Uniqueness Conditions of the Representation of a Topology in ${ Top}(X)$

Abstract

In this article motivated by unique representation of a topology $\tau$ on X, we introduce some kind of representation for $\tau$ in ${ Top}(X)$ and find the equivalent condition for a topological space to have unique representation. As the main part of the paper, we introduce the natural representation for every topology $\tau$ on a set X of the form $\bigwedge _{a\in L(X)}\mathscr {T}(a,\mathscr {W}_a)$, where $\mathscr {W}_a$ is the set of all neighborhoods and deleted neighborhoods of a and L(X) is the set of all non-isolated points of X, and we characterize the spaces whose topology has unique representation. Also it is proved that every first countable Hausdorff space has unique representation property.

1 Introduction

The work of Birkhoff [3] in 1936, concerning ${ Top}(X)$, the lattice of all topologies on a set X, which was the starting point of interest in this lattice, has played a major role in both general topology and combinatorial set theory. In contrast to the importance of this lattice, it seems its existence in the literature, took a rather long time, to catch the eyes of the interested authors, see for example, [8, 10] and the references therein, and for a more recent one see [11]. The early problems of interest in this lattice were mostly of combinatorial nature and, in particular, the complementation problems in this lattice have received much attention. Hartmanis ( in [8]) was the first who settled the latter problem in the context of finite sets and asked if $\textit{Top}(X)$ is complemented in case X is infinite. Steiner in his Ph.D. thesis has interestingly shown that $\textit{Top}(X)$, is indeed complemented (i.e., each element in $\textit{Top}(X)$ has a complement), see [10]. In contrast to distributive lattices, in which complements, if they exist, are unique, there are still many open problems concerning the number of complements of elements in $\textit{Top}(X)$, see [13]. Although these combinatorial problems were at first the main motivation, for interest in $\textit{Top}(X)$, but the importance of the latter lattice in topology itself is folklore too, for a good history concerning $\textit{Top}(X)$, see [9]. In this note, we are also interested in $\textit{Top}(X)$ from topological point of view. Clearly, $\textit{Top}(X)$ is a complete lattice, where the trivial topology and the discrete topology are 0 and 1 in $\textit{Top}(X)$, respectively. Fröhlich proved that, there exists a representation for every topology $\tau$ on X, (see [6]). In this article, for every representation of a topology $\tau$ on a set X, we introduce a natural one for $\tau$ over L(X).

Throughout this paper, L(X) stands for the set of non-isolated points of a topological space X, I(X) stands for the set of isolated points of X. By ${\mathscr {F}}$, we mean a filter on a set X, and by ${\mathscr {U}}_x$ we mean the neighborhood system of a point $x\in X$. Clearly, the set of all neighborhoods and deleted neighborhoods of x is a filter on X, we denote it by ${\mathscr {W}}_x$. Let X be a non-empty set and $\varnothing \ne A\subseteq X$. Then $\{F\subseteq X:\;A\subseteq F\}$ is a filter on X which is denoted by $\mathscr {F}(A)$; in particular if $A=\{x\}$, we use $\mathscr {F}(x)$ instead of $\mathscr {F}(\{x\})$.

Let us recall some useful definitions and theorems from [6, 10].

Definition 1.1

Suppose that $\mathscr {F}$ is a filter on a set X and $\varnothing \ne A\subseteq X$. We define $\mathscr {T}(A, \mathscr {F})=\mathscr {P}(X{\setminus }A)\cup \mathscr {F}$, where $\mathscr {P}(X{\setminus }A)$ is the collection of all subsets of $X{\setminus }A$. Clearly, $\mathscr {T}(A, \mathscr {F})$ is a topology on X. If $A=\{a_1, a_2, \ldots , a_n\}$, where $n\in \mathbb N$, then we use $\mathscr {T}(a_1, a_2, \ldots , a_n, \mathscr {F})$ instead of $\mathscr {T}(A, \mathscr {F})$.

A topology $\tau$ on a set X is called ultraspace (resp., infraspace) if the discrete topology (resp., the trivial topology) on X is the only topology finer (resp., coarser) than $\tau$. However, in this article, we prefer an ultraspace topology $\tau$ on a set X to be called an ultratopology and instead the space $(X, \tau )$ be called ultraspace. It is easy to predict the form of any infraspace and to show that any topology $\tau$ on a set X is the supremum of infraspaces coarser than $\tau$. In contrast to the forms of infraspaces, the forms of ultraspace are not as easily predicted, fortunately, Fröhlich in the next result has given us this form.

Theorem 1.2

• (a) The ultratopologies on X are exactly of the form $\mathscr {T}(x,\mathscr {F})$, where $x\in X$ and $\mathscr {F}$ is an ultrafilter on X, where $\mathscr {F}\ne \mathscr {F}(x)$.

• (b) There is a one-to-one correspondence between ultraspaces and ordered pairs $(x,\mathscr {F})$, where ${\mathscr {F}}\ne {\mathscr {F}}(x)$.

• (c) Every topology $\tau$ on X is the infimum of the ultratopologies on X which are finer than $\tau$.

Let X be an arbitrary set and $\mathscr {F}$ be an ultrafilter on X. The ultrafilter $\mathscr {F}$ is either principal (i.e., $|\cap \mathscr {F}|=1$) or non-principal (consequently, $\cap \mathscr {F}=\varnothing$). When $\mathscr {F}$ is a principal (respectively, non-principal) ultrafilter, then the ultratopology $\mathscr {T}(a,\mathscr {F})$ is called principal (respectively, non-principal) ultratopology, where $a\in X$.

Theorem 1.3

([10, Theorem 1.1]) A topology $\tau$ on X is a $T_1$-topology if and only if it is the infimum of a family of non-principal ultratopologies.

2 The Topology $\mathscr {T}(a,\mathscr {F})$

We know that if $\mathscr {F}$ is a free filter (i.e., $\bigcap \mathscr {F}=\varnothing )$ on X and $a\in X$, then the topology $\mathscr {T}(a,\mathscr {F})$ is a $T_1$-topology, but the converse is not true. For example, let $\mathscr {F}_1=\mathscr {F}(a)\cap \mathscr {F}_{\circ }$ where $\mathscr {F}_{\circ }=\{G\subseteq X:~G\ne \varnothing , X\backslash G \text { is finite}\}$. Then $\mathscr {T}(a,\mathscr {F}_1)$ is a $T_1$-topology, but $\mathscr {F}_1$ is not free.

By noting that every free ultrafilter contains $\mathscr {F}_{\circ }$ and every point in $X{\setminus }\{a\}$ is an isolated point in X with the topology $\mathscr {T}(a,\mathscr {F})$, we have the following straightforward fact which somehow shows that some cases of the above mentioned converse hold.

Proposition 2.1

Let $\mathscr {F}$ be a filter on a set X, $a\in X$, and $a\not \in \cap \mathscr {F}$. Then the following statements are equivalent:

• (1) $\mathscr {F}$ is a free filter on X.

• (2) $\mathscr {T}(a,\mathscr {F})$ on X is Hausdorff.

• (3) $\mathscr {T}(a,\mathscr {F})$ on X is $T_1$.

• (4) $\mathscr {F}_{\circ }\cap \mathscr {F}(a)\subseteq \mathscr {F}$.

Now a natural question arises: “when does the equality $\mathscr {T}(a,{\mathscr {F}}_1)=\mathscr {T}(a,{\mathscr {F}}_2)$ imply that ${\mathscr {F}}_1={\mathscr {F}}_2$?”

Theorem 2.2

Let $a\in X$ and $\tau =\mathscr {T}(a,\mathscr {F}_1)=\mathscr {T}(a,\mathscr {F}_2)$. Then the following statements are equivalent:

• (1) $\mathscr {F}_1=\mathscr {F}_2$.

• (2) Either $X\backslash \{a\}\in \mathscr {F}_1\cap \mathscr {F}_2$ or $X\backslash \{a\}\not \in \mathscr {F}_1\cup \mathscr {F}_2$.

• (3) Either $\mathscr {F}_1,\;\mathscr {F}_2\subseteq \mathscr {F}(a)$ or $\mathscr {F}_1,\;\mathscr {F}_2\nsubseteq \mathscr {F}(a)$.

• (4) $\mathscr {F}_1{\setminus }{\mathscr {U}}_a=\mathscr {F}_2\backslash {\mathscr {U}}_a$.

Proof

(1) $\Rightarrow$ (2). It is clear.

(2) $\Rightarrow$ (3). It is easy to see that for any filter ${\mathscr {F}}$, we have ${\mathscr {F}}\subseteq {\mathscr {F}}(a)$ if and only if $X{\setminus }\{a\}\notin {\mathscr {F}}$. Hence, this implication is obvious.

(3) $\Rightarrow$ (4). Assume that $F\in \mathscr {F}_1{\setminus }{\mathscr {U}}_a$. Therefore, $a\notin F$ and so ${\mathscr {F}_1}\nsubseteq {\mathscr {F}}(a)$. Hence, ${\mathscr {F}_2}\nsubseteq {\mathscr {F}}(a)$ and so $X{\setminus }\{a\}\in {\mathscr {F}}_2$. On the other hand, it is clear that $F\cup \{a\}\in \mathscr {T}(a,{\mathscr {F}}_1)$. Thus, $F\cup \{a\}\in \mathscr {T}(a,{\mathscr {F}}_2)$ and so $F\cup \{a\}\in {\mathscr {F}}_2$. Consequently, $F=(F\cup \{a\})\cap X{\setminus }\{a\}\in {\mathscr {F}}_2{\setminus }{\mathscr {U}}_a$. Therefore, ${\mathscr {F}}_1{\setminus }{\mathscr {U}}_a\subseteq {\mathscr {F}}_2{\setminus }{\mathscr {U}}_a$. Similarly, ${\mathscr {F}}_2{\setminus }{\mathscr {U}}_a\subseteq {\mathscr {F}}_1{\setminus }{\mathscr {U}}_a$ and hence ${\mathscr {F}}_1{\setminus }{\mathscr {U}}_a={\mathscr {F}}_2{\setminus }{\mathscr {U}}_a$.

(4) $\Rightarrow$ (1). Without loss of generality, it suffices to prove that ${\mathscr {F}}_1\subseteq {\mathscr {F}}_2$. Suppose that $F\in {\mathscr {F}}_1$. If $a\in F$, then we are done. Now, assuming $a\notin F$, it follows that $F\in {\mathscr {F}}_1{\setminus }{\mathscr {U}}_a={\mathscr {F}}_2{\setminus }{\mathscr {U}}_a$. Therefore, ${\mathscr {F}}_1\subseteq {\mathscr {F}}_2$. $\square$

Corollary 2.3

Let $\mathscr {F}_1$, $\mathscr {F}_2$ be two ultrafilters on a set X and $a\in X$. Then $\mathscr {T}(a,\mathscr {F}_1)=\mathscr {T}(a,\mathscr {F}_2)$ implies that $\mathscr {F}_1=\mathscr {F}_2$.

Corollary 2.4

Let $\tau =\bigwedge ^n_{i=1}\mathscr {T}(a_i,\mathscr {F}_i)=\bigwedge ^n_{i=1}\mathscr {T}(a_i,\mathscr {F}^{\prime }_i)$ be a topology on X, such that $\mathscr {F}_i, \mathscr {F'}_i\nsubseteq \mathscr {F}(a_i)$ for each $1\le i\le n$. Then $\mathscr {F}_i=\mathscr {F}^{\prime }_i$, for all $1\le i\le n$, where $n\in \mathbb N$.

Proof

By Theorem 2.2, it suffices to show that $\mathscr {T}(a_i,\mathscr {F}_i)=\mathscr {T}(a_i,\mathscr {F'}_i)$ for each $1\le i\le n$. Let $G\in \mathscr {T}(a_i,\mathscr {F}_i)$ and without loss of generality we can assume that $a_i\in G$. Since $G\in \mathscr {F}_i$, we infer that $G\backslash A\in \mathscr {F}_i$, where $A=\{a_1,\cdots , a_n\}\setminus \{a_i\}$. Clearly, $G\backslash A\in \tau$ and consequently $G\in \mathscr {T}(a_i,\mathscr {F'}_i)$. $\square$

Clearly, $\mathscr {T}(a_1,a_2,...,a_n,\mathscr {F})=\bigwedge ^n_{i=1}\mathscr {T}(a_i,\mathscr {F})$. More generally, if $A\subseteq X$ and $\mathscr {F}_i$ is a filter on X, for all $i\in I$, then $\bigwedge \limits _{x\in A}(\bigwedge \limits _{i\in I}\mathscr {T}(x,\mathscr {F}_i))=\bigwedge \limits _{i\in I}(\bigwedge \limits _{x\in A}\mathscr {T}(x,\mathscr {F}_i))=\mathscr {T}(A,\bigcap \limits _{i\in I}\mathscr {F}_i)$.

3 The Uniqueness Conditions of the Representation of a Principal Topology in Top(X)

Let us recall the definition of the principal topology in [10].

Definition 3.1

A topology $\tau$ on X is said to be a principal topology, when $\tau$ is the infimum of all principal ultratopologies on X finer than $\tau$, i.e., $\tau =\bigwedge \{\mathscr {T}(x,\mathscr {F}(y)):~ \tau \le \mathscr {T}(x,\mathscr {F}(y))\}$.

Now, we are looking for the conditions under which the representation of a principal topology $\tau$ on X is unique.

Remark 3.2

Let $\tau$ be a topology on X and $x\in X$. We define the set $B_x=\{y\in X:~\tau \le \mathscr {T}(x,\mathscr {F}(y))\}$. In [10], it is shown that if $\tau$ is a principal topology on X, then $B_x$ is open in X, for every $x\in X$. We can simply see that $y\in B_x$ if and only if $\;\mathscr {U}_x\subseteq \mathscr {U}_y$ and hence $B_x=\{y\in X:~\mathscr {U}_x\subseteq \mathscr {U}_y\}$ for all $x\in X$, also $B_x=B_y$ if and only if $\;{\mathscr {U}}_x={\mathscr {U}}_y$. Clearly, $B_x=\cap \mathscr {U}_x$ for all $x\in X$. Therefore, if $B_x$ is an open set in X, then $B_x$ is the smallest open subset containing x. Thus, by these facts and [10, Theorem 2.3], it follows that $B_x$ is open in X, for every $x\in X$ if and only ifthe topology on X is principal.

Theorem 3.3

Let $\tau$ be a topology on X. Then the following statements are equivalent:

• (1) $\tau$ is a principal topology on X.

• (2) For every $x\in L(X)$, there exists $A_x\subseteq X$ such that $x\not \in A_x$ and

  $$\begin{aligned} \tau =\bigwedge _{x\in L(X)}\mathscr {T}(x,\mathscr {F}(A_x)). \end{aligned}$$

• (3) $B_x$ is open, for each $x\in X$.

• (4) $\cap \mathscr {B}\in \tau$, for each subcollection $\mathscr {B}$ of $\tau$.

Proof

By Definition 3.1, we can simply see that $(1)\Leftrightarrow (2)$ and by Remark 3.2, $(1)\Leftrightarrow (3)$ is evident. Now it suffices to show that $(3)\Leftrightarrow (4)$. (3) $\Rightarrow$ (4). Let $\mathscr {B}$ be an arbitrary subcollection of $\tau$. We show that $\cap \mathscr {B}\in \tau$. If $\cap \mathscr {B}=\varnothing$, we are done. Let $x\in \cap \mathscr {B}$. Then $x\in B$, for all $B\in {\mathscr {B}}$. Since $B_x\subseteq B$, for all $B\in {\mathscr {B}}$, we infer that $B_x\subseteq \cap \mathscr {B}$. Thus, $x\in B_x\subseteq \cap \mathscr {B}$ for each $x\in \cap {\mathscr {B}}$, i.e., $\cap \mathscr {B}$ is open. (4) $\Rightarrow$ (3). It is evident, since $B_x=\bigcap {\mathscr {U}}_x$. $\square$

Proposition 3.4

Let $\tau$ be a topology on X. Then the following statements are equivalent:

• (1) $\tau$ is a $T_1$-topology on X.

• (2) $B_x=\{x\}$, for all $x\in X$.

• (3) For every $x,y\in X$, if $B_y\subseteq B_x$, then $y=x$.

Proof

The proof is straightforward. $\square$

Definition 3.5

Let $\tau$ be a principal topology on X and “$\le$” be a preorder relation on X defined by $a\le b: \Leftrightarrow \mathscr {U}_a\subseteq \mathscr {U}_b\;$, for all $a,b\in X$. Then we say there is a chain with length n from x to y if there are elements $x=x_0< x_1<\ldots< x_{n-1}< x_n=y$ in X, $\;n\ge 1\;$, such that $\mathscr {U}_{x_0}\subset \mathscr {U}_{x_1}\subset \mathscr {U}_{x_2}\subset \ldots \subset \mathscr {U}_{x_{n-1}}\subset \mathscr {U}_{x_n}$.

Now let us recall the following definition from [10].

Definition 3.6

A collection $\mathscr {D}$ of distinct principal ultratopologies

$$\begin{aligned} \{\mathscr {T}(x_1,\mathscr {F}(x_2)),\mathscr {T}(x_2,\mathscr {F}(x_3)),\ldots ,\mathscr {T}(x_n,\mathscr {F}(x_1))\} \end{aligned}$$

is called a cycle of order $n\in \mathbb N$. We say that $\{x_1, \ldots , x_n\}$ is a cycle (of order n) in the topological space X if $\tau \le \mathscr {T}(x_1, \mathscr {F}(x_2))\wedge \mathscr {T}(x_2, \mathscr {F}(x_3))\wedge \cdots \wedge \mathscr {T}(x_n, \mathscr {F}(x_1))$.

Suppose that $(X,\tau )$ is a principal topological space, $x\in L(X)$. Then, clearly, $[x]=\{y\in X:~B_x=B_y\}$ contains every cycle containing x. Also, it is easily observed that $\mathscr {T}(x_1, \mathscr {F}(x_2))\wedge \mathscr {T}(x_2, \mathscr {F}(x_3))=\mathscr {T}(x_1, \mathscr {F}(x_2))\wedge \mathscr {T}(x_2, \mathscr {F}(x_3))\wedge \mathscr {T}(x_1, \mathscr {F}(x_3))$ and $\mathscr {T}(x_1, \mathscr {F}(x_2))\wedge \mathscr {T}(x_2,\mathscr {F}(x_3))\wedge \mathscr {T}(x_3, \mathscr {F}(x_1))=\mathscr {T}(x_3,\mathscr {F}(x_2))\wedge \mathscr {T}(x_2, \mathscr {F}(x_1))\wedge \mathscr {T}(x_1,\mathscr {F}(x_3)).$

By the latter discussion, the following proposition is now immediate.

Proposition 3.7

Let $(X,\tau )$ be a non-discrete principal space. Then the representation $\tau =\wedge \{\mathscr {T}(x, \mathscr {F}(y)): x\in L(X), y\in X ~\mathrm{{ and }} ~\tau \le \mathscr {T}(x, \mathscr {F}(y))\}$ is unique (in the sense that the representation is irreducible) if and only if there is no cycle of order $n>2$ and no chain with length $n>1$ in the representation, i.e., if and only if $\;L(X)$ is the disjoin union of open sets which have exactly two points.

The following proposition is another unique representation for a principal topology.

Proposition 3.8

Let $(X,\tau )$ be a non-discrete principal space. Then we have $\tau =\wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {F}}(B_x))$ and if $\tau =\wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {F}}(C_x))$, where $C_x$ is an open set containing x, for every $x\in X$, then it follows that $B_x=C_x$, for each $x\in L(X)$.

Proof

Assume that $U\in \tau$, if $x\notin U$, then clearly $U\in \mathscr {T}(x, {\mathscr {F}}(B_x))$. Otherwise, suppose that $x\in U$, then clearly $x\in B_x\subseteq U$ and consequently $U\in \mathscr {T}(x, {\mathscr {F}}(B_x))$. Hence, $U\in \wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {F}}(B_x))$. Conversely, suppose that $U\in \wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {F}}(B_x))$ and $x\in U$, then by hypothesis $x\in B_x\subseteq U$ and so $x\in U^\circ$. Thus, $U\in \tau$. Therefore, $\tau =\wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {F}}(B_x))$. Now, suppose that $\tau =\wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {F}}(C_x))$, where $C_x$ is an open subset containing x for every $x\in L(X)$. Assume that $a\in L(X)$. Since $B_a$ is an open set, $B_a\in \mathscr {T}(a,{\mathscr {F}}(C_a))$. It follows that $B_a\in {\mathscr {F}}(C_a)$ and consequently $C_a\subseteq B_a$ and since $B_a$ is the smallest open subset containing a, we conclude that $B_a=C_a$. $\square$

In the above proposition, the openness of $C_x$ is necessary. For instance, let $\tau$ be the trivial topology on X, $|X|\ge 3$, $f:X\longrightarrow X$ be a bijection such that $f(x)\ne x$ for every $x\in X$ and $C_x=X\setminus \{f(x)\}$ for every $x\in X$. Then $\tau =\wedge _{x\in X}\mathscr {T}(x, {\mathscr {F}}(C_x))$, while $B_x=X\nsubseteq C_x$.

A principal topology $\tau$ on X is called the topology over cycles if $\tau =\bigwedge _{x\in L(X)}\tau [x]$ where $\tau [x]=\wedge \{\mathscr {T}(a,{\mathscr {F}}(b)):~a,b\in [x]\}$ for each $x\in L(X)$. Note that $\bigcup _{x\in X}[x]=X$ and if $[x]\cap [y]\ne \varnothing$, then $[x]=[y]$. Therefore, there exists $L_1\subseteq L(X)$ such that $\tau =\bigwedge _{x\in L_1}\tau [x]$, $[x]\cap [y]=\varnothing$ for every two distinct points $x,y\in L_1$ and $\bigcup _{x\in L_1}[x]=L(X)$. Thus, every topology over cycles is a topology over distinct cycles. Now, if $\tau$ is a non-discrete principal topology, we would like to know under what condition it is a topology over distinct cycles. To see this, we offer the following proposition.

Proposition 3.9

A non-discrete principal topology $\tau$ on X is a topology over distinct cycles if and only if the set [x] is open for each $x\in L(X)$.

Proof

($\Rightarrow$). Let $\tau =\wedge _{x\in X}\tau [x]$. It is clear that $\tau [x]=\wedge \{\mathscr {T}(a,{\mathscr {F}}(b)):~a,b\in [x]\}=\wedge _{a\in [x]}\wedge _{b\in [x]}\mathscr {T}(a,\mathscr {F}(b))=\mathscr {T}([x],{\mathscr {F}}([x]))$. Therefore, $\tau =\wedge _{x\in X}\mathscr {T}([x],{\mathscr {F}}([x]))$. Suppose that $a\in X$ and we show that $[a]\in \mathscr {T}([x],{\mathscr {F}}([x]))$ for every $x\in X$. Assume that $x\in X$; if $x\notin [a]$, then clearly $[a]\in \mathscr {T}([x],{\mathscr {F}}([x]))$. Now supposing $x\in [a]$, it follows that $[a]=[x]$ and so $[a]\in \mathscr {T}([x],{\mathscr {F}}([x]))$.

($\Leftarrow$). Since $[x]\subseteq B_x$ and $B_x$ is the smallest open subset containing x, we deduce that $[x]=B_x$. Thus, by Proposition 3.8, we are done. $\square$

We note that, if $(X,\tau )$ is a non-discrete principal space with no isolated point, such that its representation is unique, then by Propositions 3.7 and 3.8, $\tau$ is a topology over distinct cycles of order $n=2$.

4 Representation of a Topology with Respect to System of Neighborhoods

We recall that a topology $\tau$ on X is a $T_1$-topology if and only if it is the infimum of non-principal ultratopologies finer than $\tau$ and show that this representation for a $T_1$-topology $\tau \;$ can be written in the form $\bigwedge _{a\in L(X)}\mathscr {T}(a,\mathscr {W}_a)$, where $\mathscr {W}_a$ is the set of all neighborhoods and deleted neighborhoods of x.

In view of Theorem 1.2 (c), and the well-known fact that every filter is the intersection of all ultrafilters containing it, the next result is immediate.

Lemma 4.1

Let $(X,\tau )$ be a topological space.

• (1) If we put

  $$\begin{aligned} \psi _1= & {} \{\mathscr {T}(x,\mathscr {F}):~\tau \le \mathscr {T}(x,\mathscr {F}),~ x\in L(X)\;\text { and }\mathscr {F} \text { is an ultrafilter on } X \},\\ \psi _2= & {} \{\mathscr {T}(x,\mathscr {F}):~\tau \le \mathscr {T}(x,\mathscr {F}),~ x\in L(X)~and~\mathscr {F} \text { is a filter on } X \}, \end{aligned}$$

  then $\tau =\wedge \psi _1=\wedge \psi _2$.

• (2) If $(X,\tau )$ is a $T_1$-space and

  $$\begin{aligned} \phi _1= & {} \{\mathscr {T}(x,\mathscr {F}):~ \tau \le \mathscr {T}(x,\mathscr {F}),~ x\in L(X)~and~\mathscr {F} \text { is a free ultrafilter on } X \},\\ \phi _2= & {} \{\mathscr {T}(x,\mathscr {F}):~\tau \le \mathscr {T}(x,\mathscr {F}),~ x\in L(X)~ and~\mathscr {F} \text { is a free filter on } X \},\end{aligned}$$

  then $\tau =\wedge \phi _1=\wedge \phi _2$.

Now, let $(X,\tau )$ be a topological space and ${\mathscr {U}}'_x$ be the set of all deleted neighborhoods of x. Clearly, ${\mathscr {U}}_x$ is a filter on X and if we denote by ${\mathscr {W}}_x$ the filter induced by ${\mathscr {U}}'_x$, then ${\mathscr {W}}_x={\mathscr {U}}_x\cup {\mathscr {U}}'_x$ and it is the smallest filter containing ${\mathscr {U}}_x$ such that $x\notin \cap {\mathscr {W}}_x$. Furthermore, if X is a $T_1$ space, then ${\mathscr {W}}_x$ is the smallest free filter containing ${\mathscr {U}}_x$. Henceforth we use the notation ${\mathscr {W}}_x$ with this meaning.

The following proposition is now in order.

Proposition 4.2

Let $(X,\tau )$ be a topological space and $x\in L(X)$.

• (1) ${\mathscr {U}}_x\subseteq {\mathscr {F}}$ if and only if $\;\tau \le \mathscr {T}(x, {\mathscr {F}})$.

• (2) ${\mathscr {W}}_x\subseteq {\mathscr {F}}$ if and only if $\;x\notin \cap {\mathscr {F}}$ and $\tau \le \mathscr {T}(x, {\mathscr {F}})$.

• (3) If X is a $T_1$ space, then ${\mathscr {W}}_x\subseteq {\mathscr {F}}$ if and only if ${\mathscr {F}}$ is a free filter and $\tau \le \mathscr {T}(x, {\mathscr {F}})$.

Proof

(1 $\Rightarrow$). Suppose that ${\mathscr {U}}_x\subseteq {\mathscr {F}}$ and $U\in \tau$. If $x\notin U$, then clearly $U\in \mathscr {T}(x, {\mathscr {F}})$ and if $x\in U$, then it follows that $U\in {\mathscr {U}}_x\subseteq {\mathscr {F}}$ and so $U\in \mathscr {T}(x, {\mathscr {F}})$.

(1 $\Leftarrow$). Assume that $\tau \le \mathscr {T}(x, {\mathscr {F}})$ and $A\in {\mathscr {U}}_x$. Hence, $A^\circ \in \tau$ and $x\in A^\circ \subseteq A$. By hypothesis $A^\circ \in \mathscr {T}(x, {\mathscr {F}})$ and since $x\in A^\circ$, it follows that $A^\circ \in {\mathscr {F}}$ and consequently $A\in {\mathscr {F}}$.

The proofs of (2) and (3) are similar to (1). $\square$

Note that by part (3) of the above proposition, $(X,\tau )$ is a $T_1$ space if and only if $\;{\mathscr {W}}_x$ is a free filter on X for every $x\in L(X)$.

Proposition 4.3

Let $(X,\tau )$ be a topological space. The following statements hold:

• (1) $\tau =\wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {U}}_x)=\wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {W}}_x)$.

• (2) Suppose that ${\mathscr {F}}_x$ is a filter for every $x\in L(X)$. Then $\tau \le \wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {F}}_x)$ if and only if $\;{\mathscr {U}}_x\subseteq {\mathscr {F}}_x$ for every $x\in L(X)$.

• (3) If X is $T_1$ space, $\;\tau \le \wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {F}}_x)$ and $a\in L(X)$, then ${\mathscr {F}}_a$ is a free filter if and only if $\;{\mathscr {W}}_a\subseteq {\mathscr {F}}_a$.

Proof

• (1) By Lemma 4.1 and Proposition 4.2, if we put

  $$\begin{aligned} \phi _x=\{{\mathscr {F}}:~{\mathscr {F}} \text {is a filter on}\; X\text {and} \; \tau \le \mathscr {T}(x,{\mathscr {F}})\}, \end{aligned}$$

  then, since ${\mathscr {U}}_x={\mathscr {W}}_x\cap {\mathscr {F}}(x)$, we can write

  $$\begin{aligned} \tau= & {} \wedge \{\mathscr {T}(x,{\mathscr {F}}):~x\in L(X),~{\mathscr {F}}\in \phi _x\}\\= & {} \wedge \{\mathscr {T}(x,\wedge \phi _x):~x\in L(X)\}=\wedge _{x\in L(X)}\mathscr {T}(x,{\mathscr {U}}_x)\\= & {} \wedge _{x\in L(X)}\mathscr {T}(x,{\mathscr {W}}_x\cap {\mathscr {F}}(x))=\wedge _{x\in L(X)}\mathscr {T}(x,{\mathscr {W}}_x). \end{aligned}$$

• (2) Assume that $\tau \le \wedge _{x\in L(X)}\mathscr {T}(x, {\mathscr {F}}_x)$ and $A\in {\mathscr {U}}_x$. Thus, $x\in A^\circ \subseteq A$ and so $A^\circ \in \tau$, consequently, $A^\circ \in \mathscr {T}(x,{\mathscr {F}}_x)$. Hence, $A^\circ \in {\mathscr {F}}_x$ and so $A\in {\mathscr {F}}_x$. The converse is clear.

• (3) The proof is similar to (2).

$\square$

Corollary 4.4

Every $T_1$-topology on an infinite set X with no isolated points is the cofinite topology if and only if $\;\tau =\wedge _{x\in X}\mathscr {T}(x,{\mathscr {F}}_\circ )$, where ${\mathscr {F}}_\circ$ is the filter consisting of all cofinite subsets of X.

Remark 4.5

It is easy to see that if $x\in L(X)$, then $x\not \in \cap \mathscr {W}_x$ and $\mathscr {U}_x=\mathscr {W}_x\cap \mathscr {F}(x)$ and also $\mathscr {W}_x$ is a unique filter with this property. To see this, let $\mathscr {F}$ be a filter on X such that $x\not \in \cap \mathscr {F}$, $\mathscr {U}_x=\mathscr {F}\cap \mathscr {F}(x)$ and we show that $\mathscr {W}_x=\mathscr {F}$. Since $x\not \in \cap \mathscr {F}$ and $\mathscr {U}_x\subseteq \mathscr {F}$, we infer that $X\backslash \{x\}\in \mathscr {F}$ and so $\mathscr {U'}_x\subseteq \mathscr {F}$, consequently we have $\mathscr {W}_x=\mathscr {U}_x\cup \mathscr {U'}_x\subseteq \mathscr {F}$. Now we assume that $G\in \mathscr {F}$, if $x\in G$, then $G\in \mathscr {F}\cap \mathscr {F}(x)=\mathscr {U}_x$ and if $x\not \in G$ then $G\cup \{x\}\in \mathscr {F}\cap \mathscr {F}(x)=\mathscr {U}_x$ and conclude that $G\in \mathscr {U'}_x$.

Let $(X,\tau )$ be a $T_1$ topological space. By Proposition 4.3, we have $\tau =\bigwedge _{x\in L(X)}\mathscr {T}(x,\mathscr {W}_x)$ and if $\tau =\bigwedge _{x\in L(X)}\mathscr {T}(x,\mathscr {F}_x)$, where ${\mathscr {F}}_x$ is a free filter, then ${\mathscr {W}}_x\subseteq {\mathscr {F}}_x$ for every $x\in L(X)$. We are looking for conditions under which ${\mathscr {W}}_x={\mathscr {F}}_x$ for every $x\in L(X)$, i.e., the representation $\tau =\bigwedge _{x\in L(X)}\mathscr {T}(x,\mathscr {W}_x)$ is unique.

From here onwards, we assume that all topological spaces are $T_1$.

Definition 4.6

In Proposition 4.3, we have already noticed that for any space $(X, \tau )$, $\tau =\wedge _{x\in L(X)}\mathscr {T}(x,{\mathscr {W}}_x)$. This representation is called the natural representation of $\tau$. Also, we say that $\tau$ (or $(X,\tau )$) has the unique representation property whenever if $\tau =\wedge _{x\in L(X)}\mathscr {T}(x,{\mathscr {F}}_x)$, where ${\mathscr {F}}_x$ is a free filter for every $x\in L(X)$, then it can be deduced that ${\mathscr {W}}_x={\mathscr {F}}_x$ for every $x\in L(X)$.

By the above definition, the following proposition is obvious.

Proposition 4.7

If ${\mathscr {W}}_x$ is an ultrafilter for every $x\in L(X)$ (i.e., $(X,\tau )$ is a maximal space, see Definition 4.13), then $\tau$ has unique representation property.

Example 4.8

Let $\tau _{\circ }$ be the cofinite topology on an infinite set X. We show that $\tau _{\circ }$ does not have a unique representation property. It is clear that $\tau _{\circ }=\bigwedge _{x\in X}\mathscr {T}(x,\mathscr {W}_x)$, where $\mathscr {W}_x=\mathscr {F}_{\circ }$ for all $x\in X$. Now, for an arbitrary $a\in X$, assume that $A_a$, $X\backslash A_a$ are infinite subsets of X and without loss of generality, suppose that $a\in A_a$. Then the collection $\mathscr {F}_{\circ }\cup \{A_a\}$ is a filter base for a free filter $\mathscr {F}_a$ on X. It is easily seen that $\tau =[\bigwedge _{x\ne a}\mathscr {T}(x,\mathscr {F}_{\circ })]\wedge \mathscr {T}(a,\mathscr {F}_a)$, while ${\mathscr {W}}_a={\mathscr {F}}_\circ \ne {\mathscr {F}}_a$.

Theorem 4.9

Let $(X,\tau )$ be a topological space. Then $\tau$ has a unique representation property if and only if whenever $a\in A'\backslash A^\circ$, then there exist $U\in \mathscr {U}_a$ and $B\subseteq X$ such that $B\backslash B^\circ =\{a\}$ and $A\cap U\subseteq B$, where $a\in L(X)$ and $A\in \mathscr {F}(a)$.

Proof

($\Rightarrow$). Assume that there are $a\in L(X)$ and $A\in \mathscr {F}(a)$ such that $a\in A'\backslash A^\circ$ but whenever $A\cap U\subseteq B$ for every $U\in \mathscr {U}_a$, then we deduce that $B\backslash B^\circ \ne \{a\}$. Since $a\in A'\backslash A^\circ$, $\mathscr {W}_a\cup \{A\}$ generates a free filter ${\mathscr {F}}_a$ properly containing ${\mathscr {W}}_a$. Now, it suffices to show that $\tau =[\bigwedge _{x\in L(X)\backslash \{a\}}\mathscr {T}(x,\mathscr {W}_x)]\wedge \mathscr {T}(a,\mathscr {F}_a)$. To see this, let $\sigma =\delta \wedge \mathscr {T}(a,\mathscr {F}_a)$ where $\delta =\bigwedge _{x\in L(X)\backslash \{a\}}\mathscr {T}(x,\mathscr {W}_x)$. Then we have $\tau =\delta \wedge \mathscr {T}(a,\mathscr {W}_a)$ and as $\mathscr {W}_a\subseteq \mathscr {F}_a$ it follows that $\tau \le \sigma$. Now let $B\in \sigma$, thus $B\in \delta$ and $B\in \mathscr {T}(a,\mathscr {F}_a)$. If $a\not \in B$ we are done, thus we assume that $a\in B$. Since $B\in \mathscr {T}(a,\mathscr {F}_a)$, we infer that $B\in \mathscr {F}_a$. Now, we claim that $B\in \mathscr {U}_a$. To see this, let $B\in \mathscr {F}_a\backslash \mathscr {U}_a$. Then $U_{\circ }\cap A\subseteq B$, for some $U_{\circ }\in \mathscr {U}_a$. By hypothesis, we have $B\backslash B^\circ \nsubseteq \{a\}$. Hence, there is $x\in B\backslash B^\circ$ such that $x\ne a$. Since $x\not \in B^\circ$, it follows that $x\in L(X)$ and we have $B\not \in \mathscr {W}_x$ and consequently $B\not \in \mathscr {T}(x,\mathscr {W}_x)$. Therefore, $B\not \in \delta$ and this is a contradiction.

($\Leftarrow$). Let $\tau =\bigwedge _{x\in L(X)}\mathscr {T}(x,\mathscr {F}_x)$, where $\mathscr {F}_x$ is a free filter on X for all $x\in L(X)$. It suffices to show that $\mathscr {F}_x\cap \mathscr {F}(x)\subseteq \mathscr {U}_x$ for all $x\in L(X)$. We assume that $\mathscr {F}_a\cap \mathscr {F}(a)\nsubseteq \mathscr {U}_a$ for some $a\in L(X)$, then there is $A\in \mathscr {F}_a\cap \mathscr {F}(a)$ such that $A\not \in \mathscr {U}_a$. It is clear that $a\in A'\backslash A^\circ$ and therefore there exist $U\in {\mathscr {U}}_x$ and $B\subseteq X$ such that $B\backslash B^\circ =\{a\}$ and $A\cap U\subseteq B$. Now, let $x\in X\backslash \{a\}$. Then $x\not \in B\backslash B^\circ$. Therefore, either $x\not \in B$ or $x\in B^\circ$, that is $B\in \mathscr {T}(x,\mathscr {F}_x)$ and we infer that $B\in \bigwedge _{x\in L(X)\backslash \{a\}}\mathscr {T}(x,\mathscr {F}_x)$. On the other hand, since $U\in \mathscr {U}_a\subseteq {\mathscr {F}}_a$ and $A\in \mathscr {F}_a$, it follows that $B\in {\mathscr {F}}_a$ and so $B\in \mathscr {T}(a,{\mathscr {F}}_a)$. Hence, $B\in \tau$ and consequently $a\in B^\circ$, but this is a contradiction. $\square$

Corollary 4.10

Let $(X,\tau )$ be a topological space. If for all $a\in L(X)$ and for all $A\in \mathscr {F}(a)$, $a\in A'\backslash A^\circ$ implies that $a\notin (A\setminus A^\circ )'$, then $\tau$ has a unique representation property.

Proof

Let $a\in L(X)$, $A\in {\mathscr {F}}(a)$, and $a\in A'\backslash A^\circ$. Then by hypothesis, $a\notin (A\setminus A^\circ )'$. Therefore, there exists an open neighborhood U of a such that $(A\setminus A^\circ )\cap U=\{a\}$. Now, if we put $B=A\cap U$, then we have

$$\begin{aligned} B\setminus B^\circ =(A\cap U)\setminus (A\cap U)^\circ =(A\cap U)\setminus (A^\circ \cap U)=(A\setminus A^\circ )\cap U=\{a\}. \end{aligned}$$

$\square$

We cite the following definition from [1].

Definition 4.11

Let $(X,\tau )$ be a topological space. The space X is called an I-space if the subspace L(X) of X is a discrete space.

The following result is immediate for I-spaces.

Corollary 4.12

If $(X,\tau )$ is a non-discrete topological I-space, then $\tau$ has a unique representation property.

Definition 4.13

Let $(X,\tau )$ be a topological space. In [4], a point $x\in X$ is called a maximal point if whenever $G\in \tau$ and $x\in \overline{G}$, then $G\cup \{x\}\in \tau$. Also, in [4, proposition 1.2], it is shown that “a point $x\in X$ is maximal if and only if the trace of the open neighborhoods filter at x on the subspace $X{\setminus }\{x\}$ is an open ultrafilter.” We prefer renaming this concept as O-maximal point, but we use “maximal point” to call a point $x\in X$ if either x is an isolated point or ${\mathscr {W}}_x$ is an ultrafilter. We call a non-discrete topological space X an O-maximal space (a maximal space) if every point of X is an O-maximal point (a maximal point). The reader must be careful that, in previous sources, a maximal space is a crowded space $(X,\tau )$ such that whenever $\tau \le \tau _1$, then $(X,\tau _1)$ has an isolated point, while in this paper, maximal space is defined for every non-discrete space. In fact, by our definition, a non-discrete space $(X,\tau )$ is a maximal space if and only if for any topology $\tau _1$ on X properly containing $\tau$, $(X,\tau _1)$ has a new isolated point.

Proposition 4.14

Any maximal point is an O-maximal point and consequently any maximal space is an O-maximal space.

Proof

Suppose that x is a non-isolated point and G is an open subset of X such that $x\in \overline{G}$. Therefore, $U\cap G\nsubseteq \{x\}$ for every $U\in {\mathscr {W}}_x$ and since ${\mathscr {W}}_x$ is an ultrafilter, it follows that $G\in {\mathscr {W}}_x$. Hence, $G\cup \{x\}\in {\mathscr {W}}_x$ and so $G\cup \{x\}$ is an open set. $\square$

The converse of the above proposition is not true. For example, let $\tau _{\circ }$ be the cofinite topology on an infinite set X. Then it is clear that $x\in X$ is an O-maximal point of $(X.\tau _{\circ })$, while the free filter $\mathscr {W}_x=\mathscr {F}_{\circ }$ is not an ultrafilter.

The following proposition is an improvement of [4, Proposition 1.2].

Proposition 4.15

A point $x\in X$ is an O-maximal point if and only if the set of all open neighborhoods and deleted open neighborhoods of x is open ultrafilter.

Part (1) of the following proposition is an improvement of [4, Proposition 1.3].

Proposition 4.16

Suppose that $(X,\tau )$ is a topological space and $x\in U\in \tau$. Then the following statements hold.

• (1) The point x is an O-maximal point in U if and only if $\;x$ is an O-maximal point in X.

• (2) The point x is a maximal point in U if and only if $\;x$ is a maximal point in X.

Clearly, $x\in L(X)$ is a maximal point if and only if for every $A\notin {\mathscr {W}}_x$ there exists $U\in {\mathscr {W}}_x$ such that $A\cap U=\varnothing$. The following proposition is a similar result for O-maximal points.

Proposition 4.17

Let $(X,\tau )$ be a topological space. Then $x\in L(X)$ is an O-maximal point of $(X,\tau )$ if and only if for each $A\not \in \mathscr {W}_x$, there exists $U\in \mathscr {W}_x$ such that $(U\cap A)^\circ =\varnothing$.

Proof

($\Rightarrow$). Let $x\in L(X)$ be an O-maximal point of $(X,\tau )$ and $A\notin {\mathscr {W}}_x$. Thus, $A^\circ \notin {\mathscr {W}}_x$ and so $A^\circ \cup \{x\}$ is not open. Since x is an O-maximal point, it follows that $x\notin \overline{A^\circ }$. Therefore, there exists $U\in {\mathscr {U}}_x$ such that $U\cap A^\circ =\varnothing$ and so $(U\cap A)^\circ =\varnothing$. ($\Leftarrow$). Let $x\in \overline{A}$ and $A\in \tau$, it suffices to show that $A\cup \{x\}\in \tau$. To see this, we assume $B=A\cup \{x\}\not \in \tau$, where $x\not \in A$, then we have $B\not \in \mathscr {W}_x$. Thus, there is $U\in \mathscr {W}_x$ such that $(U\cap B)^\circ =\varnothing$. Since $A=B^\circ$, it follows that $U^\circ \cap A=\varnothing$ and it is easily seen that $(U^\circ \cup \{x\})\cap A=\varnothing$, i.e., $x\not \in \overline{A}$, which is a contradiction. $\square$

Corollary 4.18

Let $(X,\tau )$ be a topological space and $x\in L(X)$. Then x is a maximal point if and only if x is an O-maximal point of $(X,\tau )$ and $x\not \in (A\setminus A^\circ )'$, for all $A\not \in \mathscr {W}_x$.

Let $(X,\tau )$ be a topological space and $x\in L(X)$. We say that ${\mathscr {U}}_x$ is a maximal neighborhood system at x, if ${\mathscr {F}}(x)$ is the only neighborhood system at x properly containing ${\mathscr {U}}_x$. It is obvious that ${\mathscr {U}}_x$ is a maximal neighborhood system at x if and only if $\;{\mathscr {W}}_x$ is an ultrafilter on X. Therefore, it is easy to see that $\mathscr {U}_x$ is a maximal neighborhood system if and only if for each $A\subseteq X$, if $x\in A\backslash A^\circ$ then there exists $U\in \mathscr {U}_x$ such that $U\cap A=\{x\}$.

Proposition 4.19

Let $(X,\tau )$ be a topological space and assume $\tau$ has a unique representation property. Then for all $a\in L(X)$ the following statements are equivalent.

• (1) $\mathscr {U}_a$ is a maximal neighborhood system.

• (2) a is an O-maximal point of $(X,\tau )$.

• (3) For all $A\in \mathscr {F}(a)$, it follows that $a\not \in A'\backslash A^\circ$.

• (4) For each $A\subseteq X$ if $a\in A\backslash A^\circ$ there exists $U\in \mathscr {U}_a$ such that $U\cap A=\{a\}$.

Proof

It is clear that $1\Rightarrow 2$ and $3\Rightarrow 4\Rightarrow 1$. (2) $\Rightarrow$ (3). Let $a\in L(X)$ be an O-maximal point of $(X,\tau )$ and $A\in \mathscr {F}(a)$. Suppose, on the contrary, $a\in A'\backslash A^\circ$, then by Theorem 4.9 there exist $U\in \mathscr {U}_a$ and $B\subseteq X$ such that $B\backslash B^\circ =\{a\}$ and $A\cap U\subseteq B$. Since $B^\circ \in \tau$, $a\in \overline{B^\circ }$ and a is an O-maximal point of X, we infer that $B=B^\circ \cup \{a\}\in \tau$, which is a contradiction. $\square$

Now, we want to deal pointwise with the concept of a unique representation property and some other related concepts.

Definition 4.20

Let $(X,\tau )$ be a topological space. $a\in X$ is called an exact point of X whenever if $a\in L(X)$ and $\tau =[\bigwedge _{x\in L(X)\backslash \{a\}}\mathscr {T}(x,\mathscr {W}_x)]\wedge \mathscr {T}(a,\mathscr {F}_a)$, where $\mathscr {F}_a$ is a free filter on X, then it follows that $\mathscr {F}_a=\mathscr {W}_a$.

Clearly, a topological space X has a unique representation property if and only if every point of X is an exact point.

It is clear that every maximal point of $(X,\tau )$ is an exact point and also an O-maximal point of $(X,\tau )$, but if $a\in L(X)$ is either an exact point or an O-maximal point of $(X,\tau )$, then we cannot conclude that a is a maximal point. For example, suppose that $\tau =\mathscr {T}(a,\mathscr {F})$ where $a\in X$ and $\mathscr {F}$ is a free filter on X that is not an ultrafilter, then non-isolated point $a\in X$ is an exact point of X but not a maximal point of X. Assume that $\tau _{\circ }$ is the cofinite topology on an infinite set X, then every point $a\in X$ is an O-maximal point of $(X,\tau _{\circ })$, while $a\in X$ is not a maximal point of X.

Proposition 4.21

Let $(X,\tau )$ be a topological space. Then $a\in L(X)$ is a maximal point of X if and only if a is both an O-maximal point and an exact point of $(X,\tau )$.

Proof

Suppose that $a\in L(X)$ is a maximal point of $(X,\tau )$. It is easily seen that a is both an O-maximal point and an exact point of $(X,\tau )$. Conversely, let $a\in L(X)$ and the free filter $\mathscr {W}_a$ is not an ultrafilter on X, and then we prove that if a is an O-maximal point, then if follows that a is not an exact point of X. To see this, assume that a is an O-maximal point of $(X,\tau )$. Since $\mathscr {W}_a$ is not ultrafilter on X then there is $A\subseteq X$ such that $A\in \mathscr {F}(a)$ and $a\in A'\backslash A^\circ$. Now, suppose that $U\in \mathscr {U}_a$ and $A\cap U\subseteq B$, it suffices to show that $B\backslash B^\circ \ne \{a\}$. If $B\backslash \{a\}\nsubseteq B^\circ$ we are done. Hence, let $B\backslash \{a\}\subseteq B^\circ$. Then we have $\overline{B\backslash \{a\}}\subseteq \overline{B^\circ }$. Since $a\in A'$, it is clear that $a\in \overline{B\backslash \{a\}}$ and consequently $a\in \overline{B^\circ }$. As $a\in L(X)$ is an O-maximal point of $(X,\tau )$, it follows that $B=B^\circ \cup \{a\}\in \tau$ and so we have $B\backslash B^\circ =\varnothing \ne \{a\}$. $\square$

Definition 4.22

Let $(X,\tau )$ be a topological space. A point $x\in X$ is called an internal point of X whenever for all $A\subseteq X$ if $a\in \overline{A^\circ }$, then $a\in {\overline{A^\circ }}^{^\circ }$

Let us recall that a topological space X is called extremely disconnected if the closure of every open set is open; also X is called almost discrete if $\overline{I(X)}=X$. We can easily see that a space X is extremely disconnected if and only if every non-isolated point of X is an internal point.

Theorem 4.23

Let $(X,\tau )$ be an almost discrete topological space and $a\in L(X)$. Then the following statements are equivalent.

• (1) a is a maximal point of $(X,\tau )$.

• (2) a is an O-maximal point of $(X,\tau )$.

• (3) a is an internal point of $(X,\tau )$ and an isolated point in L(X).

Proof

(1) $\Rightarrow$ (2). It is clear.

(2) $\Rightarrow$ (3). Let $U\in \tau$ and $a\in U$. If we put $D=I(X )$, then $\overline{D}=X$ and so $\overline{D\cap U}=\overline{U}$. Since $a\in \overline{D\cap U}$ and $a\in L(X)$ is an O-maximal point of $(X,\tau )$, we infer that $B=(D\cap U)\cup \{a\}\in \tau$. Therefore, we have $B\cap L(X)=\{a\}$, i.e., a is an isolated point in L(X). Now, it suffices to prove that a is an internal point of $(X,\tau )$. To see this, we assume that $G\in \tau$ and $a\in \overline{G}$, then we show that $a\in \overline{G}^{^\circ }$. Since $a\in \overline{G}$ and a is an O-maximal point of $(X,\tau )$, it follows that $G\cup \{a\}\in \tau$ and thus $a\in G\cup \{a\}\subseteq \overline{G}^{^\circ }$.

(3) $\Rightarrow$ (1). By Proposition 4.21 it suffices to show that a is an O-maximal point and an exact point of $(X,\tau )$. First we show that $a\in L(X)$ is an O-maximal point of $(X,\tau )$. Let $G\in \tau$ and $a\in \overline{G}$. As a is an internal point of $(X,\tau )$, we have $a\in \overline{G}^{^\circ }$. Since a is an isolated point in L(X), there exists $V\in \tau$ such that $V\cap L(X)=\{a\}$ and $a\in V\subseteq \overline{G}$. Put $D_1=D\cup \{a\}$, then we have

$$\begin{aligned} a\in V=V\cap D_1\subseteq \overline{G}\cap D_1=(G\cap D)\cup \{a\}\subseteq G\cup \{a\}. \end{aligned}$$

Now, we show that a is an exact point of X. To see this, it suffices to prove that for all $A\in \mathscr {F}(a)$ if $a\in A'$ then $a\in A^\circ$. Let $a\in A\cap A'$. Since $D\cup \{a\}\in \tau$ and $a\in A'$ it is easily seen that $a\in \overline{A\cap D}$. Since $a\in L(X)$ is an O-maximal point of $(X,\tau )$, we infer that $W=(A\cap D)\cup \{a\}\in \tau$ and $a\in W\subseteq A$, i.e., $a\in A^\circ$. $\square$

Corollary 4.24

Let $(X,\tau )$ be an almost discrete topological space. Then the following statements are equivalent.

• (1) X is a maximal space.

• (2) X is an O-maximal space.

• (3) X is an extremely disconnected I-space.

Proposition 4.25

Let $(X,\tau )$ be a topological space. Then $a\in L(X)$ is an exact point of X if and only if for all $A\in \mathscr {F}(a)$, if $a\in A'\backslash A^\circ$, then there is $U\in \mathscr {U}_a$ such that for each $a\ne x\in U\cap A$ there exists $U_x\in \mathscr {U}_x$ and $W=\bigcup _{a\ne x\in A\cap U}U_x$ is not a deleted neighborhood of a.

Proof

($\Rightarrow$). Let $a\in L(X)$ be an exact point of $(X,\tau )$, $A\in \mathscr {F}(a)$, and $a\in A'\backslash A^\circ$. By Theorem 4.9, there exist $U\in \mathscr {U}_a$ and $B\subseteq X$ such that $B\backslash B^\circ =\{a\}$ and $A\cap U\subseteq B$. Hence, $A\backslash \{a\}\cap U\subseteq B^\circ$ and for each $a\ne x\in U\cap A$, there exists $U_x\in \mathscr {U}_x$ such that $x\in U_x\subseteq B$ and so we have $W=\bigcup _{a\ne x\in A\cap U}U_x\subseteq B^\circ$. Since $B^\circ$ is not a deleted neighborhood of a, it is clear that W is not a deleted neighborhood of a, too.

($\Leftarrow$). Suppose that $A\in \mathscr {F}(a)$ and $a\in A'\backslash A^\circ$. By hypothesis, there is $U\in \mathscr {U}_a$ such that for all $a\ne x\in U\cap A$ there exists $U_x\in \mathscr {U}_x$ such that $W=\bigcup _{a\ne x\in A\cap U}U_x$ is not a deleted neighborhood of a. Now, we set $B=W\cup \{a\}$ and then, clearly, $A\cap U\subseteq B$ and $B\backslash B^\circ =\{a\}$. Therefore, a is an exact point of $(X,\tau )$. $\square$

If, in the above proposition, we consider a regular space, then the proposition has a simpler form.

Theorem 4.26

Let $(X,\tau )$ be a regular space and $a\in L(X)$. Then the following statements are equivalent.

• (1) $a\in L(X)$ is an exact point of $(X,\tau )$.

• (2) For each $A\in \mathscr {F}(a)$, if $a\in A'\backslash A^\circ$ then for every $x\in A$ distinct from a, there exists $U_x\in \mathscr {U}_x$ such that $W=\bigcup _{a\ne x\in A}U_x$ is not a deleted neighborhood of $a\in L(X)$.

• (3) For each $A\in \mathscr {F}(a)$, if $a\in A'\backslash A^\circ$ then there is $B\subseteq X$ such that $B\backslash B^\circ =\{a\}$ and $A\backslash B=\varnothing$.

Proof

(1) $\Rightarrow$ (2). Let $a\in L(X)$ be an exact point of $(X,\tau )$, $A\in \mathscr {F}(a)$, and $a\in A'\backslash A^\circ$. Then by Proposition 4.25, there is $U\in \mathscr {U}_a$ such that for every $a\ne x\in U\cap A$, there exists $U_x\in \mathscr {U}_x$ where $\bigcup _{a\ne x\in A\cap U}U_x$ is not a deleted neighborhood of a. Since X is a regular space, without loss of generality, we can assume that U is a closed neighborhood of a, thus, for each $x\in A\backslash U$ there is $U_x\in \mathscr {U}_x$ such that $U_x\cap U=\varnothing$. Now, we claim that $W=\bigcup _{a\ne x\in A}U_x$ is not a deleted neighborhood of a. To see this, suppose that there is $V\in \mathscr {U}_a$ such that $V\backslash \{a\}\subseteq W$, then $(V\cap U)\backslash \{a\}\subseteq \bigcup _{a\ne x\in A\cap U}U_x$. Since $V\cap U\in \mathscr {U}_a$ it follows that $\bigcup _{a\ne x\in A\cap U}U_x$ is a deleted neighborhood of a, which is a contradiction.

(2) $\Rightarrow$ (3). Put $W=\bigcup _{a\ne x\in A}U_x$ and $B=W\cup \{a\}$, then clearly $B\backslash B^\circ =\{a\}$ and $A\backslash B=\varnothing$.

(3) $\Rightarrow$ (1). It is clear. $\square$

Proposition 4.27

Let X be a Hausdorff space with a countable base at $a\in L(X)$. Then a is an exact point of $(X,\tau )$.

Proof

Let $A\in \mathscr {F}(a)$ and $a\in A'\backslash A^\circ$. Since the Hausdorff space X has a countable base at $a\in L(X)$ and $a\in \overline{X\backslash A}$, there exists a sequence $\{x_n\}$ in $X\backslash A$ which converges to a and clearly $A\cap \{x_n\}=\varnothing$. We set $S=\{x_n:~n\in \mathbb {N}\}$ and it is easily seen that for all $a\ne x\not \in S$ there is $U_x\in \mathscr {U}_x$ such that $U_x\cap S=\varnothing$. Now, let $B=(\bigcup _{a\ne x\not \in S}U_x)\cup \{a\}$. Then it is clear that $B\backslash B^\circ =\{a\}$ and $A\cap X\subseteq B$, where $X\in \mathscr {U}_a$, thus by Theorem 4.9, a is an exact point of $(X,\tau )$. $\square$

Corollary 4.28

Every first countable Hausdorff space $(X,\tau )$ has a unique representation property.

Previously, we have seen that any cofinite topology on an infinite set X does not have a unique representation property. We conclude the paper by an example of a Hausdorff compact topological space which does not have a unique representation property.

Example 4.29

Let X be an infinite discrete space. It is easily seen that for all $p\in \beta X\backslash X$, there is no sequence in X converging to p. Now, we show that p is not an exact point of $\beta X$ for all $p\in \beta X\backslash X$. To see this, let $p\in \beta X\backslash X=A$. Clearly, $p\in A'$ and $\text {int}_{\beta X}A=\varnothing$, hence $p\in A'\backslash \text {int}_{\beta X}A$. Let $A\subseteq B$ and $B=\text {int}_{\beta X}B\cup \{p\}$, by Theorem 4.26, it suffices to show that B is open in $\beta X$. Since $A\subseteq B$, we have $A\backslash \{p\}\subseteq B\backslash \{p\}=\text {int}_{\beta X}B\backslash \{p\}=D$ and it is clear that D is open set in $\beta X$ and consequently $E=\beta X\backslash D$ is compact. We can easily see that $E\subseteq \beta X\backslash (A\setminus \{p\})=X\cup \{p\}$. We claim that the set E is finite. Suppose, on the contrary, E is infinite. Then there exists an infinite set $K=\{x_n:\;n\in \mathbb {N},\; x_n\in X\}\subseteq E$. As E is compact, K has a limit point in E. Since p is the only non-isolated point of E, it follows that $x_n\longrightarrow p$ and this is a contradiction. Therefore, E is finite and consequently D is a cofinite subset of $\beta X$. Thus $B=D\cup \{p\}$ is also cofinite and consequently B is open in $\beta X$.