# Generalized cell structures

Ana G. Hernández-Dávila  ,  Benjamín A. Itzá-Ortiz  Centro de Investigación en
Matemáticas, Universidad Autónoma del Estado de Hidalgo, Pachuca, Hidalgo,
Mexico  itza<EMAIL_ADDRESS>and  Rocío Leonel-Gómez
Facultad de Ciencias, Universidad Nacional Autónoma de México, Mexico City,
Mexico<EMAIL_ADDRESS>
###### Abstract.

Cell structures were introduced by W. Debski and E. Tymchatyn as a way to
study some classes of topological spaces and their continuous functions by
means of discrete approximations. In this work we weaken the notion of cell
structure and prove that the resulting class of topological space admitting
such a generalized cell structure includes non-regular spaces.

## Introduction

Cell structures may be thought of as devices to represent some topological
spaces by means of discrete approximations, more precisely, to describe a
space $X$ as homeomorphic to a perfect image of an inverse limit of graphs.
They were introduced in [1] for complete metric spaces and extended in [2] to
topologically complete spaces. The initial step in identifying a cell
structure consists in considering an inverse sequence of graphs, each of which
have attached a reflexive and symmetric relation. The vertices of each graph
are the elements of a set $G$ and the edges are described by reflexive and
symmetric subsets of $G\times G$, or equivalently, the set of edges of $G$ is
an entourage of the diagonal of $G$. Such an inverse sequence is said to be a
cell structure when it admits a couple of properties: one that allows the
induced natural relation on its inverse limit to be an equivalence relation
(where two threads in the inverse limit are declared to be related if every
pair of their corresponding components are related), and a second property
that allows the quotient of the inverse limit by its natural relation to be a
perfect mapping. This resulting quotient space is the said to admit, up to
homeomorphism, a cell structure.

In this paper we explore the natural question on the class of topological
space obtained by admitting a weaker version of a cell structures, which we
will call generalized cell structures, or g-cell structure for short. We are
able to show that there are topological spaces admitting such a g-cell
structure but not a cell structure. Furthermore, an explicit example of a
nonregular space admitting a g-cell structure will be provided.

Representing the structure of spaces as approximations of simpler more
understandable structures, such as graphs, is a fruitful idea used not only in
topology [5], but in a variety of other areas such as spin networks [6],
operator algebras [3], networks [7], among others.

We divide this work in two sections. In Section 1 we introduce the notion of
generalized cell structure and prove the basic results needed in the rest of
the paper. The main results are given in Section 2.

The authors gratefully acknowledge that this paper have benefited from
stimulating conversations with Carlos Islas and Juan Manuel Burgos. The first
author received support for this work from CONACyT scholarship Num. 926215.

##  1\. Preliminaries

In this section we give the definition of spaces admitting a g-cell structure
together with some preliminary results which will be useful in the next
section. For any set $G$, we will denote the diagonal of $G$ as
$\Delta_{G}=\\{(x,x)\colon x\in G\\}$. When no confusion arises, we will write
$\Delta$ rather than $\Delta_{G}$. The set of natural numbers is as usual
${\mathbb{N}}=\\{1,2,3,\ldots\\}$.

######  Definition 1.1.

We say that an order pair $(G,r)$ is a cellular graph if $G$ is a nonempty
topological space and $r\subset G\times G$ is a reflexive and symmetric
relation on $G$. The vertices of the graph, the elements of $G$, will also be
known as cells of the graph, while the elements of $r$ are the edges of the
graph.

######  Definition 1.2.

If $(G,r)$ is a cellular graph and $u\in G$ is a cell, we define the
neighborhood of $u$ to be the set

$B(u,r)=\\{v\in G\colon(u,v)\in r\\}.$

Therefore, the set $B(u,r)$ is the set of vertices of $G$ that are adjacent to
$u$ in $G$. More generally, for $A\subset G$ we denote

$B(A,r)=\bigcup_{a\in A}B(a,r).$

Following the terminology of [4, Section 8.1, pg. 426], given a cellular graph
$(G,r)$, since the relation $r$ is reflexive, then the diagonal $\Delta$ of
$G$ is a subset of $r$. On the other hand, since $r$ is symmetric it follows
that $r$ is equal to its own inverse relation $-r$, where
$-r=\\{(x,y)\colon(y,x)\in r\\}$. Conversely, given a entourage of the
diagonal $r$ of $G$, that is to say, a relation on $G$ which contains $\Delta$
and satisfies $r=-r$, it follows that $r$ is a reflexive and symmetric
relation. Thus, we may characterize a cellular graph as a pair $(G,r)$ where
$G$ is a nonempty topological space and $r$ is an entourage of the diagonal.
Finally, the composition of $r+r=2r$ of the relation $r$ with itself is
defined as $2r=\\{(x,z)\colon\exists y\in G,\,\,(x,y),(y,z)\in r\\}$. In other
words a pair of cells $(x,z)$ belongs to $2r$ if there exits a path of length
2 in the cellular graph $(G,r)$ joining them. Notice that $2r$ is still a
reflexive and symmetric relation of $G$. Furthermore, it is always true that
$B(x,r)\subset B(x,2r)$ and the opposite inclusion follows if and only if $r$
is transitive.

######  Definition 1.3.

Suppose that $\\{(G_{n},r_{n})\\}$ is a sequence of cellular graphs and let
$\\{g_{n}^{n+1}\\}_{n\in{\mathbb{N}}}$ be a family of continuous functions,
called bonding maps, $g_{n}^{n+1}\colon G_{n+1}\to G_{n}$ such that

  * •

$g_{n}^{n}$ is the identity on $G_{n}$,

  * •

$g_{n}^{l}=g_{n}^{m}\circ g_{m}^{l}$ for $n<m<l$ and

  * •

the bonding maps send edges to edges, that is,
$(g_{n}^{n+1}(x),g_{n}^{n+1}(y))\in r_{n}$ whenever $(x,y)\in r_{n+1}$.

We will then say that $\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$
is an inverse sequence of cellular graphs.

The inverse limit of an inverse sequence of cellular graphs
$\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$ will be denoted by
$G_{\infty}$ or $\displaystyle\lim_{\longleftarrow}\\{G_{i},g_{i}^{i+1}\\}$.
It is a nonempty subspace of $\prod_{n\in{\mathbb{N}}}G_{n}$ with the product
topology. Elements in $G_{\infty}$ are called threads and the mappings
$g_{i}\colon G_{\infty}\to G_{i}$ denote the restriction of the standard
projection maps $p_{i}\colon\prod_{n\in{\mathbb{N}}}G_{n}\to G_{i}$. In other
words,

$G_{\infty}=\left\\{\bar{x}=(x_{n})_{n\in{\mathbb{N}}}\colon\forall j\geq
i,\,\,g_{i}^{j}(x_{j})=x_{i}\right\\}\subset\prod_{n\in{\mathbb{N}}}G_{n}.$

The topology in $G_{\infty}$ is actually characterized by a basis, as the
following result shows. It is actually a special case of [4, Propositon
2.5.5]. We include a proof for completeness.

######  Proposition 1.4.

Let $G_{\infty}$ be the inverse limit of an inverse sequence of cellular
graphs $\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$. Then the
collection of the sets of the form

$\left\\{g_{i}^{-1}\left(A_{i}\right)\colon i\in{\mathbb{N}}\text{ and
}A_{i}\subset G_{i}\text{ is an open in the base of }G_{i}\right\\}$

defines a basis for the the topology of $G_{\infty}$.

###### Proof.

Let $\bar{x}\in G_{\infty}$ and let $U\subset G_{\infty}$ be an open set such
that $\bar{x}\in U$. By definition of the subspace topology, there exists an
open set $V\subset\prod_{n\in{\mathbb{N}}}G_{i}$ such that $U=V\cap
G_{\infty}$. Thus there exists an open set
$V_{0}=p_{j_{1}}^{-1}(U_{1})\cap\cdots\cap p_{j_{n}}^{-1}(U_{n})$ in the base
of $\prod_{n\in{\mathbb{N}}}G_{i}$ satisfying $\bar{x}\in V_{0}\subset V$. Let
$j=\max\\{j_{i}\\}$. Since the bounding maps are continuous, for every
$i\in\\{1,...,n\\}$, $\bigl{(}g_{j_{i}}^{j}\bigr{)}^{-1}(U_{i})$ is an open
set in $G_{j}$. Hence, $\bigcap\bigl{(}g_{j_{i}}^{j}\bigr{)}^{-1}(U_{i})$ is
also an open set and
$\bar{x}\in\bigcap\bigl{(}g_{j_{i}}^{j}\bigr{)}^{-1}(U_{i})$. Let $V_{j}$ be
an open set in the base of $G_{j}$ such that $x_{j}\in
V_{j}\subset\bigcap\bigl{(}g_{j_{i}}^{j}\bigr{)}^{-1}(U_{i})$. Now, since
$g_{j}^{-1}\bigl{(}\bigl{(}g_{j_{i}}^{j}\bigr{)}^{-1}(U_{i})\bigr{)}=g_{j_{i}}^{-1}(U_{i})=G_{\infty}\cap
p_{j_{i}}^{-1}(U_{i})$, we get that $\bar{x}\in g_{j}^{-1}(V_{j})\subset
G_{\infty}\cap p_{j_{i}}^{-1}(U_{i})\subset G_{\infty}\cap V=U$. ∎

######  Definition 1.5.

Let $\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$ be an inverse
sequence of cellular graphs and let $G_{\infty}$ be its inverse limit. We
define the natural relation $r$ on $G_{\infty}$ by

$r=\\{(\bar{x},\bar{y})\in G_{\infty}\times G_{\infty}\colon\forall
n,\,\,(x_{n},y_{n})\in r_{n}\\}.$

In turns out that the inverse limits of an inverse sequence of cellular graphs
is actually a cellular graph with the natural relation $r$ on $G_{\infty}$
given in Definition 1.5, as established in the following proposition.

######  Proposition 1.6.

If $G_{\infty}$ is the inverse limit of an inverse sequence of cellular graphs
then it is a cellular graph with respect its natural relation given in
Definition 1.5.

###### Proof.

Let $\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$ be an inverse
sequence of cellular graphs and let $G_{\infty}$ be its inverse limit. By
definition, $G_{\infty}$ is a topological space. We only need to check that
$r$ is reflexive and symmetric. This follows immediately since each $r_{n}$ is
reflexive and symmetric. ∎

We are now ready to define a generalized cell structure. It consists of just
one condition which is a necessary condition implied by the first original
condition for cell structures originally defined in [1]. It enables the
natural relation induced on the inverse limit of an inverse sequence of
cellular graphs to be transitive and therefore an equivalent relation.

######  Definition 1.7.

We say that an inverse sequence of cellular graphs
$\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$ is a generalized cell
structure, or a g-cell structure for short, if its natural relation $r$ given
in Definition 1.5 is an equivalence relation.

######  Proposition 1.8.

Let $\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$ be an inverse
sequence of cellular graphs and let $G_{\infty}$ be its inverse limit. Each of
the following statement implies the next.

  1. (1)

For each $i\in{\mathbb{N}}$, the relation $r_{i}$ is an equivalence relation
on $G_{i}$.

  2. (2)

For each thread $\bar{x}\in G_{\infty}$ and for each $i\in{\mathbb{N}}$ there
exists $j\geq i$ such that
$g_{i}^{j}\left(B\left(x_{j},2r_{j}\right)\right)\subset
B\left(x_{i},r_{i}\right)$.

  3. (3)

The inverse sequence $\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$
is a g-cell structure.

###### Proof.

To prove (1) implies (2), let $\bar{x}\in G_{\infty}$ and let
$i\in{\mathbb{N}}$. Take $j=i$. Since $r_{j}$ is transitive, by hypothesis,
then $B(x_{j},2r_{j})=B(x_{j},r_{j})$. Since $g_{i}^{i}$ is the identity, this
proves that $(2)$ holds.

Now, we prove (2) implies (3). By Proposition 1.6, the natural relation is
reflexive and symmetric. To check that it is a g-cell structure, the
transitivity of the natural relation is the only property missing. The proof
follows from the same argument given in the proof of [1, Lemma 3.1], which we
repeat here for completeness sake. Suppose that
$(\bar{x},\bar{y}),(\bar{y},\bar{z})\in r$ and let $i\in{\mathbb{N}}$. Then
there is $j\geq i$ such that
$g_{i}^{j}\left(B\left(x_{j},2r_{j}\right)\right)\subset
B\left(x_{i},r_{i}\right)$. Since $(x_{j},y_{j}),(y_{j},z_{j})\in r_{j}$ then
$z_{j}\in B(x_{j},2r_{j})$. Therefore $z_{i}=g_{i}^{j}(z_{j})\in
B(x_{i},r_{i})$, that is, $(x_{i},z_{i})\in r_{i}$. Thus $(\bar{x},\bar{z})\in
r$, as wanted. ∎

Condition (2) in previous Proposition 1.8 may be interpreted as follows: For
each $\bar{x}\in G_{\infty}$ and each $i\in{\mathbb{N}}$, there exits $j\geq
i$ such that $g_{i}^{j}$ collapses any path of length two beginning at the
vertex $x_{j}$ and through three different vertices to an edge beginning at
$x_{i}$.

Given an inverse sequence of cellular graphs, to prove that it is a g-cell
structure will usually be verified by checking condition (2) in Proposition
1.8.

Notice that Definition 1.7 allows the spaces $G_{n}$ to be not necessary
discrete. It would be trivial to propose a g-cell structure of any topological
space $X$ by considering the inverse sequence $G_{n}=X$, $r_{n}=\Delta$ and
the bonding maps $g_{n}^{n+1}$ are all the identity maps. Therefore, to avoid
such trivialities, it becomes important to require either that every $G_{n}$
has the discrete topology or that not all $r_{n}$ are equal to $\Delta$. The
role of the bonding maps, evidently, becomes predominant, as the following
examples show.

######  Example 1.9.

For each $i\in{\mathbb{N}}$, consider $G_{i}=[0,1]$ with the standard topology
and define

$r_{i}=\Delta\cup\left\\{(x,1-x),(1-x,x)\colon
0<x<\tfrac{1}{2}\right\\}\cup\left\\{(0,\tfrac{1}{2}),(\tfrac{1}{2},0),(\tfrac{1}{2},1),(1,\tfrac{1}{2})\right\\},$

while the bonding maps are given by

$g_{n}^{n+1}(x)=\begin{cases}x&\text{ if }0\leq x<\frac{1}{2}\\\ 1-x&\text{ if
}\frac{1}{2}\leq x\leq 1\end{cases}.$

Then $\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$ is a g-cell
structure while none of the $G_{n}$’s is a discrete spaces.

Observe that Example 1.9 also exhibits that the converse of the implication
$(1)\Rightarrow(2)$ in Proposition 1.8 is false. The following example will
show how a modification on the bonding maps may produce the loss of the cell
structure on a sequence of cellular graphs.

######  Example 1.10.

Consider the cellular graphs $\\{(G_{n},r_{n})\\}$ given in Example 1.9. Now
redefine all the bonding maps $g_{n}^{n+1}$ as the identity maps. Then the
resulting inverse sequence is no longer a cell structure. Indeed, the constant
sequences $\bar{0}=(0,0,0,\ldots)$,
$\bar{\frac{1}{2}}=(\frac{1}{2},\frac{1}{2},\frac{1}{2},\ldots)$ and
$\bar{1}=(1,1,1,\ldots)$ belong to $G_{\infty}$. Furthermore,
$(\bar{0},\bar{\frac{1}{2}})\in r$ and $(\bar{\frac{1}{2}},\bar{1})\in r$.
However, $(\bar{0},\bar{1})\not\in r$. Thus $r$ is not an equivalence
relation.

######  Definition 1.11.

Let $\left\\{\left(G_{n},r_{n}\right),g_{n}^{n+1}\right\\}$ be a g-cell
structure. Let
$G_{\infty}=\displaystyle\lim_{\longleftarrow}\\{G_{n},g_{n}^{n+1}\\}$ and let
$r$ be the induced relation on $G_{\infty}$ as given in Definition 1.5. We
denote the quotient space $G^{\ast}$ as

$G^{\ast}=G_{\infty}/r.$

We will say that a topological space $X$ admits a g-cell structure if there
exists a g-cell structure such that $X$ is homeomorphic to $G^{\ast}$.

In Example 1.9 one may show that $G_{\infty}$ is homeomorphic to
$[0,\tfrac{1}{2}]$ with the usual topology and that $(\bar{0},\bar{1})$ is the
only nontrivial relation in $r$. Thus $G_{\infty}$ is homeomorphic to $S^{1}$.
This gives a g-cell structure for $S^{1}$ alternative to cell structure
obtained in [1].

When every $r_{j}$ is an equivalence relation, one might think that the
inverse limit of $G_{j}^{\ast}=G_{j}/r_{j}$ is equal to $G^{\ast}$. This is
not true, as the following example shows.

######  Example 1.12.

Consider the discrete spaces $G_{n}=(0,\tfrac{1}{2}]\cap{\mathbb{Q}}$ together
with $r_{n}=G_{n}\times G_{n}$. Define now the bonding maps $g_{n}^{n+1}$ to
be the inclusion maps. Then $G^{\ast}$ is empty, while the inverse limit of
$G_{n}^{\ast}$ is the one point space.

The above Example 1.12 may be slightly modified by taking
$G_{n}={\mathbb{N}}$, $r_{n}=G_{n}\times G_{n}$ and all bonding maps to be the
identity. Then again $G^{\ast}={\mathbb{N}}$ is different from the inverse
limit of the one-point space $G_{n}^{\ast}$.

######  Proposition 1.13.

Let $\bigl{\\{}(G_{n},r_{n}),g_{n}^{n+1}\bigr{\\}}_{n\in{\mathbb{N}}}$ be a
g-cell structure. The quotient map $\pi\colon G_{\infty}\to G^{\ast}$ is
closed if and only if for every $\bar{x}\in G_{\infty}$ and every open
neighbourhood $A\subset G_{\infty}$ of $B(\bar{x},r)$ there exists $U$ in the
base of $G_{\infty}$ such that $\bar{x}\in U$ and $B\left(U,r\right)\subset
A$.

###### Proof.

Let $\bar{x}\in G_{\infty}$ and let $A\subset G_{\infty}$ be an open subset in
$G_{\infty}$ such that $B(\bar{x},r)\subset A$. Define

$\displaystyle O$ $\displaystyle=\\{\bar{z}\in
G_{\infty}\colon\pi(\bar{z})\cap A\not=\emptyset\\}=\\{\bar{z}\in
G_{\infty}:B(\bar{z},r)\subset A\\}.$

If $\pi$ is a closed map, by [4, Proposition 2.4.9], the set $O$ is an open
set and since $\bar{x}\in O$, by Proposition 1.4, there exists $U$ in the base
of $G_{\infty}$ such that $\bar{x}\in U\subset O$. By definition of $O$, for
each $\bar{y}\in U$, $B(\bar{y},r)\subset A$. Thus, $B(U,r)\subset A$. For the
converse, if $\bar{z}\in O$ then, by hypothesis, there exists a basis element
$U$ such that $\bar{z}\in U$ and $B\left(U,r\right)\subset A$. Let $\bar{y}\in
U$ then $B(\bar{y},r)\subset B(U,r)\subset A$ and thus $\bar{y}\in O$.
Consequently $U\subset O$ and so $O$ is an open set. By [4, Proposition
2.4.9], $\pi$ is a closed map. ∎

In Propositon 1.13, if $G_{n}$ has the discrete topology for all $n$, we
obtained a condition similar to [1, Proposition 3.8] for cell structures as
the following result shows.

######  Corollary 1.14.

Let $\bigl{\\{}(G_{n},r_{n}),g_{n}^{n+1}\bigr{\\}}_{n\in{\mathbb{N}}}$ be a
g-cell structure where every $G_{n}$ has the discrete topology. If for every
$\bar{x}\in G_{\infty}$ and every open neighbourhood $A\subset G_{\infty}$ of
$B(x,r)$ there exists $j\in{\mathbb{N}}$ such that
$B(g_{j}^{-1}(x_{j}),r)\subset A$ then the collection

$\\{G^{\ast}\setminus\pi\left(G_{\infty}\setminus\mathcal{V}\right)\colon\mathcal{V}\text{
is open in }G_{\infty}\\}$

is a basis of open sets for the topology of $G^{\ast}$.

###### Proof.

Let $[\bar{x}]\in G^{*}$ and let $U\subset G^{*}$ be an open set containing
$[\bar{x}]$. Since $\pi^{-1}(U)$ is an open set in $G_{\infty}$, then by
Proposition 1.4, for each $\bar{y}\in B(\bar{x},r)$ there exists $i_{\bar{y}}$
such that $g_{i_{\bar{y}}}^{-1}(y_{i_{\bar{y}}})\subset\pi^{-1}(U)$. Let
$\mathcal{V}=\bigcup\left\\{g_{i_{\bar{y}}}^{-1}(y_{i_{\bar{y}}}):\bar{y}\in
B(\bar{x},r)\text{ and
}g_{i_{\bar{y}}}^{-1}(y_{i_{\bar{y}}})\subset\pi^{-1}(U)\right\\}$, then
$\mathcal{V}$ is an open set $G_{\infty}$ such that
$B(\bar{x},r)\subset\mathcal{V}\subset\pi^{-1}(U)$. By Proposition 1.13,
$\pi(G_{\infty}\backslash\mathcal{V})$ is a closed set and thus
$\mathrm{G^{*}}\backslash\pi(G_{\infty}\backslash\mathcal{V})$ is an open set.
On the other hand we have $G_{\infty}\backslash\mathcal{V}\subset
G_{\infty}\backslash B(\bar{x},r)$. This mean that
$B(\bar{x},r)\cap(G_{\infty}\backslash\mathcal{V})=\emptyset$ and so
$[\bar{x}]\notin\pi(G_{\infty}\backslash\mathcal{V})$. Thus
$[\bar{x}]\in\mathrm{G^{*}}\backslash\pi(G_{\infty}\backslash\mathcal{V})$.
Finally, since $G_{\infty}\backslash\pi^{-1}(U)\subset
G_{\infty}\backslash\mathcal{V}$, then
$\pi(G_{\infty}\backslash\pi^{-1}(U))\subset\pi(G_{\infty}\backslash\mathcal{V})$,
so that $G^{\ast}\setminus\pi(G_{\infty}\backslash\mathcal{V})\subset
G^{\ast}\setminus\pi(G_{\infty}\backslash\pi^{-1}(U))$. Now, since
$\pi(\pi^{-1}(G^{*}\backslash U))=G^{*}\backslash U$ and
$G_{\infty}\setminus\pi^{-1}(U)=\pi^{-1}(G^{\ast}\setminus U)$ then
$\pi(G_{\infty}\backslash\pi^{-1}(U))=G^{*}\backslash U$. We conclude that
$G^{*}\backslash\pi(G_{\infty}\backslash\mathcal{V})\subset U$. ∎

######  Lemma 1.15.

Let $\bigl{\\{}(G_{i},r_{i}),g_{i}^{i+1}\bigr{\\}}_{i\in{\mathbb{N}}}$ be a
g-cell structure where each $G_{i}$ satisfies that for each $x\in G_{i}$ and
for each open set $U$ containing $B(x,r_{i})$, there exists an open set $O$
such that $x\in O$ and $B(O,r)\subset U$. Let $i\in{\mathbb{N}}$ and let
$\bar{x}=(x_{1},x_{2},x_{3},...)\in G_{\infty}$. Then
$A_{x_{i},U}=\bigl{\\{}[\bar{z}]\in G^{*}:z_{i}\in U\text{ and there exists
}j>i\text{ such that }(z_{j},w_{j})\notin r_{j},\forall\bar{w}\in
g_{i}^{-1}\left(B(U,r_{i}\right)\backslash U)\bigr{\\}}$ is an open set in
$G^{*}$ and $\pi(\bar{x})\in A_{x_{i},U}$.

###### Proof.

We will show that the condition in $A_{x_{i},U}$ is independent of the
representative of the class. Let $\bar{z}\in G_{\infty}$ such that $z_{i}\in
U$ and $j>i$ satisfying $(z_{j},y_{j})\notin r_{j}$ for all $\bar{y}\in
g_{i}^{-1}(B(U,r_{i})\backslash U)$. Let $\bar{w}\in G_{\infty}$ such that
$(\bar{w},\bar{z})\in r$. Since $z_{i}\in U$, there exists an open set $O$
such that $z_{i}\in O$ and $B(O,r)\subset U$. Hence, since $w_{i}\in B(O,r)$,
then $w_{i}\in U$. By the definition of g-cell structure, there exits $k>j$
such that $g_{j}^{k}(B(z_{k},2r_{k}))\subset B(z_{j},r_{j})$ and suppose that
there exists $\bar{y}\in g_{i}^{-1}(B(U,r_{i}))\backslash U$ such that
$(y_{k},w_{k})\in r_{k}$. Then $y_{k}\in B(z_{k},2r_{k})$, which implies
$(z_{j},y_{j})\in r_{j}$. This is a contradiction on the choosing of $j$.
Thus, for all $\bar{y}\in g_{i}^{-1}(B(U,r_{i})\backslash U)$,
$(y_{k},w_{k})\notin r_{k}$, this mean that $\bar{w}$ satisfies the condition
in $A_{x_{i},U}$.

Now, we will prove that $A_{x_{i},U}$ is an open set. Let
$\bar{z}\in\pi^{-1}(A_{x_{i},U})$, then $z_{i}\in U$ and there exists $j>i$
such that $(z_{j},y_{j})\notin r_{j}$ for all $\bar{y}\in
g_{i}^{-1}(B(U,r_{i})\backslash U)$. Since $z_{i}\in U$, then there exists an
open set $O$ such that $z_{i}\in O$ and $B(O,r_{i})\subset U$. Then
$\bar{z}\in g_{i}^{-1}(O)$. We will show that $g_{i}^{-1}(O)\subset
A_{x_{i},U}$. Let $\bar{u}\in g_{i}^{-1}(O)$, then $u_{i}\in O\subset U$.
Suppose that there exists $\bar{y}\in g_{i}^{-1}(B(U,r_{i})\backslash U)$ such
that $(y_{j},u_{j})\in r_{j}$. Then $(y_{i},u_{i})\in r_{i}$. Hence, $y_{i}\in
B(u_{i},r_{i})\subset B(O,r_{i})\subset U$. This is a contradiction because
$y_{i}\in B(U,r_{i})\backslash U$. Consequently, $(y_{j},u_{j})\notin r_{j}$
for all $\bar{y}\in g_{i}^{-1}(B(U,r_{i})\backslash U)$. Then $[\bar{u}]\in
A_{x_{i},U}$ which implies that $\bar{u}\in\pi^{-1}(A_{x_{i},U})$. Hence,
$\pi^{-1}(A_{x_{i},U})$ is an open set, so that $A_{x_{i},U}$ is an open set
in $G^{*}$. ∎

##  2\. Results

In this section we present the main results of this work. Since we do not
require that for each thread $\bar{x}$ and each $i\in N$ there is $j\geq i$
such that $g_{i}^{j}(B(x_{i},r_{i})$ is finite (a condition that implies that
the quotient map $G_{\infty}\mapsto G^{\ast}$ to be a perfect map), we begin
by giving alternative condition on the g-cell structure which implies
$G^{\ast}$ to be Hausdorff or normal.

######  Theorem 2.1.

Let $\bigl{\\{}(G_{i},r_{i}),g_{i}^{i+1}\bigr{\\}}_{i\in{\mathbb{N}}}$ be a
g-cell structure where each $G_{i}$ is a Hausdorff space and satisfies that
for each $x\in G_{i}$ and for each open set $U$ containing $B(x,r_{i})$, there
exists an open set $O$ such that $x\in O$ and $B(O,r_{i})\subset U$. Then the
quotient space $G^{*}$ is a Hausdorff space.

###### Proof.

Let $\pi(\bar{x})$, $\pi(\bar{y})$ in $G^{*}$ such that
$\pi(\bar{x})\neq\pi(\bar{y})$ and $\bar{x},\bar{y}\in G_{\infty}$. Then
$(\bar{x},\bar{y})\notin r$ and thus there exists $i\in{\mathbb{N}}$ such that
$(x_{i},y_{i})\notin r_{i}$. Since that $G_{i}$ is a Hausdorff space there
exists disjoint open sets $U$ and $V$ such that $x_{i}\in U$ and $y_{i}\in V$.
Define the following sets:

$A_{x_{i},U}=\bigl{\\{}[\bar{z}]\in G^{*}:z_{i}\in U\text{ and }\exists
j>i\text{ such that }(z_{j},w_{j})\notin r_{j},\forall\bar{w}\in
g_{i}^{-1}\left(B(U,r_{i})\backslash U\right)\bigr{\\}},$
$B_{y_{i},V}=\bigl{\\{}[\bar{z}]\in G^{*}:z_{i}\in V\text{ and }\exists
j>i\text{ such that }(z_{j},w_{j})\notin r_{j},\forall\bar{w}\in
g_{i}^{-1}\left(B(V,r_{i})\backslash V\right)\bigr{\\}}.$

By the Lemma 1.15, $A_{x_{i},U}$ and $B_{y_{i},V}$ are open sets in $G^{*}$
containing $\pi(\bar{x})$ and $\pi(\bar{y})$, respectively. We will show that
$A_{x_{i},U}$ and $B_{y_{i},V}$ are disjoint sets. Suppose that there exists
$[\bar{z}]\in G^{*}$ such that $[\bar{z}]\in A_{x_{i}}\cap B_{y_{i}}$, then
there exists $\bar{w}$ and $\bar{u}$ in $[\bar{z}]$ such that $\bar{w}$
satisfies the condition in $A_{x_{i},U}$ and $\bar{u}$ satisfies the condition
in $B_{y_{i},V}$. Since $(w_{i},u_{i})\in r_{i}$, then $w_{i}\in B(V,r_{i})$.
But, $U\cap V=\emptyset$ then $w_{i}\in B(V,r_{i})\backslash V$ and so
$\bar{w}\in g_{i}^{-1}(B(V,r_{i})\setminus V$. Hence, $(w_{j},u_{j})\notin
r_{j}$ for some $j$. This is a contradiction because $\bar{u}$ and $\bar{w}$
are in the same equivalence class. ∎

######  Corollary 2.2.

If $\left\\{\left(G_{i},r_{i}\right),g_{i}^{i+1}\right\\}$ is a g-cell
structure where each $G_{i}$ is a discrete space then $G^{\ast}$ is a
Hausdorff space.

###### Proof.

Since $G_{i}$ is a discrete space, then it is a Hausdorff space. Also, if
$x\in G_{i}$ and $U$ is an open set containing $B(x,r_{i})$, then $O=\\{x\\}$
is an open set containing $x$ and $B(O,r_{i})\subset U$. By the Theorem 2.1,
$G^{*}$ is a Hausdorff space. ∎

######  Theorem 2.3.

Let $X$ be a space admitting a g-cell structure. Then $X$ is normal provided
its g-cell structure satisfies the following properties: (i) Every $G_{n}$ is
a metric space. (ii) For every $\bar{x}\in G_{\infty}$ and for every open
neighborhood $A\subset G_{\infty}$ of $B(\bar{x},r)$ there exists $U$ in the
base of $G_{\infty}$ such that $\bar{x}\in U$ and $B(U,r)\subset A$.

###### Proof.

By Proposition 1.13, the quotient map $\pi:G_{\infty}\to G^{*}$ is closed.
Since every $G_{n}$ is metric then $\prod_{n\in{\mathbb{N}}}G_{n}$ is also
metric, [4, Theorem 4.2.2, pg. 259]. Therefore $G_{\infty}$ is a normal space
and applying [4, Theorem 1.5.20, pg. 46] we get that $G^{*}$ is a normal space
and thus $X$ is a normal space. ∎

######  Theorem 2.4.

Let $\\{X_{k}\\}_{k\in{\mathbb{N}}}$ be a sequence of topological spaces each
admitting a cellular structure and without isolated points. Then the wedge sum
$X=\bigvee_{k\in{\mathbb{N}}}X_{k}$ does not admit a cell structure but does
admit a g-cell structure.

###### Proof.

For each $k\in{\mathbb{N}}$, let us denote $p_{k}\in X_{k}$ the distinguished
base point in $X_{k}$ and let
$q:\bigsqcup_{k\in{\mathbb{N}}}X_{k}\to\bigvee_{k\in{\mathbb{N}}}X_{k}$ the
quotient map which identifies all the points $p_{k}$ into a single point
$x_{0}$. Since each $X_{k}$ is a metrizable space (in fact completely
metrizable by [1, Theorem 3.6]), let $\rho_{k}$ be a metric in $X_{k}$, for
each $k\in{\mathbb{N}}$. Suppose that there exists a countable neighborhood
base $\\{u_{i}\\}_{i\in{\mathbb{N}}}$ of the point $x_{0}$. Since
$q^{-1}(u_{k})$ is an open set in $\sqcup_{k\in{\mathbb{N}}}X_{k}$, then
$q^{-1}(u_{k})\cap X_{k}$ is an open set in $X_{k}$. Also, $q^{-1}(u_{k})\cap
X_{k}$ is a non-degenerate set because $x_{0}\in q^{-1}(u_{k})\cap X_{k}$ and
$X_{k}$ is a topological space without isolated points. Then there exists
$y_{k}\in q^{-1}(u_{k})\cap X_{k}\backslash\\{p_{k}\\}$. Put
$d_{k}=\rho_{k}(p_{k},y_{k})$ and denote $B_{k}=\\{y_{k}\in
X_{k}\colon\rho_{k}(y_{k},p_{k})<d_{k}\\}$ the open ball with center $p_{k}$
and radius $d_{k}$. By definition of the mapping $q$ we have
$q^{-1}\left(q\left(\sqcup B_{k}\right)\right)=\sqcup B_{k}$, then by
definition of quotient topology, we have that $q\left(\sqcup B_{k}\right)$ is
an open set in $X=\bigvee_{k\in{\mathbb{N}}}X_{k}$ which contains $x_{0}$. We
claim that $q(\sqcup B_{k})$ does not contain any of the elements of the base
$\\{u_{i}\\}_{i\in{\mathbb{N}}}$. Indeed, if $u_{m}\subset q(\sqcup B_{k})$,
then $q^{-1}(u_{m})\subset q^{-1}(q(\sqcup B_{k}))=\sqcup B_{k}$. Hence,
$y_{m}\in q^{-1}(u_{m})\cap X_{m}\subset B_{m}$, a contradiction. Thus,
$x_{0}$ does no admit a countable neighborhood base. Then $X$ is not a first
countable space. We conclude that $X$ is not a metrizable space. Thus X does
not admit a cell structure by [1, Theorem 3.6].

For each $k\in{\mathbb{N}}$, let
$(\ast)_{k}=\\{(G_{i}^{k},r_{i}^{k}),(g_{k})_{i}^{i+1}\\}$ be a cell structure
for $X_{k}$ built as in [1, Theorem 4.3]. The cell structure $(\ast)_{k}$
determines a inverse limit which will be denoted by $G_{\infty}^{k}$ and
$G_{k}^{*}$ will denote the space determined by the cell structure. The
$i^{th}$ projection map restricted to $G_{\infty}^{k}$ will be denoted by
$(g_{k})_{i}$. The quotient map of $G_{\infty}^{k}$ onto $G_{k}^{*}$ will be
denoted by $\pi_{k}$ while $\varphi_{k}$ will denote the homeomorphism of
$G_{k}^{*}$ onto $X_{k}$.

For each $k\in{\mathbb{N}}$, fix
$\bar{x}_{k}=(x_{1}^{k},x_{2}^{k},\ldots)\in(\phi_{k}\circ\pi_{k})^{-1}(p_{k})$.
Furthermore, since $(*)_{k}$ is a cell structure for $X_{k}$, we may choose
inductively $j_{i}^{k}>j_{i-1}^{k}$ the least natural number such
$(g_{k})_{j_{i-1}^{k}}^{j_{i}^{k}}\bigl{(}B\bigl{(}x_{j_{i}^{k}}^{k},2r_{j_{i}^{k}}^{k}\bigr{)}\bigr{)}\subset
B\bigl{(}x_{j_{i-1}^{k}}^{k},r_{j_{i-1}^{k}}^{k}\bigr{)}$, where
$j_{0}^{k}=1$. We define the following sets:

  * •

$G_{i}=\bigcup_{k\in\mathbb{N}}G_{j_{i}^{k}}^{k}$

  * •

$A_{i}=\bigcup_{k\in\mathbb{N}}r_{j_{i}^{k}}^{k}$

  * •

$B_{i}=\bigcup_{k,t}B\left((g_{k})_{j_{i}^{k}}(\varphi_{k}\circ\pi_{k})^{-1}(p_{k}),r_{j_{i}^{k}}^{k}\right)\times
B\left((g_{t})_{j_{i}^{t}}(\varphi_{t}\circ\pi_{t})^{-1}(p_{t}),r_{j_{i}^{t}}^{t}\right)$

For each $i\in\mathbb{N}$, we will consider on $G_{i}$ the relation
$R_{i}=A_{i}\cup B_{i}$. Notice that each $R_{i}$ is a reflexive relation,
because $A_{i}$ contains to the diagonal of $G_{i}\times G_{i}$. Also, $R_{i}$
is a symmetric relation, since $A_{i}$ and $B_{i}$ are symmetric. Thus, for
each $i\in\mathbb{N}$, $(G_{i},R_{i})$ is a cellular graph. We define, for
each $i\in{\mathbb{N}}$, the bonding maps as follow: let $f_{i}^{i}$ be the
identity on $G_{i}$ and $f_{i}^{i+1}:G_{i+1}\to G_{i}$ defined by

$f_{i}^{i+1}(x)=(g_{k})_{j_{i}^{k}}^{j_{i+1}^{k}}(x),\text{ }x\in
G_{j_{i+1}^{k}}^{k}.$

We will prove that $\bigl{(}f_{i}^{i+1}(x),f_{i}^{i+1}(y)\bigr{)}\in R_{i}$
whenever $(x,y)\in R_{i+1}$.

  * •

If $(x,y)\in A_{i+1}$, then $(x,y)\in r_{j_{i+1}^{k}}^{k}$ for some
$k\in\mathbb{N}$. Since $(g_{k})_{j_{i}^{k}}^{j_{i+1}^{k}}$ is a composition
of bonding maps, we get that $\left(f_{i}^{i+1}(x),f_{i}^{i+1}(y)\right)\in
r_{j_{i}^{k}}^{k}$. Therefore,
$\bigl{(}f_{i}^{i+1}(x),f_{i}^{i+1}(y)\bigr{)}\in A_{i}\subset R_{i}$.

  * •

If $(x,y)\in B_{i+1}$, there exist $l,k\in\mathbb{N}$ such that $x\in
B\bigl{(}(g_{k})_{j_{i+1}^{k}}(\varphi_{k}\circ\pi_{k})^{-1}(p_{k}),r_{j_{i+1}^{k}}^{k}\bigr{)}$
and $y\in
B\bigl{(}(g_{l})_{j_{i+1}^{l}}(\varphi_{l}\circ\pi_{l})^{-1}(p_{l}),r_{j_{i+1}^{l}}^{l}\bigr{)}$.
Thus, there is $z=(g_{k})_{j_{i+1}^{k}}(\bar{z})$ such that
$\bar{z}=(\varphi_{k}\circ\pi_{k})^{-1}(p_{k})$ and $(x,z)\in
r_{j_{i+1}^{k}}^{k}$. This implies that
$f_{i}^{i+1}(x)=(g_{k})_{j_{i}^{k}}^{j_{t+1}^{k}}(x)$ and
$f_{i}^{i+1}(z)=(g_{k})_{j_{i}^{k}}^{j_{t+1}^{k}}(z)$, therefore
$\bigl{(}f_{i}^{i+1}(x),f_{i}^{i+1}(z)\bigr{)}\in r_{j_{i}^{k}}^{k}$. Also,
since
$f_{i}^{i+1}(z)=(g_{k})_{j_{i}^{k}}^{j_{i+1}^{k}}\bigl{(}(g_{k})_{j_{i+1}^{k}}(\bar{z})\bigr{)}=(g_{k})_{j_{i}^{k}}(\bar{z})$,
we have $f_{i}^{i+1}(x)\in
B\bigl{(}(g_{k})_{j_{i}^{k}}(\varphi_{k}\circ\pi_{k})^{-1}(p_{k}),r_{j_{i}^{k}}^{k}\bigr{)}$.
In the same way, we get that $f_{i}^{i+1}(y)\in
B\bigl{(}(g_{l})_{j_{i}^{l}}(\varphi_{l}\circ\pi_{l})^{-1}(p_{l}),r_{j_{i}^{k}}^{l}\bigr{)}$.
Hence, we conclude that $\bigl{(}f_{i}^{i+1}(x),f_{i}^{i+1}(y)\bigr{)}\in
B_{i}\subset R_{i}$.

Therefore $\bigl{\\{}\bigl{(}G_{i},R_{i}\bigr{)},f_{i}^{i+1}\bigr{\\}}$ is a
inverse sequence of cellular graphs and we will denote $G_{\infty}$ its
inverse limit. Now to show that the condition of g-cell structure is
satisfied, let $\bar{x}\in G_{\infty}$ and let $i\in\mathbb{N}$. Then, for
some $t\in\mathbb{N}$,
$\bar{x}\in\lim\limits_{\longleftarrow}\\{F_{i}^{t},f_{i}^{i+1}\\}:=F_{\infty}^{t}$,
where we set, for each $i\in\mathbb{N}$, $F_{i}^{t}=G_{j_{i}^{t}}^{t}$. Notice
that $G_{\infty}^{t}=F_{\infty}^{t}$, because the set
$\\{j_{s}^{t}:s\in\mathbb{N}\\}$ is cofinal in $\mathbb{N}$. Then, there is
$\bar{x}^{\prime}=(x_{1}^{\prime},x_{2}^{\prime},\ldots,x_{i}^{\prime},\ldots)\in
G_{\infty}^{t}$ which corresponds to
$\bar{x}=\left(x_{j_{1}^{t}}^{t},x_{j_{2}^{t}}^{t},\ldots,x_{j_{i}^{t}}^{t},\ldots\right)\in\lim\limits_{\longleftarrow}\\{F_{i}^{t},f_{i}^{i+1}\\}$.
For $i^{\prime}=j_{i}^{t}\in\mathbb{N}$, let $k>j_{i+2}^{t}>i$ such that
$(g_{t})_{i^{\prime}}^{j_{k}^{t}}(B(x^{\prime}_{j_{k}^{t}},2r_{j_{k}^{t}}^{t}))\subset
B(x^{\prime}_{i^{\prime}}\,{},r_{i^{\prime}}^{t})$. We will show that
$f_{i}^{k}(B(x_{k},2R_{k}))\subset B(x_{i},R_{i})$. Let $y\in
B(x_{k},2R_{k})$. Notice that $2R_{k}=2A_{k}\cup
2B_{k}\cup(A_{k}+B_{k})\cup(B_{k}+A_{k})$.

  * •

If $(x_{k},y)\in 2A_{k}$, then there exists $z\in G_{k}$ such that
$(x_{k},z)\in A_{k}$ and $(z,y)\in A_{k}$. Since $x_{k}\in G_{j_{k}^{t}}^{t}$,
then $(x_{k},z)\in r_{j_{k}^{t}}^{t}$ and necessarily $z\in
G_{j_{k}^{t}}^{t}$. Thus $(z,y)\in r_{j_{k}^{t}}^{t}$. This implies that $y\in
B\bigl{(}x_{k},2r_{j_{k}^{t}}^{t}\bigr{)}=B\bigl{(}x^{\prime}_{j_{k}^{t}},2r_{j_{k}^{t}}^{t}\bigr{)}$.
Therefore $f_{i}^{k}(y)=(g_{t})_{i^{\prime}}^{j_{k}^{t}}(y)\in
B(x^{\prime}_{i^{\prime}},r_{i^{\prime}}^{t})=B(x^{\prime}_{j_{i}^{t}},r_{j_{i}^{t}}^{t})=B(x_{i},r_{j_{i}^{t}}^{t})\subset
B(x_{i},A_{i})$. We get that $(f_{i}^{k}(y),f_{i}^{k}(x_{k}))\in A_{i}\subset
R_{i}$.

  * •

If $(x_{k},y)\in 2B_{k}$, since $B_{k}$ is transitive, we have $(x_{k},y)\in
B_{k}$ and so we obtain the result.

  * •

If $(x_{k},y)\in A_{k}+B_{k}$, there exists $z\in G_{k}$ such that
$(x_{k},z)\in A_{k}$ and $(z,y)\in B_{k}$. Since $(x_{k},z)\in A_{k}$, then
$(x_{k},z)\in r_{j_{k}^{t}}^{t}$. Hence
$\left((g_{t})_{j_{i+1}^{t}}^{{j_{k}^{t}}}(x_{k}),(g_{t})_{j_{i+1}^{t}}^{{j_{k}^{t}}}(z)\right)\in
r_{j_{i+1}^{t}}^{t}$. On the other hand, we have that $z\in G_{j_{k}^{t}}^{t}$
and $y\in G_{j_{k}^{s}}^{s}$ for some $s\in\mathbb{N}$. Since $(z,y)\in
B_{k}$, then $(g_{t})_{j_{i+2}^{t}}^{j_{k}^{t}}(z)\in
B\left((g_{t})_{j_{i+2}^{t}}(\varphi_{t}\circ\pi_{t})^{-1}(p_{t}),r_{j_{i+2}^{t}}^{t}\right)$
and $(g_{s})_{j_{i}^{s}}^{j_{k}^{s}}(y)\in$
$B\left((g_{s})_{j_{i}^{s}}(\varphi_{s}\circ\pi_{s})^{-1}(p_{s}),r_{j_{i}^{s}}^{s}\right)$.
Now, $((g_{t})_{j_{i+2}^{t}}^{j_{k}^{t}}(z),w)\in r_{j_{i+2}^{t}}^{t}$ for
some $w\in(g_{t})_{j_{i+2}^{t}}(\varphi_{t}\circ\pi_{t})^{-1}(p_{t})$ and
therefore $(w,(g_{t})_{j_{i+2}^{t}}(\bar{x}_{t}))\in r_{j_{i+2}^{t}}^{t}$.
This mean that $(g_{t})_{j_{i+2}^{t}}^{j_{k}^{t}}(z)\in
B((g_{s})_{j_{i+2}^{t}}(\bar{x}_{t})),2r_{j_{i+2}^{t}}^{t})$ and by the
hypothesis, $(g_{t})_{j_{i+1}^{t}}^{j_{k}^{t}}(z)\in
B((g_{t})_{j_{i+1}^{t}}(\bar{x}_{t})),r_{j_{i+1}^{t}}^{t})$. We obtain
$(g_{t})_{j_{i+1}^{t}}^{{j_{k}^{t}}}(x_{k})\in
B\left((g_{t})_{j_{i+1}^{t}}(\bar{x}_{t}),2r_{j_{i+1}^{t}}^{t}\right)$. Also,
$g_{j_{i}^{t}}^{j_{i+1}^{t}}\left(B\left((g_{t})_{j_{i+1}^{t}}(\bar{x}_{t}),2r_{j_{i+1}^{t}}^{t}\right)\right)\subset
B\left((g_{t})_{j_{i}^{t}}(\bar{x}_{t}),r_{j_{i}^{t}}\right)$. Then
$f_{i}^{k}(x_{k})=g_{j_{i}^{t}}^{j_{k}^{t}}(x_{k})\in
B\left((g_{t})_{j_{i}^{t}}(\bar{x}_{t}),r_{j_{i}^{t}}^{t}\right)$. We conclude
that $f_{i}^{k}(x_{k})\in
B\left((g_{t})_{j_{i}^{t}}(\bar{x}_{t}),r_{j_{i}^{t}}^{t}\right)\subset
B\left((g_{t})_{j_{i}^{t}}(\varphi_{t}\circ\pi_{t})^{-1}(p_{t}),r_{j_{i}^{t}}^{t}\right)$.
Since $f_{i}^{k}(y)=(g_{s})_{j_{i}^{s}}^{j_{k}^{s}}(y)\in
B\left|((g_{s})_{j_{i}^{s}}(\bar{x}_{s}),r_{j_{i}^{s}}^{s}\right)\subset
B\left((g_{s})_{j_{i}^{s}}(\varphi_{s}\circ\pi_{s})^{-1}(p_{s}),r_{j_{i}^{s}}^{s}\right)$,
we get that $\left(f_{i}^{k}(x_{k}),f_{i}^{k}(y)\right)\in B_{i}\subset
R_{i}$.

  * •

If $(x_{k},y)\in B_{k}+A_{k}$, there exists $z\in G_{k}$ such that
$(x_{k},z)\in B_{k}$ and $(z,y)\in A_{k}$. Since $(z,y)\in A_{k}$, there
exists $s\in\mathbb{N}$ such that $(z,y)\in r_{j_{k}^{s}}^{s}$. Hence
$((g_{s})_{j_{i+1}^{s}}^{{j_{k}^{s}}}(y),(g_{s})_{j_{i+1}^{s}}^{{j_{k}^{s}}}(z))\in
r_{j_{i+1}^{s}}^{s}$. In the other hands, we have that $z\in
G_{j_{k}^{s}}^{s}$ and $x_{k}\in G_{j_{k}^{t}}^{t}$. Since $(x_{k},z)\in
B_{k}$, then $(g_{s})_{j_{i+2}^{s}}^{j_{k}^{s}}(z)\in
B((g_{s})_{j_{i+2}^{s}}(\varphi_{s}\circ\pi_{s})^{-1}(p_{s}),r_{j_{i+2}^{s}}^{s})$
and $(g_{t})_{j_{i}^{t}}^{j_{k}^{t}}(x_{k})\in
B((g_{t})_{j_{i}^{t}}(\varphi_{t}\circ\pi_{t})^{-1}(p_{t}),r_{j_{i}^{t}}^{t})$.
Thus $((g_{s})_{j_{i+2}^{s}}^{j_{k}^{s}}(z),w)\in r_{j_{i+2}^{s}}^{s}$ where
$w\in(g_{s})_{j_{i+2}^{s}}(\varphi_{s}\circ\pi_{s})^{-1}(p_{s})$. Hence,
$(w,(g_{s})_{j_{i+2}^{s}}(\bar{x}_{s}))\in r_{j_{i+2}^{s}}^{s}$. This implies
that $(g_{s})_{j_{i+2}^{s}}^{j_{k}^{s}}(z)\in
B((g_{s})_{j_{i+2}^{s}}(\bar{x}_{s})),2r_{j_{i+2}^{s}}^{s})$ and by the
hypothesis, we get that $(g_{s})_{j_{i+1}^{s}}^{j_{k}^{s}}(z)\in
B((g_{s})_{j_{i+1}^{s}}(\bar{x}_{s})),r_{j_{i+1}^{s}}^{s})$. Notice that
$(g_{s})_{j_{i+1}^{s}}^{{j_{k}^{s}}}(y)\in
B((g_{s})_{j_{i+1}^{s}}(\bar{x}_{s}),2r_{j_{i+1}^{s}}^{s})$. Also, we have
$g_{j_{i}^{s}}^{j_{i+1}^{s}}(B((g_{s})_{j_{i+1}^{s}}(\bar{x}_{s}),2r_{j_{i+1}^{s}}^{s}))\subset
B((g_{s})_{j_{i}^{s}}(\bar{x}_{s}),r_{j_{i}^{s}}^{s})$. Therefore
$f_{i}^{k}(y)=g_{j_{i}^{s}}^{j_{k}^{s}}(y)\in
B((g_{s})_{j_{i}^{s}}(\bar{x}_{s}),r_{j_{i}^{s}}^{s})$. We conclude that
$f_{i}^{k}(y)\in B((g_{s})_{j_{i}^{s}}(\bar{x}_{s}),r_{j_{i}^{s}}^{s})\subset
B((g_{s})_{j_{i}^{s}}(\varphi_{s}\circ\pi_{s})^{-1}(p_{s}),r_{j_{i}^{s}}^{s})$.
Since $f_{i}^{k}(x_{k})=(g_{t})_{j_{i}^{t}}^{j_{k}^{t}}(x_{k})\in
B((g_{t})_{j_{i}^{t}}(\bar{x}_{t}),r_{j_{i}^{t}}^{t})\subset
B((g_{t})_{j_{i}^{t}}(\varphi_{t}\circ\pi_{t})^{-1}(p_{t}),r_{j_{i}^{t}}^{t})$,
we get that $(f_{i}^{k}(x_{k}),f_{i}^{k}(y))\in B_{i}\subset R_{i}$.

We now define the following sets:

$A=\\{(\bar{x},\bar{y})\in G_{\infty}\times G_{\infty}:(x_{i},y_{i})\in
A_{i},\text{ for each }i\in\mathbb{N}\\}$

and

$B=\\{(\bar{x},\bar{y})\in G_{\infty}\times G_{\infty}:(x_{i},y_{i})\in
B_{i},\text{ for each }i\in\mathbb{N}\\}.$

Notice that since $A$ is the induced relation on $G_{\infty}$ by the reflexive
and symmetric relation $A_{i}$ on $G_{i}$, for each $i$, we have that $A$ is
an equivalence relation. So we will consider the quotient space $G_{\infty}/A$
(which is in fact homeomorphic to $\sqcup X_{k}$) and we will introduce and
equivalence relation on $G_{\infty}/A$ as follows:

$\displaystyle
B^{\prime}=\bigl{\\{}\bigl{(}[\bar{x}]_{A},[\bar{y}]_{A}\bigr{)}\in
G_{\infty}/A\times G_{\infty}/A:(\bar{x},\bar{y})\in B\bigr{\\}}\cup\Delta.$

We will prove that the definition of $B^{\prime}$ does not depend on the class
representative. It will suffice to show that given $\bar{u}\in[\bar{x}]_{A}$
such that for each $i\in{\mathbb{N}}$ we have $x_{i}\in
B\bigl{(}(g_{k})_{j_{i}^{k}}(\varphi_{k}\circ\pi_{k})^{-1}(p_{k}),r_{j_{i}^{k}}^{k}\bigr{)}$
for some $k\in\mathbb{N}$, implies that for each $i\in{\mathbb{N}}$ we also
have $u_{i}\in
B\bigl{(}(g_{k})_{j_{i}^{k}}(\varphi_{k}\circ\pi_{k})^{-1}(p_{k}),r_{j_{i}^{k}}^{k}\bigr{)}$
for some $k\in\mathbb{N}$. For this purpose, fix an arbitrary
$i\in{\mathbb{N}}$. Since $x_{i}\in G_{i}$, there exists $k\in{\mathbb{N}}$
such that $x_{i}\in G_{j_{i}^{k}}^{k}$. Then $(u_{i},x_{i})\in
r_{j_{i}^{k}}^{k}$ and $(u_{i+1},x_{i+1})\in r_{j_{i+1}^{k}}$. Also, by
assumption, there exists $\bar{w}\in G_{\infty}$ such that
$\pi_{k}(\bar{w})=\varphi_{k}^{-1}(p_{k})$ and $(x_{i+2},w_{j_{i+2}^{k}})\in
r_{j_{i+2}^{k}}^{k}$. Thus $x_{i+2}\in
B\bigl{(}x_{j_{i+2}^{k}}^{k},2r_{j_{i+2}^{k}}^{k}\bigr{)}$ and this implies
that $\bigl{(}x_{i+1},x_{j_{i+1}^{k}}^{k}\bigr{)}\in r_{j_{i+1}^{k}}^{k}$.
Hence, $u_{i+1}\in B\bigl{(}x_{j_{i+1}^{k}}^{k},2r_{j_{i+1}^{k}}^{k}\bigr{)}$
and so, $(u_{i},x_{j_{i}^{k}}^{k})\in r_{j_{i}^{k}}^{k}$. We get that
$u_{i}\in
B\bigl{(}(g_{k})_{j_{i}^{k}}(\varphi_{k}\circ\pi_{k})^{-1}(p_{k}),r_{j_{i}^{k}}^{k}\bigr{)}$,
as wanted. Thus $B^{\prime}$ is well defined. Notice that $B^{\prime}$ is an
equivalence relation. Consider the funtion
$\varphi:(G_{\infty}/A)/B^{\prime}\to G_{\infty}/R$ defined as
$\varphi\bigl{(}\bigl{[}[\bar{x}]_{A}\bigr{]}_{B^{\prime}}\bigr{)}=[\bar{x}]_{R}$.
We will show tha $\varphi$ is well defined.

${G_{\infty}}$${G_{\infty}/A}$${(G_{\infty}/A)/B^{\prime}}$${G_{\infty}/R}$$\scriptstyle{\pi_{A\cup
B}}$$\scriptstyle{\pi_{A}}$$\scriptstyle{\pi_{B^{\prime}}}$$\scriptstyle{\varphi}$
Figure 1. The spaces $G_{\infty}/R$ and $(G_{\infty}/A)/B$ are homeomorphic

Let $[\bar{y}]_{A}\in\bigl{[}[\bar{x}]_{A}\bigr{]}_{B^{\prime}}$. If
$\bigl{(}[\bar{x}]_{A},[\bar{y}]_{A}\bigr{)}\in B^{\prime}\backslash\Delta$,
then $(\bar{x},\bar{y})\in B\subset R$. If $[\bar{x}]_{A}=[\bar{y}]_{A}$, then
$(x_{i},y_{i})\in A_{i}$ and thus $(\bar{x},\bar{y})\in A\subset R$. It
follows that the diagram in Figure 1 commutes. Hence $\varphi$ is continuous.
We now show that $\varphi$ is injective. Let
$\bigl{(}[\bar{x}]_{A},[\bar{y}]_{A}\bigr{)}\notin B^{\prime}$, then
$(\bar{x},\bar{y})\notin A$, since otherwise $[\bar{x}]_{A}=[\bar{y}]_{A}$ and
so $\bigl{(}[\bar{x}]_{A},[\bar{y}]_{A}\bigr{)}\in\Delta\subset B^{\prime}$, a
contradiction. But also $(\bar{x},\bar{y})\notin B$ since otherwise
$(\bar{x},\bar{y})\in B$ implies by definition that
$\bigl{(}[\bar{x}]_{A},[\bar{y}]_{A}\bigr{)}\in B^{\prime}$, a contradiction.
We conclude that $[\bar{x}]_{R}\neq[\bar{y}]_{R}$ and thus $\varphi$ is
injective. It is clear that $\varphi$ is surjective and its inverse
$[\bar{x}]_{R}\mapsto\bigl{[}[\bar{x}]_{A}\bigr{]}_{B^{\prime}}$ is also
continuous by an analogous argument using a commutative diagram. Thus
$\varphi$ is a homeomorphism

∎

In fact the space $X$ in Theorem 2.4 is a normal space. In the next result we
prove that more is true: there is a non-regular space admitting a g-cell
structure.

First, we build the g-cell structure which will determine a non-regular space.

######  Definition 2.5.

Consider the following sets with the discrete topology.

$G_{1}=\\{a_{1}\\}\cup\left\\{b_{1}^{k}:k\in{\mathbb{N}}\right\\},\text{
}G_{2}=\\{a_{2}\\}\cup\left\\{b_{2}^{k}:k\in{\mathbb{N}}\right\\}\cup
C_{1}^{2}\cup\left\\{d_{2}^{k}:k\in{\mathbb{N}}\right\\}$

where $C_{1}^{2}=\left\\{c^{1}_{2,1},c^{2}_{2,1}\right\\}$ and let $L_{2}=1$.
In general, for each $i>2$ let $L_{i}=L_{i-1}+i-1$,
$C_{k}^{i}=\left\\{c^{1}_{i,k},c^{2}_{i,k}\right\\}$ with $1\leq k\leq L_{i}$
and

$G_{i}=\bigl{\\{}a_{i}\bigr{\\}}\cup\bigl{\\{}b_{i}^{k}:k\in{\mathbb{N}}\bigr{\\}}\cup\left(\bigcup_{k=1}^{L_{i}}C_{k}^{i}\right)\cup\bigl{\\{}d_{i}^{k}:k\in{\mathbb{N}}\bigr{\\}}.$

Now, for each $i\in{\mathbb{N}}$,

$r_{i}=\bigl{\\{}\bigl{(}a_{i},b_{i}^{k}\bigr{)},\bigl{(}b_{i}^{k},a_{i}\bigr{)}:k\geq
i\bigr{\\}}\cup\bigl{\\{}\bigl{(}b_{i}^{k},b_{i}^{n}\bigr{)}:k,n\geq
i\bigr{\\}}\cup\bigl{\\{}(a,b):a,b\in C_{k}^{i},k=1,...,L_{i}\bigr{\\}}\\\
\cup\bigl{\\{}\bigl{(}d_{i}^{(i-1)j+k},b_{i}^{k}\bigr{)},\bigl{(}b_{i}^{k},d_{i}^{(i-1)j+k}\bigr{)}:k<i,j\in{\mathbb{N}}\cup\\{0\\}\bigr{\\}},$

$g_{i}^{i}:G_{i}\to G_{i}$ is the identity function on $G_{i}$ and
$g_{i}^{i+1}:G_{i+1}\to G_{i}$ is defined as

$g_{i}^{i+1}(a)=\begin{cases}a_{i}&\text{if }a=a_{i+1}\hfill(1)\\\
b_{i}^{k}&\mbox{if }a=b_{i+1}^{k}\hfill(2)\\\ a_{i}&\text{if
}a=c^{1}_{i+1,1}\hfill(3)\\\ b_{i}^{i}&\text{if }a=c^{2}_{i+1,1}\hfill(4)\\\
c^{r}_{i,k-1}&\text{if }a=c^{r}_{i+1,k},r=1,2,k=2,...,L_{i}+1\hfill(5)\\\
d_{i}^{k-(L_{i}+1)}&\text{if
}a=c^{1}_{i+1,k},k=L_{i}+2,...,L_{i+1}\hfill(6)\\\
b_{i}^{k-(L_{i}+1)}&\text{if
}a=c^{2}_{i+1,k},k=L_{i}+2,...,L_{i+1}\hfill(7)\\\
d_{i}^{(i-1)(j+1)+n}&\text{if }a=d_{i+1}^{k},k=i\cdot j+n\text{ for
}0<n<i,j\in{\mathbb{N}}\cup\\{0\\}\phantom{p}\hfill(8)\\\ a_{i}&\text{if
}a=d_{i+1}^{k},k=i\cdot j,j\in{\mathbb{N}}\cup\\{0\\}\hfill(9)\end{cases}$

First we will prove the following lemma.

######  Lemma 2.6.

The sequence
$\left\\{\bigl{(}G_{i},r_{i}\bigr{)},g_{i}^{i+1}\right\\}_{i\in{\mathbb{N}}}$
given in Definition 2.5 is an inverse sequence of cellular graphs.
Furthermore, given $\bar{w}=(w_{1},w_{2},w_{3},...)\in G_{\infty}$ such that
$w_{i}=d_{i}^{k}$ for some $k,i\in{\mathbb{N}}$, then there exists
$j_{0}\in{\mathbb{N}}$ such that $w_{j_{0}}\in C_{s}^{j_{0}}$, for some
$s\in{\mathbb{N}}$.

###### Proof.

By definition, $r_{i}$ is a reflexive and symmetric relation for each
$i\in\mathbb{N}$. Also, by definition of the bonding functions, we get that
$\\{(G_{i},r_{i}),g_{i}^{i+1}\\}_{i\in\mathbb{N}}$ satisfies the first two
conditions of inverse system. We will show that the third condition is
satistied. Let $a,b\in G_{i+1}$ such that $(a,b)\in r_{i+1}$.

  * •

If $a=a_{i+1}$ and $b=b_{i+1}^{k}$ with $k\geq i$, then $g_{i}^{i+1}(a)=a_{i}$
and $g_{i}^{i+1}(b)=b_{i}^{k}$. Hence, $(g_{i}^{i+1}(a),g_{i}^{i+1}(b))\in
r_{i}$.

  * •

If $a=b_{i+1}^{k}$ and $b=b_{i+1}^{n}$, then
$g_{i}^{i+1}(b_{i+1}^{k})=b_{i}^{k}$ and $g_{i}^{i+1}(b_{i+1}^{n})=b_{i}^{n}$.
Therefore, $(g_{i}^{i+1}(a),g_{i}^{i+1}(b))\in r_{i}$.

  * •

If $a,b\in C_{k}^{i+1}$ we can suppose that $a=c^{1}_{i+1,k}$ and
$b=c^{2}_{i+1,k}$. If $k=1$, then $g_{i}^{i+1}(a)=a_{i}$ and
$g_{i}^{i+1}(b)=b_{i}^{i}$. Thus, $(g_{i}^{i+1}(a),g_{i}^{i+1}(b))\in r_{i}$ .
Now, if $k\in\\{2,...,L_{i}+1\\}$, we get that $g_{i}^{i+1}(a)=c^{1}_{i,k-1}$
and $g_{i}^{i+1}(b)=c^{2}_{i,k-1}$. Hence, $(g_{i}^{i+1}(a),g_{i}^{i+1}(b))\in
r_{i}$. By last, if $k\in\\{L_{i}+2,...,L_{i+1}\\}$, then
$g_{i}^{i+1}(a)=d_{i}^{k-(L_{i}+1)}$ and $g_{i}^{i+1}(b)=b_{i}^{k-(L_{i}+1)}$.
Therefore $(g_{i}^{i+1}(a),g_{i}^{i+1}(b))\in r_{i}$.

  * •

Suppose that $b=b_{i+1}^{k},\text{ with }k<i+1,a=d_{i+1}^{i\cdot j+k}\text{
with }j\in\mathbb{N}\cup\\{0\\}$. If $i\cdot j+k=i\cdot r$ for some
$r\in\mathbb{N}$, then $g_{i}^{i+1}(a)=a_{i}$ and $g_{i}^{i+1}(b)=b_{i}^{k}$.
So, we get that $(g_{i}^{i+1}(a),g_{i}^{i+1}(b))\in r_{i}$. Now, if $i\cdot
j+k=i\cdot r+n$ with $0<n<i$, then $g_{i}^{i+1}(a)=d_{i}^{(i-1)(j+1)+n}$ and
$g_{i}^{i+1}(b)=b_{i}^{k}$. Since that $n<i$, we conclude that
$(g_{i}^{i+1}(a),g_{i}^{i+1}(b))\in r_{i}$.

Consequently, we get that $\\{(G_{i},r_{i}),g_{i}^{i+1}\\}$ is an inverse
system of cellular graphs. Let $k\leq i-1$, notice that
$w_{i+1}=c^{1}_{i+1,k+L_{i}+1}$, then we can take $j_{0}=i+1$ and
$s=k+L_{j_{0}-1}+1$ and we get that $w_{j}\in C_{k+L_{j_{0}-1}+1}^{j}$. Now,
we suppose that $k>i-1$. Then we can factorize $k$ as $k=(i-1)(j+1)+n$ for
some $j\in{\mathbb{N}}\cup\\{0\\}$ and $0<n\leq i-1$. By definition of the
function $g_{i}^{i+1}$ we get that $w_{i+1}=d_{i+1}^{i\cdot j+n}$. Now, we
consider the preimage of $w_{i+1}$ under the function $g_{i+1}^{i+2}$. We have
two cases again. If $i\cdot j+n\leq i$, we get that $w_{i+2}=c^{1}_{i+2,i\cdot
j+n+L_{i+1}+1}$. Hence, we can take $j_{0}=i+2$ and $s=i\cdot
j+n+L_{j_{0}-1}+1$. This implies that $w_{j_{0}}\in C_{i\cdot
j+n+L_{j_{0}-1}+1}^{j_{0}}$. If $i\cdot j+n>i$, then
$w_{i+2}=d_{i+2}^{(i+1)(j-1)+n}$. In general, if $w_{i+t}$ is not in some
$C_{k}^{i}$, then $w_{i+t}=d_{i+t}^{(i+t-1)(j-t+1)+n}$. Taking $m=j+1$, we
have $w_{i+m}=d_{i+m}^{n}$. Since $0<n<i-1\leq i+m-1$, we get that
$w_{i+m+1}=c^{i}_{i+m+1,n+L_{i+m}+1}$. Now, we can take $j_{0}=i+m+1$ and
$s=n+L_{j_{0}-1}+1$ and we get that $w_{j_{0}}\in
C_{n+L_{j_{0}-1}+1}^{j_{0}}$. Consequently, for each $\bar{w}\in G_{\infty}$
such that $w_{i}=d_{i}^{k}$ with $k,i\in{\mathbb{N}}$, there exists
$j\in{\mathbb{N}}$ such that $w_{j}\in C_{s}^{j}$, for some
$s\in{\mathbb{N}}$. ∎

We will also need the following.

######  Lemma 2.7.

The sequence
$\left\\{\bigl{(}G_{i},r_{i}\bigr{)},g_{i}^{i+1}\right\\}_{i\in{\mathbb{N}}}$
given in Definition 2.5 is a g-cell structure.

###### Proof.

By Lemma 2.6, the system given in Definition 2.5 is an inverse system. It
remains to prove that the condition for a g-cell structure is satisfied. Let
$\bar{w}=(w_{1},w_{2},w_{3},...)\in G_{\infty}$ and $i\in{\mathbb{N}}$. If
$w_{i}=a_{i}$ or $w_{i}=b_{i}^{k}$ with $k\geq i$, then
$B(w_{i},r_{i})=\\{a_{i}\\}\cup\\{b_{i}^{k}:k\geq i\\}$ and
$B(w_{i},2r_{i})=\\{a_{i}\\}\cup\\{b_{i}^{k}:k\geq i\\}$. Now, if
$w_{i}=b_{i}^{k}$, with $k<i$, we get that
$B(w_{i},r_{i})=\\{b_{i}^{k}\\}\cup\\{d_{i}^{(i-1)j+k}:j\in{\mathbb{N}}\cup\\{0\\}\\}$
and
$B(w_{i},2r_{i})=\\{b_{i}^{k}\\}\cup\\{d_{i}^{(i-1)j+k}:j\in{\mathbb{N}}\cup\\{0\\}\\}$.
If $w_{i}\in C_{k}^{i}$ for some $k\in{\mathbb{N}}$, then
$B(w_{i},r_{i})=C_{k}^{i}$ and $B(w_{i},2r_{i})=C_{k}^{i}$. Hence, if
$w_{i}\in\\{a_{i}\\}\cup\\{b_{i}^{k}:k\in{\mathbb{N}}\\}\cup\left(\bigcup_{k=1}^{L_{i}}C_{k}^{i}\right)$,
then $B(w_{i},r_{i})=B(w_{i},2r_{i})$. Finally, if $\bar{w}\in G_{\infty}$
such that $w_{i}=d_{i}^{k}$ for some $k,i\in{\mathbb{N}}$. Then, by the Lemma
2.6, there exists $j_{0}\in{\mathbb{N}}$ such that $w_{j_{0}}\in
C_{s}^{j_{0}}$ for some $s\in{\mathbb{N}}$. Therefore, we have that
$B(w_{j_{0}},r_{j_{0}})=B(w_{j_{0}},2r_{j_{0}})$. We conclude that for each
$i\in{\mathbb{N}}$ and $\bar{w}\in G_{\infty}$, there exists
$j\in{\mathbb{N}}$ such that $g_{i}^{j}(B(w_{j},2r_{j}))\subset
B(w_{i},r_{i})$. Consequently,
$\\{(G_{i},r_{i}),g_{i}^{i+1}\\}_{i\in{\mathbb{N}}}$ is a g-cell structure. ∎

We are now ready to prove our last result.

######  Theorem 2.8.

There are spaces admitting a g-cell structure which are not regular.

###### Proof.

Consider the inverse sequence of cell graphs
$\left\\{\bigl{(}G_{i},r_{i}\bigr{)},g_{i}^{i+1}\right\\}_{i\in{\mathbb{N}}}$
given in Definition 2.5. By Lemma 2.7 it is a g-cell structure. Let
$\bar{a}=(a_{1},a_{2},a_{3},...)$,
$A=\left\\{(b_{1}^{k},b_{2}^{k},b_{3}^{k},...)\colon
k\in{\mathbb{N}}\right\\}$ and $B=\pi(A)$. We will show that $B$ is a closed
set in $G^{*}$ and it does not contain $\pi(\bar{a})$. Let
$b_{k}=(b_{1}^{k},b_{2}^{k},b_{3}^{k},...)\in A$ for some $k\in{\mathbb{N}}$.
By definition of the relation $r_{k+1}$, the element $a_{k+1}$ is related with
$b_{k+1}^{t}$ if and only if $t\geq k+1$. Hence, $(b_{k+1}^{k},a_{k+1})\notin
r_{k+1}$ this implies that $(b_{k},\bar{a})\notin r$. Since $k$ was chosen
arbitrarily, $\bar{a}$ is not related with some element of $A$. Therefore,
$\pi(\bar{a})\notin B$.

Let $\bar{x}=(x_{1},x_{2},x_{3},...)\in\pi^{-1}(G^{*}\backslash B)$. We shall
show that $\pi^{-1}(G^{*}\backslash B)\subset G_{\infty}\backslash A$. Notice
that if $\bar{b}\in\pi^{-1}(G^{*}\backslash B)$, then $\pi(\bar{b})\in
G^{*}\backslash B$ and thus $\pi(\bar{b})\notin B=\pi(A)$. It result that
$\bar{b}\notin A$ and this implies that $\bar{b}\in G_{\infty}\backslash A$.
Therefore, $\bar{x}\notin A$. Now, we will show that there exists
$k\in{\mathbb{N}}$ such that for every $n\in{\mathbb{N}}$, $x_{k}\neq
b_{k}^{n}$. Suppose that for every $k\in{\mathbb{N}}$ there exists
$n\in{\mathbb{N}}$ such that $x_{k}=b_{k}^{n}$. In particular for $k=1$, there
exists $n_{1}$ such that $x_{1}=b_{1}^{n_{1}}$. Since that
$x_{1}=g_{1}^{2}(x_{2})$ and for $k=2$ there exists $n_{2}$ such that
$x_{2}=b_{2}^{n_{2}}$, then $x_{2}=b_{2}^{n_{1}}$. Applying induction, we
suppose that $x_{t}=b_{t}^{n_{1}}$ for some $t\in{\mathbb{N}}$. Since
$x_{t}=g_{t}^{t+1}(x_{t+1})$ and for $k=t+1$ there exists $n_{t+1}$ such that
$x_{t+1}=b_{t+1}^{n_{t+1}}$, we get that $x_{t+1}=b_{t+1}^{n_{1}}$. Then
$x_{t}=b_{t}^{n_{1}}$ for every $t\in{\mathbb{N}}$. Hence, $\bar{x}\in A$
which is a contradiction. So, there exists $k\in{\mathbb{N}}$ such that for
each $n\in{\mathbb{N}}$, $x_{k}\neq b_{k}^{n}$.

We shall show that $g_{k}^{-1}(x_{k})$ is an open set and $\bar{x}\in
g_{k}^{-1}(x_{k})\subset\pi^{-1}(G^{*}\backslash B)$. Let
$\bar{w}=(w_{1},w_{2},w_{3},...)\in g_{k}^{-1}(x_{k})$, then $w_{k}=x_{k}$.
Now, if $\bar{w}=\bar{a}$, then $\bar{w}\in\pi^{-1}(G^{*}\backslash B)$
because $\pi(\bar{a})\in G^{*}\backslash B$. Suppose that
$\bar{w}\neq\bar{a}$, then $w_{k}=d_{k}^{n}$ or $w_{k}=c_{k,m}^{r}$ with
$n,m\in{\mathbb{N}}$ and $r\in\\{1,2\\}$. First we will consider the case
which $\bar{w}$ satisfies $w_{k}=d_{k}^{n}$ for some $n\in{\mathbb{N}}$. By
the Lemma 2.6, there exists $l>k$ such that $w_{l}=c_{l,t}^{r}$ for some
$t\in{\mathbb{N}}$ and $r\in\\{1,2\\}$. Therefore, by definition of $r_{l}$,
$(w_{l},b_{l}^{s})=(c_{l,t}^{r},b_{l}^{s})\notin r_{l}$ for every
$s\in{\mathbb{N}}$. Then $(\bar{w},b_{s})\notin r$ for each $b_{s}\in A$ this
implies that $\pi(\bar{w})\notin B$. Hence,
$\bar{w}\in\pi^{-1}(G^{*}\backslash B)$. Now, we will consider $\bar{w}$ such
that $w_{k}=c_{k,m}^{r}$ for some $m\in{\mathbb{N}}$ and $r\in\\{1,2\\}$. Then
$(c_{k,m}^{r},b_{k}^{s})\notin r_{l}$ for every $s\in{\mathbb{N}}$. Similarly
to the first case, we get that $\bar{w}\in\pi^{-1}(G^{*}\backslash B)$. We
conclude that $g_{k}^{-1}(x_{k})$ is an open set such that $\bar{x}\in
g_{k}^{-1}(x_{k})\subset\pi^{-1}(G^{*}\backslash B)$. Consequently
$\pi^{-1}(G^{*}\backslash B)$ is open this implies that $G^{*}\backslash B$ is
open in $G^{*}$. Therefore, $B$ is closed in $G^{*}$.

To show that $G^{*}$ is not regular, we will prove that there are not disjoint
open sets $U$ and $V$ in $G^{*}$ such that $\pi(\bar{a})\in U$ and $B\subset
V$. Let $U$, $V$ open sets in $G^{*}$ such that $\pi(\bar{a})\in U$ and
$B\subset V$. Then $\pi^{-1}(U)$ and $\pi^{-1}(V)$ are open sets in
$G_{\infty}$ which satisfy $\pi^{-1}(\pi(\bar{a}))\subset\pi^{-1}(U)$ and
$\pi^{-1}(B)\subset\pi^{-1}(V)$. Also, $\pi^{-1}(\pi(\bar{a}))=\\{\bar{a}\\}$.
Since $\bar{a}\in\pi^{-1}(U)$ and $\pi^{-1}(U)$ is open, there exists a basic
open $g_{j}^{-1}(u_{j})$ such that $\bar{a}\in
g_{j}^{-1}(u_{j})\subset\pi^{-1}(U)$. This implies that $a_{j}=u_{j}$ and thus
$g_{j}^{-1}(a_{j})\subset\pi^{-1}(U)$. By definition of $r_{j}$, we have that
$(a_{j},b_{j}^{j})\in r_{j}$. Consider the thread
$b_{j}=(b_{1}^{j},b_{2}^{j},b_{3}^{j},...,b_{j-1}^{j},b_{j}^{j},...)$. Then
$b_{j}\in A$ and thus, $b_{j}\in\pi^{-1}(B)\subset\pi^{-1}(V)$. Since
$\pi^{-1}(V)$ is open, for $b_{j}$ there exists a basic open
$g_{i}^{-1}(b_{i}^{j})$ such that $b_{j}\in
g_{i}^{-1}(b_{i}^{j})\subset\pi^{-1}(V)$. If $\bar{z}\in
g_{i+1}^{-1}(b_{i+1}^{j})$, then $z_{i+1}=b_{i+1}^{j}$ this implies that
$z_{i}=g_{i}^{i+1}(b_{i+1}^{j})=b_{i}^{j}$. Hence, $\bar{z}\in
g_{i}^{-1}(b_{i}^{j})$ and $g_{i+1}^{-1}(b_{i+1}^{j})\subset
g_{i}^{-1}(b_{i}^{j})$. Without losing generality, we suppose that $i>j$.
Remember that $L_{i+1}=L_{i}+i$, then $L_{i}+2\leq j+L_{i}+1\leq L_{i+1}$.
Applying the row 7 of the bonding function $g_{i}^{i+1}$, we get that
$g_{i}^{i+1}(c^{2}_{i+1,j+(L_{i}+1)})=b_{i}^{j}$. Now, we consider the thread

$\bar{c}=\bigl{(}b_{1}^{j},b_{2}^{j},b_{3}^{j},...,b_{j}^{j},...,b_{i}^{j},c^{2}_{i+1,j+(L_{i}+1)},c^{2}_{i+2,j+(L_{i}+1)+1},c^{2}_{i+3,j+(L_{i}+1)+2},...\bigr{)}.$

Then $\bar{c}\in g_{i}^{-1}(b_{i}^{j})$ and thus, $\bar{c}\in\pi^{-1}(V)$.
Define the thread $\bar{d}$ as

$\bar{d}=\bigl{(}a_{1},...,g_{i-1}^{i+1}(c^{1}_{i+1,j+(L_{i}+1)}),g_{i}^{i+1}(c^{1}_{i+1,j+(L_{i}+1)}),c^{1}_{i+1,j+(L_{i}+1)},c^{1}_{i+2,j+(L_{i}+1)+1},...\bigr{)}.$

By definition of $r_{i+k}$ with $k\in{\mathbb{N}}$, we have that
$\bigl{(}c^{2}_{i+k,j+(L_{i}+k)},c^{1}_{i+k,j+(L_{i}+k)}\bigr{)}\in r_{i+k}$.
Then $\bigl{(}\bar{d},\bar{c}\bigr{)}\in r$ and therefore
$\bar{d}\in\pi^{-1}(V)$. Since $i>j$, we get that $L_{i}+1\leq j+L_{i}+1\leq
L_{i}+i=L_{i+1}$ and applying the row 6 of the bonding function $g_{i}^{i+1}$,
it result that $g_{i}^{i+1}\bigl{(}c^{1}_{i+1,j+(L_{i}+1)}\bigr{)}=d_{i}^{j}$.
If $j=i-1$, by row 9 of the function $g_{i-1}^{i}$, we have that
$g_{i-1}^{i}\bigl{(}d_{i}^{j}\bigr{)}=a_{i-1}=a_{j}$. In other case ($j<i-1$),
applying the row 8 of the function $g_{i-1}^{i}$,
$g_{i-1}^{i}\bigl{(}d_{i}^{j}\bigr{)}=d_{i-1}^{i-2+j}$. Considering now the
function $g_{i-2}^{i-1}$, we have two cases again. If $j=i-2$, then
$g_{i-2}^{i-1}\bigl{(}d_{i-1}^{i-2+j}\bigr{)}=a_{i-2}=a_{j}$ and in the case
$j<i-2$, $g_{i-2}^{i-1}\bigl{(}d_{i-1}^{i-2+j}\bigr{)}=d_{i-2}^{i-3+j}$. Since
$i>j>1$, there exists $0<l<i$ such that $j=i-l$. If we continue with this
process until $j=i-l$, we will get that
$g_{i-l}^{i-l+1}\bigl{(}d_{i-l+1}^{i-l+j}\bigr{)}=a_{i-l}=a_{j}$. Therefore
$g_{j}^{i}\bigl{(}d_{i}^{j}\bigr{)}=a_{j}$. Consequently, $\bar{d}\in
g_{j}^{-1}(a_{j})$ and thus $\bar{d}\in\pi^{-1}(U)$. So, we get that
$\pi^{-1}(U)\cap\pi^{-1}(V)\neq\emptyset$. This mean that there exists
$\bar{z}\in G_{\infty}$ such that $\pi(\bar{z})\in U$ y $\pi(\bar{z})\in V$.
Hence, $\pi(\bar{z})\in U\cap V$and this implies that $U\cap V\neq\emptyset$.
Therefore, if $U$ and $V$ are open sets in $G^{*}$ such that $\pi(\bar{a})\in
U$ and $B\subset V$, then $U\cap V\neq\emptyset$. We conclude that $G^{*}$ is
not regular. ∎

Evidently, there is a net version of Definition 1.7, as was observed in [2]
for cell structures. Using this generalized definition one may check that
uniform spaces $(X,\mathcal{U})$ have a nontrivial g-cell structure
$\\{(G_{u},u)\\}_{u\in\mathcal{U}}$ where $G_{u}=X$ and $g_{u}^{u}=I_{X}$ for
all $u,v\in\mathcal{U}$. It would be interesting to determine what is the
class of topological spaces admitting a g-cell structure.

## References

  * [1] W. Debski and E.D. Tymchantyn.  Cell structures and completely metrizable spaces and their mappings.  Colloq. Math, 147:181–194, 2017.
  * [2] W. Debski and E.D. Tymchantyn.  Cell structures and topologically complete spaces.  Topol. Appl., 239:293–307, 2018.
  * [3] G. Elliott.  On the Classification of Inductive Limits of Sequences of Semisimple Finite-Dimensional Algebras.  J. Algebra, 38:29–44, 1976.
  * [4] R. Engelking.  General Topology.  Sigma Series in Pure Mathematics, 1989.
  * [5] S. Lefschetz.  On compact spaces.  Ann. Math., 32:521–538, 1931.
  * [6] C. Nash.  Topology and physics—a historical essay.  In I.M. James, editor, History of Topology, chapter 12, pages 359–415. North Holland, Amsterdam, 1999.
  * [7] M. Newman.  Networks.  Oxford University Press, Oxford, 2018.


∎

11institutetext:  1School of Physical Sciences, National Institute of Science
Education and Research. 22institutetext: P.O. Jatni, Khurda, Odisha, India.
PIN- 752050
22email<EMAIL_ADDRESS>
2 22email<EMAIL_ADDRESS>
*Corresponding Author

# Quantum Multiplayer Colonel Blotto Game

Joydeep Naskar1 Alan C. Maioli 2∗

(Received: 01 August 2020)

###### Abstract

In this work we successfully present a quantum version of the multiplayer
Colonel Blotto game. We find that players with access to the quantum
strategies has a advantage over the classical ones. The payoff is invariant
under the order of the operator’s strategies.

###### Keywords:

Colonel Blotto Game Multiplayer Quantum Games

##  1 Introduction

Quantum game theory is a branch of quantum information which overlaps with
game theory. The last one is an active field of mathematics with a plethora of
works and applications. Nowadays, physicists refer games from game theory as
classical games. One can rewrite a classical game using state vectors and
operators to depict strategies and measurement to obtain the payoff. Along
these lines, one can add new operators without classic analogue, and with
entanglement and superposition Meyer  new features appear. This process is
often called quantization of a game. Therefore, any quantum game must preserve
all features from its classical counterpart. Since 1999 several classical
games was brought to the quantum realm due to the intrinsic probabilistic
nature of quantum theory and game theory (for mixed strategies), such as
Prisoners’ Dilemma Eisert , Hawk-Dove Ahmad ; Tomassini , Quantum duel
Flitney_2004 ; Schmidt_2012  and recently Colonel Blotto
maioli2019quantization . The quantum games are investigated in order to find
new properties santos2020entanglement ; dhiman2020implementation ;
anand2020solving ; faleiro2020quantum ; du2003multiprisoner  or deep
connections with game theory brunner2013connection .
Our purpose is to present a quantum version of multiplayer Colonel Blotto
game. Therefore we use a similar procedure made by the Maioli et al
maioli2019quantization , however there is a subtle difference near the end of
the game. In other words, we use the density matrix formalism in order to
perform the measurements and obtain the payoff. This choice has the advantage
to separate the subsystems via partial trace. One can find similar use of the
density matrix formalism in quantum games in the reference guo2008survey , it
also contains an excellent description of quantum games. In the construction
by Maioli et al in maioli2019quantization , 2 players, namely Blotto and enemy
shared a common qubit for representation of their soldiers. This led to a
philosophical discussion introduced by naskarcomment2020 . In the following
construction, we have resolved that conundrum.
A classical description of the game is given at section (2). In this work we
successfully present a quantum version of the multiplayer Colonel Blotto game
on section (3), and this is corroborated by the examples on subsection (4.1).
Therefore, we investigate the payoff’s alteration due to the existence of
quantum strategies at subsection (4.2).

##  2 The Classical Game

The Colonel Blotto game can be formulated in different waysgenblotto2020 ;
discblotto2008 ; contblotto1950 ; jayantimultiplayerblotto2020 . In this
section we present a classical version of the multiplayer Colonel Blotto game,
a similar version of the game can be found at jayantimultiplayerblotto2020 .
We consider the number of players $N>2$ and it has $n$ number of battlefields.
Player 1 is defined as Blotto, while other players are the enemies, player 2
is enemy 1 and so on. Any strategy chosen by player $j$ is presented as a
vector,

$\mathbf{x}_{j}=(x_{1}^{j},x_{2}^{j},...,x_{n}^{j}),$ (1)

where $x_{k}^{j}$ is the number of soldiers the player $j$ allocate on the
battlefield $k$. We are considering a continuous game contblotto1950 , so the
numbers $x_{k}^{j}$ not need to be integer, then the set of all the possible
strategies for player $j$

$\mathbb{S}_{j}=\bigg{\\{}\mathbf{x}_{j}\in\mathbb{R}^{+n}\bigg{|}\sum_{k=1}^{n}x_{k}^{j}=X_{j}\bigg{\\}},$
(2)

where $\mathbb{R}^{+n}$ stands for the n-dimensional non-negative real set and
$X_{j}$ is the total number of soldiers. Then each player choose a strategy
and the payoff of the $j^{th}$ player can be calculated

$\$_{j}(X)=\sum_{k=1}^{n}sgn\left(x_{k}^{j}-x_{k,j}^{max}\right),$ (3)

where $x_{k,j}^{max}$ is the highest number of soldiers among all the others
players

$x_{k,j}^{max}=\max\\{x_{k}^{1},x_{k}^{2},...,x_{k}^{j-1},x_{k}^{j+1},...,x_{k}^{N}\\},$
(4)

and

$sgn(x)=\left\\{\begin{array}[]{@{}ll@{}}+1,&\text{if}\ x\geq 0\\\
0,&\text{if}\ x=0\\\ -1,&\text{if}\ x<0\end{array}\right..$ (5)

##  3 The Quantum Game

In order to define the quantum description of the game we choose the Hilbert
space to be a direct product of $k$ soldier spaces, each corresponding to a
player and the battlefield space

$\mathbb{H}=\mathbb{H}_{S}^{1}\otimes\mathbb{H}_{S}^{2}\otimes...\otimes\mathbb{H}_{S}^{N}\otimes\mathbb{H}_{T},$
(6)

We use 2-dimensional soldier spaces $\mathbb{H}_{S}^{j}$ for any player $j$
and a n-dimensional battlefield space $\mathbb{H}_{T}$. The game begins in an
initial state $\ket{\psi_{i}}$ that belongs to the Hilbert space $\mathbb{H}$.
We represent our 2-dimensional space basis vectors as $\ket{0}_{j}$ and
$\ket{1}_{j}$, where

$\ket{0}_{j}=\begin{pmatrix}1\\\ 0\end{pmatrix},\ \ \ \ \
|1\rangle_{j}=\begin{pmatrix}0\\\ 1\end{pmatrix},$ (7)

and the battlefield’s basis $\\{|T_{i}\rangle\\}$ with $i=1,2...n$ where,

$|T_{1}\rangle=\begin{pmatrix}1\\\ 0\\\ \vdots\\\ 0\end{pmatrix},\ \ \ \ \
|T_{2}\rangle=\begin{pmatrix}0\\\ 1\\\ \vdots\\\ 0\end{pmatrix},\ \ \ \dots\ \
\ |T_{n}\rangle=\begin{pmatrix}0\\\ 0\\\ \vdots\\\ 1\end{pmatrix}.$ (8)

The initial state of the system $\Psi_{i}$ is written as

$\ket{\Psi_{i}}=\ket{0}_{1}\otimes\ket{0}_{2}\otimes...\otimes\ket{0}_{N}\otimes\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\ket{T_{k}},$
(9)

which we write in notation

$\ket{\Psi_{i}}=\prod^{\otimes
N}_{j=1}\ket{0}_{j}\otimes\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\ket{T_{k}}.$ (10)

Blotto’s games contain an usual condition which is $X_{B}\geq X_{j}$, i. e.
Blotto has the highest total number of soldiers. The number of soldiers Blotto
has i.e, $X_{B}$ is public information known to all players. To allocate
troops, the $j^{th}$ player applies the unitary operator,

$\hat{U}_{j}=\sum_{k=1}^{n}\mathbb{I}_{1}\otimes...\otimes\mathbb{I}_{j-1}\otimes\hat{R}(\lambda^{j}_{k})\otimes\mathbb{I}_{j+1}\otimes...\otimes\mathbb{I}_{N}\otimes\hat{\Pi}_{k},$
(11)

where $j^{th}$ is the player, $\hat{\Pi}_{k}$ is the projector
$|T_{k}\rangle\langle T_{k}|$, the operator $\hat{R}(\lambda^{j}_{k})$ is the
rotation matrix

$\hat{R}(\lambda^{j}_{k})=\begin{pmatrix}\cos(\lambda^{j}_{k})&-\sin(\lambda^{j}_{k})\\\
\sin(\lambda^{j}_{k})&\cos(\lambda^{j}_{k})\end{pmatrix},$ (12)

and $\lambda^{j}_{k}$ is the angle which the vector $\ket{0}_{j}$ will be
rotated. The rotation angle $\lambda^{j}_{k}$ is defined as

$\lambda^{j}_{k}=\frac{\pi}{2}\frac{x^{j}_{k}}{X_{B}},$ (13)

where $x^{j}_{k}$ stands for the number of $j^{th}$ player’s troops in the
$k^{th}$ battlefield. Therefore, each player rotates the $\ket{0}_{j}$ towards
$\ket{1}_{j}$. This definition is important to depict the classical game,
whoever has the highest projection on to $\ket{1}_{j}$ win a specific
battlefield.The constraints are

$\sum_{k=1}^{n}x^{j}_{k}=X_{j}.$ (14)

This ensures that the rotation angle $\lambda^{k}_{j}\in[0,\frac{\pi}{2}]$. To
introduce quantum strategies, we generalise the allocation operator for
$j^{th}$ player to

$\hat{U}_{j}=\sum_{k=1}^{n}\mathbb{I}_{1}\otimes...\otimes\mathbb{I}_{j-1}\otimes\hat{Q}(\lambda^{j}_{k},\phi^{j}_{k})\otimes\mathbb{I}_{j+1}\otimes...\otimes\mathbb{I}_{N}\otimes\hat{\Pi}_{k},$
(15)

where

$\hat{Q}(\lambda^{j}_{k},\phi^{j}_{k})=\begin{pmatrix}e^{i\phi^{j}_{k}}\cos(\lambda^{j}_{k})&-\sin(\lambda^{j}_{k})\\\
\sin(\lambda^{j}_{k})&e^{-i\phi^{j}_{k}}\cos(\lambda^{j}_{k})\end{pmatrix},$
(16)

where $\phi^{j}_{k}$ is a phase that represent a quantum resource and when
$\phi^{j}_{k}=0$ one can easily verify
$\hat{Q}(\lambda^{j}_{k},0)=\hat{R}(\lambda^{j}_{k})$. To introduce
entanglement in our initial state, we use an operator ${\hat{J}}$. In order to
preserve the structure of a quantum game,

$\ket{\Psi_{f}}=\hat{J}^{\dagger}\hat{U}_{N}...\hat{U}_{1}\hat{J}\ket{\Psi_{i}},$
(17)

the ${\hat{J}}$ must commute with all classical strategies,

$\left[\hat{J},\sum_{k=1}^{n}\mathbb{I}_{1}\otimes...\otimes\mathbb{I}_{j-1}\otimes\hat{R}(\lambda^{k}_{j})\otimes\mathbb{I}_{j+1}\otimes...\otimes\mathbb{I}_{N}\otimes\hat{\Pi}_{k}\right]=0.$
(18)

A common procedure is to define the entanglement operator as

$\hat{J}=e^{i\gamma\hat{A}/2},$ (19)

where $\gamma\in\left[0,\pi/2\right]$ is a parameter that introduce the
entanglement. Therefore, if the operator $\hat{A}$ satisfy $\hat{A}^{2n}=1$,
then

$\hat{J}=\cos(\gamma/2)\hat{\mathbb{I}}+i\sin(\gamma/2)\hat{A}.$ (20)

We choose $\hat{A}$

$\hat{A}=(-1)^{N}\begin{pmatrix}0&1\\\
-1&0\end{pmatrix}_{1}\otimes...\otimes\begin{pmatrix}0&1\\\
-1&0\end{pmatrix}_{N}\otimes\begin{pmatrix}\pm i&0&\ldots&0\\\ 0&\pm
i&\ldots&0\\\ \vdots&\vdots&\ddots&\vdots\\\ 0&0&\ldots&\pm i\\\
\end{pmatrix}_{n\times n},$ (21)

i.e,

$\hat{A}=(-1)^{N}\prod^{\otimes N}_{j=1}\begin{pmatrix}0&1\\\
-1&0\end{pmatrix}_{j}\otimes\begin{pmatrix}\pm i&0&\ldots&0\\\ 0&\pm
i&\ldots&0\\\ \vdots&\vdots&\ddots&\vdots\\\ 0&0&\ldots&\pm i\\\
\end{pmatrix}_{n\times n},$ (22)

which have an important condition: in the last matrix, at least one of the
elements must have a different sign when compared to the others. All the
players play sequentially and the final state obtained at equation (17), then
we construct the density matrix

$\rho=\ket{\Psi_{f}}\bra{\Psi_{f}}.$ (23)

To find the payoff of the $j^{th}$ player, partial trace is taken

$\rho_{j}=tr_{X}\rho,$ (24)

where X runs over all players except j, i.e, $X=\\{1,2,...,j-1,j+1,...,N\\}$.
Therefore we perform a measurement

$M_{k,j}=tr\left[\left(\ket{1}_{j}\bra{1}_{j}\otimes\ket{T_{k}}\bra{T_{k}}\right)\rho_{j}\right].$
(25)

For a given battlefield k, we define the quantity

$M_{k,j}^{max}=\max\\{M_{k,1},M_{k,2},...,M_{k,j-1},M_{k,j+1},...M_{k,N}\\},$
(26)

for all n battlefields. The idea behind the definition of equation (26) is to
obtain the highest value of the measurements $M_{k,i}$ for all $i$ players
except $i=j$. This quantity is useful to define the payoff for the $j^{th}$
player

$\$_{j}=\sum_{k\neq j}^{n}sgn\left[M_{k,j}-M_{k,j}^{max}\right].$ (27)

The idea behind the definition of the payoff is to compare the measurement
made by player $j$ with the highest value among all the others players.

##  4 Results

In this section we have shown examples in order to elucidate some features of
the quantum game. In subsection 4.1 we show a situation where all players have
access only to classical strategies. In other hand, we allow enemy 3 to
perform a quantum move in subsection 4.2.

###  4.1 Classical strategy

Consider an example with 3 players respectively fighting over 2 territories.
Blotto has 6 soldiers, enemy 1 has 4 soldiers and enemy 2 has 3 soldiers. Let
Blotto choose the strategy (3,3), enemy 1 choose (3,1) and enemy 2 choose
(0,3) as shown in Table (1).

|  Battlefield $\rightarrow$
---
Players$\downarrow$
Battlefield 1 | Battlefield 2
Blotto | 3 | 3
enemy 1 | 3 | 1
enemy 2 | 0 | 3
Table 1: The allocation of soldiers for all players over 2 battlefields. Each
number of soldier allocated to each battlefield has a correspondence with an
angle, presented at equation (13).

In battlefield 1, both Blotto and enemy 1 tied whereas in Battlefield 2, both
Blotto and enemy 2 tied. Clearly, the payoffs for Blotto, enemy 1 and enemy 2
are $0$, $-1$ and $-1$ respectively. We will now play the same strategies in
quantum formalism. Blotto has maximum number of soldiers $X_{B}=6$. Thus the
angles to be rotated in the corresponding qubits of the initial state are
$(\frac{\pi}{4},\frac{\pi}{4})$, $(\frac{\pi}{4},\frac{\pi}{12})$ and
$(0,\frac{\pi}{4})$ by Blotto, enemy 1 and enemy 2 respectively. We tabulate
the measurements in Table (2). Therefore, the payoffs can be calculated with
the help of equation (27) which are $0$, $-1$ and $-1$, for Blotto, enemy 1
and enemy 2, respectively. Through this example, we verified that the
classical payoffs agree with their counterpart classical strategies in the
quantum game. In addition, one can calculate the measurements $M_{k,j}$ via

$M_{k,j}=\frac{1}{n}\sin^{2}{\lambda^{k}_{j}},$ (28)

when all players are restricted to play with classical strategies

$\phi^{k}_{j}=0\quad\forall\quad j\leq N\quad{\rm and}\quad k\leq n.$ (29)

|  Battlefield $\rightarrow$
---
Players $\downarrow$
Battlefield 1 | Battlefield 2
Blotto | $\frac{1}{4}$ | $\frac{1}{4}$
enemy 1 | $\frac{1}{4}$ | $\frac{1}{2}\sin^{2}{\frac{\pi}{12}}$
enemy 2 | 0 | $\frac{1}{4}$
Table 2: The values of the measurements for the strategies presented at table
1

###  4.2 A quantum move

In this subsection we shown some examples where the enemy 2 has access to
quantum strategy. Through the following cases we consider the maximum
entanglement which is $\gamma=\pi/2$.

####  4.2.1 First example

First we analyse the case which both Blotto and enemy 1 choose the same
strategy presented at table (1). However enemy 2 will choose the strategy
$x_{1}^{3}=0$, $x_{2}^{3}=3$, $\phi_{2}^{3}=0$ and $\phi_{1}^{3}$ for several
values. The payoffs are plotted at figure (1) for each $\phi_{1}^{3}$ that
belongs to the domain $\left[0,\pi/2\right]$, for all players. The enemy 2 can
increase its own payoff if he choose a quantum strategy with
$\phi_{1}^{3}>\pi/4$.

Figure 1: Plot of the payoff of all players. The blue dot-dashed line
corresponds to the Blotto’s payoff, the dotted black line to the enemy’s 1
payoff and the red full line to the enemy’s 2 payoff. In this case the quantum
resource on the second battlefield is $\phi_{2}^{3}=0$. One can see that enemy
3 can have positive payoff for some values of $\phi_{1}^{3}$.

####  4.2.2 Second example

Another example, which is illustrated at figure (2), contains the same values
used on the first example, sub-subsection (4.2.1). However, we let
$\phi_{2}^{3}$ vary. We observe no change of players’ payoff for any values
between $0<\phi_{2}^{3}<\pi/2$. In other words, the payoffs has a
discontinuity jump at $\phi_{2}^{3}=0$, as one can see after the comparison
between the figures (1) and (2). Therefore, considering both previous
scenarios, it is better for the enemy 2 to choose values of quantum resources
$0<\phi_{2}^{3}<\pi/2$ and $\phi_{1}^{3}>\pi/4$, which gives $\$_{3}=+2$ .

Figure 2: Plot of the payoff of all players. The blue dot-dashed line
corresponds to the Blotto’s payoff, the dotted black line to the enemy’s 1
payoff and the red full line to the enemy’s 2 payoff.

##  5 Conclusion

We have presented a quantum version of the multiplayer Colonel Blotto game.
Through our work we see that the payoff do not alter under the change of the
order of the operators. This happens because the operators’ strategies of
different players commute with each other. We also find an great advantage for
players who has access to quantum strategies.

###### Acknowledgements.

Part of this work is supported by INSPIRE, Dept. of Science & Technology,
Govt. of India.

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