Y)$, then either $Z ext X$
or $Z ext\,Y$.
\end{itemize}
\end{itemize}
Our proposition reads
\bigskip
\noindent {\bf Proposition 5.6.3. ([33])}\hfill\break
Under (RI11) \ $i(X\wedge Y) =_{\mu} (iX) \wedge (iY)$.
\bigskip
\begin{proof}
First, let us check that $i(X\wedge Y)\,{\it ingr\/}\,(iX)\wedge (iY)$.
Assume,
$$
Z\neq_{\mu} N\;\;\;\mbox{and}\;\;\; Z\,{\it ingr\/}\; i(X\wedge Y).
$$
As $i(X\wedge Y)=_{\mu}$ {\it class\/} $(T: NTP(T,X\wedge Y))$ it follows
by (RI6) that there exist $P\neq_{\mu} N,~W$
\end{proof}
\vspace*{2mm}
\begin{center}
\unitlength=0.60mm
\special{em:linewidth 0.4pt}
\linethickness{0.4pt}
\begin{picture}(77.00,84.00)
\put(44.00,89.00){\vector(-3,-2){26.78}}
\put(51.00,88.00){\vector(4,-3){21.22}}
\put(18.00,57.00){\vector(4,-3){23.11}}
\put(73.00,61.00){\vector(-1,-1){20.22}}
\put(44.00,94.00){\makebox(0,0)[cc]{$i(X \wedge Y)$}}
\put(10.00,63.00){\makebox(0,0)[cc]{$Z$}}
\put(100.00,64.00){\makebox(0,0)[cc]{$W: NTP(W, X\wedge Y)$}}
\put(97.00,49.00){\makebox(0,0)[cc]{$W {\it ingr\/} X$}}
\put(97.00,40.00){\makebox(0,0)[cc]{$W {\it ingr\/} Y$}}
\put(46.00,32.00){\makebox(0,0)[cc]{$P$}}
\end{picture}
\end{center}
\vspace*{-17mm}
\noindent and thus (by monotonicity) $W {\it ingr\/}\; i X$, $W {\it ingr\/}\; i Y$
(as $i(X\wedge Y)\,ingr\,iX, iY)$
so that $W {\it ingr\/} (iX) \wedge (iY)$ and hence $P {\it ingr\/}
(iX) \wedge (iY)$.
\noindent By (RI5), $i(X\wedge Y)\,{\it ingr\/}\,(iX) \wedge (iY)$\newline
Next, we prove that:
$$
(iX) \wedge (iY)\,{\it ingr\/}\,i(X\wedge Y).
$$
Assume to this end that $Z$\, {\it ingr\/}\,$(iX) \wedge (iY)$.
\bigskip
\noindent {\bf Claim:}
$$
Z\,{\it ingr\/}\,i(X \wedge Y)
$$
Suppose not: as $Z {\it ingr\/} (X \wedge Y)$, these exists $T\neq_{\mu}
N$ such that:
\[
\begin{array}{l}
EC(T,Z)\wedge EC(T, X \wedge Y). \;\;\mbox{As}\;\; EC(T, X\wedge Y)\; \;\;
\mbox{we have}:\\
C(T, X\wedge Y) \;\; \mbox{hence}\;\; C(T,X), C(T,Y) \;\;
\mbox{and}\;\; T ext\,(X\wedge Y).
\end{array}
\]
\noindent By (RI11), either $T\,ext\,X$ or $T\, ext\,Y$ let e.g. $T\, ext\, X$,
then $EC(T,X)\,$ hence $EC(T,Z)$ and $EC(T,X)$ and this means that
$\neg NTP (Z,X)$ hence not $Z\,{\it ingr\/}\,iX$, a contradiction.
Hence $Z\,{\it ingr\/}\,(i(X \wedge Y))$. It follows by (RI5) that
$$
\bigl[(iX)\wedge (iY)\bigr]\,{\it ingr\/}\, \bigl[i(x\wedge Y)\bigr]\quad\mbox{and}
$$
finally
$$(iX \wedge iY) =_{\mu} i(X\wedge Y)$$
\vspace*{-6mm}
\begin{flushright}
$\Box$
\end{flushright}
Let us sum up our results in the final
\bigskip
\noindent {\bf Proposition 5.6.4. ([33])}\hbox{}\hfill
{\it
\vspace*{-1.2mm}
\begin{itemize}
\item [(a)] If a rough inclusion $\mu$ satisfies (RI6-RI10),
then the operator $i$ becomes a \v{C}ech interior operator;
\vfil\eject
\item [(b)] If $\mu$ satisfies in addition (RI11), then the
\v{C}ech interior operator $i$ becomes a topological interior
operator;
\item [(c)] in both cases (a) and (b), the restriction of $i$ to
non-null objects coincides with the quasi - topological
operator of~[1].
\end{itemize}
As a result, \v{C}ech topological spaces (case (a)) or
topological spaces (case (b)) generated from the rough inclusion
$\mu$ contain as subsets, quasi - topological spaces of [1].
}
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\end{document}