# Annihilator graph of the ring $C_{\mathscr{P}}(X)$

Pratip Nandi Department of Pure Mathematics, University of Calcutta, 35,
Ballygunge Circular Road, Kolkata 700019, West Bengal, India
<EMAIL_ADDRESS>,  Sudip Kumar Acharyya Department of Pure
Mathematics, University of Calcutta, 35, Ballygunge Circular Road, Kolkata
700019, West Bengal, India<EMAIL_ADDRESS>and  Atasi Deb Ray Department
of Pure Mathematics, University of Calcutta, 35, Ballygunge Circular Road,
Kolkata 700019, West Bengal, India<EMAIL_ADDRESS>
###### Abstract.

In this article, we introduce the annihilator graph of the ring
$C_{\mathscr{P}}(X)$, denoted by $AG(C_{\mathscr{P}}(X))$ and observe the
effect of the underlying Tychonoff space $X$ on various graph properties of
$AG(C_{\mathscr{P}}(X))$. $AG(C_{\mathscr{P}}(X))$, in general, lies between
the zero divisor graph and weakly zero divisor graph of $C_{\mathscr{P}}(X)$
and it is proved that these three graphs coincide if and only if the
cardinality of the set of all $\mathscr{P}$-points, $X_{\mathscr{P}}$ is $\leq
2$. Identifying a suitable induced subgraph of $AG(C_{\mathscr{P}}(X))$,
called $G(C_{\mathscr{P}}(X))$, we establish that both
$AG(C_{\mathscr{P}}(X))$ and $G(C_{\mathscr{P}}(X))$ share similar graph
theoretic properties and have the same values for the parameters, e.g.,
diameter, eccentricity, girth, radius, chromatic number and clique number. By
choosing the ring $C_{\mathscr{P}}(X)$ where $\mathscr{P}$ is the ideal of all
finite subsets of $X$ such that $X_{\mathscr{P}}$ is finite, we formulate an
algorithm for coloring the vertices of $G(C_{\mathscr{P}}(X))$ and thereby get
the chromatic number of $AG(C_{\mathscr{P}}(X))$. This exhibits an instance of
coloring infinite graphs by just a finite number of colors. We show that any
graph isomorphism $\psi:AG(C_{\mathscr{P}}(X))\rightarrow
AG(C_{\mathscr{Q}}(Y))$ maps $G(C_{\mathscr{P}}(X))$ isomorphically onto
$G(C_{\mathscr{Q}}(Y))$ as a graph and a graph isomorphism
$\phi:G(C_{\mathscr{P}}(X))\rightarrow G(C_{\mathscr{Q}}(Y))$ can be extended
to a graph isomorphism $\psi:AG(C_{\mathscr{P}}(X))\rightarrow
AG(C_{\mathscr{Q}}(Y))$ under a mild restriction on the function $\phi$.
Finally, we show that atleast for the rings $C_{\mathscr{P}}(X)$ with finitely
many $\mathscr{P}$-points, so far as the graph properties are concerned, the
induced subgraph $G(C_{\mathscr{P}}(X))$ is a good substitute for
$AG(C_{\mathscr{P}}(X)$.

###### Key words and phrases:

triangulated, hypertriangulated, complemented, chromatic number, girth,
dominating number, induced subgraph and graph isomorphism

###### 2010 Mathematics Subject Classification:

Primary 54C40; Secondary 05C69

The first author thanks the CSIR, New Delhi – 110001, India, for financial
support

##  1\. Introduction

Let $C(X)$ be the ring of all real valued continuous functions defined on a
Tychonoff space $X$ and $\mathscr{P}$, an ideal of closed sets in $X$ in the
following sense: if $A\in\mathscr{P}$ and $B\in\mathscr{P}$, then $A\cup
B\in\mathscr{P}$ and if $A\in\mathscr{P}$ and $C\subset A$ with $C$, closed in
$X$, then $C\in\mathscr{P}$. In [4], we have studied the zero divisor graph of
the ring $C_{\mathscr{P}}(X)=\\{f\in C(X):cl_{X}(X\setminus
Z(f))\in\mathscr{P}\\}$, where $Z(f)=\\{x\in X:f(x)=0\\}$ is the zero set of
$f$. In this article, we introduce the annihilator graph of the same ring
which we denote by $AG(C_{\mathscr{P}}(X))$ and observe how the topology of
the underlying space helps to interpret its graph properties and vice-versa.
Though Badawi investigated the annihilator graph of a commutative ring in [6]
and achieved the values of typical graph parameters through the properties of
the ring, our study of annihilator graph of $C_{\mathscr{P}}(X)$ reveals the
interactions among the properties of the ring $C_{\mathscr{P}}(X)$, the graph
$AG(C_{\mathscr{P}}(X))$ and the topological behaviour of the space $X$.

It is well known that the annihilator graph of any commutative ring $R$
contains, as a subgraph, the zero divisor graph of $R$ [6]. We show here that
the annihilator graph of $C_{\mathscr{P}}(X)$ is contained in its weakly zero
divisor graph. We also establish that the cardinality of the set of all
$\mathscr{P}$ points of the topological space $X$ completely determines when
and only when all the three graphs coincide; where the $\mathscr{P}$ points
are those $x\in X$ such that $cl_{X}V\in\mathscr{P}$, for some neighborhood
$V$ of $x$. In section 3, certain graph parameters, e.g., the diameter,
radius, girth, length of the smallest cycle containing two given vertices, are
calculated and condition(s) under which the annihilator graph of
$C_{\mathscr{P}}(X)$ is triangulated, hypertriangulated and uniquely
complemented, are determined.

A subgraph $H$ of a given graph $G$ is called an induced subgraph if two
vertices $a,b$ of $H$ are adjacent in $G$ whenever they are adjacent in $H$.
In Section 4 of this article we identify an induced subgraph of
$AG(C_{\mathscr{P}}(X))$, denoted by $G(C_{\mathscr{P}}(X))$ which exhibit a
similar behaviour as a graph, so far as typical graph properties are
concerned. For example, both have same diameter, girth, radius, eccentricity,
clique and chromatic number. At times, handling $G(C_{\mathscr{P}}(X))$ comes
out more convenient, as it has less number of edges than those of
$AG(C_{\mathscr{P}}(X))$. In Section 5, knowing that there are only finitely
many isolated points in $X$, we device an algorithm for coloring the vertices
of $G(C_{F}(X))$ (where $C_{F}(X)$ is the ring obtained by choosing
$\mathscr{P}$ as the ideal of all finite subsets of $X$) and show how this
coloring of $G(C_{F}(X))$ determines a coloring for $AG(C_{F}(X))$, although
it is an infinite graph.

The last section of this article discusses how far the role of
$G(C_{\mathscr{P}}(X))$ can replace the role of $AG(C_{\mathscr{P}}(X))$ as a
graph. We answer in affirmative that any graph isomorphism
$AG(C_{\mathscr{P}}(X))\rightarrow AG(C_{\mathscr{Q}}(Y))$ maps
$G(C_{\mathscr{P}}(X))$ isomorphically onto $G(C_{\mathscr{Q}}(Y))$. Perhaps
it is more intriguing whether any graph isomorphism
$G(C_{\mathscr{P}}(X))\rightarrow G(C_{\mathscr{Q}}(Y))$ can be extended to a
graph isomorphism $AG(C_{\mathscr{P}}(X))\rightarrow AG(C_{\mathscr{Q}}(Y))$.
We have shown in Section 6 that under some mild conditions, this is true. We
also show that $C_{\mathscr{P}}(X)$ with $|X_{\mathscr{P}}|$ (= cardinality of
$X_{\mathscr{P}}$) $=$ finite, is indeed an example for which the graph
properties of $G(C_{\mathscr{P}}(X))$ (and hence, $AG(C_{\mathscr{P}}(X))$)
completely determines the ring properties of $C_{\mathscr{P}}(X)$ and
conversely.

##  2\. Prerequisites and Technical terms related to graphs and
$C_{\mathscr{P}}(X)$

The distance between two distinct vertices $f$ and $g$ in
$AG(C_{\mathscr{P}}(X))$ (respectively, $G(C_{\mathscr{P}}(X))$), denoted by
$d_{AG}(f,g)$ (respectively, $d_{G}(f,g)$), is the length of the shortest path
from $f$ to $g$. In what follows, the set of vertices of any graph $H$ is
denoted by $V(H)$. The diameter of a graph $H$ is defined by:
$diam(H)=max\\{d_{H}(f,g):f,g\in V(H)\\}$. The eccentricity $ecc(f)$ of $f\in
V(H)$ is defined by: $ecc(f)=max\\{d_{H}(f,g):g\in V(H)\\}$. An $f\in V(H)$ is
said to be in the center of $H$ if $ecc(f)\leq ecc(g)$ holds for each $g\in
V(H)$ and in this case $ecc(f)$ is called the radius of the graph. The center
and radius of a graph $H$ is denoted respectively by $C(H)$ and $rad(H)$. The
girth of $H$, denoted by $gr(H)$, is the length of the smallest cycle in this
graph. A graph $H$ is called triangulated (respectively, hypertriangulated) if
each vertex (respectively, each edge) of this graph is a vertex (respectively,
an edge) of a triangle. The smallest length of a cycle containing two distinct
vertices $f$ and $g$ in $AG(C_{\mathscr{P}}(X))$ is denoted by $c(f,g)$.
A subset $D$ of $V(H)$ is called a dominating set in $H$ if for each $f\in
V(H)\setminus D$, there exists $g\in D$ such that $f$ and $g$ are adjacent. A
dominating set $D$ in $H$ is called a total dominating set if for each $f\in
V(H)$, there exists $g\in D$ such that $f$ and $g$ are adjacent. The
dominating number (respectively, total dominating number) of the graph $H$ is
defined as $dt(H)=\min\\{|D|:D\text{ is a dominating set in }H\\}$
(respectively, $dt_{t}(H)=\min\\{|D|:D\text{ is a total dominating set in
}H\\}$). It is evident that $dt(H)\leq dt_{t}(H)$ and for a simple graph $H$,
$dt_{t}(H)\geq 2$. A complete subgraph of a graph $H$ is any subset $K$ of $H$
such that each pair of distinct vertices in $K$ are adjacent. The clique
number of $H$ is defined as $cl(H)=\sup\\{|K|:K\ is\ a\ complete\ subgraph\
ofH\\}$. A coloring of a graph is a labeling of the vertices of the graph with
colors such that no two adjacent vertices have the same color. More precisely,
for a cardinal number $\Lambda$ (finite or infinite), a $\Lambda$-coloring of
$H$ is a map $\psi:V(H)\to[0,\Lambda)$ with the following condition: whenever
$f,g\in V(H)$ and $f$ and $g$ are adjacent, $\psi(f)\neq\psi(g)$. The
chromatic number of $H$ is defined as follows:
$\chi(H)=min\\{\Lambda:\text{there exists a }\Lambda\text{-coloring of }H\\}$.
For more information related to graph theoretic terms, the reader is referred
to the book [7].

######  Definition 2.1.

[4] $X$ is called locally $\mathscr{P}$ at a point $x\in X$, if there exists
an open neighborhood $V$ of $x$ in $X$ such that $cl_{X}V\in\mathscr{P}$. $X$
is said to be locally $\mathscr{P}$ if it is locally $\mathscr{P}$ at each
point on it.

Let $X_{\mathscr{P}}=\\{x\in X:X\text{ is locally }\mathscr{P}\text{ at
}x\\}$. Then it is easy to see that $X_{\mathscr{P}}$ is an open set in $X$.
Also, $X$ is locally $\mathscr{P}$ if and only if $X_{\mathscr{P}}=X$.

As a special case of $C_{\mathscr{P}}(X)$, we get the ring $C(X)$. In fact, if
$\mathscr{P}$ is the ideal of all closed subsets of $X$, we get
$C_{\mathscr{P}}(X)=C(X)$ with $X_{\mathscr{P}}=X$. If $\mathscr{P}$ is the
ideal of all compact subsets of $X$ then $C_{\mathscr{P}}(X)=C_{K}(X)$ and
$X_{\mathscr{P}}$ is the set of all points at which $X$ is locally compact. On
the other hand, if $\mathscr{P}$ is the ideal of all finite subsets of $X$
then $C_{\mathscr{P}}(X)=C_{F}(X)=\\{f\in C(X):X\setminus Z(f)\text{ is
finite}\\}$ and $X_{\mathscr{P}}$ is the set of all isolated points in $X$,
denoted later on by $K_{X}$.

######  Lemma 2.2.

[4] Let $x\in X_{\mathscr{P}}$ and $G$ be a neighborhood of $x$ in $X$. Then
there exists $f\in C_{\mathscr{P}}(X)$ such that $x\in X\setminus Z(f)\subset
cl_{X}(X\setminus Z(f))=X\setminus int_{X}Z(f)\subset G$.

######  Lemma 2.3.

Let $K\subset X_{\mathscr{P}}$ be a compact set and $G$, a neighborhood of $K$
in $X$. Then there exists $f\in C_{\mathscr{P}}(X)$ such that $K\subset
X\setminus Z(f)\subset cl_{X}(X\setminus Z(f))\subset G$.

###### Proof.

Let $x\in K$. By Lemma 2.2, there exists $f_{x}\in C_{\mathscr{P}}(X)$ such
that $x\in X\setminus Z(f_{x})\subset cl_{X}(X\setminus Z(f_{x}))\subset G$.
$K$ being compact, the open cover $\\{X\setminus Z(f_{x}):x\in K\\}$ has a
finite subcover; i.e., there exist $x_{1},x_{2},\ldots,x_{n}\in K$ such that
$K\subset\bigcup\limits_{i=1}^{n}(X\setminus
Z(f_{x_{i}}))=X\setminus(\bigcap_{i=1}^{n}Z(f_{x_{i}}))=X\setminus Z(f)$ where
$f=\sum_{i=1}^{n}f_{x_{i}}^{2}\in C_{\mathscr{P}}(X)$. Consequently, $K\subset
X\setminus Z(f)\subset cl_{X}(X\setminus
Z(f))=\bigcup\limits_{i=1}^{n}cl_{X}(X\setminus Z(f_{x_{i}}))\subset G$. ∎

It has been proved in our earlier paper [see Theorem 2.3, [4]] that
$Z(C_{\mathscr{P}}(X))^{*}=\\{f\in
C_{\mathscr{P}}(X)\setminus\\{0\\}:int_{X}Z(f)\cap
X_{\mathscr{P}}\neq\emptyset\\}$. The following is a couple of easy
observations, that we frequently use in our discussion.

######  Observation 2.4.

(i) Every point in the co-zero set of $f\in C_{\mathscr{P}}(X)$ is a
$\mathscr{P}$-point.
(ii) $fg=0$ if and only if $X\setminus Z(g)\subseteq int_{X}Z(f)$, for any
$f,g\in C_{\mathscr{P}}(X)$.

For more information on the ring $C_{\mathscr{P}}(X)$, the reader is referred
to see the articles [1] and [2].

##  3\. Annihilator graph of $C_{\mathscr{P}}(X)$

We begin this section by introducing the annihilator graph of
$C_{\mathscr{P}}(X)$ whose vertices are the non-zero zero divisors of
$C_{\mathscr{P}}(X)$. The adjacency of two vertices of this graph is
reformulated in terms of their zero sets and the $\mathscr{P}$-points of the
space $X$.

######  Definition 3.1.

[6] The annihilator graph $AG(R)$, of a commutative ring $R$, is a simple
graph whose set of vertices is $Z(R)^{*}=Z(R)\setminus\\{0\\}$ and two
distinct vertices $a,\ b$ are adjacent if $ann(a)\cup ann(b)\neq ann(a.b)$;
here $ann(a)$ stands for the ideal $\\{r\in R:ra=0\\}$

The following lemma translates a purely algebraic statement into an equivalent
topological terms.

######  Lemma 3.2.

Let $f,g\in Z(C_{\mathscr{P}}(X))^{*}$. Then $int_{X}Z(f)\cap
X_{\mathscr{P}}\subset int_{X}Z(g)\cap X_{\mathscr{P}}$ if and only if
$ann(f)\subset ann(g)$.

###### Proof.

Let $int_{X}Z(f)\cap X_{\mathscr{P}}\subset int_{X}Z(g)\cap X_{\mathscr{P}}$
and $h\in ann(f)$. Then $f.h=0$ and so, by using Observation 2.4, $X\setminus
Z(h)\subset int_{X}Z(f)\cap X_{\mathscr{P}}\subset int_{X}Z(g)\cap
X_{\mathscr{P}}$, so that $g.h=0$, i.e., $h\in ann(g)$.
Conversely let $ann(f)\subset ann(g)$ and $x\in int_{X}Z(f)\cap
X_{\mathscr{P}}$. Then by Lemma 2.2, there exists $h\in C_{\mathscr{P}}(X)$
such that $x\in X\setminus Z(h)\subset int_{X}Z(f)\cap X_{\mathscr{P}}$.
Clearly, $h\neq 0$ and $f.h=0$. Now $f.h=0\implies h\in ann(f)\subset
ann(g)\implies g.h=0$. Therefore, $x\in X\setminus Z(h)\subset int_{X}Z(g)\cap
X_{\mathscr{P}}$. So, $int_{X}Z(f)\cap X_{\mathscr{P}}\subset int_{X}Z(g)\cap
X_{\mathscr{P}}$. ∎

In light of Lemma 3.2, the rule for two vertices of $AG(C_{\mathscr{P}}(X))$
to be adjacent, takes the following form:

######  Theorem 3.3.

Two vertices $f,\ g$ in $AG(C_{\mathscr{P}}(X))$ are adjacent if and only if
$int_{X}Z(f)\cap X_{\mathscr{P}}\not\subset int_{X}Z(g)\cap X_{\mathscr{P}}$
and $int_{X}Z(g)\cap X_{\mathscr{P}}\not\subset int_{X}Z(f)\cap
X_{\mathscr{P}}$.

###### Proof.

Let $f,\ g$ be adjacent vertices in $AG(C_{\mathscr{P}}(X))$. Then $ann(f)\cup
ann(g)\subsetneqq ann(f.g)$ and so there exists $h\in
ann(f.g)\setminus[ann(f)\cup ann(g)]\implies f.g.h=0$ but $f.h\neq 0$ and
$g.h\neq 0$. Since $f.h\neq 0$, $(X\setminus Z(f))\cap(X\setminus
Z(h))\neq\emptyset$. Choose $x\in(X\setminus Z(f))\cap(X\setminus Z(h))$. By
Observation 2.4, $x\in X_{\mathscr{P}}$. Similarly, $y\in X_{\mathscr{P}}$
such that $y\in(X\setminus Z(g))\cap(X\setminus Z(h))$. Now,
$f.g.h=0\implies(X\setminus Z(f))\cap(X\setminus Z(h))=X\setminus Z(fh)\subset
int_{X}Z(g)$ and so, $x\in int_{X}Z(g)\cap X_{\mathscr{P}}$. Similarly, $y\in
int_{X}Z(f)\cap X_{\mathscr{P}}$. Therefore, $x\in[int_{X}Z(g)\cap
X_{\mathscr{P}}]\setminus int_{X}Z(f)$ and $y\in[int_{X}Z(f)\cap
X_{\mathscr{P}}]\setminus int_{X}Z(g)$. So, $int_{X}Z(f)\cap
X_{\mathscr{P}}\not\subset int_{X}Z(g)\cap X_{\mathscr{P}}$ and
$int_{X}Z(g)\cap X_{\mathscr{P}}\not\subset int_{X}Z(f)\cap X_{\mathscr{P}}$.
Suppose $f,\ g$ are non-adjacent vertices in $AG(C_{\mathscr{P}}(X))$. Then by
definition, $ann(f)\cup ann(g)=ann(f.g)$. Since the union of two ideals
$ann(f)$ and $ann(g)$ is an ideal $ann(f.g)$, it follows that either
$ann(f)\subset ann(g)$ or $ann(g)\subset ann(f)$. Then by Lemma 3.2, either
$int_{X}Z(f)\cap X_{\mathscr{P}}\subset int_{X}Z(g)\cap X_{\mathscr{P}}$ or
$int_{X}Z(g)\cap X_{\mathscr{P}}\subset int_{X}Z(f)\cap X_{\mathscr{P}}$. ∎

The following theorem endorses that annihilator graph $AG(C_{\mathscr{P}}(X))$
is indeed a subgraph of the weakly zero divisor graph
$W\Gamma(C_{\mathscr{P}}(X))$. A weakly zero divisor graph $W\Gamma(R)$ of a
commutative ring $R$ is a simple graph with $Z(R)^{*}$ as the set of vertices
and two distinct vertices $a,b$ are adjacent if there exists $c,d\in R$ such
that $c\in ann(a)\setminus\\{0\\}$, $d\in ann(b)\setminus\\{0\\}$ and $c.d=0$
[9].

######  Theorem 3.4.

$AG(C_{\mathscr{P}}(X))$ is a subgraph of $W\Gamma(C_{\mathscr{P}}(X))$.

###### Proof.

Let $f,\ g$ be two adjacent vertices in $AG(C_{\mathscr{P}}(X))$. By Theorem
3.3, $(int_{X}Z(f)\cap X_{\mathscr{P}})\cap(X\setminus Z(g))\neq\emptyset$ and
$(int_{X}Z(g)\cap X_{\mathscr{P}})\cap(X\setminus Z(f))\neq\emptyset$.
Therefore, by Lemma 2.2, there exist $h_{1},\ h_{2}\in C_{\mathscr{P}}(X)$
such that $\emptyset\neq X\setminus Z(h_{1})\subset cl_{X}(X\setminus
Z(h_{1}))\subset(int_{X}Z(f)\cap X_{\mathscr{P}})\cap(X\setminus Z(g))$ and
$\emptyset\neq X\setminus Z(h_{2})\subset cl_{X}(X\setminus
Z(h_{2}))\subset(int_{X}Z(g)\cap X_{\mathscr{P}})\cap(X\setminus Z(f))$. So,
$h_{1},\ h_{2}\neq 0$ and $f.h_{1}=0=g.h_{2}$. i.e., $h_{1},\ h_{2}\in
Z(C_{\mathscr{P}}(X))^{*}$. Since $(int_{X}Z(g)\cap
X_{\mathscr{P}})\cap(X\setminus Z(f))\cap(int_{X}Z(f)\cap
X_{\mathscr{P}})\cap(X\setminus Z(g))=\emptyset$, it follows that $(X\setminus
Z(h_{1}))\cap(X\setminus Z(h_{2}))=\emptyset$ and so, $h_{1}.h_{2}=0$. i.e.,
$h_{1}\in ann(f)\setminus\\{0\\}$, $h_{2}\in ann(g)\setminus\\{0\\}$ and
$h_{1}.h_{2}=0$, proving that $f,\ g$ are adjacent in
$W\Gamma(C_{\mathscr{P}}(X))$. ∎

For if $|X_{\mathscr{P}}|\leq 1$, the ring $C_{\mathscr{P}}(X)$ is either an
integral domain or the trivial ring $0$ and so, $Z(C_{\mathscr{P}}(X))^{*}$ is
an empty set. Therefore, in case $|X_{\mathscr{P}}|\leq 1$ , all the three
graphs are empty graphs, producing
$\Gamma(C_{\mathscr{P}}(X))=AG(C_{\mathscr{P}}(X))=W\Gamma(C_{\mathscr{P}}(X))$.
Observing that $AG(C_{\mathscr{P}}(X))$ has the same set of vertices as the
other two graphs under consideration and that it lies between
$\Gamma(C_{\mathscr{P}}(X))$ and $W\Gamma(C_{\mathscr{P}}(X))$, we may refine
Theorem 2.8 of [4] by allowing the possibility of a graph to become an empty
graph, as follows:

######  Theorem 3.5.

$\Gamma(C_{\mathscr{P}}(X))=AG(C_{\mathscr{P}}(X))=W\Gamma(C_{\mathscr{P}}(X))$
if and only if $|X_{\mathscr{P}}|\leq 2$.

Henceforth, in view of Theorem 3.5, we carry on our discussion on
$AG(C_{\mathscr{P}}(X))$ with the assumption that the Tychonoff space $X$ is
such that $|X_{\mathscr{P}}|\geq 3$.

The following theorem provides conditions under which a pair of vertices of
$AG(C_{\mathscr{P}}(X))$ admits of a third vertex adjacent to both of them.

######  Theorem 3.6.

Let $f,\ g\in Z(C_{\mathscr{P}}(X))^{*}$.

  1. (i)

If $f$ and $g$ are not adjacent in $\Gamma(C_{\mathscr{P}}(X))$ then there
exists a vertex $h\in Z(C_{\mathscr{P}}(X))^{*}$ such that $h$ is adjacent to
both $f$ and $g$ in $AG(C_{\mathscr{P}}(X))$.

  2. (ii)

If $f$ and $g$ are adjacent in $\Gamma(C_{\mathscr{P}}(X))$, then there exists
a vertex $h\in Z(C_{\mathscr{P}}(X))^{*}$ such that $h$ is adjacent to both
$f$ and $g$ in $AG(C_{\mathscr{P}}(X))$ if and only if $|int_{X}Z(f)\cap
X_{\mathscr{P}}|\geq 2$ and $|int_{X}Z(g)\cap X_{\mathscr{P}}|\geq 2$.

###### Proof.

(i) Follows from Lemma 2.1(6) and Lemma 2.3 of [6].
(ii) If possible let $|int_{X}Z(f)\cap X_{\mathscr{P}}|=1$. Since $f$ and $g$
are adjacent in $\Gamma(C_{\mathscr{P}}(X))$, $fg=0$ holds. So, $X\setminus
Z(g)\subset int_{X}Z(f)\cap X_{\mathscr{P}}=\\{x\\}$ (say). Then
$int_{X}Z(g)=X_{\mathscr{P}}\setminus\\{x\\}$ and so, $int_{X}Z(f)\cap
int_{X}Z(g)\cap X_{\mathscr{P}}=\emptyset$. Since $h$ is adjacent to $f$ in
$AG(C_{\mathscr{P}}(X))$, by Theorem 3.3, $\\{x\\}=int_{X}Z(f)\cap
X_{\mathscr{P}}$ $\not\subset int_{X}Z(h)\cap X_{\mathscr{P}}$ $\implies
x\notin int_{X}Z(h)\cap X_{\mathscr{P}}\implies int_{X}Z(h)\cap
X_{\mathscr{P}}\subset X_{\mathscr{P}}\setminus\\{x\\}=int_{X}Z(g)\cap
X_{\mathscr{P}}$. By Theorem 3.3 again, $h$ is not adjacent to $g$ in
$AG(C_{\mathscr{P}}(X))$, which is a contradiction.
Conversely, let $|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$ and
$|int_{X}Z(g)\cap X_{\mathscr{P}}|\geq 2$. If $int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}\neq\emptyset$ then by Lemma 2.2, there exists $h\in
C_{\mathscr{P}}(X)$ such that $\emptyset\neq X\setminus Z(h)\subset
int_{X}Z(f)\cap int_{X}Z(g)\cap X_{\mathscr{P}}$. So, $h\neq 0$ and
$f.h=0=g.h$. i.e., $h\in Z(C_{\mathscr{P}}(X))^{*}$ is adjacent to both $f,g$
in $\Gamma(C_{\mathscr{P}}(X))$ and hence, in $AG(C_{\mathscr{P}}(X))$. Let
$int_{X}Z(f)\cap int_{X}Z(g)\cap X_{\mathscr{P}}=\emptyset$. If we choose
$x,y\in int_{X}Z(f)\cap X_{\mathscr{P}}$ and $z,w\in int_{X}Z(g)\cap
X_{\mathscr{P}}$ then $x,y,z,w$ are all distinct. By Lemma 2.2, there exists
$h_{x}\in C_{\mathscr{P}}(X)$ such that $x\in X\setminus Z(h_{x})\subset
X\setminus int_{X}Z(h_{x})\subset[int_{X}Z(f)\cap
X_{\mathscr{P}}]\setminus\\{y\\}\implies\\{y\\}\cup X\setminus[int_{X}Z(f)\cap
X_{\mathscr{P}}]\subset int_{X}Z(h_{x})$ and so $y,z,w\in int_{X}Z(h_{x})\cap
X_{\mathscr{P}}$ and $x\in X\setminus Z(h_{x})$. Similarly we can find
$h_{z}\in C_{\mathscr{P}}(X)$ such that $x,y,w\in int_{X}Z(h_{z})\cap
X_{\mathscr{P}}$ and $z\in X\setminus Z(h_{z})$. Let $h=h_{x}^{2}+h_{z}^{2}\in
C_{\mathscr{P}}(X)$ and so, $int_{X}Z(h)=int_{X}Z(h_{x})\cap int_{X}Z(h_{z})$.
Therefore, $y,w\in int_{X}Z(h)\cap X_{\mathscr{P}}$ and $x,z\in X\setminus
Z(h)$. Clearly, $h\in Z(C_{\mathscr{P}}(X))^{*}$. $x\in[int_{X}Z(f)\cap
X_{\mathscr{P}}]\setminus int_{X}Z(h)$ and $w\in[int_{X}Z(h)\cap
X_{\mathscr{P}}]\setminus int_{X}Z(f)\implies f,h$ are adjacent in
$AG(C_{\mathscr{P}}(X))$, by Theorem 3.3. Similarly, we can show that $g,h$
are adjacent in $AG(C_{\mathscr{P}}(X))$. Thus $h\in
Z(C_{\mathscr{P}}(X))^{*}$ is adjacent to both $f,g$ in
$AG(C_{\mathscr{P}}(X))$. ∎

The next few results directly follow from Theorem 3.6:

######  Corollary 3.7.

For $f,g\in Z(C_{\mathscr{P}}(X))^{*}$,

$d_{AG}(f,g)=\begin{cases}1&\text{ whenever }f,g\text{ are adjacent}\\\
2&\text{ otherwise}\end{cases}$

and hence, $diam(AG(C_{\mathscr{P}}(X)))=2$.

######  Corollary 3.8.

For each $f\in Z(C_{\mathscr{P}}(X))^{*}$, $ecc(f)=2$ and hence,
$rad(AG(C_{\mathscr{P}}(X)))=2$ and
$C(AG(C_{\mathscr{P}}(X)))=Z(C_{\mathscr{P}}(X))^{*}$.

###### Proof.

For each $f$, as $2f$ and $f$ are non-adjacent, $ecc(f)=2$ and hence, each $f$
is in the center. ∎

######  Corollary 3.9.

An edge $f-g$ in $AG(C_{\mathscr{P}}(X))$ is an edge of a triangle in
$AG(C_{\mathscr{P}}(X))$ if and only if either $f-g$ is not an edge in
$\Gamma(C_{\mathscr{P}}(X))$ or else, $|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq
2$ and $|int_{X}Z(g)\cap X_{\mathscr{P}}|\geq 2$.

A topological phenomenon of $X$ produces a sufficient condition for
$AG(C_{\mathscr{P}}(X))$ to be hypertriangulated, as seen in the next theorem.

######  Theorem 3.10.

$AG(C_{\mathscr{P}}(X))$ is hypertriangulated if $X_{\mathscr{P}}$ does not
contain any isolated point of $X$.

###### Proof.

If $X_{\mathscr{P}}$ does not contain any isolated point then
$|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$ for every $f\in
Z(C_{\mathscr{P}}(X))^{*}$. By Corollary 3.9, every edge is an edge of a
triangle in $AG(C_{\mathscr{P}}(X))$ and so, $AG(C_{\mathscr{P}}(X))$ is
hypertriangulated. ∎

The following example establishes that the converse of Theorem 3.10 is not
always true.

######  Example 3.11.

Consider the completely regular Hausdorff space $X=\mathbb{N}^{*}$, the one
point compactification of $\mathbb{N}$. If $\mathscr{P}$ is the collection of
all finite sets of $X$ then $C_{\mathscr{P}}(X)=C_{F}(X)$. Clearly,
$X_{\mathscr{P}}=\mathbb{N}$. If $f\in Z(C_{\mathscr{P}}(X))^{*}$ then
$X\setminus Z(f)$ is finite and therefore, $Z(f)$ is a clopen set in $X$ such
that $X_{\mathscr{P}}\cap Z(f)=X_{\mathscr{P}}\cap int_{X}Z(f)$ is infinite.
i.e., $|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$, for all $f\in
Z(C_{\mathscr{P}}(X))$. Hence, $AG(C_{\mathscr{P}}(X))$ is hypertriangulated.

However, if $X_{\mathscr{P}}$ contains an isolated point of $X$ and
$AG(C_{\mathscr{P}}(X))$ is hypertriangulated then $X_{\mathscr{P}}$ is never
a clopen subset of $X$. For if $X_{\mathscr{P}}$ is clopen and $x\in
X_{\mathscr{P}}$ is an isolated point of $X$, taking $f=1_{x}$ and
$g=1_{X_{\mathscr{P}}\setminus\\{x\\}}$ we get $f.g=0$. Since both
$X_{\mathscr{P}}$ and $\\{x\\}$ are clopen sets in $X$, $f,g\in
C_{\mathscr{P}}(X)$ and therefore, $f$ and $g$ are adjacent in
$AG(C_{\mathscr{P}}(X))$. Since $int_{X}Z(g)\cap X_{\mathscr{P}}=\\{x\\}$, by
Theorem 3.6(ii), no vertex is adjacent to both $f$ and $g$ in
$AG(C_{\mathscr{P}}(X))$. Consequently, $f-g$ in $AG(C_{\mathscr{P}}(X))$ is
not an edge of a triangle, i.e., $AG(C_{\mathscr{P}}(X))$ is not
hypertriangulated.

We record the results related to hypertriangulatedness of
$AG(C_{\mathscr{P}}(X))$ for special choices of $\mathscr{P}$.

######  Theorem 3.12.

  1. (1)

$AG(C(X))$ is hypertriangulated if and only if $X$ has no isolated point.

  2. (2)

$AG(C_{K}(X))$ is hypertriangulated if and only if $X$ has no isolated point.

  3. (3)

$AG(C_{F}(X))$ is hypertriangulated if and only if the number of isolated
points of $X$ is infinite.

###### Proof.

  1. (1)

If $\mathscr{P}$ is the ideal of all closed subsets of $X$ then
$C_{\mathscr{P}}(X)=C(X)$ and $X_{\mathscr{P}}=X$ (= clopen in $X$). So,
$AG(C(X))$ is hypertriangulated if and only if $X$ has no isolated point.

  2. (2)

If $\mathscr{P}$ is the ideal of all compact subsets of $X$ then
$C_{\mathscr{P}}(X)=C_{K}(X)$ and $X_{\mathscr{P}}$ is the set of all points
at which $X$ is locally compact. Therefore, for a locally compact space $X$,
$AG(C_{K}(X))$ is hypertriangulated if and only if $X$ has no isolated point.

  3. (3)

If $\mathscr{P}$ is the ideal of all finite subsets of $X$ then
$C_{\mathscr{P}}(X)=C_{F}(X)=\\{f\in C(X):X\setminus Z(f)\text{ is finite}\\}$
and $X_{\mathscr{P}}$ is the set of all isolated points in $X$. It is to be
noted that $C_{\mathscr{P}}(X)$ is a non-zero ring if and only if $X$ contains
atleast one isolated point. So, if the number of isolated points of $X$ is
finite then certainly $X_{\mathscr{P}}$ is a clopen subset of $X$ and hence,
$AG(C_{F}(X))$ is never hypertriangulated. Let the number of isolated points
of $X$ be infinite. Then proceeding as in Example 3.11, $AG(C_{F}(X))$ is
hypertriangulated.

∎

######  Theorem 3.13.

$AG(C_{\mathscr{P}}(X))$ is always triangulated.

###### Proof.

Let $f\in Z(C_{\mathscr{P}}(X))^{*}$. Then $int_{X}Z(f)\cap
X_{\mathscr{P}}\neq\emptyset$. If $|int_{X}Z(f)\cap X_{\mathscr{P}}|=1$, say
$int_{X}Z(f)\cap X_{\mathscr{P}}=\\{x\\}$, then as $|X_{\mathscr{P}}|\geq 3$,
we can find distinct $y,z$ in $X_{\mathscr{P}}\setminus\\{x\\}$. By Lemma 2.3,
there exists $g,h\in C_{\mathscr{P}}(X)$ such that $\\{x,y\\}\subset
X\setminus Z(g)\subset X\setminus int_{X}Z(g)\subset
X_{\mathscr{P}}\setminus\\{z\\}$ and $\\{x,z\\}\subset X\setminus Z(h)\subset
X\setminus int_{X}Z(h)\subset X_{\mathscr{P}}\setminus\\{y\\}$. Then $g,h\neq
0$, $z\in int_{X}Z(g)\cap X_{\mathscr{P}}$ and $y\in int_{X}Z(h)\cap
X_{\mathscr{P}}\implies g,h\in Z(C_{\mathscr{P}}(X))^{*}$. Now
$x\in[int_{X}Z(f)\cap X_{\mathscr{P}}]\setminus int_{X}Z(g)$ and
$y\in[int_{X}Z(g)\cap X_{\mathscr{P}}]\setminus int_{X}Z(f)\implies f,g$ are
adjacent in $AG(C_{\mathscr{P}}(X))$. Similarly, $g,h$ and $f,g$ are adjacent
in $AG(C_{\mathscr{P}}(X))$. Thus $f-g-h-f$ constitutes a triangle with $f$ as
a vertex in $AG(C_{\mathscr{P}}(X))$.
If $|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$ then by Theorem 2.22 of [4], $f$
is a vertex of a triangle in $\Gamma_{\mathscr{P}}(X)$ and hence, a vertex of
a triangle in $AG(C_{\mathscr{P}}(X))$. ∎

Before calculating eccentricity, girth, radius and length of shortest cycle
containing two specific vertices, we jot down two very elementary facts which
are amply clear and need no further clarification:
In $AG(C_{\mathscr{P}}(X))$,

  1. (i)

$f$ and $g$ are adjacent if and only if $m_{1}f$ and $m_{2}g$ are adjacent,
for all positive integers $m_{1},m_{2}$.

  2. (ii)

$m_{1}f$ and $m_{2}f$, for any positive integers $m_{1},m_{2}$, are not
adjacent.

In view of these elementary facts, the following are immediate:

######  Theorem 3.14.

In $AG(C_{\mathscr{P}}(X))$,

  1. (i)

Any edge $f-g$ is an edge of a square and so, $3\leq c(f,g)\leq 4$.
Furthermore, for any $f,g\in Z(C_{\mathscr{P}}(X))^{*}$,

    1. (a)

if $f,g$ are adjacent in $AG(C_{\mathscr{P}}(X))$ with $f.g\neq 0$, then
$c(f,g)=3$;

    2. (b)

if $f,g$ are adjacent in $AG(C_{\mathscr{P}}(X))$ with $f.g=0$ then
$c(f,g)=\begin{cases}3&\text{ if }|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq
2\text{ and }|int_{X}Z(g)\cap X_{\mathscr{P}}|\geq 2\\\ 4&\text{ if
}|int_{X}Z(f)\cap X_{\mathscr{P}}|=1\text{ or }|int_{X}Z(g)\cap
X_{\mathscr{P}}|=1\end{cases}$

    3. (c)

if $f,g$ are non-adjacent in $AG(C_{\mathscr{P}}(X))$ then $c(f,g)=4$.

  2. (ii)

$gr(AG(C_{\mathscr{P}}(X)))=3$.

###### Proof.

(i) Since $f-g$ is an edge in $AG(C_{\mathscr{P}}(X))$, it follows that
$f-g-2f-2g-f$ completes a square. Consequently, $3\leq c(f,g)\leq 4$.

  1. (a)

$f,g$ are adjacent in $AG(C_{\mathscr{P}}(X))$ with $f.g\neq 0$ means that
$f,g$ are not adjacent in $\Gamma(C_{\mathscr{P}}(X))$ and hence, by Theorem
3.6(i), there exists a vertex $h\in Z(C_{\mathscr{P}}(X))^{*}$ such that $f-h-
g-f$ constitutes a cycle of length 3 in $AG(C_{\mathscr{P}}(X))$. i.e.,
$c(f,g)=3$.

  2. (b)

By Theorem 3.6(ii) there exists a vertex $h\in Z(C_{\mathscr{P}}(X))^{*}$ such
that $f-h-g-f$ constitutes a cycle of length 3 in $AG(C_{\mathscr{P}}(X))$ if
and only if $|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$ and $|int_{X}Z(g)\cap
X_{\mathscr{P}}|\geq 2$. So, in this case,
$c(f,g)=\begin{cases}3&\text{ if }|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq
2\text{ and }|int_{X}Z(g)\cap X_{\mathscr{P}}|\geq 2\\\ 4&\text{ if
}|int_{X}Z(f)\cap X_{\mathscr{P}}|=1\text{ or }|int_{X}Z(g)\cap
X_{\mathscr{P}}|=1\end{cases}$

  3. (c)

If $f,g$ are non-adjacent in $AG(C_{\mathscr{P}}(X))$ then by Theorem 3.6(i),
there exists a vertex $h\in Z(C_{\mathscr{P}}(X))^{*}$ such that $h$ is
adjacent to both of $f$ and $g$, constituting a square $f-h-g-2h-f$.
Therefore, $c(f,g)=4$.

(ii) Follows from the fact that $AG(C_{\mathscr{P}}(X))$ is triangulated. ∎

######  Corollary 3.15.

Each chord-less cycle in $AG(C_{\mathscr{P}}(X))$ is of length $3$ or $4$.

The crux of Theorem 3.14 may be represented diagrammatically as follows:

$f$$g$$f$$g$$f.g\neq 0\text{ and }f,g\text{ are adjacent}$$f.g=0\text{ and
}|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$and $|int_{X}Z(g)\cap
X_{\mathscr{P}}|\geq 2$$f$$g$$f$$g$$f.g=0\text{ and }|int_{X}Z(f)\cap
X_{\mathscr{P}}|=1$$f,g\text{ are non-adjacent}$or $|int_{X}Z(g)\cap
X_{\mathscr{P}}|=1$

The following theorem refines Theorem 3.14 in absence of any isolated point of
$X$ in $X_{\mathscr{P}}$.

######  Theorem 3.16.

If $X_{\mathscr{P}}$ contains no isolated point of $X$, then for any $f,g\in
Z(C_{\mathscr{P}}(X))^{*}$,

$c(f,g)=\begin{cases}3&\text{ if }f,g\text{ are adjacent}\\\ 4&\text{
otherwise}\end{cases}$

###### Proof.

If $X_{\mathscr{P}}$ contains no isolated point of $X$ then $|int_{X}Z(f)\cap
X_{\mathscr{P}}|\geq 2$ for all $f\in Z(C_{\mathscr{P}}(X))^{*}$. So, by
Theorem 3.14, the result holds. ∎

Example 3.11 guarantees that the converse of Theorem 3.16 doesn’t always hold.
More generally, in any completely regular Hausdorff space $X$ with infinitely
many isolated points, as $X\setminus Z(f)$, for each $f\in(Z(C_{F}(X)))^{*}$,
is a finite subset of isolated points, $int_{X}Z(f)\cap
X_{\mathscr{P}}=Z(f)\cap X_{\mathscr{P}}$ is infinite. As a consequence,
$|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$ holds for all
$f\in(Z(C_{F}(X)))^{*}$. So, by Theorem 3.14, in spite of the presence of
infinitely many isolated points of $X$, for every pair
$f,g\in(Z(C_{F}(X)))^{*}$,

$c(f,g)=\begin{cases}3&\text{ if }f,g\text{ are adjacent}\\\ 4&\text{
otherwise}\end{cases}$

In the next theorem we show that a sufficient condition for the converse of
Theorem 3.16 to hold is that $X_{\mathscr{P}}$ is a clopen (not necessarily
proper) subset of $X$.

######  Theorem 3.17.

If $X_{\mathscr{P}}$ is a clopen (not necessarily proper) subset of $X$ and
for any $f,g\in Z(C_{\mathscr{P}}(X))^{*}$,

$c(f,g)=\begin{cases}3&\text{ if }f,g\text{ are adjacent}\\\ 4&\text{
otherwise}\end{cases}$

then $X_{\mathscr{P}}$ does not contain any isolated point of $X$.

###### Proof.

let $X_{\mathscr{P}}$ be a clopen subset of $X$. If $x\in X_{\mathscr{P}}$ is
an isolated point of $X$ then choosing $f=1_{x}\in Z(C_{\mathscr{P}}(X))^{*}$
and $g=1_{X_{\mathscr{P}}\setminus\\{x\\}}\in Z(C_{\mathscr{P}}(X))^{*}$ we
find that $f.g=0$ and $int_{X}Z(g)\cap X_{\mathscr{P}}=\\{x\\}$. Then by
Theorem 3.14, $c(f,g)=4$ though $f,g$ are adjacent in
$AG(C_{\mathscr{P}}(X))$, contradicting the hypothesis. ∎

Interpreting Theorem 3.16and Theorem 3.17 for special choices of
$\mathscr{P}$, we get an explicit formula for $c(f,g)$ in terms of the
adjacency of the vertices $f$ and $g$ in $C(X)$ as well as in $C_{K}(X)$.

######  Theorem 3.18.

Let $X$ be a completely regular Hausdorff space.

  1. (1)

$X$ has no isolated point if and only if for any $f,g\in Z(C(X))^{*}$,

$c(f,g)=\begin{cases}3&\text{ if }f,g\text{ are adjacent}\\\ 4&\text{
otherwise}\end{cases}$

  2. (2)

If $X$ is locally compact then $X$ has no isolated point if and only if for
any $f,g\in Z(C_{K}(X))^{*}$,

$c(f,g)=\begin{cases}3&\text{ if }f,g\text{ are adjacent}\\\ 4&\text{
otherwise}.\end{cases}$

Two vertices $a,b$ in a graph $G$ are called orthogonal, denoted by $a\perp
b$, if $a,b$ are adjacent in $G$ and no other vertex of $G$ is adjacent to
both of $a$ and $b$. The following theorem sets a criterion for existence of a
pair of orthogonal vertices in $AG(C_{\mathscr{P}}(X))$ in terms of the
topological behaviour of $X$ and $X_{\mathscr{P}}$.

######  Theorem 3.19.

For any $f,g\in Z(C_{\mathscr{P}}(X))^{*}$, $f\perp g$ in
$AG(C_{\mathscr{P}}(X))$ if and only if $f.g=0$ and either $|int_{X}Z(f)\cap
X_{\mathscr{P}}|=1$ or $|int_{X}Z(g)\cap X_{\mathscr{P}}|=1$.

###### Proof.

If $f\perp g$ then $f,g$ are adjacent in $AG(C_{\mathscr{P}}(X))$ and no other
vertex is adjacent to both $f,g$ in $AG(C_{\mathscr{P}}(X))$. By Theorem
3.6(i), $f.g\neq 0$ is impossible. So, $f.g=0$ and therefore by Theorem
3.6(ii), either $|int_{X}Z(f)\cap X_{\mathscr{P}}|=1$ or $|int_{X}Z(g)\cap
X_{\mathscr{P}}|=1$.
Converse follows from Theorem 3.6(ii) and the definition of $f\perp g$. ∎

######  Theorem 3.20.

$AG(C_{\mathscr{P}}(X))$ does not contain any orthogonal pair of vertices if
$X_{\mathscr{P}}$ has no isolated point of $X$. The converse holds if
$X_{\mathscr{P}}$ is a clopen (not necessarily proper) subset of $X$.

###### Proof.

$AG(C_{\mathscr{P}}(X))$ contains an orthogonal pair of vertices $f,g$ then by
Theorem 3.19, either $|int_{X}Z(f)\cap X_{\mathscr{P}}|$ or $|int_{X}Z(g)\cap
X_{\mathscr{P}}|$ is a singleton, say $\\{x\\}$. Since $X_{\mathscr{P}}$ is
open in $X$, it follows that $x\in X_{\mathscr{P}}$ is an isolated point of
$X$, contradicting our hypothesis.
Conversely let $X_{\mathscr{P}}$ be a clopen subset of $X$ and $x\in
X_{\mathscr{P}}$ be an isolated point of $X$. By choosing $f=1_{x}\in
Z(C_{\mathscr{P}}(X))^{*}$ and $g=1_{X_{\mathscr{P}}\setminus\\{x\\}}\in
Z(C_{\mathscr{P}}(X))^{*}$, it follows from Theorem 3.19 that $f\perp g$ in
$AG(C_{\mathscr{P}}(X))$. ∎

Once again, Example 3.11 endorses that abundance of isolated points in
$X_{\mathscr{P}}$ does not ensure the existence of a pair of orthogonal
vertices in $AG(C_{\mathscr{P}}(X))$.

A graph $G$ is said to be complemented if for each vertex $a$ in $G$, there
exists a vertex $b$ in $G$ such that $a\perp b$. A complemented graph $G$ is
called uniquely complemented if $a\perp b$ and $a\perp c$ implies that $b$ and
$c$ are adjacent to the same set of vertices in $G$. It is well known that a
graph may be complemented without being uniquely complemented. In the next
theorem we prove that for $AG(C_{\mathscr{P}}(X))$, the terms ‘complemented’
and ‘uniquely complemented’ are synonymous.

######  Theorem 3.21.

If $AG(C_{\mathscr{P}}(X))$ is complemented then it is uniquely complemented.

###### Proof.

Let $f\perp g$ and $f\perp h$ in $AG(C_{\mathscr{P}}(X))$ for $f,g,h\in
Z(C_{\mathscr{P}}(X))^{*}$. By Theorem 3.19, $f.g=0=f.h$ and either
$|int_{X}Z(f)\cap X_{\mathscr{P}}|=1$ or $|int_{X}Z(g)\cap
X_{\mathscr{P}}|=1=|int_{X}Z(h)\cap X_{\mathscr{P}}|$.
If $int_{X}Z(f)\cap X_{\mathscr{P}}=\\{x\\}$ then $f.g=0\implies X\setminus
Z(g)\subset int_{X}Z(f)\cap X_{\mathscr{P}}=\\{x\\}$, so that,
$int_{X}Z(g)\cap X_{\mathscr{P}}=X_{\mathscr{P}}\setminus\\{x\\}$. Similarly,
$f.h=0\implies int_{X}Z(h)\cap
X_{\mathscr{P}}=X_{\mathscr{P}}\setminus\\{x\\}$. Hence, $int_{X}Z(g)\cap
X_{\mathscr{P}}=int_{X}Z(h)\cap X_{\mathscr{P}}$.
If $|int_{X}Z(g)\cap X_{\mathscr{P}}|=1=|int_{X}Z(h)\cap X_{\mathscr{P}}|$
then assuming $int_{X}Z(g)\cap X_{\mathscr{P}}=\\{x\\}$ and $int_{X}Z(h)\cap
X_{\mathscr{P}}=\\{y\\}$, we arrive at the following : $f.g=0\implies
X\setminus Z(f)\subset int_{X}Z(g)\cap X_{\mathscr{P}}=\\{x\\}$ and
$f.h=0\implies X\setminus Z(f)\subset int_{X}Z(h)\cap
X_{\mathscr{P}}=\\{y\\}$. Therefore, $X\setminus Z(f)=x=y$ and hence,
$int_{X}Z(g)\cap X_{\mathscr{P}}=int_{X}Z(h)\cap X_{\mathscr{P}}$.
So, in any case, $int_{X}Z(g)\cap X_{\mathscr{P}}=int_{X}Z(h)\cap
X_{\mathscr{P}}$. By Theorem 3.3, it is evident that $g,h$ are adjacent to the
same set of vertices in $AG(C_{\mathscr{P}}(X))$. ∎

If $|X_{\mathscr{P}}|=2$ then
$AG(C_{\mathscr{P}}(X))(=\Gamma(C_{\mathscr{P}}(X)))$ becomes a complete
bipartite graph and hence, it is uniquely complemented. For if
$|X_{\mathscr{P}}|<2$, $AG(C_{\mathscr{P}}(X))$ remains (trivially) uniquely
complemented as well. So, being confined to our blanket assumption that
$|X_{\mathscr{P}}|\geq 3$, we prove next that $AG(C_{\mathscr{P}}(X))$ is
almost never a (uniquely) complemented graph.

######  Theorem 3.22.

$AG(C_{\mathscr{P}}(X))$ is uniquely complemented if and only if
$|X_{\mathscr{P}}|=3$.

###### Proof.

In view of Theorem 3.21, it is enough to show that the graph is complemented
if and only if $|X_{\mathscr{P}}|=3$.
Let $X_{\mathscr{P}}=\\{x,y,z\\}$. So, $f\in Z(C_{\mathscr{P}}(X))^{*}$
implies that either $|X\setminus Z(f)|=1$ or $2$.
If $|X\setminus Z(f)|=1$, say $X\setminus Z(f)=\\{x\\}$, then
$g=1_{\\{y,z\\}}\in Z(C_{\mathscr{P}}(X))^{*}$ such that $f.g=0$ and
$|int_{X}Z(g)\cap X_{\mathscr{P}}|=1$. By Theorem 3.19, $f\perp g$ in
$AG(C_{\mathscr{P}}(X))$.
If $|X\setminus Z(f)|=2$, say $X\setminus Z(f)=\\{x,y\\}$, then
$|int_{X}Z(f)\cap X_{\mathscr{P}}|=1$. By choosing $g=1_{z}$ it is easy to see
that $g\in Z(C_{\mathscr{P}}(X))^{*}$ with $f.g=0$. By Theorem 3.19 again,
$f\perp g$ in $AG(C_{\mathscr{P}}(X))$.
For if $|X_{\mathscr{P}}|\geq 4$ there are atleast four distinct vertices
$x,y,z,w\in X_{\mathscr{P}}$. By Lemma 2.3, there exist $f_{x},f_{y}\in
Z(C_{\mathscr{P}}(X))^{*}$ such that $\\{y,z,w\\}\subset X\setminus
Z(f_{x})\subset X\setminus int_{X}Z(f_{x})\subset X\setminus\\{x\\}$ and
$\\{x,z,w\\}\subset X\setminus Z(f_{y})\subset X\setminus
int_{X}Z(f_{y})\subset X\setminus\\{y\\}$. Then $x\in int_{X}Z(f_{x})\cap
X_{\mathscr{P}}$ and $y\in int_{X}Z(f_{y})\cap X_{\mathscr{P}}$. Let
$f=f_{x}.f_{y}\in C_{\mathscr{P}}(X)$. Then $x,y\in int_{X}Z(f)\cap
X_{\mathscr{P}}\implies f\in Z(C_{\mathscr{P}}(X))^{*}$ and $z,w\in X\setminus
Z(f)$. If possible let $g\in Z(C_{\mathscr{P}}(X))^{*}$ be such that $f\perp
g$ in $AG(C_{\mathscr{P}}(X))$. By Theorem 3.19, $f.g=0$ and $|int_{X}Z(g)\cap
X_{\mathscr{P}}|=1$ (as $|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$). But
$f.g=0\implies X\setminus Z(f)\subset int_{X}Z(g)\cap X_{\mathscr{P}}\implies
z,w\in int_{X}Z(g)\cap X_{\mathscr{P}}$ i.e., $|int_{X}Z(g)\cap
X_{\mathscr{P}}|\geq 2$ and so, we arrive at a contradiction. Hence, no vertex
is orthogonal to such $f$ in $AG(C_{\mathscr{P}}(X))$. i.e.,
$AG(C_{\mathscr{P}}(X))$ is not complemented when $|X_{\mathscr{P}}|\geq 4$. ∎

In particular, special choices of $\mathscr{P}$ reveals when $C(X)$ and
$C_{K}(X)$ are (uniquely) complemented, as recorded in the following
Corollary:

######  Corollary 3.23.

  1. (1)

$AG(C(X))$ is uniquely complemented if and only if $X$ has atmost three
points.

  2. (2)

$AG(C_{K}(X))$ is uniquely complemented if and only if $X$ is locally compact
at atmost three points.

######  Lemma 3.24.

Let $f,g\in Z(C_{\mathscr{P}}(X))^{*}$ such that $f.g=0$ and $int_{X}Z(f)\cap
int_{X}Z(g)\cap X_{\mathscr{P}}=\emptyset$. Then for some $h\in
C_{\mathscr{P}}(X)$, $[int_{X}Z(f)\cup int_{X}Z(g)]\cap X_{\mathscr{P}}\subset
int_{X}Z(h)\cap X_{\mathscr{P}}$ implies $h=0$.

###### Proof.

$f.g=0\implies X\setminus Z(f)\subset int_{X}Z(g)\cap X_{\mathscr{P}}$ and
$X\setminus Z(g)\subset int_{X}Z(f)\cap X_{\mathscr{P}}$. Therefore,
$X\setminus Z(f)\cup X\setminus Z(g)\subset[int_{X}Z(g)\cap
X_{\mathscr{P}}]\cup[int_{X}Z(f)\cap X_{\mathscr{P}}]=[int_{X}Z(f)\cup
int_{X}Z(g)]\cap X_{\mathscr{P}}\subset int_{X}Z(h)\cap
X_{\mathscr{P}}\implies X\setminus[int_{X}Z(h)\cap X_{\mathscr{P}}]\subset
Z(f)\cap Z(g)$. Thus $X\setminus Z(h)\subset X\setminus[int_{X}Z(h)\cap
X_{\mathscr{P}}]\subset Z(f)\cap Z(g)\cap X_{\mathscr{P}}\implies X\setminus
Z(h)\subset int_{X}[Z(f)\cap Z(g)\cap X_{\mathscr{P}}]=int_{X}Z(f)\cap
int_{X}Z(g)\cap X_{\mathscr{P}}=\emptyset\implies Z(h)=X$, i.e., $h=0$. ∎

######  Theorem 3.25.

$dt(AG(C_{\mathscr{P}}(X)))=2$ if and only if there exists a pair of vertices
$f,g$ in $AG(C_{\mathscr{P}}(X))$ such that $f.g=0$ and $int_{X}Z(f)\cap
int_{X}Z(g)\cap X_{\mathscr{P}}=\emptyset$.

###### Proof.

Let $dt(AG(C_{\mathscr{P}}(X)))=2$ and $\\{f,g\\}$ be a dominating set in
$AG(C_{\mathscr{P}}(X))$. Since $2f$ is not adjacent to $f$ and
$2f\notin\\{f,g\\}$, $2f$ is adjacent to $g$ and so $f$ is adjacent to $g$ in
$AG(C_{\mathscr{P}}(X))$. By Theorem 3.3, we get

(3.1) $\begin{split}int_{X}Z(f)\cap X_{\mathscr{P}}\not\subset int_{X}Z(g)\cap
X_{\mathscr{P}}\text{ and }\\\ int_{X}Z(g)\cap X_{\mathscr{P}}\not\subset
int_{X}Z(f)\cap X_{\mathscr{P}}\end{split}$

If possible let $f.g\neq 0$ and $x\in X\setminus Z(f)\cap X\setminus Z(g)$.
Then by Lemma 2.2, there exists $h\in C_{\mathscr{P}}(X)$ such that $x\in
X\setminus Z(h)\subset X\setminus int_{X}Z(h)\subset X\setminus Z(f)\cap
X\setminus Z(g)\implies h\neq 0$ and $Z(f)\cup Z(g)\subset int_{X}Z(h)\implies
h\in Z(C_{\mathscr{P}}(X))^{*}$ and $h$ is not adjacent to both $f,g$ in
$AG(C_{\mathscr{P}}(X))$. By Equation 3.1 and the fact that $Z(f)\cup
Z(g)\subset int_{X}Z(h)$, we get $h\neq f$ and $h\neq g$, i.e.,
$h\notin\\{f,g\\}$ which contradicts that $\\{f,g\\}$ is a dominating set in
$AG(C_{\mathscr{P}}(X))$.
If possible let $int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}\neq\emptyset$. Choosing $h=f^{2}+g^{2}\in C_{\mathscr{P}}(X)$,
$int_{X}Z(h)=int_{X}Z(f)\cap int_{X}Z(g)\implies int_{X}Z(h)\cap
X_{\mathscr{P}}\neq\emptyset\implies h\in Z(C_{\mathscr{P}}(X))^{*}$ and
$h\neq f$, $h\neq g$ by Equation 3.1. Since $int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}=int_{X}Z(h)\cap X_{\mathscr{P}}$, $h\notin\\{f,g\\}$ is not
adjacent to both $f,g$ in $AG(C_{\mathscr{P}}(X))$, which gives a
contradiction.
Conversely, let $f,g\in Z(C_{\mathscr{P}}(X))^{*}$ be such that $f.g=0$ and
$int_{X}Z(f)\cap int_{X}Z(g)\cap X_{\mathscr{P}}=\emptyset$. Since for each
$h\in Z(C_{\mathscr{P}}(X))^{*}$, $ecc(h)=2$ implies that
$dt(AG(C_{\mathscr{P}}(X)))\geq 2$. We now show that $\\{f,g\\}$ is a
dominating set in $AG(C_{\mathscr{P}}(X))$. If possible let $h$ be not
adjacent to both $f$ and $g$ in $AG(C_{\mathscr{P}}(X))$. Then by Theorem 3.3,
[either $int_{X}Z(f)\cap X_{\mathscr{P}}\subset int_{X}Z(h)\cap
X_{\mathscr{P}}$ or $int_{X}Z(h)\cap X_{\mathscr{P}}\subset int_{X}Z(f)\cap
X_{\mathscr{P}}$] and [either $int_{X}Z(g)\cap X_{\mathscr{P}}\subset
int_{X}Z(h)\cap X_{\mathscr{P}}$ or $int_{X}Z(h)\cap X_{\mathscr{P}}\subset
int_{X}Z(g)\cap X_{\mathscr{P}}$].
If $int_{X}Z(f)\cap X_{\mathscr{P}}\subset int_{X}Z(h)\cap X_{\mathscr{P}}$
and $int_{X}Z(g)\cap X_{\mathscr{P}}\subset int_{X}Z(h)\cap X_{\mathscr{P}}$
then $[int_{X}Z(f)\cup int_{X}Z(g)]\cap X_{\mathscr{P}}\subset int_{X}Z(h)\cap
X_{\mathscr{P}}$ and hence by Lemma 3.24, we get $h=0$ which is not possible
as $h\in Z(C_{\mathscr{P}}(X))^{*}$.
If $int_{X}Z(h)\cap X_{\mathscr{P}}\subset int_{X}Z(f)\cap X_{\mathscr{P}}$
and $int_{X}Z(h)\cap X_{\mathscr{P}}\subset int_{X}Z(g)\cap X_{\mathscr{P}}$
then $int_{X}Z(h)\cap X_{\mathscr{P}}\subset int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}=\emptyset$ which contradicts that $h\in
Z(C_{\mathscr{P}}(X))^{*}$.
If $int_{X}Z(h)\cap X_{\mathscr{P}}\subset int_{X}Z(f)\cap X_{\mathscr{P}}$
and $int_{X}Z(g)\cap X_{\mathscr{P}}\subset int_{X}Z(h)\cap X_{\mathscr{P}}$
then $int_{X}Z(g)\cap X_{\mathscr{P}}\subset int_{X}Z(f)\cap
X_{\mathscr{P}}\implies int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}=int_{X}Z(g)\cap X_{\mathscr{P}}$. Therefore $int_{X}Z(g)\cap
X_{\mathscr{P}}=\emptyset$ which contradicts that $g\in
Z(C_{\mathscr{P}}(X))^{*}$.
If $int_{X}Z(f)\cap X_{\mathscr{P}}\subset int_{X}Z(h)\cap X_{\mathscr{P}}$
and $int_{X}Z(h)\cap X_{\mathscr{P}}\subset int_{X}Z(g)\cap X_{\mathscr{P}}$
then $int_{X}Z(f)\cap X_{\mathscr{P}}\subset int_{X}Z(g)\cap X_{\mathscr{P}}$
and we get a similar contradiction.
Hence, $\\{f,g\\}$ is a dominating set in $AG(C_{\mathscr{P}}(X))$, proving
$dt(AG(C_{\mathscr{P}}(X)))=2$. ∎

######  Theorem 3.26.

$dt(AG(C_{\mathscr{P}}(X)))=2$ if and only if $\Gamma(C_{\mathscr{P}}(X))$ is
not hypertriangulated.

###### Proof.

By Theorem $2.28$ in [4], $\Gamma(C_{\mathscr{P}}(X))$ is hypertriangulated if
and only if for any edge $f-g$ in $\Gamma(C_{\mathscr{P}}(X))$,
$int_{X}Z(f)\cap int_{X}Z(g)\cap X_{\mathscr{P}}\neq\emptyset$. Let
$\Gamma(C_{\mathscr{P}}(X))$ be not hypertriangulated. Then there is an edge
$f-g$ in $\Gamma(C_{\mathscr{P}}(X))$ such that $int_{X}Z(f)\cap
int_{X}Z(g)\cap X_{\mathscr{P}}=\emptyset$. Since $f-g$ is an edge in
$\Gamma(C_{\mathscr{P}}(X))$, $f.g=0$. Hence, by Theorem 3.25,
$dt(AG(C_{\mathscr{P}}(X)))=2$. Conversely, if $dt(AG(C_{\mathscr{P}}(X)))=2$
then there exists $f,g\in Z(C_{\mathscr{P}}(X))^{*}$ such that $f.g=0$ and
$int_{X}Z(f)\cap int_{X}Z(g)\cap X_{\mathscr{P}}=\emptyset$. Then
$f.g=0\implies f-g$ is an edge in $\Gamma(C_{\mathscr{P}}(X))$ and
$int_{X}Z(f)\cap int_{X}Z(g)\cap X_{\mathscr{P}}=\emptyset\implies f-g$ is not
an edge of a triangle in $\Gamma(C_{\mathscr{P}}(X))$. Therefore,
$\Gamma(C_{\mathscr{P}}(X))$ is not hypertriangulated. ∎

The following Corollary is an immediate consequence of Theorem 3.26 and
Theorem $2.29$ in [4].

######  Corollary 3.27.

If $cl_{X}(X_{\mathscr{P}})\notin\mathscr{P}$, then
$dt(AG(C_{\mathscr{P}}(X)))>2$.

######  Theorem 3.28.

If $X_{\mathscr{P}}$ is clopen and properly contains a clopen subset of $X$
then

$dt(AG(C_{\mathscr{P}}(X)))=2.$

###### Proof.

Since $X_{\mathscr{P}}$ is a disconnected subspace of $X$, $X_{\mathscr{P}}$
contains a non-empty proper clopen subset, say $K$. Since $X_{\mathscr{P}}$ is
clopen in $X$, $K$ is clopen in $X$ also. Consider $f=1_{K}$ and
$g=1_{X_{\mathscr{P}}\setminus K}$. Since both $K$ and
$X_{\mathscr{P}}\setminus K$ are clopen in $X$, $f,g\in C_{\mathscr{P}}(X)$
and $f.g=0$. Also $int_{X}Z(f)\cap X_{\mathscr{P}}=X_{\mathscr{P}}\setminus K$
and $int_{X}Z(f)\cap X_{\mathscr{P}}=K\implies int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}=\emptyset$ and $f,g\in Z(C_{\mathscr{P}}(X))^{*}$. So, by
Theorem 3.25, $dt(AG(C_{\mathscr{P}}(X)))=2$. ∎

The converse of Theorem 3.28 is not true in general. For example let
$X=\mathbb{R}$ and $\mathscr{P}$ be the ideal of all closed sets in $X$. Then
$C_{\mathscr{P}}(X)=C(X)$ and $X_{\mathscr{P}}=\mathbb{R}$ which is not a
disconnected subspace of $X$. However, $dt(AG(C_{\mathscr{P}}(X)))=2$, because
if we consider $f(x)=\begin{cases}0&x\leq 0\\\ x&x\geq 0\end{cases}$ and
$g(x)=\begin{cases}x&x\leq 0\\\ 0&x\geq 0\end{cases}$, then $f,g\in
Z(C(X))^{*}$ with $f.g=0$ and $int_{X}Z(f)\cap int_{X}Z(g)=\emptyset$. So, by
Theorem 3.25, $dt(AG(C(\mathbb{R})))=2$.

In the following theorem, the dominating numbers of the annihilator graphs of
$C(X)$, $C_{K}(X)$ and $C_{F}(X)$ are recorded.

######  Theorem 3.29.

For a topological space $X$,

  1. (1)

$dt(AG(C(X)))=2$, if $X$ is disconnected.
If $X$ is connected then $dt(AG(C(X)))=2$ if and only if $X$ is not a middle
$P$-space.

  2. (2)

$dt(AG(C_{K}(X)))=2$, if $X$ is a locally compact and disconnected space.

  3. (3)

$dt(AG(C_{F}(X)))=\begin{cases}2,&\textnormal{ if }K_{X}\textnormal{ is
finite}\cr\aleph_{0},&\textnormal{ otherwise}\end{cases}$.

###### Proof.

  1. (1)

If $X$ is disconnected then $\Gamma(C(X))$ is not hypertriangulated and
therefore, by Theorem 3.26, $dt(AG(C(X)))=2$.
If $X$ is connected then $X$ is not a middle $P$-space if and only if
$\Gamma(C(X))$ is not hypertriangulated [5]. So, the result follows from
Theorem 3.26.

  2. (2)

If $X$ is locally compact then $X=X_{\mathscr{P}}$ and so, by disconnectedness
of $X$, we get by Theorem 3.28 that $dt(AG(C_{K}(X)))=2$.

  3. (3)

If $K_{X}$, the set of isolated points of $X$ is finite then it is a
disconnected subset of $X$ containing proper clopen subsets. Therefore, by
Theorem 3.28, $dt(AG(C_{F}(X)))=2$.
Let $dt(AG(C_{F}(X)))$ be infinite. Choosing a countably infinite subset $K$
of $K_{X}$, we set $W=\\{1_{x}\in Z(C_{F}(X))^{*}:x\in K\\}$. We claim that
$W$ is a dominating set in $AG(C_{F}(X))$. If possible, let there be $f\in
Z(C_{F}(X))^{*}\setminus W$ such that $f$ is not adjacent to any $1_{x}$,
$x\in K$. i.e., For each $x\in K$, either $Z(f)\subseteq Z(1_{x})$ or
$Z(1_{x})\subseteq Z(f)$. In any case, $K\subseteq X\setminus Z(f)$ which
leads to a contradiction, as $X\setminus Z(f)$ is finite. Hence,
$dt(AG(C_{F}(X)))\leq|K|=\aleph_{0}$.
It is now enough to show that no finite set of vertices is a dominating set in
$AG(C_{F}(X))$. Let $W^{\prime}=\\{f_{i}\in Z(C_{F}(X))^{*}:i=1,2,...,n\\}$.
Choose $x\in K_{X}\setminus\left(\bigcup\limits_{i=1}^{n}X\setminus
Z(f_{i})\right)$. Such a choice of $x$ is possible, as
$\bigcup\limits_{i=1}^{n}X\setminus Z(f_{i})$ is a finite subset of $K_{X}$
and $K_{X}$ is infinite. Considering $f=1_{A}$ where
$A=\\{x\\}\cup\left(\bigcup\limits_{i=1}^{n}X\setminus Z(f_{i})\right)$ we
observe that $f\in Z(C_{F}(X))^{*}$, however, $f$ is not adjacent to any
$f_{i}\in W^{\prime}$.
Hence, $dt(AG(C_{F}(X)))=\aleph_{0}$.

∎

The observation that the vertices $f$ and $2f$ are non-adjacent in
$AG(C_{\mathscr{P}}(X))$ leads to the result : $dt(AG(C_{\mathscr{P}}(X)))=2$
if and only if $dt_{t}(AG(C_{\mathscr{P}}(X)))=2$. In the next theorem, we
generalize this result if $dt(AG(C_{\mathscr{P}}(X)))<c$(= the cardinality of
the continuum).

######  Theorem 3.30.

If $dt(AG(C_{\mathscr{P}}(X)))<c$ then
$dt(AG(C_{\mathscr{P}}(X)))=dt_{t}(AG(C_{\mathscr{P}}(X)))$.

###### Proof.

Let $H$ be a dominating set in $AG(C_{\mathscr{P}}(X))$ such that
$|H|=dt(AG(C_{\mathscr{P}}(X)))$ $<c$. If $H$ is not a total dominating set of
$AG(C_{\mathscr{P}}(X))$, there exists $f\in H$ such that $f$ is not adjacent
to any vertices of $H$. So, $r.f$ is not adjacent to any vertices of $H$, for
each $r\in\mathbb{R}\setminus\\{0\\}$. As a result
$\\{r.f:r\in\mathbb{R}\setminus\\{0\\}\\}\subset H$ which gives $|H|\geq c$.
Therefore, $H$ is a total dominating set of $AG(C_{\mathscr{P}}(X))$ and
hence, $dt_{t}(AG(C_{\mathscr{P}}(X)))\leq|H|=dt(AG(C_{\mathscr{P}}(X)))$. The
reverse implication follows from the definition. ∎

##  4\. An induced subgraph of $C_{\mathscr{P}}(X)$

In this section, we define an equivalence relation on the set of vertices of
$AG(C_{\mathscr{P}}(X))$ and fetch exactly one element from each equivalence
class to constitute a set of vertices for a subgraph of it. The adjacency
relation on the new set of vertices is defined accordingly so that it becomes
an induced subgraph $G(C_{\mathscr{P}}(X))$ of $AG(C_{\mathscr{P}}(X))$. As
proposed in the introduction, we observe that the graph properties of
$G(C_{\mathscr{P}}(X))$ are mostly analogous to those of
$AG(C_{\mathscr{P}}(X))$.
A relation $\sim$ on $C_{\mathscr{P}}(X)$ is defined as follows: For all
$f,g\in C_{\mathscr{P}}(X)$,

$f\sim g\textnormal{ if and only if }int_{X}Z(f)\cap
X_{\mathscr{P}}=int_{X}Z(g)\cap X_{\mathscr{P}}.$

$\sim$ being an equivalence relation on $C_{\mathscr{P}}(X)$, partitions the
set in disjoint equivalence classes. In what follows, $[f]$ denotes the
equivalence class containing $f$. The following observations are useful in the
subsequent development of this paper:

  1. (a)

No two members of $[f]$ are adjacent in $AG(C_{\mathscr{P}}(X))$, for each
$f\in C_{\mathscr{P}}(X)$.

  2. (b)

$[0]=\\{0\\}$. For if $g\in[0]$, $int_{X}Z(g)\cap
X_{\mathscr{P}}=X_{\mathscr{P}}$ and $X\setminus Z(g)\subset X_{\mathscr{P}}$
which imply $g=0$.

  3. (c)

For any $f$, $g\in Z(C_{\mathscr{P}}(X))^{*}$, $f$ is adjacent to $g\implies$
$f$ is adjacent to $h$, for all $h\in[g]$.

Choosing exactly one element from each equivalence class, a subset $S$ of
$C_{\mathscr{P}}(X)$ is constructed. The induced subgraph obtained by taking
the set of vertices as $V(C_{\mathscr{P}}(X))=\\{f\in
S\setminus\\{0\\}:int_{X}Z(f)\cap X_{\mathscr{P}}\neq\emptyset\\}\subset
Z(C_{\mathscr{P}}(X))^{*}$, is denoted by $G(C_{\mathscr{P}}(X))$. By
definition of induced subgraph, for any $f,g\in V$, $f$ is adjacent to $g$ in
$G(C_{\mathscr{P}}(X))$ if and only if $f$ is adjacent to $g$ in
$AG(C_{\mathscr{P}}(X))$, i.e., $int_{X}Z(g)\cap X_{\mathscr{P}}\not\subset
int_{X}Z(f)\cap X_{\mathscr{P}}$ and $int_{X}Z(f)\cap
X_{\mathscr{P}}\not\subset int_{X}Z(g)\cap X_{\mathscr{P}}$. Further,

  1. (a)

$f=g$ in $V(C_{\mathscr{P}}(X))$ if and only if $int_{X}Z(f)\cap
X_{\mathscr{P}}=int_{X}Z(g)\cap X_{\mathscr{P}}$.

  2. (b)

For each $f\in Z(C_{\mathscr{P}}(X))^{*}$ there exists a unique $f^{\prime}\in
V(C_{\mathscr{P}}(X))$ such that $[f^{\prime}]=[f]$.

A useful result based on these observations is recorded in the following lemma
for future references.

######  Lemma 4.1.

For each edge $f-g$ in $AG(C_{\mathscr{P}}(X))$ there exists a unique edge
$f^{\prime}-g^{\prime}$ in $G(C_{\mathscr{P}}(X))$ such that
$[f^{\prime}]=[f]$ and $[g^{\prime}]=[g]$.

Before discussing various graph parameters of the induced subgraph of
$AG(C_{\mathscr{P}}(X))$, we cite a few illustrative examples of
$G(C_{\mathscr{P}}(X))$ and visualize them when $|X_{\mathscr{P}}|$ is finite.

######  Example 4.2.

Let $|X_{\mathscr{P}}|=2$. If $X_{\mathscr{P}}=\\{x,y\\}$ then any vertex $f$
in $AG(C_{\mathscr{P}}(X))$ ($=\Gamma(C_{\mathscr{P}}(X))$) takes the form
$r.1_{x}$ or $r.1_{y}$, for some $r\in\mathbb{R}\setminus\\{0\\}$, where
$1_{A}$ denotes the characteristic function on $A\subseteq X$. Therefore,
$Z(C_{\mathscr{P}}(X))^{*}=\\{r.1_{x}:r\in\mathbb{R}\setminus\\{0\\}\\}\cup\\{r.1_{y}:r\in\mathbb{R}\setminus\\{0\\}\\}$
and hence $AG(C_{\mathscr{P}}(X))$ is a complete bipartite graph as shown
below:

$\\{r.1_{x}:r\in\mathbb{R}\setminus\\{0\\}\\}$$\\{r.1_{y}:r\in\mathbb{R}\setminus\\{0\\}\\}$

Choosing $V=\\{1_{x},1_{y}\\}$, we obtain $G(C_{\mathscr{P}}(X))$ as the
complete graph $K_{2}$ with two vertices, as the following:

$1_{x}$$1_{y}$$K_{2}$

######  Example 4.3.

Let $|X_{\mathscr{P}}|=3$. Then $f\in Z(C_{\mathscr{P}}(X))^{*}$ if and only
if either $|X\setminus Z(f)|=1$ or $2$. If $X_{\mathscr{P}}=\\{x,y,z\\}$ then
$X\setminus Z(f)$ is one of $\\{x\\}$, $\\{y\\}$, $\\{z\\}$, $\\{x,y\\}$,
$\\{x,z\\}$, $\\{y,z\\}$. Therefore, $AG(C_{\mathscr{P}}(X))$ is a $6$-partite
graph of the following form :

$\forall f,X\setminus Z(f)=\\{x\\}$$\forall f,X\setminus Z(f)=\\{y\\}$$\forall
f,X\setminus Z(f)=\\{z\\}$$\forall f,X\setminus Z(f)=\\{x,z\\}$$\forall
f,X\setminus Z(f)=\\{x,y\\}$$\forall f,X\setminus Z(f)=\\{y,z\\}$

Considering
$V=\\{1_{x},1_{y},1_{z},1_{\\{x,y\\}},1_{\\{x,z\\}},1_{\\{y,z\\}}\\}$, the
induced subgraph $G(C_{\mathscr{P}}(X))$ takes the form as shown below:

$1_{x}$$1_{y}$$1_{z}$$1_{\\{y,z\\}}$$1_{\\{x,z\\}}$$1_{\\{x,y\\}}$

Scrutinising the pattern of getting $G(C_{\mathscr{P}}(X))$ from
$AG(C_{\mathscr{P}}(X))$ in the above examples, one expects that if
$|X_{\mathscr{P}}|=\ finite$ then $V=\\{1_{A}\ :\ \emptyset\neq A\subsetneqq
X_{\mathscr{P}}\\}$ serves as the set of vertices of $G(C_{\mathscr{P}}(X))$
and two vertices $1_{A},1_{B}$ are adjacent in $G(C_{\mathscr{P}}(X))$ if and
only if $A\not\subset B$ and $B\not\subset A$. This can be deduced as a
special case of a more general example (where $X_{\mathscr{P}}$ need not be
finite) as cited below.

######  Example 4.4.

Consider $\mathcal{P}$ as the ideal of all finite subsets of $X$. Then
$C_{\mathscr{P}}(X)=C_{F}(X)$ and $X_{\mathscr{P}}=K_{X}$, the set of all
isolated points of $X$. If $f\in C_{F}(X)$ then $X\setminus Z(f)$ is finite
and therefore, $[f]=[1_{X\setminus Z(f)}]$. So, considering
$V=\\{1_{A}:\emptyset\neq A\subsetneqq K_{X}\text{ and }A\text{ is
finite}\\}$, the induced subgraph $G(C_{F}(X))$ of $AG(C_{F}(X))$ is achieved.
Clearly, two vertices $1_{A},1_{B}$ are adjacent in $G(C_{F}(X))$ if and only
if $A\not\subset B$ and $B\not\subset A$.

It is relevant to inquire when the induced subgraph is finite. The following
theorem asserts that the finiteness of $G(C_{\mathscr{P}}(X))$ completely
depends upon the cardinality of $X_{\mathscr{P}}$.

######  Theorem 4.5.

$G(C_{\mathscr{P}}(X))$ is a finite graph if and only if $X_{\mathscr{P}}$ is
a finite set.
Moreover, if $X_{\mathscr{P}}$ is finite, then
$|G(C_{\mathscr{P}}(X))|=2^{|X_{\mathscr{P}}|}-2$ and
$deg(f)=2^{|X_{\mathscr{P}}|}-2^{|X\setminus Z(f)|}-2^{|X_{\mathscr{P}}\cap
Z(f)|}+1$ for each $f\in V(C_{\mathscr{P}}(X))$.

###### Proof.

Let $x,y\in X_{\mathscr{P}}$. Then by Lemma 2.2, there exist $f_{x},f_{y}\in
C_{\mathscr{P}}(X)$ such that $x\in X\setminus Z(f_{x})\subset X\setminus
int_{X}Z(f_{x})\subset X\setminus\\{y\\}$ and $y\in X\setminus Z(f_{y})\subset
X\setminus int_{X}Z(f_{y})\subset X\setminus\\{x\\}$. Then $y\in
int_{X}Z(f_{x})\cap X_{\mathscr{P}}$, $x\in X\setminus Z(f_{x})$ and $x\in
int_{X}Z(f_{y})\cap X_{\mathscr{P}}$, $y\in X\setminus Z(f_{y})$. So,
$f_{x},f_{y}\in Z(C_{\mathscr{P}}(X))^{*}$ such that $[f_{x}]\neq[f_{y}]$. Let
$f^{\prime}_{x},f^{\prime}_{y}\in V(C_{\mathscr{P}}(X))$ be such that
$[f^{\prime}_{x}]=[f_{x}]$ and $[f^{\prime}_{y}]=[f_{y}]$. Then $x\mapsto
f^{\prime}_{x}$ defines an injective function from $X_{\mathscr{P}}$ to
$G(C_{\mathscr{P}}(X))$. Hence,
$|G(C_{\mathscr{P}}(X))|\geq|X_{\mathscr{P}}|$.
If $X_{\mathscr{P}}$ is infinite then $G(C_{\mathscr{P}}(X))$ is an infinite
graph. In other words, if $G(C_{\mathscr{P}}(X))$ is a finite graph then
$X_{\mathscr{P}}$ is a finite set.
Conversely, if $X_{\mathscr{P}}$ is finite then by a special case of Example
4.4, we get $V=\\{1_{A}:\emptyset\neq A\subsetneqq X_{\mathscr{P}}\\}$ and so,
$|V|=|\mathcal{P}(X_{\mathscr{P}})|-2$, which is finite. Therefore,
$G(C_{\mathscr{P}}(X))$ is a finite graph.
Also,
$|G(C_{\mathscr{P}}(X))|=|\mathcal{P}(X_{\mathscr{P}})|-2=2^{|X_{\mathscr{P}}|}-2$.
Let $f\in V$. Then $f=1_{A}$ where $A=X\setminus Z(f)$.If $1_{B}\in V$ is not
adjacent to $1_{A}$ then either $B\subset A$ or $A\subset B$. Now $\\{1_{B}\in
V:B\subset A\\}=\mathcal{P}(A)\setminus\\{\emptyset\\}\implies|\\{1_{B}\in
V:B\subset A\\}|=|\mathcal{P}(A)\setminus\\{\emptyset\\}|=2^{|A|}-1$ and
$\\{1_{B}\in V:A\subsetneqq B\\}=\\{A\cup
C:C\in\mathcal{P}(X_{\mathscr{P}}\setminus
A)\setminus\\{\emptyset,X_{\mathscr{P}}\setminus A\\}\\}\implies|\\{1_{B}\in
V:A\subsetneqq B\\}|=|\mathcal{P}(X_{\mathscr{P}}\setminus
A)\setminus\\{\emptyset,X_{\mathscr{P}}\setminus
A\\}|=2^{|X_{\mathscr{P}}\setminus A|}-2$. Therefore, $\\{1_{B}\in
V:1_{B}\text{ is not adjacent to }1_{A}\text{ in
}G(C_{\mathscr{P}}(X))\\}=\\{1_{B}\in V:B\subsetneqq A\\}\sqcup\\{1_{B}\in
V:A\subsetneqq B\\}\implies|\\{1_{B}\in V:1_{B}\text{ is not adjacent to
}1_{A}\text{ in
}G(C_{\mathscr{P}}(X))\\}|=(2^{|A|}-1)+(2^{|\mathcal{P}(X_{\mathscr{P}}\setminus
A)|}-2)=2^{|A|}+2^{|X_{\mathscr{P}}\setminus A|}-3$. Hence,

$\displaystyle deg(f)=deg(1_{A})$ $\displaystyle=$ $\displaystyle|\\{1_{B}\in
V\ :\ 1_{B}\text{ is adjacent to }1_{A}\text{ in }G(C_{\mathscr{P}}(X))\\}|$
$\displaystyle=$ $\displaystyle|V|-|\\{1_{B}\in V\ :\ 1_{B}\text{ is not
adjacent to}1_{A}\text{ in }G(C_{\mathscr{P}}(X))\\}|$ $\displaystyle=$
$\displaystyle(2^{|X_{\mathscr{P}}|}-2)-(2^{|A|}+2^{|X_{\mathscr{P}}\setminus
A|}-3)$ $\displaystyle=$ $\displaystyle
2^{|X_{\mathscr{P}}|}-2^{|A|}-2^{|X_{\mathscr{P}}\setminus A|}+1.$

Since $A=X\setminus Z(f)$, $X_{\mathscr{P}}\setminus A=X_{\mathscr{P}}\cap
Z(f)$ and consequently, $deg(f)=2^{|X_{\mathscr{P}}|}-2^{|X\setminus
Z(f)|}-2^{|X_{\mathscr{P}}\cap Z(f)|}+1$, for each $f\in
V(C_{\mathscr{P}}(X))$. ∎

We now establish that $AG(C_{\mathscr{P}}(X))$ and $G(C_{\mathscr{P}}(X))$
have same values for certain important graph parameters, namely, diameter,
eccentricity and girth.

######  Theorem 4.6.

$d_{G}(f,g)=d_{AG}(f,g)$ for each $f,g\in V(C_{\mathscr{P}}(X))$.

###### Proof.

Let $f,g\in V(C_{\mathscr{P}}(X))$ be not adjacent in $V(C_{\mathscr{P}}(X))$.
Then $f,g$ are not adjacent in $AG(C_{\mathscr{P}}(X))$ and so,
$d_{AG(C_{\mathscr{P}}(X))}(f,g)=2$ by Corollary 3.7, i.e., there exists $h\in
Z(C_{\mathscr{P}}(X))^{*}$ which is adjacent to both $f,g$ in
$AG(C_{\mathscr{P}}(X))$ . From Theorem 3.3, it follows that $int_{X}Z(h)\cap
X_{\mathscr{P}}\neq int_{X}Z(f)\cap X_{\mathscr{P}}$ and $int_{X}Z(h)\cap
X_{\mathscr{P}}\neq int_{X}Z(g)\cap X_{\mathscr{P}}$, i.e., $[h]\neq[f]$ and
$[h]\neq[g]$ and so, we can consider $h\in V(C_{\mathscr{P}}(X))$. Thus $h$ is
adjacent to both $f,g$ in $G(C_{\mathscr{P}}(X))\implies d_{G}(f,g)=2$. Thus
$d_{G}(X)(f,g)=d_{AG}(f,g)$ for each $f,g\in V(C_{\mathscr{P}}(X))$. ∎

######  Corollary 4.7.

$diam(G(C_{\mathscr{P}}(X)))=diam(AG(C_{\mathscr{P}}(X)))=2$

######  Theorem 4.8.

$ecc_{G}(f)=2=ecc_{AG}(f)$ for each $f\in V(C_{\mathscr{P}}(X))$.

###### Proof.

Let $f\in V(C_{\mathscr{P}}(X))$. Then $X\setminus Z(f)\neq\emptyset$.
If $|X\setminus Z(f)|=1$, say $X\setminus Z(f)=\\{x\\}$ then $int_{X}Z(f)\cap
X_{\mathscr{P}}=X_{\mathscr{P}}\setminus\\{x\\}$.
Since $|X_{\mathscr{P}}|\geq 3$, there exist distinct $y,z\in
X_{\mathscr{P}}\setminus\\{x\\}$. By Lemma 2.3, we find $g\in
C_{\mathscr{P}}(X)$ such that $\\{x,y\\}\subset X\setminus Z(g)\subset
X\setminus int_{X}Z(g)\subset X\setminus\\{z\\}$. Then $z\in int_{X}Z(g)\cap
X_{\mathscr{P}}$ and $x,y\notin Z(g)\implies g\in Z(C_{\mathscr{P}}(X))^{*}$
and $int_{X}Z(g)\cap X_{\mathscr{P}}\subsetneqq int_{X}Z(f)\cap
X_{\mathscr{P}}$, i.e., $[f]\neq[g]$. By Lemma 4.1, there exists
$g^{\prime}\in V(C_{\mathscr{P}}(X))$ such that $[g^{\prime}]=[g]$. Then
$f,g^{\prime}\in V(C_{\mathscr{P}}(X))$ with $int_{X}Z(g^{\prime})\cap
X_{\mathscr{P}}\subsetneqq int_{X}Z(f)\cap X_{\mathscr{P}}\implies
f,g^{\prime}$ are not adjacent in $G(C_{\mathscr{P}}(X))$.
Let $|X\setminus Z(f)|\geq 2$ and choose distinct $x,y\in X\setminus Z(f)$.
By Lemma 2.2, there exists $g\in C_{\mathscr{P}}(X)$ such that $x\in
X\setminus Z(g)\subset X\setminus int_{X}Z(g)\subset
X\setminus[Z(f)\cup\\{y\\}]$. Then $g\neq 0$ and $Z(f)\cup\\{y\\}\subset
int_{X}Z(g)\implies g\in Z(C_{\mathscr{P}}(X))^{*}$ and $int_{X}Z(f)\cap
X_{\mathscr{P}}\subsetneqq int_{X}Z(g)\cap X_{\mathscr{P}}$, i.e.,
$[f]\neq[g]$. As done earlier, considering $g\in V(C_{\mathscr{P}}(X))$ we
find that $f,g$ are not adjacent in $G(C_{\mathscr{P}}(X))$.
Therefore, in any case, for each $f\in V(C_{\mathscr{P}}(X))$, there exists
$g\in V(C_{\mathscr{P}}(X))$ such that $g$ is not adjacent to $f$ in
$G(C_{\mathscr{P}}(X))$. So, $ecc_{G}(f)\geq 2$. As
$diam(G(C_{\mathscr{P}}(X)))=2$, it follows that $ecc_{G}(f)=2$
($=ecc_{AG}(f)$), for each $f\in V(C_{\mathscr{P}}(X))$. ∎

######  Corollary 4.9.

  1. (i)

$C(G(C_{\mathscr{P}}(X)))=V(C_{\mathscr{P}}(X))$

  2. (ii)

$rad(G(C_{\mathscr{P}}(X)))=2=rad(AG(C_{\mathscr{P}}(X)))$.

######  Theorem 4.10.

Let $f,g\in V(C_{\mathscr{P}}(X))$.

  1. (1)

If $f.g\neq 0$ then there always exists a vertex in $V(C_{\mathscr{P}}(X))$
which is adjacent to both $f,g$ in $G(C_{\mathscr{P}}(X))$.

  2. (2)

If $f.g=0$ then there exists a vertex in $V(C_{\mathscr{P}}(X))$ which is
adjacent to both $f,g$ in $G(C_{\mathscr{P}}(X))$ if and only if
$|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$ and $|int_{X}Z(g)\cap
X_{\mathscr{P}}|\geq 2$.

###### Proof.

  1. (1)

If $f.g\neq 0$ then by Theorem 3.6, there always exists a vertex $h$ in
$Z(C_{\mathscr{P}}(X))^{*}$ such that $h$ is adjacent to both $f$ and $g$ in
$AG(C_{\mathscr{P}}(X))$. So, $[h]\neq[f]$ and $[h]\neq[g]$. Let
$h^{\prime}\in V(C_{\mathscr{P}}(X))$ such that $[h^{\prime}]=[h]$. Clearly,
$h^{\prime}$ is adjacent to both $f,g$ in $G(C_{\mathscr{P}}(X))$.

  2. (2)

Let $f.g=0$. If $|int_{X}Z(f)\cap X_{\mathscr{P}}|\geq 2$ and
$|int_{X}Z(g)\cap X_{\mathscr{P}}|\geq 2$, then by Theorem 3.6(ii), there
exists a vertex $h$ in $Z(C_{\mathscr{P}}(X))^{*}$ which is adjacent to both
$f,g$ in $AG(C_{\mathscr{P}}(X))$. Let $h^{\prime}\in V(C_{\mathscr{P}}(X))$
such that $[h^{\prime}]=[h]$. Then, $h$ is adjacent to both $f,g$ in
$G(C_{\mathscr{P}}(X))$. Conversely let $h$ be adjacent to both $f,g$ in
$G(C_{\mathscr{P}}(X))$. Then $h$ is adjacent to both $f,g$ in
$AG(C_{\mathscr{P}}(X))$ also and hence by Theorem 3.6(ii), $|int_{X}Z(f)\cap
X_{\mathscr{P}}|\geq 2$ and $|int_{X}Z(g)\cap X_{\mathscr{P}}|\geq 2$.

∎

######  Theorem 4.11.

$G(C_{\mathscr{P}}(X))$ is hypertriangulated if and only if
$AG(C_{\mathscr{P}}(X))$ is hypertriangulated.

###### Proof.

Let $G(C_{\mathscr{P}}(X))$ be hypertriangulated and $f-g$ be an edge in
$AG(C_{\mathscr{P}}(X))$. Then by Lemma 4.1 there exists a unique edge
$f^{\prime}-g^{\prime}$ in $G(C_{\mathscr{P}}(X))$ such that
$[f^{\prime}]=[f]$ and $[g^{\prime}]=[g]$. Since $G(C_{\mathscr{P}}(X))$ is
hypertriangulated, there exists $h\in V(C_{\mathscr{P}}(X))$ such that $h$ is
adjacent to both $f^{\prime},g^{\prime}$ in $G(C_{\mathscr{P}}(X))$. So, $h$
is adjacent to both $f,g$ in $AG(C_{\mathscr{P}}(X))$ and hence $f-g-h-f$ is a
triangle in $AG(C_{\mathscr{P}}(X))$, proving that $AG(C_{\mathscr{P}}(X))$ is
hypertriangulated.
Conversely, let $f-g$ be an edge in $G(C_{\mathscr{P}}(X))$. Then $f-g$ is an
edge in $AG(C_{\mathscr{P}}(X))$. So, there exists $h\in
Z(C_{\mathscr{P}}(X))^{*}$ such that $f-g-h-f$ is a triangle in
$AG(C_{\mathscr{P}}(X))$. Consider $h^{\prime}\in V(C_{\mathscr{P}}(X))$ such
that $[h^{\prime}]=[h]$. Then $f-g-h^{\prime}-f$ is a triangle in
$G(C_{\mathscr{P}}(X))$. Therefore, $G(C_{\mathscr{P}}(X))$ is
hypertriangulated. ∎

######  Theorem 4.12.

$G(C_{\mathscr{P}}(X))$ is always triangulated.

###### Proof.

Let $f\in V(C_{\mathscr{P}}(X))$. Since $AG(C_{\mathscr{P}}(X))$ is always
triangulated and $f\in Z(C_{\mathscr{P}}(X))^{*}$, there exists a triangle in
$AG(C_{\mathscr{P}}(X))$ containing $f$ as a vertex, say $f-g-h-f$ for some
$g,h\in Z(C_{\mathscr{P}}(X))^{*}$. Let $g^{\prime},h^{\prime}\in
V(C_{\mathscr{P}}(X))$ be such that $[g^{\prime}]=[g]$ and $[h^{\prime}]=[h]$.
Hence, $f-g^{\prime}-h^{\prime}-f$ is a triangle in $G(C_{\mathscr{P}}(X))$,
proving that $G(C_{\mathscr{P}}(X))$ is triangulated. ∎

######  Corollary 4.13.

$gr(G(C_{\mathscr{P}}(X)))=gr(AG(C_{\mathscr{P}}(X)))$.

The following theorem follows immediately from Theorem 4.10 :

######  Theorem 4.14.

For $f,g\in V(C_{\mathscr{P}}(X))$, $f\perp_{G}g$ if and only if $f.g=0$ and
either $|int_{X}Z(f)\cap X_{\mathscr{P}}|=1$ or $|int_{X}Z(g)\cap
X_{\mathscr{P}}|=1$.

######  Theorem 4.15.

For $f,g\in Z(C_{\mathscr{P}}(X))^{*}$, there exist unique
$f^{\prime},g^{\prime}\in V(C_{\mathscr{P}}(X))$ such that $f\perp_{AG}g$ if
and only if $f^{\prime}\perp_{G}g^{\prime}$.

###### Proof.

For $f,g\in Z(C_{\mathscr{P}}(X))^{*}$, there exist unique
$f^{\prime},g^{\prime}\in V(C_{\mathscr{P}}(X))$ such that $[f^{\prime}]=[f]$
and $[g]=[g^{\prime}]$. $[f^{\prime}]=[f]\implies$ $int_{X}Z(f)\cap
X_{\mathscr{P}}=int_{X}Z(f^{\prime})\cap X_{\mathscr{P}}$ and similarly,
$[g]=[g^{\prime}]\implies$ $int_{X}Z(g)\cap
X_{\mathscr{P}}=int_{X}Z(g^{\prime})\cap X_{\mathscr{P}}$. Then $f.g=0\iff
X\setminus Z(f)\subset int_{X}Z(g)\cap
X_{\mathscr{P}}=int_{X}Z(g^{\prime})\cap X_{\mathscr{P}}\iff
f.g^{\prime}=0\iff X\setminus Z(g^{\prime})\subset int_{X}Z(f)\cap
X_{\mathscr{P}}=int_{X}Z(f^{\prime})\cap X_{\mathscr{P}}\iff
f^{\prime}.g^{\prime}=0$. Therefore,

$\begin{split}f\perp_{AG}g&\iff f.g=0\text{ and either }|int_{X}Z(f)\cap
X_{\mathscr{P}}|=1\\\ &\qquad\qquad\quad\qquad\text{ or }|int_{X}Z(g)\cap
X_{\mathscr{P}}|=1\\\ &\iff f^{\prime}.g^{\prime}=0\text{ and either
}|int_{X}Z(f^{\prime})\cap X_{\mathscr{P}}|=1\\\
&\qquad\qquad\quad\qquad\text{ or }|int_{X}Z(g^{\prime})\cap
X_{\mathscr{P}}|=1\\\ &\iff
f^{\prime}\perp_{G(C_{\mathscr{P}}(X))}g^{\prime},\text{ by Theorem
\ref{Th5.10}}\end{split}$

∎

######  Corollary 4.16.

For $f,g\in V(C_{\mathscr{P}}(X))$, $f\perp_{G}g$ if and only if
$f\perp_{AG}g$.

######  Theorem 4.17.

If $G(C_{\mathscr{P}}(X))$ is complemented then it is uniquely complemented.

###### Proof.

Let $f\perp_{G}g$ and $f\perp_{G}h$, for some $f,g,h\in
V(C_{\mathscr{P}}(X))$. By Theorem 4.14, $f.g=0=f.h$ and either
$|int_{X}Z(f)\cap X_{\mathscr{P}}|=1$ or $|int_{X}Z(g)\cap
X_{\mathscr{P}}|=1=|int_{X}Z(h)\cap X_{\mathscr{P}}|$.
If $|int_{X}Z(f)\cap X_{\mathscr{P}}|=1$, say $int_{X}Z(f)\cap
X_{\mathscr{P}}=\\{x\\}$, then $f.g=0\implies X\setminus Z(g)\subset
int_{X}Z(f)\cap X_{\mathscr{P}}=\\{x\\}$, so that $int_{X}Z(g)\cap
X_{\mathscr{P}}=X_{\mathscr{P}}\setminus\\{x\\}$. Similarly, $f.h=0\implies
int_{X}Z(h)\cap X_{\mathscr{P}}=X_{\mathscr{P}}\setminus\\{x\\}$. So,
$int_{X}Z(g)\cap X_{\mathscr{P}}=int_{X}Z(h)\cap X_{\mathscr{P}}$.
If $|int_{X}Z(g)\cap X_{\mathscr{P}}|=1=|int_{X}Z(h)\cap X_{\mathscr{P}}|$
then assuming $int_{X}Z(g)\cap X_{\mathscr{P}}=\\{x\\}$ and $int_{X}Z(h)\cap
X_{\mathscr{P}}=\\{y\\}$ we get $X\setminus Z(f)=\\{x\\}$ and $X\setminus
Z(f)=\\{y\\}$ respectively from $f.g=0$ and $f.h=0$. Therefore, $x=y$ and so,
$int_{X}Z(g)\cap X_{\mathscr{P}}=int_{X}Z(h)\cap X_{\mathscr{P}}$.
i.e., $int_{X}Z(g)\cap X_{\mathscr{P}}=int_{X}Z(h)\cap X_{\mathscr{P}}$ holds
in any case. So, any $k\in V(C_{\mathscr{P}}(X))$ is adjacent to $g$ in
$G(C_{\mathscr{P}}(X))$ if and only if $k$ is adjacent to $h$ in
$G(C_{\mathscr{P}}(X))$. i.e., $g,h$ are adjacent to the same set of vertices
in $G(C_{\mathscr{P}}(X))$ which proves that $G(C_{\mathscr{P}}(X))$ is
uniquely complemented. ∎

######  Theorem 4.18.

$G(C_{\mathscr{P}}(X))$ is uniquely complemented if and only if
$AG(C_{\mathscr{P}}(X))$ is uniquely complemented.

###### Proof.

In view of Theorem 3.21 and 4.17, it is enough to show that
$G(C_{\mathscr{P}}(X))$ is complemented if and only if
$AG(C_{\mathscr{P}}(X))$ is complemented.
Let $G(C_{\mathscr{P}}(X))$ be complemented and $f\in
Z(C_{\mathscr{P}}(X))^{*}$. Let $f^{\prime}\in V(C_{\mathscr{P}}(X))$ be such
that $[f^{\prime}]=[f]$. Since $G(C_{\mathscr{P}}(X))$ is complemented, there
exists $g\in V(C_{\mathscr{P}}(X))$ such that $f\perp_{G}g$ and hence by
Corollary 4.16, $f\perp_{AG}g$, proving that $AG(C_{\mathscr{P}}(X))$ is
complemented.
Conversely, let $f\in V(C_{\mathscr{P}}(X))$. Since $AG(C_{\mathscr{P}}(X))$
is complemented, there exists $g\in Z(C_{\mathscr{P}}(X))^{*}$ such that
$f\perp_{AG}g$. By Theorem 4.17, $f\perp_{AG}g^{\prime}$ where $g^{\prime}\in
V(C_{\mathscr{P}}(X))$ and $[g^{\prime}]=[g]$. Therefore,
$G(C_{\mathscr{P}}(X))$ is complemented. ∎

######  Lemma 4.19.

For each dominating set $W$ of $AG(C_{\mathscr{P}}(X))$,
$W^{\prime}=\\{f^{\prime}\in V(C_{\mathscr{P}}(X)):[f^{\prime}]=[f]\text{ for
some }f\in W\\}$ is a dominating set of $G(C_{\mathscr{P}}(X))$ with
$|W^{\prime}|\leq|W|$.

###### Proof.

Clearly, $W\cap V(C_{\mathscr{P}}(X))\subset W^{\prime}$. Let $f\in
V(C_{\mathscr{P}}(X))\setminus W^{\prime}$. Then it follows that $f\in
Z(C_{\mathscr{P}}(X))^{*}\setminus W$. $W$ being a dominating set in
$AG(C_{\mathscr{P}}(X))$, there exists $g\in W$ such that $g$ is adjacent to
$f$ in $AG(C_{\mathscr{P}}(X))$. Choose $g^{\prime}\in V(C_{\mathscr{P}}(X))$
such that $[g^{\prime}]=[g]$. Then $g^{\prime}\in W^{\prime}$ and
$f,g^{\prime}$ are adjacent in $G(C_{\mathscr{P}}(X))$, proving $W^{\prime}$
to be a dominating set of $G(C_{\mathscr{P}}(X))$. That $|W^{\prime}|\leq|W|$
is evident. ∎

######  Theorem 4.20.

$dt(G(C_{\mathscr{P}}(X)))\leq dt(AG(C_{\mathscr{P}}(X)))$.

###### Proof.

Let $W$ be a dominating set in $AG(C_{\mathscr{P}}(X))$. From Lemma 4.19,
there exists a dominating set $W^{\prime}$ in $G(C_{\mathscr{P}}(X))$ such
that $|W^{\prime}|\leq|W|$. So,
$dt(G(C_{\mathscr{P}}(X)))\leq|W^{\prime}|\leq|W|$. $W$ being an arbitrary
dominating set in $AG(C_{\mathscr{P}}(X))$ it then follows that
$dt(G(C_{\mathscr{P}}(X)))\leq dt(AG(C_{\mathscr{P}}(X)))$. ∎

######  Theorem 4.21.

$dt(G(C_{\mathscr{P}}(X)))=2$ if and only if there exist $f,g\in
V(C_{\mathscr{P}}(X))$ with $f.g=0$ and $int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}=\emptyset$.

###### Proof.

Let there be $f,g\in V(C_{\mathscr{P}}(X))$ obeying the given conditions. As
$f,g\in Z(C_{\mathscr{P}}(X))^{*}$, by Theorem 3.25,
$dt(AG(C_{\mathscr{P}}(X)))=2$. Also by Theorem 4.8, $ecc(f)=2$ for each $f\in
V(C_{\mathscr{P}}(X))$ and therefore, $dt(G(C_{\mathscr{P}}(X)))>1$. Clubbing
all these facts with Theorem 4.20, we get $dt(G(C_{\mathscr{P}}(X)))=2$.
Conversely, let $dt(G(C_{\mathscr{P}}(X)))=2$ and $\\{f,g\\}\subset
V(C_{\mathscr{P}}(X))$ constitute a dominating set of $G(C_{\mathscr{P}}(X))$.
We claim that $f,g$ are adjacent in $G(C_{\mathscr{P}}(X))$. If not then
either $int_{X}Z(f)\cap X_{\mathscr{P}}\subsetneqq int_{X}Z(g)\cap
X_{\mathscr{P}}$ or $int_{X}Z(g)\cap X_{\mathscr{P}}\subsetneqq
int_{X}Z(f)\cap X_{\mathscr{P}}$. Without loss of generality, we assume that
$int_{X}Z(f)\cap X_{\mathscr{P}}\subsetneqq int_{X}Z(g)\cap X_{\mathscr{P}}$.
For $g\in V(C_{\mathscr{P}}(X))$, $X\setminus Z(g)\neq\emptyset$.
If $|X\setminus Z(g)|\geq 2$, say $x,y\in X\setminus Z(g)$, then by Lemma 2.2,
there exists $h\in C_{\mathscr{P}}(X)$ such that $x\in X\setminus Z(h)\subset
X\setminus int_{X}Z(h)\subset X_{\mathscr{P}}\setminus[Z(g)\cup\\{y\\}]$. Then
$Z(g)\cup\\{y\\}\subset int_{X}Z(h)$ and $h\neq 0$. Since
$y\in[int_{X}Z(h)\cap X_{\mathscr{P}}]\setminus Z(g)$, it follows that
$int_{X}Z(g)\cap X_{\mathscr{P}}\subsetneqq int_{X}Z(h)\cap X_{\mathscr{P}}$.
Let $h^{\prime}\in V(C_{\mathscr{P}}(X))$ with $[h^{\prime}]=[h]$. Therefore,
$int_{X}Z(f)\cap X_{\mathscr{P}}\subsetneqq int_{X}Z(g)\cap
X_{\mathscr{P}}\subsetneqq int_{X}Z(h^{\prime})\cap X_{\mathscr{P}}\implies
h^{\prime}\in V(C_{\mathscr{P}})\setminus\\{f,g\\}$ and $h$ is not adjacent to
$f,g$; this contradicts our hypothesis that$\\{f,g\\}$ is a dominating set of
$G(C_{\mathscr{P}}(X))$.
If $|X\setminus Z(g)|=1$, say $X\setminus Z(g)=\\{x\\}\implies int_{X}Z(g)\cap
X_{\mathscr{P}}=X_{\mathscr{P}}\setminus\\{x\\}$ then assume if possible that
$X\setminus Z(f)=\\{y\\}$. Then $int_{X}Z(f)\cap
X_{\mathscr{P}}=X_{\mathscr{P}}\setminus\\{y\\}$. Therefore $int_{X}Z(f)\cap
X_{\mathscr{P}}\subsetneqq int_{X}Z(g)\cap X_{\mathscr{P}}\implies
X_{\mathscr{P}}\setminus\\{y\\}\subsetneqq X_{\mathscr{P}}\setminus\\{x\\}$
which is not possible. Therefore, $|X\setminus Z(f)|\geq 2$ and choose any
$y,z\in X\setminus Z(f)$. By Lemma 2.3, there exists $h\in C_{\mathscr{P}}(X)$
such that $\\{x,y\\}\subset X\setminus Z(h)\subset X\setminus
int_{X}Z(h)\subset X\setminus[Z(f)\cup\\{z\\}]$. Therefore,
$Z(f)\cup\\{z\\}\subset int_{X}Z(h)\subset X\setminus\\{x,y\\}$ and $h\neq 0$.
Let $h^{\prime}\in V(C_{\mathscr{P}}(X))$ be such that $[h^{\prime}]=[h]$.
Then $int_{X}Z(f)\cap X_{\mathscr{P}}\subsetneqq int_{X}Z(h^{\prime})\cap
X_{\mathscr{P}}\subsetneqq
X_{\mathscr{P}}\setminus\\{x\\}=int_{X}Z(h^{\prime})\cap X_{\mathscr{P}}$. So,
$h^{\prime}\in V(C_{\mathscr{P}})\setminus\\{f,g\\}$ is not adjacent to $f,g$,
leading to a contradiction.
Hence, $f,g$ are adjacent in $G(C_{\mathscr{P}}(X))$ so that

(4.1) $\begin{split}int_{X}Z(f)\cap X_{\mathscr{P}}\not\subset int_{X}Z(g)\cap
X_{\mathscr{P}}\text{ and }\\\ int_{X}Z(g)\cap X_{\mathscr{P}}\not\subset
int_{X}Z(f)\cap X_{\mathscr{P}}\end{split}$

Our next claim is $f.g=0$. If possible let $f.g\neq 0$ and choose $x\in
X\setminus Z(f)\cap X\setminus Z(g)$. By Lemma 2.2, there exists $h\in
C_{\mathscr{P}}(X)$ such that $x\in X\setminus Z(h)\subset X\setminus
int_{X}Z(h)\subset X\setminus Z(f)\cap X\setminus Z(g)\implies h\neq 0$ and
$Z(f)\cup Z(g)\subset int_{X}Z(h)$. Clearly, $h^{\prime}\in
V(C_{\mathscr{P}}(X))$ with $[h^{\prime}]=[h]$, is not adjacent to both $f,g$
in $G(C_{\mathscr{P}}(X))$. By Equation 4.1 and the fact that $Z(f)\cup
Z(g)\subset int_{X}Z(h^{\prime})$, we get $h^{\prime}\notin\\{f,g\\}$, which
contradicts that $\\{f,g\\}$ is a dominating set in $G(C_{\mathscr{P}}(X))$.
Finally, to show that $int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}=\emptyset$. If possible let $int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}\neq\emptyset$ and choose $x\in int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}$. Let $h=f^{2}+g^{2}\in C_{\mathscr{P}}(X)$. Then
$int_{X}Z(h)=int_{X}Z(f)\cap int_{X}Z(g)\implies int_{X}Z(h)\cap
X_{\mathscr{P}}\neq\emptyset\implies h\in Z(C_{\mathscr{P}}(X))^{*}$. If
$h^{\prime}\in V(C_{\mathscr{P}}(X))$ is such that $[h^{\prime}]=[h]$ then
$int_{X}Z(h^{\prime})=int_{X}Z(f)\cap int_{X}Z(g)$. So, by Equation 4.1,
$h^{\prime}\neq f$, $h^{\prime}\neq g$. Since $int_{X}Z(f)\cap int_{X}Z(g)\cap
X_{\mathscr{P}}=int_{X}Z(h^{\prime})\cap X_{\mathscr{P}}$, $h^{\prime}$ is not
adjacent to both $f,g$ in $G(C_{\mathscr{P}}(X))$ and we arrive at a
contradiction. ∎

An immediate consequence is recorded in the following corollary.

######  Corollary 4.22.

$dt(G(C_{\mathscr{P}}(X)))=2$ if and only if
$dt_{t}(G(C_{\mathscr{P}}(X)))=2$.

######  Lemma 4.23.

Every total dominating set in $G(C_{\mathscr{P}}(X))$ is also a total
dominating set in $AG(C_{\mathscr{P}}(X))$.

###### Proof.

Let $W$ be a total dominating set in $G(C_{\mathscr{P}}(X))$ and $f\in
Z(C_{\mathscr{P}}(X))^{*}$. Then there exists $f^{\prime}\in
V(C_{\mathscr{P}}(X))$ such that $[f^{\prime}]=[f]$. Since $W$ is a total
dominating set in $G(C_{\mathscr{P}}(X))$, there exists $g\in W$ such that
$f^{\prime},g$ are adjacent in $G(C_{\mathscr{P}}(X))$ and hence $f$ and $g$
are adjacent in $AG(C_{\mathscr{P}}(X))$. Therefore, $W$ is a total dominating
set in $AG(C_{\mathscr{P}}(X))$. ∎

######  Theorem 4.24.

$dt_{t}(G(C_{\mathscr{P}}(X)))=dt_{t}(AG(C_{\mathscr{P}}(X)))$.

###### Proof.

Let $H$ be a total dominating set in $AG(C_{\mathscr{P}}(X))$. Consider
$H^{\prime}=\\{f^{\prime}\in V(C_{\mathscr{P}}(X)):[f^{\prime}]=[f]\text{ for
some }f\in H\\}$. Let $f\in V(C_{\mathscr{P}}(X))$. Then $f\in
Z(C_{\mathscr{P}}(X))^{*}$ and hence there exists $g\in H$ such that $f,g$ are
adjacent in $AG(C_{\mathscr{P}}(X))$. Therefore $f$ and $g^{\prime}$ are
adjacent in $G(C_{\mathscr{P}}(X))$ where $g^{\prime}\in H^{\prime}$. It
follows that $H^{\prime}$ is a total dominating set in
$AG(C_{\mathscr{P}}(X))$. Therefore,
$dt_{t}(G(C_{\mathscr{P}}(X)))\leq|H^{\prime}|\leq|H|$, proving
$dt_{t}(G(C_{\mathscr{P}}(X)))\leq dt_{t}(AG(C_{\mathscr{P}}(X)))$.
Conversely, let $W$ be a total dominating set in $G(C_{\mathscr{P}}(X))$. Then
by Lemma 4.23, $W$ is a total dominating set in $AG(C_{\mathscr{P}}(X))$ and
hence $dt_{t}(AG(C_{\mathscr{P}}(X)))\leq|W|$. Hence,
$dt_{t}(AG(C_{\mathscr{P}}(X)))\leq dt_{t}(G(C_{\mathscr{P}}(X)))$ ∎

The next corollary follows from Theorem 4.20 and Theorem 4.24

######  Corollary 4.25.

$dt(G(C_{\mathscr{P}}(X)))=dt_{t}(G(C_{\mathscr{P}}(X)))$ if and only if
$dt(G(C_{\mathscr{P}}(X)))=dt(AG(C_{\mathscr{P}}(X)))$
$=dt_{t}(AG(C_{\mathscr{P}}(X)))=dt_{t}(G(C_{\mathscr{P}}(X)))$.

We cite an example where
$dt(G(C_{\mathscr{P}}(X)))=dt_{t}(G(C_{\mathscr{P}}(X)))$.

######  Example 4.26.

Let $X$ be a completely regular Hausdorff space with atleast two isolated
points and $\mathscr{P}$ be the ideal of all finite sets of $X$. Then
$C_{\mathscr{P}}(X)=C_{F}(X)$ and $X_{\mathscr{P}}$ is the set of all isolated
points of $X$. So, $|X_{\mathscr{P}}|\geq 2$. As seen before,
$V(C_{\mathscr{P}}(X))=\\{1_{A}:\emptyset\neq A\subsetneqq
X_{\mathscr{P}}\\}$.
If $X_{\mathscr{P}}$ is finite then $X_{\mathscr{P}}$ is clopen and so,
proceeding as in Theorem 4.21,
$W^{\prime}=\\{1_{x},1_{X_{\mathscr{P}}\setminus\\{x\\}}\\}$, for some $x\in
X_{\mathscr{P}}$ is a total dominating set of $G(C_{\mathscr{P}}(X))$. So
$dt_{t}(G(C_{\mathscr{P}}(X)))=2$ and hence
$dt(G(C_{\mathscr{P}}(X)))=dt_{t}(G(C_{\mathscr{P}}(X)))$, by Corollary 4.22.
If $X_{\mathscr{P}}$ is infinite then set $W^{\prime}=\\{1_{x}:x\in K\\}$, for
some countable infinite subset $K$ of $X_{\mathscr{P}}$. Proceeding as in
Theorem 3.29, it is easy to see that, $W^{\prime}$ is a total dominating set
in $G(C_{\mathscr{P}}(X))$. Clearly,
$dt_{t}(G(C_{\mathscr{P}}(X)))\leq|W^{\prime}|=|K|=\aleph_{0}$. If $W$ is any
dominating set of $G(C_{\mathscr{P}}(X))$ then it is enough to show that, $W$
is never finite. If possible let $W=\\{f_{i}\in
V(C_{\mathscr{P}}(X)):i=1,2,\ldots,n\\}$. Choosing any $x\in
X_{\mathscr{P}}\setminus(\bigcup\limits_{i=1}^{n}(X\setminus Z(f_{i})))$ (such
a point exists as $X_{\mathscr{P}}$ is infinite), we construct a finite subset
$A=\\{x\\}\cup\bigcup\limits_{i=1}^{n}(X\setminus Z(f_{i}))$ of
$X_{\mathscr{P}}$. Then $X\setminus Z(f_{i})\subsetneqq A=X\setminus Z(1_{A})$
for each $i=1,2,\ldots,n$. So $1_{A}\in V(C_{\mathscr{P}}(X))\setminus W$ and
$1_{A}$ is not adjacent to any $f_{i}$, $i=1,2,...,n$, contradicting the
assumption that $W$ is a dominating set. Thus
$dt(G(C_{\mathscr{P}}(X)))\geq\aleph_{0}\geq dt_{t}(G(C_{\mathscr{P}}(X)))$.
Hence $dt(G(C_{\mathscr{P}}(X)))=dt_{t}(G(C_{\mathscr{P}}(X)))$.

So, from Theorem 3.29 and the last example, we get

$dt(G(C_{F}(X)))=\begin{cases}2&\text{when }X\text{ has finitely many isolated
points }\\\ \aleph_{0}&\text{otherwise}\end{cases}$

######  Theorem 4.27.

$cl(G(C_{\mathscr{P}}(X)))=cl(AG(C_{\mathscr{P}}(X)))$.

###### Proof.

As $G(C_{\mathscr{P}}(X))$ is a subgraph $AG(C_{\mathscr{P}}(X))$,
$cl(G(C_{\mathscr{P}}(X)))\leq cl(AG(C_{\mathscr{P}}(X)))$.
Let $H$ be any complete subgraph of $AG(C_{\mathscr{P}}(X))$. Consider
$H^{\prime}=\\{f^{\prime}\in V(C_{\mathscr{P}}(X)):[f^{\prime}]=[f]\text{ for
some }f\in H\\}$. Completeness of $H^{\prime}$ is immediate from the
completeness of $H$. Also, for any $f\in AG(C_{\mathscr{P}}(X))$, there exists
a unique $f^{\prime}\in G(C_{\mathscr{P}}(X))$ with $[f]=[f^{\prime}]$ defines
a function $\psi:H\rightarrow H^{\prime}$ given by $f\mapsto f^{\prime}$
whenever $[f]=[f^{\prime}]$. As $H$ is complete, it follows that $\psi$ is
one-one. The construction of $H^{\prime}$ itself indicates that $\psi$ is
onto.
So, $|H|=|H^{\prime}|\leq cl(G(C_{\mathscr{P}}(X)))$ implies that
$cl(AG(C_{\mathscr{P}}(X)))\leq cl(G(C_{\mathscr{P}}(X)))$. ∎

######  Theorem 4.28.

$\chi(G(C_{\mathscr{P}}(X)))=\chi(AG(C_{\mathscr{P}}(X)))$.

###### Proof.

Certainly, $\chi(G(C_{\mathscr{P}}(X)))\leq\chi(AG(C_{\mathscr{P}}(X)))$.
Let $\chi(G(C_{\mathscr{P}}(X)))=\Lambda$. If $f\in Z(C_{\mathscr{P}}(X))^{*}$
is such that $f\in V(C_{\mathscr{P}}(X))$ then it is already colored. So, we
assume $f\notin V(C_{\mathscr{P}}(X))$. For each $f$ there exists a unique
$f^{\prime}\in V(C_{\mathscr{P}}(X))$ such that $[f^{\prime}]=[f]$. As
$f,f^{\prime}$ are non-adjacent in $AG(C_{\mathscr{P}}(X))$, we color $f$ by
the color of $f^{\prime}$. We claim that the coloring is consistent. If $f,g$
in $AG(C_{\mathscr{P}}(X))$ have the same color, say $f^{\prime}$, then by the
rule of coloring, $[f]=[f^{\prime}]=[g]$ which implies that $f$ and $g$ are
non-adjacent, proving consistency, as desired. So,
$\chi(AG(C_{\mathscr{P}}(X)))\leq\chi(G(C_{\mathscr{P}}(X)))$. ∎

##  5\. An algorithm for coloring $AG(C_{F}(X))$

In this section, we show that the chromatic number and the clique number of
$AG(C_{F}(X))$ are same and this value is completely determined by the
cardinality of the set of all isolated points of $X$. Moreover, if $|K_{X}|$
is finite then $G(C_{F}(X))$ is a finite graph and we are successful in
devising an algorithm to color the vertices of $G(C_{F}(X))$. As a
consequence, obtain the chromatic number of $AG(C_{F}(X))$, even though it is
an infinite graph.

######  Theorem 5.1.

If $|K_{X}|$ is infinite then $|V(C_{F}(X))|=|K_{X}|$.

###### Proof.

From Theorem 4.5, taking $\mathscr{P}$ as the ideal of all finite subsets of
$X$, it follows that $|V(C_{F}(X))|$ is infinite. Clearly,
$|K_{X}|=|\\{1_{x}:x\in K_{X}\\}|\leq|V(C_{F}(X))|$. $\\{X\setminus Z(f):f\in
V(C_{F}(X))\\}$ being a family of nonempty finite subsets of $K_{X}$,
$\\{X\setminus Z(f):f\in
V(C_{F}(X))\\}\subset\bigcup\limits_{n=1}^{\infty}(K_{X})^{n}=|K_{X}|$ (as
$|K_{X}|$ is infinite) so that $|\\{X\setminus Z(f):f\in
V(C_{F}(X))\\}|\leq|K_{X}|$. For distinct $f,g\in V(C_{F}(X))$, $X\setminus
Z(f)\neq X\setminus Z(g)\implies|V(C_{F}(X))|=|\\{X\setminus Z(f):f\in
V(C_{F}(X))\\}|$, i.e., $|V(C_{F}(X))|\leq|K_{X}|$. ∎

######  Theorem 5.2.

If $|K_{X}|$ is infinite then $cl(G(C_{F}(X)))=|K_{X}|=\chi(G(C_{F}(X)))$.

###### Proof.

Since $cl(G(C_{F}(X)))\leq|V(C_{F}(X))|$ it follows from Theorem 5.1 that
$cl(G(C_{F}(X)))\leq|K_{X}|$. Observing that the subgraph $H$ of $G(C_{F}(X))$
whose set of vertices is $\\{1_{x}:x\in K_{X}\\}$, is complete, we get
$cl(G(C_{F}(X)))\geq|H|=|K_{X}|$. Hence, $cl(G(C_{F}(X)))=|K_{X}|$.
Also, $cl(G(C_{F}(X)))\leq\chi(G(C_{F}(X)))\leq V(C_{F}(X))$. By Theorem 5.1
the result follows. ∎

######  Theorem 5.3.

If $|K_{X}|$ is finite then
$cl(G(C_{F}(X)))\geq\binom{|K_{X}|}{[\frac{|K_{X}|}{2}]}$, where
$\binom{n}{r}=\frac{n!}{(n-r)!r!}$ and $[m]$ denotes the positive integer less
than or equal to $m$.

###### Proof.

Let $|K_{X}|=\\{x_{1},x_{2},\ldots,x_{n}\\}$. We define an order ‘$<$’ on
$K_{X}$, so that $x_{1}<x_{2}<\ldots<x_{n}$. For each $k\leq n$, define
$\mathscr{A}_{k}=\\{A\subset K_{X}:|A|=k\\}$.
We make a convention that for any $A\in\mathscr{A}_{k}$, whenever we write
$A=\\{a_{1},a_{2},\ldots,a_{k}\\}$, it is understood that
$a_{1}<a_{2}<\ldots<a_{k}$.
Define an order $<$ on $\mathscr{A}_{k}$ by $A<B$ if and only if either
($a_{1}<b_{1}$) or (there exists some $i\in\\{2,\ldots,n\\}$ such that
$a_{i}<b_{i}$ and $a_{j}=b_{j}$ for all $j<i$). It is clear that for each
$k=1,2,\ldots,n$, $(\mathscr{A}_{k},<)$ is a totally ordered set with the
least element $A^{1}=\\{x_{1},x_{2},\ldots,x_{k}\\}$. For each $k$, $H_{k}$,
with set of vertices $V(H_{k})=\\{1_{A}:A\in\mathscr{A}_{k}\\}$ is a complete
subgraph of $G(C_{F}(X))$. It is further observed that
$|\mathscr{A}_{k}|=\binom{|K_{X}|}{k}$. Therefore,
$|H_{k}|=\binom{|K_{X}|}{k}$, for each $k=1,2,\ldots,n$.
Among the complete subgraphs $H_{k}$ of $G(C_{F}(X))$,
$H_{[\frac{|K_{X}|}{2}]}$ is maximal with number of vertices =
$\binom{|K_{X}|}{[\frac{|K_{X}|}{2}]}$. Hence,
$cl(G(C_{F}(X)))\geq\binom{|K_{X}|}{[\frac{|K_{X}|}{2}]}$. ∎

######  Theorem 5.4.

If $|K_{X}|$ is finite then
$\chi(G(C_{F}(X)))=\binom{|K_{X}|}{[\frac{|K_{X}|}{2}]}$.

###### Proof.

Let $|K_{x}|=n$. Proceeding as in Theorem 5.3, we obtain the complete
subgraphs $H_{k}$ of $G(C_{F}(X))$, among which $H_{[\frac{n}{2}]}$ is maximal
with number of vertices $=\binom{n}{[\frac{n}{2}]}$. So, we start with
coloring $H_{[\frac{n}{2}]}$ by $\binom{n}{[\frac{n}{2}]}$-many distinct
colors.
Before writing the coloring scheme, we make a few conventions and introduce
some notations that will be used in the algorithm for coloring the vertices of
$G(C_{F}(X))$.

  1. (1)

$A^{1}$ denotes the first element of each $\mathscr{A}_{k}$. So,
$A^{1}=\\{x_{1},x_{2},\ldots,x_{k}\\}$. (From the context we understand which
$\mathscr{A}_{k}$ is under consideration).

  2. (2)

$A^{l}$ denotes the last element of each $\mathscr{A}_{k}$. So,
$A^{l}=\\{x_{n-k+1},x_{n-k+2},\ldots,x_{n}\\}$.

  3. (3)

$curr$ is a variable, used to denote some member of $\mathscr{A}_{k}$, for any
$k$.

  4. (4)

$next$ is a variable, used to denote the immediate successor of $curr$.

  5. (5)

$prev$ is a variable, used to denote a subset of $\mathscr{A}_{k}$, for any
$k$.

  6. (6)

$1_{A}\sim 1_{B}$ reads as “Color $1_{A}$ using the color of $1_{B}$”.

  7. (7)

$x\leftarrow y$ means “Assign the value $y$ to the variable $x$”.

  8. (8)

Since the elements of $A$ are ordered, for any $A\in\mathscr{A}_{k}$, $A_{-i}$
designates the set obtained from $A$ by deleting its $i$-th element. Clearly,
$A\in\mathscr{A}_{k}$ implies that $A_{-i}\in\mathscr{A}_{k-1}$.

  9. (9)

For any $A\in\mathscr{A}_{k-1}$, define $A_{+j}=A\cup\\{x_{j}\\}$. It is to
note that $A_{+j}\in\mathscr{A}_{k}$ if and only if $x_{j}\notin A$.

It is quite clear that if $A^{1}$ is the first element of $\mathscr{A}_{k}$
then ${A^{1}}_{-k}$ is the first element of $\mathscr{A}_{k-1}$.
Step 1: Starting from $k=[\frac{n}{2}]$, we color $H_{k-1}$ by using the
colors of $H_{k}$, adopting the following algorithm:

  1. 1.

$k\leftarrow[\frac{n}{2}]-1$

  2. 2.

$curr\leftarrow A^{1}$

  3. 3.

$1_{curr}\sim 1_{curr_{+k}}$

  4. 4.

$prev\leftarrow\\{curr_{+k}\\}$

  5. 5.

$curr\leftarrow next$

  6. 6.

IF $j\in\\{1,2,\ldots,n\\}$ is the smallest such that
$curr_{+j}\in\mathscr{A}_{k}$ and $curr_{+j}\neq B$, for all $B\in prev$ THEN
$1_{curr}\sim 1_{curr_{+j}}$ and $prev\leftarrow prev\cup\\{curr_{+j}\\}$

  7. 7.

IF $curr=A^{l}$ THEN GOTO Step 8 ELSE GOTO Step 5

  8. 8.

$k\leftarrow k-1$

  9. 9.

CONTINUE Step 2 to Step 8 TILL $k=1$

  10. 10.

STOP when $k=1$.

We first check that the scheme described above exhausts coloring of all the
vertices of $H_{k-1}$ with the colors used for $H_{k}$. The first element of
$H_{k-1}$ is colored by the first element of the $H_{k}$. It is enough to show
that the last element of $H_{k-1}$ gets a color.
We first show that $1_{A}\sim 1_{B}$ where
$A=\\{x_{1},x_{2},\ldots,x_{k-3},x_{n-1},x_{n}\\}$ and $B=$
$\\{x_{1},x_{2},\ldots,$ $x_{k-3},x_{k-2},x_{n-1},x_{n}\\}$.
Using the scheme of coloring described above, it is not hard to see that the
vertex corresponding to $\\{x_{1},x_{2},\ldots,x_{k-3},x_{k-2},x_{n-1}\\}$ is
colored by the vertex corresponding to
$\\{x_{1},x_{2},\ldots,x_{k-2},x_{k-1},x_{n-1}\\}$ and
$\\{x_{1},x_{2},\ldots,$ $x_{k-3},x_{k-2},x_{n}\\}$ by
$\\{x_{1},x_{2},\ldots,$ $x_{k-2},x_{k-1},x_{n}\\}$. Also,
$B_{-i}=\\{x_{1},x_{2},\ldots,x_{i-1},x_{i+1},\ldots,x_{n-1},x_{n}\\}>A$ as
$x_{i+1}>x_{i}$. Hence, $1_{B_{-i}}$ ($i=1,2,\ldots,k-2$) are not colored by
$1_{B}$ and therefore, $1_{B}$ is available for $1_{A}$.
Proceeding similarly, we get that $1_{A}\sim 1_{B}$ where
$A=\\{x_{1},x_{n-k+3},\ldots,x_{n-1},x_{n}\\}$ and
$B=\\{x_{1},x_{2},x_{n-k+3},\ldots,x_{n-1},x_{n}\\}$.
So, for $A^{l}=\\{x_{n-k+2},x_{n-k+3},\ldots,x_{n-1},x_{n}\\}$,
$C=\\{x_{1},x_{n-k+2},\ldots,x_{n-1},x_{n}\\}$ is available, so that
$1_{A^{l}}\sim 1_{C}$ (because, none of $1_{C_{-i}}$ ($i\neq 1$) are colored
by $C$, as seen above).
Step 2: Starting from $k=[\frac{n}{2}]$, we color $H_{k+1}$ by using the
colors of $H_{k}$, adopting the following algorithm:

  1. 1.

$k\leftarrow[\frac{n}{2}]+1$

  2. 2.

$curr\leftarrow A^{1}$

  3. 3.

$1_{curr}\sim 1_{curr_{-k}}$

  4. 4.

$prev\leftarrow\\{curr_{-k}\\}$

  5. 5.

$curr\leftarrow next$

  6. 6.

IF $j$ is the largest such that $curr_{-j}\neq B$, for all $B\in prev$ THEN
$1_{curr}\sim 1_{curr_{-j}}$ and $prev\leftarrow prev\cup\\{curr_{-j}\\}$

  7. 7.

IF $curr=A^{l}$ THEN GOTO Step 8 ELSE GOTO Step 5

  8. 8.

$k\leftarrow k+1$

  9. 9.

CONTINUE Step 2 TO Step 8 TILL $k=n$

  10. 10.

STOP when $k=n$.

We claim that the colors (i.e., vertices of $H_{k-1}$) do not exhaust before
all the vertices of $H_{k}$ are colored. Let $A\in\mathscr{A}_{k}$ and for
each $j$, $A_{-j}\in prev$. Let $B^{j}\in\mathscr{A}_{k}$ be such that
$1_{B^{j}}\sim 1_{A_{-j}}$, for $j=1,2,\ldots k$. Also, each $B^{j}$ has been
colored before $A$ implies that $B^{j}<A$, for all $j=1,2,\ldots k$. If
$A=\\{a_{1},a_{2},\ldots,a_{k}\\}$ then
$B^{j}=\\{a_{1},\ldots,a_{j-1},a_{j+1},\ldots,a_{k}\\}\cup\\{b_{j}\\}$ where
$b_{j}\in K_{X}$. Clearly, $b_{j}\neq a_{i}$, for each $i\neq j$ and
$b_{j}\neq a_{j}$ as $A\neq B^{j}$. Hence, there are atleast $2k$ distinct
elements (i.e., $a_{1},\ldots,a_{k},b_{1},\ldots,b_{k}$) exist in $K_{X}$.
But, $[\frac{n}{2}]+1$ being the minimum value of $k$ in the above algorithm,
$2k\geq 2([\frac{n}{2}]+1)>n=|K_{X}|$, which is impossible. Therefore, using a
color of $H_{k-1}$ it is possible to color all the vertices of $H_{k}$, for
each $k=[\frac{n}{2}]+1$, …, n.
Proceeding as in the previous case, we observe that if $1_{A}\in H_{i}$ and
$1_{B}\in H_{j}$ ($i<j$) have the same color then (as $i<j$) $A\subsetneqq B$.
Finally, we check whether the coloring is consistent. For that matter, we
choose any $1_{A}$ and $1_{B}$ from $V(C_{F}(X))$ such that they have the same
color. Then $1_{A}\in H_{i}$ and $1_{B}\in H_{j}$ for some $i\neq j$. In other
words, $A\in\mathscr{A}_{i}$ and $B\in\mathscr{A}_{j}$. Without loss of
generality, we assume that $i<[\frac{n}{2}]$ and $j>[\frac{n}{2}]$. Then
proceeding as before, we get some $1_{C},1_{D}\in H_{[\frac{n}{2}]}$ such that
$1_{A}$ and $1_{C}$ have the same color and $1_{B}$ and $1_{D}$ have the same
color. So, $A\subsetneqq C$ and $D\subsetneqq B$. As $1_{A}$ and $1_{B}$ have
the same color, it follows that $1_{C}$ and $1_{D}$ also have the same color,
which is possible when and only when $C=D$. Therefore, $A\subsetneqq
C=D\subsetneqq B$; i.e., $1_{A}$ and $1_{B}$ are non-adjacent. ∎

######  Corollary 5.5.

If $|K_{X}|$ is finite,
$cl(G(C_{F}(X)))=\chi(G(C_{F}(X)))=\binom{|K_{X}|}{[\frac{|K_{X}|}{2}]}$.

###### Proof.

Since $cl(G(C_{F}(X)))\leq\chi(G(C_{F}(X)))$, the result follows from Theorem
5.4 and Theorem 5.3. ∎

The following example illustrates the coloring scheme of Theorem 5.4.

######  Example 5.6.

Let $X$ be a Tychonoff space with $K_{X}=\\{x_{1},x_{2},x_{3},x_{4},x_{5}\\}$.
Then $|K_{X}|=5$.

$H_{3}:$$H_{2}:$$H_{1}:$$H_{4}:$$1_{x_{1}}$$1_{x_{2}}$$1_{x_{3}}$$1_{x_{4}}$$1_{x_{5}}$$1_{\\{x_{1},x_{2}\\}}$$1_{\\{x_{1},x_{3}\\}}$$1_{\\{x_{1},x_{4}\\}}$$1_{\\{x_{1},x_{5}\\}}$$1_{\\{x_{2},x_{3}\\}}$$1_{\\{x_{2},x_{4}\\}}$$1_{\\{x_{2},x_{5}\\}}$$1_{\\{x_{3},x_{4}\\}}$$1_{\\{x_{3},x_{5}\\}}$$1_{\\{x_{4},x_{5}\\}}$$1_{\\{x_{1},x_{2},x_{3}\\}}$$1_{\\{x_{1},x_{2},x_{4}\\}}$$1_{\\{x_{1},x_{2},x_{5}\\}}$$1_{\\{x_{1},x_{3},x_{4}\\}}$$1_{\\{x_{1},x_{3},x_{5}\\}}$$1_{\\{x_{1},x_{4},x_{5}\\}}$$1_{\\{x_{2},x_{3},x_{4}\\}}$$1_{\\{x_{2},x_{3},x_{5}\\}}$$1_{\\{x_{2},x_{4},x_{5}\\}}$$1_{\\{x_{3},x_{4},x_{5}\\}}$$1_{\\{x_{1},x_{2},x_{3},x_{4}\\}}$$1_{\\{x_{1},x_{2},x_{3},x_{5}\\}}$$1_{\\{x_{1},x_{2},x_{4},x_{5}\\}}$$1_{\\{x_{1},x_{3},x_{4},x_{5}\\}}$$1_{\\{x_{2},x_{3},x_{4},x_{5}\\}}$

Combining Theorem 4.27, Theorem 4.28, Theorem 5.2 and Corollary 5.5, we get

######  Theorem 5.7.

$\chi(AG(C_{F}(X)))=cl(AG(C_{F}(X)))=\begin{cases}\binom{|K_{X}|}{[\frac{|K_{X}|}{2}]},&\textnormal{
if }|K_{X}|\textnormal{ is finite}\cr|K_{X}|,&\textnormal{
otherwise}\end{cases}$

##  6\. On the induced graph isomorphisms

In the earlier sections, we have seen that $G(C_{\mathscr{P}}(X))$ and
$AG(C_{\mathscr{P}}(X))$ behave in a similar fashion so far as
triangulatedness, hypertriangulatedness and complementedness are concerned and
both have the same diameter, girth, eccentricity, clique number and chromatic
number. In this section, we investigate whether a graph isomorphism
$G(C_{\mathscr{P}}(X))\rightarrow G(C_{\mathscr{Q}}(Y))$ completely determines
a graph isomorphism $AG(C_{\mathscr{P}}(X))\rightarrow AG(C_{\mathscr{Q}}(Y))$
and get a partial answer to this query. However, we show that for
$C_{\mathscr{P}}(X)$ with $|X_{\mathscr{P}}|=$ finite, the answer is complete.

In what follows, $X,Y$ stand for two completely regular Hausdorff spaces and
$\mathscr{P},\mathscr{Q}$ are ideals of closed sets in $X$ and $Y$
respectively. The equivalence relations on $C_{\mathscr{P}}(X)$ and
$C_{\mathscr{Q}}(Y)$ are denoted respectively by $\sim_{X}$, $\sim_{Y}$ and
their corresponding equivalence classes by $[f]_{X}$ and $[g]_{Y}$, ($f\in
C_{\mathscr{P}}(X)$, $g\in C_{\mathscr{Q}}(Y))$ respectively.

We first show that the restriction of an isomorphism
$\psi:AG(C_{\mathscr{P}}(X))\rightarrow AG(C_{\mathscr{Q}}(Y))$ on
$G(C_{\mathscr{P}}(X))$ takes $G(C_{\mathscr{P}}(X))$ isomorphically onto
$G(C_{\mathscr{Q}}(Y))$. Before that we need a lemma :

######  Lemma 6.1.

Let $\psi:AG(C_{\mathscr{P}}(X))\to AG(C_{\mathscr{Q}}(Y))$ be a graph
isomorphism and $f,g\in Z(C_{\mathscr{P}}(X))^{*}$. Then $int_{X}Z(f)\cap
X_{\mathscr{P}}=int_{X}Z(g)\cap X_{\mathscr{P}}$ if and only if
$int_{Y}Z(\psi(f))\cap Y_{\mathscr{Q}}=int_{Y}Z(\psi(g))\cap Y_{\mathscr{Q}}$.

###### Proof.

Let $int_{X}Z(f)\cap X_{\mathscr{P}}=int_{X}Z(g)\cap X_{\mathscr{P}}$, for
some $f,g\in AG(C_{\mathscr{P}}(X))$. Then $f,g$ are not adjacent in
$AG(C_{\mathscr{P}}(X))$ and so, $\psi(f),\psi(g)$ are not adjacent in
$AG(C_{\mathscr{Q}}(Y))$. Without loss of generality, assume that
$int_{Y}Z(\psi(f))\cap Y_{\mathscr{Q}}\subset int_{Y}Z(\psi(g))\cap
Y_{\mathscr{Q}}$.
For if $y\in[int_{Y}Z(\psi(g))\cap Y_{\mathscr{Q}}]\setminus Z(\psi(f))$, by
Lemma 2.2, there exists $h\in C_{\mathscr{Q}}(Y)$ such that $y\in Y\setminus
Z(h)\subset Y\setminus int_{Y}Z(h)\subset[int_{Y}Z(\psi(g))\cap
Y_{\mathscr{Q}}]\setminus Z(\psi(f))$. Clearly, $h\neq 0$, $Z(\psi(f))\subset
int_{Y}Z(h)$ and $\psi(g)h=0$. Now it is a routine check that
$\psi(g)h=0\implies\psi(f)$ and $h$ are adjacent and hence, contradicts the
assumption. So, $int_{Y}Z(\psi(g))\cap Y_{\mathscr{Q}}\subset Z(\psi(f))\cap
Y_{\mathscr{Q}}$.
The converse part is clear. ∎

######  Theorem 6.2.

If $\psi:AG(C_{\mathscr{P}}(X))\to AG(C_{\mathscr{Q}}(Y))$ is a graph
isomorphism then $\psi|_{G(C_{\mathscr{P}}(X))}$ maps $G(C_{\mathscr{P}}(X))$
isomorphically onto $G(C_{\mathscr{Q}}(Y))$.

###### Proof.

Let $\psi:AG(C_{\mathscr{P}}(X))\to AG(C_{\mathscr{Q}}(Y))$ be a graph
isomorphism and $f_{X}\in V(C_{\mathscr{P}}(X))$. Then $\psi(f_{X})\in
Z(C_{\mathscr{Q}}(Y))^{*}$. By Lemma 4.1, there exists a unique $f_{Y}\in
V(C_{\mathscr{Q}}(Y))$ such that $[f_{Y}]_{Y}=[\psi(f_{X})]_{Y}$; i.e.,
$int_{Y}Z(f_{Y})\cap Y_{\mathscr{Q}}=int_{Y}Z(\psi(f_{X}))\cap
Y_{\mathscr{Q}}$. Consider the map $\phi:V(C_{\mathscr{P}}(X))\to
V(C_{\mathscr{Q}}(Y))$ given by $\phi(f_{X})=f_{Y}$, where $f_{Y}\in
V(C_{\mathscr{Q}}(Y))$ such that $int_{Y}Z(f_{Y})\cap
Y_{\mathscr{Q}}=int_{Y}Z(\psi(f_{X}))\cap Y_{\mathscr{Q}}$.
If $f_{Y}\in V(C_{\mathscr{Q}}(Y))$ then $\psi^{-1}(f_{Y})\in
Z(C_{\mathscr{P}}(X))^{*}$. Choosing $f_{X}\in V(C_{\mathscr{P}}(X))$ in such
a way that $int_{X}Z(f_{X})\cap X_{\mathscr{P}}=int_{X}Z(\psi^{-1}(f_{Y}))\cap
X_{\mathscr{P}}$. By Lemma 6.1, it is easy to see that $\phi(f_{X})=f_{Y}$. As
a result, $\phi$ is a bijective map and it is easy to see that
$\phi:G(C_{\mathscr{P}}(X))\to G(C_{\mathscr{Q}}(Y))$ is a graph isomorphism.
∎

######  Theorem 6.3.

Let $\phi:G(C_{\mathscr{P}}(X))\to G(C_{\mathscr{Q}}(Y))$ be a graph
isomorphism satisfying the condition $|[f]_{X}|=|[\phi(f)]_{Y}|$, for all
$f\in V(C_{\mathscr{P}}(X))$ then $\phi$ can be extended to a graph
isomorphism between $AG(C_{\mathscr{P}}(X))$ and $AG(C_{\mathscr{Q}}(Y))$.

###### Proof.

For each $f\in Z(C_{\mathscr{P}}(X))^{*}$ there exists $f^{\prime}\in
V(C_{\mathscr{P}}(X))$ such that $[f^{\prime}]_{X}=[f]_{X}$. So,
$\phi(f^{\prime})\in V(C_{\mathscr{Q}}(Y))$ and hence by the hypothesis,
$|[f^{\prime}]_{X}|=|[\phi(f^{\prime})]_{Y}|$ which implies that
$|[f]_{X}|=|[\phi(f^{\prime})]_{Y}|$ and so, there exists a bijection between
$|[f]_{X}|$ and $|[\phi(f^{\prime})]_{Y}|$, say
$\psi_{f^{\prime}}:[f]_{X}\to[\phi(f^{\prime})]_{Y}$. Since an equivalence
relation on a set yields a partition, pasting these bijections we get a
bijective map $\psi$ from $Z(C_{\mathscr{P}}(X))^{*}$ to
$Z(C_{\mathscr{Q}}(Y))^{*}$.
Let $f,g\in Z(C_{\mathscr{P}}(X))^{*}$ be adjacent in
$AG(C_{\mathscr{P}}(X))$. Then there are $f^{\prime},g^{\prime}\in
V(C_{\mathscr{P}}(X))$ such that $[f^{\prime}]_{X}=[f]_{X}$ and
$[g^{\prime}]_{X}=[g]_{X}$. $f^{\prime},g^{\prime}$ are adjacent in
$G(C_{\mathscr{P}}(X))\implies\phi(f^{\prime}),\phi(g^{\prime})$ are adjacent
in $G(C_{\mathscr{Q}}(Y))$. Now
$\psi(f)=\psi_{f^{\prime}}(f)\in[\phi(f^{\prime})]_{Y}\implies
int_{Y}Z(\psi(f))\cap Y_{\mathscr{Q}}=int_{Y}Z(\phi(f^{\prime}))\cap
Y_{\mathscr{Q}}$ and
$\psi(g)=\psi_{g^{\prime}}(g)\in[\phi(g^{\prime})]_{Y}\implies
int_{Y}Z(\psi(g))\cap Y_{\mathscr{Q}}=int_{Y}Z(\phi(g^{\prime}))\cap
Y_{\mathscr{Q}}$. It follows that $\psi(f),\psi(g)$ are adjacent in
$AG(C_{\mathscr{Q}}(Y))$. Similarly, if $f,g$ are adjacent in
$AG(C_{\mathscr{Q}}(Y))$ then $\psi^{-1}(f),\psi^{-1}(g)$ are adjacent in
$AG(C_{\mathscr{P}}(X))$ proving that $\psi:AG(C_{\mathscr{P}}(X))\to
AG(C_{\mathscr{Q}}(Y))$ is a graph isomorphism. ∎

The following example guarantees the existence of a ring $C_{\mathscr{P}}(X)$
obeying the condition of Theorem 6.3

######  Example 6.4.

Let $X,Y$ be two completely regular Hausdorff spaces and
$\mathscr{P},\mathscr{Q}$ be two ideal of closed sets on $X,Y$ respectively
such that $|X_{\mathscr{P}}|\geq 2,|Y_{\mathscr{Q}}|\geq 2$ are finite. (For
example, taking $X,Y$ as completely regular Hausdorff spaces with atleast two
isolated points and
$\mathscr{P}=\\{\\{x\\},\\{y\\},\\{z\\},\\{x,y\\},\\{y,z\\},\\{x,z\\},\\{x,y,z\\}\\}$
and $\mathscr{Q}=\\{\\{a\\},\\{b\\},\\{a,b\\}\\}$, where $x,y,z\in X$ are
isolated points in $X$ and $a,b\in Y$ are isolated points in $Y$, we obtain
$X_{\mathscr{P}}=\\{x,y,z\\}$ and $Y_{\mathscr{Q}}=\\{a,b\\}$).
If $f\in C_{\mathscr{P}}(X)$ then $X\setminus Z(f)\subset
X_{\mathscr{P}}\implies X\setminus Z(f)$ is finite and hence, $Z(f)$ is a
clopen subset of $X$. So, $X\setminus
Z(f)=X_{\mathscr{P}}\setminus[int_{X}Z(f)\cap X_{\mathscr{P}}]$. Let $f\in
Z(C_{\mathscr{P}}(X))^{*}$. Then $int_{X}Z(f)\cap
X_{\mathscr{P}}\neq\emptyset\iff X\setminus Z(f)$ is a non-empty proper subset
of $X_{\mathscr{P}}$ and therefore, $Z(C_{\mathscr{P}}(X))^{*}=\\{f\in
C(X):\emptyset\neq X\setminus Z(f)\subsetneqq X_{\mathscr{P}}\\}$.
If $f\in Z(C_{\mathscr{P}}(X))^{*}$ and $X\setminus
Z(f)=\\{x_{1},x_{2},...x_{n}\\}$ where $1\leq n<|X_{\mathscr{P}}|$, then $f$
can be written as $f=\sum\limits_{i=1}^{n}f(x_{i})1_{x_{i}}$. Now,
$g\in[f]_{X}\iff X\setminus Z(g)=X\setminus
Z(f)\implies[f]_{X}=\\{\sum\limits_{i=1}^{n}r_{i}1_{x_{i}}:r_{i}\in\mathbb{R}\setminus\\{0\\}\text{
for each }i=1,2,...n\\}$ and therefore, $|[f]_{X}|$ comes out as
$2^{\aleph_{0}}$. Thus, for each $f\in Z(C_{\mathscr{P}}(X))^{*}$,
$|[f]_{X}|=2^{\aleph_{0}}$. Similarly, for each $f\in
Z(C_{\mathscr{Q}}(Y))^{*}$, $|[f]_{Y}|=2^{\aleph_{0}}$. If
$\phi:G(C_{\mathscr{P}}(X))\to G(C_{\mathscr{Q}}(Y))$ is a graph isomorphism
then for all $f\in V(\mathscr{P}(X))$,
$|[f]_{x}|=2^{\aleph_{0}}=|[\phi(f)]_{Y}|$. Hence, $G(C_{\mathscr{P}}(X))$ and
$G(C_{\mathscr{Q}}(Y))$ are graph isomorphic $\implies AG(C_{\mathscr{P}}(X))$
and $AG(C_{\mathscr{Q}}(Y))$ are graph isomorphic by Theorem 6.3.

######  Remark 6.5.

It is also to note that $|X_{\mathscr{P}}|=|Y_{\mathscr{Q}}|$ : $\phi$ is a
bijection between $V(C_{\mathscr{P}}(X))$ and
$V(C_{\mathscr{Q}}(Y))\implies|V(C_{\mathscr{P}}(X))|=|V(C_{\mathscr{Q}}(Y))|\implies
2^{|X_{\mathscr{P}}|}-2=2^{|Y_{\mathscr{Q}}|}-2$, by Theorem 4.5. So, we find
a bijective map between $X_{\mathscr{P}}$ and $Y_{\mathscr{Q}}$. The proof of
Theorem 6.18 in [4] indicates that such a bijective map leads to a ring
isomorphism between the rings $C_{\mathscr{P}}(X)$ and $C_{\mathscr{Q}}(Y)$.

Finally, in view of the result that the annihilator graphs of two isomorphic
commutative rings are graph isomorphic, we make the following observation:

######  Observation 6.6.

If $X$ and $Y$ are two completely regular Hausdorff spaces and $\mathscr{P}$,
$\mathscr{Q}$ are ideals of closed sets on $X,Y$ respectively such that
$X_{\mathscr{P}}$ and $Y_{\mathscr{Q}}$ are finite sets with
$|X_{\mathscr{P}}|\geq 2$ and $|Y_{\mathscr{Q}}|\geq 2$ then the following
statements are equivalent:

  1. (1)

$G(C_{\mathscr{P}}(X))$ and $G(C_{\mathscr{Q}}(Y))$ are graph isomorphic.

  2. (2)

$AG(C_{\mathscr{P}}(X))$ and $AG(C_{\mathscr{Q}}(Y))$ are graph isomorphic.

  3. (3)

$C_{\mathscr{P}}(X)$ and $C_{\mathscr{Q}}(Y)$ are ring isomorphic.

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