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%Figure 1: Example of a boundary.

\centerline{{\bf Finite Topology Concepts and Neighbourhoods}}

\bigskip
Let $X$ be a general set.  If a subset $U(x)$ of $X$ is determined for each element $x$
in $X$, we call the pair $(X, U(\cdot))$ a finite topological space.  The subset $U(x)$ is called
a neighbourhood of $x$.  In most cases, we assume $x \in U(x)$ (in this case, we call
$X$ {\it filled\/}), but it is not necessary in all cases.  $U(x)$ is interpreted as a set of points
neighbouring to $x$.

Using the notion of a neighbourhood, we can define many concepts.

\medskip\noindent
{\bf Definition (Boundary).}  A {\it boundary\/} $A^\partial$ of a subset $A$ of $X$ is defined as:
$$A^\partial = \{ x : \hbox{\it $U(x) \cap A \ne \emptyset$ and $U(x) \cap A^c \ne \emptyset$}\},$$
where $A^c$ is the complement of $A$ and $\emptyset$ is the empty set.

\medskip
In the field of image processing, the process of obtaining $A^\partial$ from $A$ is called an
extraction of the outline of $A$, which is shown in Figure 1.

We can divide $A^\partial$ into the following two parts -- an inner boudary $A^{\partial_i}$ and an
outer boundary $A^{\partial_o}$ which are defined as:
$$\hbox{\it$A^{\partial_i} = A \cap A^\partial$ and $A^{\partial_o} = A^c \cap A^\partial$.}$$
Of course, we see from these definitions that $A^\partial = A^{\partial_i} \cup A^{\partial_o}$.

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\medskip\noindent
{\bf Definition (Interior).}  An {\it interior\/} $A^i$ of a subset $A$ of $X$ is defined as:
$$A^i = \{ x : x \in A \Rightarrow U(x) \subseteq A \}.$$

In image processing, the process of obtaining $A^i$ is called a contraction of $A$.

\medskip\noindent
{\bf Definition (Closure).}  A {\it closure\/} $A^b$ of a subset $A$ of $X$ is defined as:
$$A^b = \{ x : U(x) \cap A \ne \emptyset \}.$$

This concept is referred to as ``expansion'' in image processing.

\medskip\noindent
{\bf Definition (Isolated Points).}  A set of {\it isolated points\/} $A^s$ of a subset $A$ of $X$ is
defined as:
$$A^s = \{ x : \hbox{\it $x \in A$ and $(U(x) \setminus x) \cap A = \emptyset$}\}.$$

A point in $A^s$ is referred to as ``noise'' in image processing.  Conversely, we define
a continuous part $A^n$ of $A$ as $A^n = A \setminus A^s$, which can also be written as:
$$A^n = \{ x : \hbox{\it $x \in A$ and $(U(x) \setminus x) \cap A \ne \emptyset$}\}.$$
$A^n$ is called a ``noise elimination'' of $A$ in image processing.

\medskip\noindent
{\bf Definition (Inflation).}  An {\it inflation\/} $A^f$ of a subset $A$ of $X$ is defined as:
$$A^f = \{ x : (\exists y)(\hbox{$y \in A$, $x \in U(y)$})\} = \bigcup_{y \in A} U(y).$$

This concept seems similar to ``closure'', but they do not always coincide.  We
present the following definition.

\medskip\noindent
{\bf Definition (Symmetry).}  A neighbourhood $U(\cdot)$ of a finite topological space
$(X, U(\cdot))$ is called {\it symmetric\/} if and only if:
$$\hbox{\it For all $x, y \in X$, $y \in U(x)$ implies $x \in U(y)$.}$$

The neighbourhoods (a), (b), and (c) shown in Figure 2 are symmetric, but
neighbourhoods (d), (e) and (f) are not.

By using the concept of symmetricity, we can give the following lemma.

\medskip\noindent
{\bf Lemma 2.1.}  If a neighbourhood $U(\cdot)$ is symmetric, then for all subsets $A$:
$$A^b = A^f.$$

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%Figure 2: Symmetricity of neighbourhoods.

\medskip\noindent
{\bf Proof.}  Let us look at the following formulae.
$$\eqalign{x \in A^f &\Leftrightarrow \hbox{\it for some $y \in A$, $x \in U(y)$}\cr
 &\Leftrightarrow \hbox{\it for some $y \in A$, $y \in U(x)$ {\rm ({\it by symmetricity\/})}}\cr
 &\Leftrightarrow U(x) \cap A \ne \emptyset\cr
 &\Leftrightarrow x \in A^b.\cr}$$
Thus, we get the result.

\hfill Q.E.D.

\medskip
The following two definitions are analogies of very popular concepts of openness
and closedness in continuous topologies.

\medskip\noindent
{\bf Definition (Open Set).}  A subset $G$ of $X$ is {\it open\/} if and only if:
$$G = G^i.$$

\medskip\noindent
{\bf Definition (Closed Set).}  A subset $F$ of $X$ is {\it closed\/} if and only if:
$$F = F^b.$$

We present here a second type of closure and interior.

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%Figure 3: $A$ is connected by neighbourhood $U_9$, but not by $U_5$.

\medskip\noindent
{\bf Definition (f-closure and f-interior).}  An {\it f-closure\/} $A^{f_b}$ and an {\it f-interior\/} $A^{f_i}$
of a subset $A$ of $X$ is defined as:
$$A^{f_b} = \bigcap \{ F : \hbox{\it $A \subseteq F$, $F$ is closed\/} \},$$
$$A^{f_i} = \bigcup \{ G : \hbox{\it $G \subseteq A$, $G$ is open\/} \}.$$

We can show that an f-closure is gotten by a repetition of an ordinary closure
(see Lemma 2.1 in Reference [1]) and an f-interior is gotten by a repetition of an
ordinary interior.

\medskip\noindent
{\bf Definition (Connected Set).}  A subset $A$ of $X$ is connected, if and only if, for
any $B$, $C$ in $X$:
$$\hbox{\it $A = B \cup C$, $B \ne \emptyset$, $C \ne \emptyset$, and $B \cap C = \emptyset$ implies $B^b \cap C \ne \emptyset$.}$$

This concept coincides with the usual intuitive concept of connectivity.  In Fig%-
ure 3, the image $A$ is connected by the neighbourhood $U_9$, but not by the neigh%-
bourhood $U_5$.

For connectivity, we present the following theorem.

\medskip\noindent
{\bf Theorem 2.1.}  Let $X$ be filled and finite (i.e., containing only a finite number of
points).  Then, a subset $A$ of $X$ is connected if and only if for every $x \in A$:
$$( \cdots ((\{x\}^b \cap A)^b \cap A)^b \cdots )^b \supseteq A.$$
That is, by a finite process of taking closures of $A$, $A$ is covered.

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\medskip\noindent
{\bf Proof.  $\Rightarrow$)}  We assume that for some $x \in A$:
$$( \cdots ((\{x\}^b \cap A)^b \cap A)^b \cdots )^b \not\supseteq A.$$
For convenience, we denote an intersection of $A$ and an n-th closure as $P_n$ ($= P_{n-1}^b \cap
A$, $P_1 = \{x\}^b \cap A$).  As $X$ is finite, we can assume that $P_{n+1} = P_n$.  If we let
$$\hbox{$B = P_n$, $C = A \setminus B$ ($= A \setminus P_n$),}$$
then $B \cup C = B \cup (A \setminus B) = A \cup B = A$ and $C \ne \emptyset$ (as $P_n \not\supseteq A$).  Since $X$ is filled,
$x \in P_n$.  Hence, $x \in B = P_n \ne \emptyset$.  $B \cap C = \emptyset$ is clear.  But,
$$B^b \cap C = P_n^b \cap C = P_n \cap (A \setminus P_n) = \emptyset.$$
Therefore, $X$ is not connected.

\medskip\noindent
{\bf $\Leftarrow$)}  Let $B$ and $C$ be non-void subsets of $A$ such that $B \cap C = \emptyset$ and $B^b \cap C = \emptyset$.
Then, there exists an element $x$ in $B$, and we can construct a set $P_n$ as a procedure
described previously and $P_{n+1} = P_n$.

Let us show that $(B^b \cap A)^b \cap C = \emptyset$.  If the left hand side is not empty, there
exists some element $z$ in it such that:
$$\hbox{\it $z \in C$ and $z \in (B^b \cap A)^b$.}$$
Thus, $U(z) \cap B^b \cap A \ne \emptyset$, where $B^b \cap A = B^b \cap (B \cup C) = (B^b \cap B) \cup (B^b \cap C) =
B \cup B^b \cap C$.  Hence, $U(z) \cap B^b \cap C \ne \emptyset$ (as $B \cap U(z) = \emptyset$ is clear), which contradicts
with $B^b \cap C = \emptyset$.  Thus, we can set $B = B^b \cap A$.  If $P_i \subseteq B^b \cap A$, then $P_{i+1} \subseteq B^b \cup A$
because:
$$P_{i+1} = P_i^b \cap A \subseteq (B^b \cap A)^b \cap A = B^b \cap A.$$
Therefore, we obtain the result:
$$P_n \subseteq B^b \cap A = B^b \cap (B \cup C) = B,$$
i.e.,
$$P_n \not\supseteq A.$$

\hfill Q.E.D.

\medskip
The following facts are easily derived:

\medskip
1. $((A^c)^i)^c = A^b$, $((A^c)^b)^c = A^i$.
\medskip
2. $A^\partial = A^b \cap (A^c)^b$, $(A^c)^\partial = A^\partial$.
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\medskip
3. If $x \in A^s$, then $X \not\in (A \setminus \{x\})^b$.
\medskip
4. If $A^s \ne \emptyset$, then $A$ is not connected.
\medskip
5. If $X$ is filled, then $A \subseteq A^b$, $A^i \subseteq A$.
\medskip
6. If $A$ is open, then $A^c$ is closed.  Conversely, if $A$ is closed, then $A^c$ is open.

\bye
