On Some Generalized Topologies Satisfying all Separation Axioms

Abstract

The properties of generalized topological spaces generated by some kind of density operators are presented. Moreover, the separation axioms of such generalized topological spaces are investigated. The examples of generalized topological spaces satisfing all separation axioms are included.

1 Introduction and Preliminaries

Let X be a non-empty set. The family of all subsets of X will be denoted by $2^X$. We shall say that a $\sigma$-ideal ${\mathcal {J}}\subset 2^X$ is admissible if ${\mathcal {J}}$ is a proper $\sigma$-ideal of sets such that $\bigcup {\mathcal {J}}=X$. For any $A, B\in 2^X$ the symbol $A \triangle B$ will stand for the set $(A{\setminus } B) \cup (B{\setminus } A)$. A cardinality of a set A will be denoted by ${{\,\textrm{card}\,}}(A)$.

In the paper a notion of a generalized topological space, introduced in Császár [1] by Császár, will be used. We shall say that a family $\Gamma \subset 2^X$ is a generalized topology in X if $\emptyset \in \Gamma$ and $\bigcup _{t\in T}G_t\in \Gamma$ whenever $\{G_t:t\in T\}\subset \Gamma$. The pair $(X,\Gamma )$ is called a generalized topological space. If $X\in \Gamma$ then we shall say that $(X,\Gamma )$ is a strong generalized topological space.

In the theory of generalized topological spaces almost all notions (e.g. an interior of a set, a closure of a set, a boundary of a set, a compact set, separated sets) are defined as in standard topological spaces (see [1, 2]). The interior, the closure and the boundary of $A\subset X$ will be denoted by ${{\,\textrm{int}\,}}_\Gamma (A)$, ${{\,\textrm{cl}\,}}_\Gamma (A)$ and ${{\,\textrm{Fr}\,}}_{\Gamma }(A)$, respectively. Separation axioms for a generalized topological space are defined as in the case of the classical topological space Császár [3]. Moreover, the definitions of Lindelöf space, first countable and second countable space can be adapted from the classical topological space. In addition, using methods analogous to those used in the case of a classical topological space, it can be shown

Lemma 1

Let $\langle X, \Gamma \rangle$ be a Hausdorff strong generalized topological space. If every two separated sets can be separated by $\Gamma$-open sets then the space $\langle X, \Gamma \rangle$ is a hereditarily normal space.

In the third part of the paper we will focus on a measurable space i.e. a triple $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$, where X is a non-empty set, ${\mathcal {S}}$ is a $\sigma$-algebra of subsets of X and ${\mathcal {J}}\subset {\mathcal {S}}$ is an admissible $\sigma$-ideal of sets. A measurable hull of a set $A\subset X$ is any set $B\in {\mathcal {S}}$ such that $A\subset B$ and for any $C\subset B{\setminus } A$ if $C\in {\mathcal {S}}$ then $C\in {\mathcal {J}}$. The set B described above is called an ${\mathcal {S}}$-measurable hull of a set A. We shall say that $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ has the hull property if any set $A\subset X$ has an ${\mathcal {S}}$-measurable hull. The space $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ satisfies the countable chain condition CCC if every disjoint family of sets from ${\mathcal {S}}{\setminus } {\mathcal {J}}$ is at most countable. It is well known that if $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ satisfies CCC then it has the hull property.

We shall say that a set $E\subset B\subset X$ is full in B if for every set $A\in {\mathcal {S}}$ such that $E\subset A$ we have that $B{\setminus } A\in {\mathcal {J}}$.

Property 2

Let $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ have the hull property and $A\subset X$. A set $B\subset A$ is full in A if and only if every ${\mathcal {S}}$-measurable hull of A is an ${\mathcal {S}}$-measurable hull of B.

Proof

Let $B\subset A$ be full in A. Let $H\supset A$ be an ${\mathcal {S}}$-measurable hull of A. We shall show that H is an ${\mathcal {S}}$-measurable hull of B. Let $C\supset B$ be any ${\mathcal {S}}$-measurable hull of B. Thus $H{\setminus }(H\cap C)\subset (H{\setminus } A)\cup (A{\setminus } (H\cap C))$. Since $B\subset H\cap C\in {\mathcal {S}}$ and B is full in A, we obtain $A{\setminus } (H\cap C)\in {\mathcal {J}}$. Therefore $(H{\setminus } A)\cap (H{\setminus } (H\cap C))\in {\mathcal {S}}$ and $(H{\setminus } A)\cap (H{\setminus } (H\cap C))\subset H{\setminus } A$ so $(H{\setminus } A)\cap (H{\setminus } (H\cap C)) \in {\mathcal {J}}$. In consequence we have that $(H{\setminus } (H\cap C))\in {\mathcal {J}}$. From this and the fact that $H\cap B$ is an ${\mathcal {S}}$-measurable hull of B we obtain that H is an ${\mathcal {S}}$-hull of B.

Now, assume that every ${\mathcal {S}}$-measurable hull of A is an ${\mathcal {S}}$-measurable hull of B. We shall show that B is full in A. Let $C\in {\mathcal {S}}$ and $B\subset C$. We shall prove that $A{\setminus } C\in {\mathcal {J}}$. Let H be an ${\mathcal {S}}$-measurable hull of A. Clearly, H is an ${\mathcal {S}}$-measurable hull of B, so $H{\setminus } (C\cap H)\in {\mathcal {J}}$. Thus $A{\setminus } C\subset H{\setminus } C=H{\setminus } (C\cap H)\in {\mathcal {J}}$ and the proof is finished. $\square$

2 $\Phi ^*$ Operator and ${\mathcal {T}}^*$-Generalized Topology

Let ${\mathcal {J}}_*\subset 2^X$ be an admissible $\sigma$-ideal and $\Phi ^*:2^X \rightarrow 2^X$ be an operator having the following properties:

$1^*$:

$\Phi ^*(\emptyset )=\emptyset$ and $\Phi ^*(X)=X$,

$2^*$:

for any $A, B\subset X$ if $A\subset B$ then $\Phi ^*(A)\subset \Phi ^*(B)$,

$3^*$:

for any $A, B\subset X$ if $A\triangle B\in {\mathcal {J}}_*$ then $\Phi ^*(A)=\Phi ^*(B)$,

$4^*$:

$A{\setminus } \Phi ^*(A)\in {\mathcal {J}}_*$ for any $A\subset X$.

For example it is easy to see that the operator $\Phi _1^*:2^X \rightarrow 2^X$ defined in the following way: $\Phi _1^*(A)=X$ if $A\not \in {\mathcal {J}}_*$ and $\Phi _1^*(A)=\emptyset$ if $A\in {\mathcal {J}}_*$ satisfies properties $1^*$–$4^*$. Recall that if $\lambda ^*$ denotes the outer Lebesgue measure on ${\mathbb {R}}$ then $x_0\in {\mathbb {R}}$ is an outer density point of a set $A\subset {\mathbb {R}}$ iff $\lim \limits _{h\rightarrow 0^+}\frac{\lambda ^*(A\cap [x_0-h, x_0+h])}{2\,h}=1$. One can easily show that the operator $\Phi _d^*:2^{{\mathbb {R}}}\rightarrow 2^{{\mathbb {R}}}$ given by the formula

$$\begin{aligned} \Phi _d^*(A)=\{x\in {\mathbb {R}}: x\;\text {is an outer density point of}\; A\} \end{aligned}$$

for $A\subset {\mathbb {R}}$ satisfies properties $1^*$–$4^*$ (see [8]).

Let us consider the family

$$\begin{aligned} {\mathcal {T}}^*=\{A\subset X: A\subset \Phi ^*(A)\}. \end{aligned}$$

Note that for the operator $\Phi _1^*$ defined above, we get that ${\mathcal {T}}^*_1=\{A\subset X: A\subset \Phi ^*_1(A)\}=\{A: A\not \in {\mathcal {J}}_*\}\cup \{\emptyset \}$ and the family ${\mathcal {T}}^*_1$ does not have to be a topology. We see at once that ${\mathcal {T}}^*_1$ is not a topology if and only if there exist sets $A, B\subset X$ such that $A\not \in {\mathcal {J}}_*$, $B\not \in {\mathcal {J}}_*$ and $A\cap B\in {\mathcal {J}}_*{\setminus }\{\emptyset \}$. Paper Hejduk and Loranty [8] demonstrates that ${\mathcal {T}}^*_d=\{A\subset X: A\subset \Phi ^*_d(A)\}$ is not also a topology.

One can prove immediately the following properties.

Property 3

The family ${\mathcal {T}}^*$ is a strong generalized topology and it is not equal to $2^X$.

Proof

Obviously $\emptyset$ and X belong to ${\mathcal {T}}^*$. Moreover, if $\{A_t\}_{t\in T}\subset {\mathcal {T}}^*$ then condition $2^*$ implies that $A_t\subset \Phi ^*(\bigcup _{t\in T}A_t)$ for any $t\in T$. Therefore $\bigcup _{t\in T} A_t\subset \Phi ^*(\bigcup _{t\in T}A_t)$ and, in consequence $\bigcup _{t\in T}A_t\in {\mathcal {T}}^*$. Obviously $\{x\}\not \in {\mathcal {T}}^*$ for any $x\in X$, so ${\mathcal {T}}^*\not =2^X$. $\square$

Property 4

If $W\in {\mathcal {T}}^*$ and $A\in {\mathcal {J}}_*$, then $W{\setminus } A\in {\mathcal {T}}^*$. Moreover, if $W\in {\mathcal {T}}^*{\setminus }\{\emptyset \}$, then $W\not \in {\mathcal {J}}_*$.

Proof

Let $W\in {\mathcal {T}}^*$, $A\in {\mathcal {J}}_*$ and $V=W{\setminus } A \subset W$. Thus, by Condition $3^*$, we obtain that $V\subset W\subset \Phi ^*(W)=\Phi ^*(V)$, which gives that $V\in {\mathcal {T}}^*$.

Let $W\in {\mathcal {T}}^*{\setminus }\{\emptyset \}$. Suppose, contrary to our claim that $W\in {\mathcal {J}}_*$. Condition $3^*$ implies that $W\subset \Phi ^*(W)=\emptyset$, which is impossible. $\square$

Property 5

For any $A\subset X$ we have that $A\cap \Phi ^*(A)\in {\mathcal {T}}^*$.

Proof

Clearly, $A=(A\cap \Phi ^*(A)) \cup (A{\setminus } \Phi ^*(A))$ for any $A\subset X$. Conditions $3^*$ and $4^*$ give that $\Phi ^*(A)=\Phi ^*(A\cap \Phi ^*(A))$ and, in consequence, we obtain that $A\cap \Phi ^*(A)\subset \Phi ^*(A\cap \Phi ^*(A))$, which means that $A\cap \Phi ^*(A)\in {\mathcal {T}}^*$. $\square$

Property 6

The smallest $\sigma$-algebra $\sigma ({\mathcal {T}}^*\cup {\mathcal {J}}_*)$ generated by the family ${\mathcal {T}}^*\cup {\mathcal {J}}_*$ is equal to $2^X$.

Proof

Let $A\subset X$. Clearly, $A=(A\cap \Phi ^*(A))\cup A{\setminus } \Phi ^*(A))$. Condition $4^*$ and Property 5 give that $A\in \sigma ({\mathcal {T}}^*\cup {\mathcal {J}}_*)$. $\square$

Property 7

For any $A\subset X$ we have that ${{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(A)= A\cap \Phi ^*(A)$. Moreover, ${{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(A)=\emptyset$ iff $A\in {\mathcal {J}}_*$.

Proof

Property 5 implies that $A\cap \Phi ^*(A)\subset {{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(A)$. Now, we prove that ${{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(A)\subset A\cap \Phi ^*(A)$. Let $x\in {{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(A)$. There exists a set $W\in {\mathcal {T}}^*$ such that $x\in W\subset A$. Thus, by Condition $2^*$, $W\subset \Phi ^*(W)\subset \Phi ^*(A)$ and, in consequence, $x\in W\subset A\cap \Phi ^*(A)$. Therefore ${{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(A)\subset A\cap \Phi ^*(A)$ and the proof of the first part of the theorem is finished. Clearly, if $A\in {\mathcal {J}}_*$, then $\Phi _*(A)=\emptyset$ by condition $1^*$ and $3^*$, so ${{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(A)=\emptyset$. Assume now that ${{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(A)=\emptyset$ and suppose that $A\not \in {\mathcal {J}}_*$. Obviously, $A=(A\cap \Phi _*(A))\cup (A{\setminus }\Phi _*(A))$ and, by Condition $4^*$, $A{\setminus }\Phi _*(A)\in {\mathcal {J}}_*$. Thus $A\cap \Phi _*(A)\not \in {\mathcal {J}}_*$, so ${{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(A)=A\cap \Phi _*(A)\not =\emptyset$. This contradiction ends the proof. $\square$

Property 8

For any $A\subset X$ we have that

• (a) ${{\,\textrm{cl}\,}}_{{\mathcal {T}}_*}(A)=A\cup (X{\setminus } \Phi ^*(X{\setminus } A))$;

• (b) ${{\,\textrm{Fr}\,}}_{{\mathcal {T}}_*}(A)\in {\mathcal {J}}_*$;

• (c) ${{\,\textrm{cl}\,}}_{{\mathcal {T}}^*}(A){\setminus } A= (X{\setminus } A){\setminus } \Phi ^*(X{\setminus } A) \in {\mathcal {J}}_*$.

Proof

Let $A\subset X$. By Property 7 we obtain immediately ${{\,\textrm{cl}\,}}_{{\mathcal {T}}_*}(A)=X{\setminus } {{\,\textrm{int}\,}}_{{\mathcal {T}}^*}(X{\setminus } A)= X{\setminus } ((X{\setminus } A)\cap \Phi ^*(X{\setminus } A))=A\cup (X{\setminus } \Phi ^*(X{\setminus } A))$.

Moreover, ${{\,\textrm{Fr}\,}}_{{\mathcal {T}}_{*}}(A)={{\,\textrm{cl}\,}}_{{\mathcal {T}}_{*}}(A){\setminus } {{\,\textrm{int}\,}}_{{\mathcal {T}}_{*}}(A)=[A\cup (X{\setminus } \Phi ^*(X{\setminus } A))]{\setminus } (A\cap \Phi ^*(A))=[A\cup (X{\setminus } \Phi ^*(X{\setminus } A))] \cap [(X{\setminus } A)\cup (X{\setminus }\Phi ^*(A))] =[A\cap (X{\setminus } \Phi ^*(A))] \cup [(X{\setminus } A)\cap (X{\setminus } \Phi ^*(X{\setminus } A))] \cup X{\setminus } (\Phi ^*(X{\setminus } A)\cup \Phi ^*(A)) \subset (A{\setminus }\Phi ^*(A))\cup ((X{\setminus } A) {\setminus } \Phi ^*(X{\setminus } A))\cup [A{\setminus } ((\Phi ^*(X{\setminus } A)\cup \Phi ^*(A))] \cup [(X{\setminus } A){\setminus } (\Phi ^*(X{\setminus } A)\cup \Phi ^*(A))]\subset (A{\setminus }\Phi ^*(A))\cup ((X{\setminus } A) {\setminus } \Phi ^*(X{\setminus } A))\in {\mathcal {J}}_*$.

By Property 7 and Condition $4^*$, we have that ${{\,\textrm{cl}\,}}_{{\mathcal {T}}^*}(A){\setminus } A= (X{\setminus } {{\,\textrm{int}\,}}_{T^*}(X{\setminus } A)){\setminus } A= (X{\setminus } ((X{\setminus } A)\cap \Phi ^*(X{\setminus } A))){\setminus } A= (A\cup (X{\setminus }\Phi ^*(X{\setminus } A))){\setminus } A= (X{\setminus } A){\setminus } \Phi ^*(X{\setminus } A))\in {\mathcal {J}}_*$ $\square$

In the case of a topological space, the notion of a nowhere dense set may be introduced by different equivalent definitions. In the case of a generalized topological space, these definitions can lead to different notions. In Korczak-Kubiak et al. [9] one can find two notions connected with nowhere density in generalized topological space $(X, \Gamma )$. We say that a set $A\subset X$ is $\Gamma$-nowhere dense if ${{\,\textrm{int}\,}}_{\Gamma }({{\,\textrm{cl}\,}}_{\Gamma }(A))=\emptyset$. A set $A\subset X$ is $\Gamma$-strongly nowhere dense if for $V\in \Gamma {\setminus }\{\emptyset \}$ there exists $U\in \Gamma {\setminus }\{\emptyset \}$ such that $U\subset V$ and $A\cap U=\emptyset$. A set is $\Gamma$-meager (-strong meager) if it is a countable union of $\Gamma$-nowhere dense (-strongly nowhere dense) sets. It is easy to see that if A is $\Gamma$-strongly nowhere dense then it is $\Gamma$-nowhere dense. The converse theorem is not true in general (see [9]). However, using Properties 7 and 4, it can be proved, as in article Hejduk and Loranty [7], that

Theorem 9

Let $A\subset X$. The following conditions are equivalent:

• (i) for any $W\in {\mathcal {T}}^*{\setminus }\{\emptyset \}$ there exists $V\in {\mathcal {T}}^*{\setminus }\{\emptyset \}$ such that $V\subset W$ and $V\cap A=\emptyset$,

• (ii) ${{\,\textrm{int}\,}}_{{\mathcal {T}}^*}({{\,\textrm{cl}\,}}_{{\mathcal {T}}^*}(A))=\emptyset$,

• (iii) $A\in {\mathcal {J}}_*$.

Thus, in the space $\langle X, {\mathcal {T}}^*\rangle$ the notions of a $\Gamma$-nowhere dense set and a $\Gamma$-strongly nowhere dense set are equivalent. We have

Corollary 10

In the space $\langle X, {\mathcal {T}}^*\rangle$ all families: the family of strong nowhere dense sets, the family of nowhere dense sets, the family of meager sets and the family of strong meager sets coincide with the family of ${\mathcal {J}}_*$.

Moreover, using the methods presented in Hejduk and Loranty [7] one can prove the following properties.

Property 11

A set $A\subset X$ is ${\mathcal {T}}^*$-compact if and only if A is is finite.

Property 12

The space $\langle X, {\mathcal {T}}^*\rangle$ neither fulfills the first nor the second axiom of countability and is not separable.

Property 13

If ${\mathcal {J}}$ contains an uncountable set then $\langle X, {\mathcal {T}}^*\rangle$ is not a Lindelöf space.

3 Separation Axioms for a GTS Generated by an Operator Connected with Measurable Hull

From now, we will consider a measurable space $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$, where ${\mathcal {S}}$ is a $\sigma$-algebra of subsets of X and ${\mathcal {J}}\subset {\mathcal {S}}$ is a admissible $\sigma$-ideal.

Let $\Phi :{\mathcal {S}}\rightarrow 2^X$ be any operator satisfying the following properties:

$1^\circ$:

$\Phi (\emptyset )=\emptyset$ and $\Phi (X)=X$,

$2^\circ$:

for any $A, B\in {\mathcal {S}}$ if $A\subset B$ then $\Phi (A)\subset \Phi (B)$,

$3^\circ$:

for any $A, B\in {\mathcal {S}}$ if $A\triangle B\in {\mathcal {J}}$ then $\Phi (A)=\Phi (B)$,

$4^\circ$:

$A{\setminus } \Phi (A)\in {\mathcal {J}}$ for any $A\in {\mathcal {S}}$.

It is easy to see that if $B_1$ and $B_2$ are ${\mathcal {S}}$-measurable hulls of a set $A\subset X$ then $B_1\triangle B_2\in {\mathcal {J}}$, so $\Phi (B_1)=\Phi (B_2)$. Therefore, if $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ has a hull property and we put for any $A\subset X$

$$\begin{aligned} \Phi _h^*(A)=\Phi (B), \end{aligned}$$

where B is a ${\mathcal {S}}$-measurable hull of A, we obtain an operator $\Phi _h^*:2^X\rightarrow 2^X$. It is easy to see that the operator $\Phi _h^*$ satisfies conditions $1^*$–$4^*$. Indeed, conditions $1^*$ and $4^*$ are obvious. To prove condition $2^*$ let us consider sets $A, B\subset X$ such that $A\subset B$. Let $A_1$ and $B_1$ be ${\mathcal {S}}$-measurable hulls of A and B, respectively. Clearly $A_1\cap B_1$ is ${\mathcal {S}}$-measurable hull of A. Thus $\Phi ^*_h(A)=\Phi (A_1\cap B_1)\subset \Phi (B_1)=\Phi _h^*(B)$. Now, let $A, B\subset X$ be such that $A\triangle B=C\in {\mathcal {J}}$. Let $A_1$ and $B_1$ be ${\mathcal {S}}$-measurable hulls of A and B, respectively. Obviously, $A=C \triangle B\subset C\cup B\subset C\cup B_1$ and $\Phi _h^*(A)=\Phi (A_1)=\Phi (A_1 \cap (B_1\cap C))=\Phi (A_1\cap B_1)\subset \Phi (B_1)=\Phi _h^*(B)$. Similarly, we can show that$\Phi _h^*(B)\subset \Phi _h^*(A)$, which completes the proof of condition $3^*$.

Thus the family

$$\begin{aligned} {\mathcal {T}}_h^*=\{A\subset X: A\subset \Phi _h^*(A)\} \end{aligned}$$

is a strong generalized topology and it is not equal to $2^X$. Moreover, it is easy to prove the following property:

Property 14

If $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ has a hull property, $A, B\subset X$ and C is an ${\mathcal {S}}$-measurable hull of B then $\Phi _h^*(A\cup B)=\Phi _h^*(A\cup C)$.

Proof

Let $C_1$ be an ${\mathcal {S}}$-measurable hull of $A\cup C$. We will show that $C_1$ is an ${\mathcal {S}}$-measurable hull of $A\cup B$. Let $E\subset C_1{\setminus } (A\cup B)$ and $E\in {\mathcal {S}}$. Obviously, $E=(E\cap C) \cup (E{\setminus } C)$ and $E\cap C, E{\setminus } C \in {\mathcal {S}}$. Moreover, $E\cap C\subset C{\setminus } B$ and $E{\setminus } C\subset C_1{\setminus } (A\cup C)$, so $(E\cap C) \cup (E{\setminus } C) \in {\mathcal {J}}$. Thus $\Phi _h^*(A\cup B)=\Phi (C_1)=\Phi _h^*(A\cup C)$. $\square$

3.1 A Non-separation Property

We shall say that a space $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ has a non-separation property with respect to ${\mathcal {S}}$ iff for any set $E\not \in {\mathcal {J}}$ there exist disjoint subsets of E which cannot be separated by sets from ${\mathcal {S}}$. Let us note that the spaces $({\mathbb {R}}, {\mathcal {L}}, {\mathbb {L}})$ and $({\mathbb {R}}, {\mathcal {B}}a, {\mathcal {I}})$, where ${\mathbb {R}}$ is the set of reals, ${\mathcal {L}}$—the $\sigma$-algebra of Lebesgue measurable sets, ${\mathbb {L}}$—the $\sigma$-ideal of sets with the Lebesgue measure zero, ${\mathcal {B}}a$—the $\sigma$-algebra of sets with Baire property and ${\mathcal {I}}$—the $\sigma$-ideal of meager sets, have a non-separation property with respect to ${\mathcal {L}}$ and ${\mathcal {B}}a$, respectively (see [4, 5]).

Theorem 15

[5]. Let $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ have a non-separation property with respect to ${\mathcal {S}}$ and satisfy CCC. Every infinite set $C\subset X$ can be decomposed into an infinite family of sets, each of which is full in C. In particular, any set $C\not \in {\mathcal {J}}$ can be decomposed into subsets $C_1$ and $C_2$ full in C.

From the above and Property 2 we obtain immediately

Corollary 16

If $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ has a non-separation property with respect to ${\mathcal {S}}$ and satisfies CCC, then every infinite set $C\subset X$ can be decomposed into subsets $C_1$ and $C_2$ such that every ${\mathcal {S}}$-measurable hull of C is simultaneously an ${\mathcal {S}}$-measurable hull of $C_1$ and $C_2$.

Theorem 17

If $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ has a non-separation property with respect to ${\mathcal {S}}$ and satisfy CCC, then the space $\langle X, {\mathcal {T}}_h^*\rangle$ is totally disconnected.

Proof

Let $A\subset X$ and ${{\,\textrm{card}\,}}(A)\ge 2$. Fix $x,y\in A$ such that $x\not =y$. Corollary 16 gives that there exist disjoint sets $X_1$ and $X_2$ such that $X=X_1\cup X_2$ and X is an ${\mathcal {S}}$-measurable hull of $X_1$ and $X_2$. Then the sets $(X_1{\setminus } \{x\})\cup \{y\}$ and $(X_2{\setminus } \{y\})\cup \{x\}$ are disjoint, clopen and their union is equal to X. Putting $A_1=((X_1{\setminus } \{x\})\cup \{y\})\cap A$ and $A_2=((X_2{\setminus } \{y\})\cup \{x\})\cap A$ we obtain that $A_1$, $A_2$ are non-empty sets, $A= A_1\cup A_2$ and $A_1$, $A_2$ are separated sets. In consequence, we have that A is a disconnected set. $\square$

3.2 The Separation Axioms

From now, we will assume that $\langle X, {\mathcal {S}}, {\mathcal {J}}\rangle$ has a non-separation property with respect to ${\mathcal {S}}$ and satisfies CCC. We will focus on the separation axioms of the space $\langle X, {\mathcal {T}}_h^*\rangle$.

Theorem 18

The space $\langle X, {\mathcal {T}}_h^*\rangle$ is a Hausdorff space.

Proof

Let $x\not =y$. Corollary 16 implies that there exist disjoint sets $C_1$ and $C_2$ such that $X=C_1\cup C_2$ and X is an ${\mathcal {S}}$-measurable hull of $C_1$ and $C_2$. Obviously $C_1, C_2\in {\mathcal {T}}_h^*$ because $\Phi _h^*(C_1)=\Phi _h^*(C_2)=\Phi (X)=X$.

If $x\in C_1$ and $y\in C_2$ (or $y\in C_1$ and $x\in C_2$) then putting $V_1=C_1$ and $V_2=C_2$ we obtained that $V_1, V_2\in {\mathcal {T}}_h^*$, $V_1\cap V_2=\emptyset$, $x\in V_1$ and $y\in V_2$ (or $y\in V_1$ and $x\in V_2$).

If $x, y\in C_1$ then putting $V_1=C_1{\setminus } \{y\}$ and $V_2=C_2\cup \{y\}$ we obtain that $V_1, V_2\in {\mathcal {T}}_h^*$ because $\Phi _h^*(V_1)=\Phi _h^*(C_1{\setminus } \{y\})=\Phi _h^*(C_1)=X\supset V_1$ and $\Phi _h^*(V_2)=\Phi _h(C_2\cup \{y\})=\Phi _h^*(C_2)=X\supset V_1$. Moreover, we have that $x\in V_1$, $y\in V_2$ and $V_1\cap V_2=\emptyset$. If $x, y\in C_2$ the proof is similar. $\square$

Remark

Let us note that the sets $V_1$ and $V_2$ constructed in the above proof are clopen sets.

Theorem 19

The space $\langle X, {\mathcal {T}}_h^*\rangle$ is a hereditarily normal space.

Proof

By Theorem 18 we obtain that $\langle X, {\mathcal {T}}_h^*\rangle$ is a Hausdorff space.

Let $A_1, A_2$ be separated sets in X. Put $C=X{\setminus } (A_1\cup A_2)$.

If $C\in {\mathcal {J}}$ then $V_1=(X{\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2)){\setminus } C\in {\mathcal {T}}_h^*$ and $V_2=(X{\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_1)){\setminus } C\in {\mathcal {T}}_h^*$. Clearly $A_1\subset V_1$, $A_2\subset V_2$ and $V_1\cap V_2=\emptyset$.

If $C\not \in {\mathcal {J}}$, then Theorem 16 gives that there are disjoint sets $C_1, C_2\subset C$ such that $C=C_1\cup C_2$ and every ${\mathcal {S}}$-measurable hull of C is an ${\mathcal {S}}$-measurable hull of $C_1$ and $C_2$. Putting $V_1=(A_1\cup C_1) {\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2)$ and $V_2=(A_2\cup C_2) {\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_1)$ we obtain that $A_1\subset V_1$, $A_2\subset V_2$ and $V_1\cap V_2=\emptyset$. Moreover, $V_1, V_2 \in {\mathcal {T}}_h^*$. Indeed, note first that Property 8 implies that $B={{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2){\setminus } A_2 \in {\mathcal {J}}$. Moreover, if H is an ${\mathcal {S}}$-measurable hull of C then, by Property 14,

$$\begin{aligned} \Phi _h^*(V_1)= & {} \Phi _h^*((A_1\cup C_1) {\setminus } {{\,\text {cl}\,}}_{{\mathcal {T}}_h^*}(A_2))=\Phi _h^*((A_1\cup C_1) {\setminus } (B\cup A_2))\\ {}= & {} \Phi _h^*((A_1\cup C_1) \cap (X{\setminus } B)\cap X{\setminus } A_2))= \Phi _h^*((A_1\cup C_1) {\setminus } B)\\ {}= & {} \Phi _h^*(A_1\cup C_1) = \Phi _h^*(A_1\cup H)=\Phi _h^*(A_1\cup C). \end{aligned}$$

Simultaneously,

$$\begin{aligned} \Phi _h^*(A_1\cup C)= & {} \Phi _h^*((A_1\cup C) {\setminus } B)= \Phi _h^*((A_1\cup C) {\setminus } ({{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2){\setminus } A_2)) \\= & {} \Phi _h^*(((A_1\cup C) {\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2))\cup ((A_1\cup C_1)\cap A_2))\\= & {} \Phi _h^*((A_1\cup C) {\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2)). \end{aligned}$$

We show that $V_1\subset \Phi _h^*(V_1)$. Suppose that there exists a point $x_0\in V_1{\setminus } \Phi _h^*(V_1)$. Thus $x_0\in {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(X{\setminus } ((A_1\cup C){\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2)))$. Indeed, suppose contrary to our claim, that $x_0\not \in {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(X{\setminus } ((A_1\cup C){\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2)))$. Therefore one could find a set $W\in {\mathcal {T}}_h^*$ such that $x_0\in W\subset (A_1\cup C) {\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2)$, so $x_0\in W\subset \Phi _h^*(W)\subset \Phi _h^*((A_1\cup C) {\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2))= \Phi _h^*(V_1)$, which is impossible.

Thus

$$\begin{aligned} x_0\in & {} {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(X{\setminus } ((A_1\cup C){\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2)))={{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}((X{\setminus } (A_1\cup C))\cup {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2))\\= & {} {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}((X{\setminus } (X{\setminus } A_2))\cup {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2))={{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2\cup {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2))={{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2), \end{aligned}$$

which contradicts the fact that $x_0\in V_1=(A_1\cup C_1) {\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2)$. Finally, we obtain that $V_1\subset \Phi _h^*(V_1)$, so $V_1\in {\mathcal {T}}_h^*$. By the similar arguments, one can prove that $V_2\in {\mathcal {T}}_h^*$. Lemma 1 implies now, that $\langle X, {\mathcal {T}}_h^*\rangle$ is a hereditarily normal space. $\square$

Corollary 20

The space $\langle X, {\mathcal {T}}_h^*\rangle$ is a normal space.

Remark 21

The analysis of the above proof allows us to conclude that if the sets $A_1$ and $A_2$ are closed and disjoint then the sets $V_1=(A_1\cup C_1) {\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_2)=A_1\cup C_1$ and $V_2=(A_2\cup C_2) {\setminus } {{\,\textrm{cl}\,}}_{{\mathcal {T}}_h^*}(A_1)=A_2\cup C_2$ are clopen sets, because $V_1=X{\setminus } V_2$.

Theorem 22

The space $\langle X, {\mathcal {T}}_h^*\rangle$ is a completely regular space.

Proof

Let $F\subset X$ be a ${\mathcal {T}}_h^*$-closed set and $x_0\not \in F$. Thus from Corollary 20 and Remark 21 it follows that there exist disjoint clopen sets $V_1$ and $V_2$ such that $x_0\in V_1$ and $F\subset V_2$. Thus the function $f(x)=\chi _{V_1}(x)$ for $x\in X$ is ${\mathcal {T}}_h^*$-continuous, $f(x_0)=1$ and $f(F)\subset \{0\}$. $\square$

Lemma 23

In the space $\langle X, {\mathcal {T}}_h^*\rangle$ every set $A\subset X$ is ${\mathcal {T}}_h^*$-$G_{\delta }$-sets.

Proof

Let $A\subset X$ and $C=X{\setminus } A$. If $C\in {\mathcal {J}}$ then Property 4 gives that $A\in {\mathcal {T}}_h^*$, so A is a ${\mathcal {T}}_h^*$-$G_{\delta }$-set. If $C\not \in {\mathcal {J}}$ then Corollary 16 implies that there are disjoint sets $C_1, C_2\subset C$ such that $C=C_1\cup C_2$ and every ${\mathcal {S}}$-measurable hull H of C is an ${\mathcal {S}}$-measurable hull of $C_1$ and $C_2$. By Property 14, we obtain that

$$\begin{aligned} \Phi _h^*(A\cup C_1)=\Phi _h^*(A\cup H)= \Phi _h^*(A\cup C)=X, \end{aligned}$$

so $A\cup C_1\in {\mathcal {T}}_h^*$. Similarly, we can show that $A\cup C_2\in {\mathcal {T}}_h^*$. Therefore, $A=(A\cup C_1)\cap (A\cup C_2)$ is a ${\mathcal {T}}_h^*$-$G_{\delta }$-set. $\square$

From the above we obtain immediately

Theorem 24

The space $\langle X, {\mathcal {T}}_h^*\rangle$ is a perfectly normal space.

4 Examples

Example 25

Let $\lambda$ be the Lebesgue measure, $\lambda ^*$ be the outer Lebesgue measure, $A\subset {\mathbb {R}}$ and

$$\begin{aligned} \Phi _{0}^*(A)=\left\{ x\in {\mathbb {R}}: \lim \limits _{h\rightarrow 0^+} \frac{\lambda ^*(A\cap [x_0-h,x_0+h])}{2h}=1\right\} . \end{aligned}$$

For any $A\subset {\mathbb {R}}$ we have

$$\begin{aligned} \Phi _0^*(A)=\Phi _0(H) \end{aligned}$$

where H is a measurable hull of A and $\Phi _0:{\mathcal {L}}\rightarrow 2^{{\mathbb {R}}}$ is defined in the following way:

$$\begin{aligned} \Phi _{0}(B)=\left\{ x\in {\mathbb {R}}: \lim \limits _{h\rightarrow 0^+} \frac{\lambda (B\cap [x_0-h,x_0+h])}{2h}=1\right\} , \end{aligned}$$

for any $B\in {\mathcal {L}}$.

Example 26

Let $\lambda$ be the Lebesgue measure, $\lambda ^*$ be the outer Lebesgue measure, $A\subset {\mathbb {R}}$ and

$$\begin{aligned} \Phi _{1}^*(A)=\left\{ x\in {\mathbb {R}}: \limsup \limits _{h\rightarrow 0^+} \frac{\lambda ^*(A\cap [x_0-h,x_0+h])}{2h}=1\right\} . \end{aligned}$$

For any $A\subset {\mathbb {R}}$ we have

$$\begin{aligned} \Phi _1^*(A)=\Phi _1(H) \end{aligned}$$

where H is a measurable hull of A and $\Phi _1:{\mathcal {L}}\rightarrow 2^{{\mathbb {R}}}$ is defined in the following way:

$$\begin{aligned} \Phi _{1}(B)=\left\{ x\in {\mathbb {R}}: \limsup \limits _{h\rightarrow 0^+} \frac{\lambda (B\cap [x_0-h,x_0+h])}{2h}=1\right\} , \end{aligned}$$

for any $B\in {\mathcal {L}}$.

Theorem 27

Let $i\in \{0,1\}$. The operator $\Phi _i^*$ satisfies conditions $1^*$–$4^*$ and the family

$$\begin{aligned} {\mathcal {T}}^*_{i}=\{A\subset {\mathbb {R}}: A\subset \Phi ^*_{i}(A)\} \end{aligned}$$

is a strong generalized topology on ${\mathbb {R}}$ containing the classical density topology ${\mathcal {T}}_d$ and the space $\langle {\mathbb {R}}, {\mathcal {T}}_{i}^*\rangle$ satisfies all separation axioms.

Remark 28

In particular, the space $\langle {\mathbb {R}}, {\mathcal {T}}_{d}^*\rangle$, where ${\mathcal {T}}_{d}^*=\{A\subset {\mathbb {R}}: A\subset \Phi ^*_d(A)\}$ (see [8]), satisfies all separation axioms.

Example 29

Let $\lambda$ be the Lebesgue measure, $\lambda ^*$ be the outer Lebesgue measure, $A\subset {\mathbb {R}}$ and $\alpha \in [0, 1)$

$$\begin{aligned} \Phi _{1\alpha }^*(A)=\left\{ x\in {\mathbb {R}}: \lim \limits _{h\rightarrow 0^+} \frac{\lambda ^*(A\cap [x_0-h,x_0+h])}{2h}>\alpha \right\} . \end{aligned}$$

For any $A\subset {\mathbb {R}}$ we have

$$\begin{aligned} \Phi _{1\alpha }^*(A)=\Phi _{1\alpha }(H) \end{aligned}$$

where H is a measurable hull of A and $\Phi _{1\alpha }:{\mathcal {L}}\rightarrow 2^{{\mathbb {R}}}$ is defined in the following way:

$$\begin{aligned} \Phi _{1\alpha }(B)=\left\{ x\in {\mathbb {R}}: \lim \limits _{h\rightarrow 0^+} \frac{\lambda (B\cap [x_0-h,x_0+h])}{2h}>\alpha \right\} , \end{aligned}$$

for any $B\in {\mathcal {L}}$.

Example 30

Let $\lambda$ be the Lebesgue measure, $\lambda ^*$ be the outer Lebesgue measure, $A\subset {\mathbb {R}}$ and $\alpha \in [0, 1)$

$$\begin{aligned} \Phi _{2\alpha }^*(A)=\left\{ x\in {\mathbb {R}}: \liminf \limits _{h\rightarrow 0^+} \frac{\lambda ^*(A\cap [x_0-h,x_0+h])}{2h}>\alpha \right\} . \end{aligned}$$

For any $A\subset {\mathbb {R}}$ we have

$$\begin{aligned} \Phi _{2\alpha }^*(A)=\Phi _{2\alpha }(H) \end{aligned}$$

where H is a measurable hull of A and $\Phi _{2\alpha }:{\mathcal {L}}\rightarrow 2^{{\mathbb {R}}}$ is defined in the following way:

$$\begin{aligned} \Phi _{2\alpha }(B)=\left\{ x\in {\mathbb {R}}: \liminf \limits _{h\rightarrow 0^+} \frac{\lambda (B\cap [x_0-h,x_0+h])}{2h}>\alpha \right\} , \end{aligned}$$

for any $B\in {\mathcal {L}}$.

Example 31

Let $\lambda$ be the Lebesgue measure, $\lambda ^*$ be the outer Lebesgue measure, $A\subset {\mathbb {R}}$ and $\alpha \in [0, 1)$

$$\begin{aligned} \Phi _{3\alpha }^*(A)=\left\{ x\in {\mathbb {R}}: \limsup \limits _{h\rightarrow 0^+} \frac{\lambda ^*(A\cap [x_0-h,x_0+h])}{2h}>\alpha \right\} . \end{aligned}$$

For any $A\subset {\mathbb {R}}$ we have

$$\begin{aligned} \Phi _{3\alpha }^*(A)=\Phi _{3\alpha }(H) \end{aligned}$$

where H is a measurable hull of A and $\Phi _{3\alpha }:{\mathcal {L}}\rightarrow 2^{{\mathbb {R}}}$ is defined in the following way:

$$\begin{aligned} \Phi _{3\alpha }(B)=\left\{ x\in {\mathbb {R}}: \limsup \limits _{h\rightarrow 0^+} \frac{\lambda (B\cap [x_0-h,x_0+h])}{2h}>\alpha \right\} , \end{aligned}$$

for any $B\in {\mathcal {L}}$.

Theorem 32

Let $i\in \{1,2,3\}$ and $\alpha \in [0,1)$. The operator $\Phi _{i\alpha }^*$ satisfies conditions $1^*$–$4^*$ and the family

$$\begin{aligned} {\mathcal {T}}^*_{i\alpha }=\{A\subset {\mathbb {R}}: A\subset \Phi ^*_{i\alpha }(A)\} \end{aligned}$$

is a strong generalized topology on ${\mathbb {R}}$ such that ${\mathcal {T}}_d\subset {\mathcal {T}}^*_{i\alpha }$ and the space $\langle {\mathbb {R}}, {\mathcal {T}}_{i\alpha }^*\rangle$ satisfies all separation axioms.

Remark 33

Let $\beta \in (0, 1]$. If in the above examples we replace $\alpha$ by $\beta$ and the symbol > by the symbol $\ge$ we obtain three new operators $\Phi ^*_{i\beta }$ for $i\in \{1,2,3\}$.

Theorem 34

Let $i\in \{1,2,3\}$ and $\beta \in (0, 1]$. The operator $\Phi _{i\beta }^*$ satisfies conditions $1^*$–$4^*$ and the family

$$\begin{aligned} {\mathcal {T}}^*_{i\beta }=\{A\subset {\mathbb {R}}: A\subset \Phi ^*_{i\beta }(A)\} \end{aligned}$$

is a strong generalized topology on ${\mathbb {R}}$ such that ${\mathcal {T}}_d\subset {\mathcal {T}}^*_{i\beta }$ and the space $\langle {\mathbb {R}}, {\mathcal {T}}_{i\beta }^*\rangle$ satisfies all separation axioms.

Example 35

Let $\Phi : {\mathcal {B}}a\rightarrow {\mathcal {B}}a$ be the lower density operator on $\langle {\mathbb {R}}, {\mathcal {B}}a, {\mathcal {I}}\rangle$. There are numerous of such operators. For example well known the ${\mathcal {I}}$-density operator (see [11]), simple density operator $\Phi _{s{\mathcal {I}}}$ (see [12]), density operator $\Phi _{{\mathcal {I}}(J)}$ (see [10]), $\langle s\rangle$-density operator $\Phi _{{\mathcal {I}}\langle s\rangle }$ being a generalization of ${\mathcal {I}}$-density operator (see [6]). Equality

$$\begin{aligned} \Phi ^*(A)=\Phi (B) \end{aligned}$$

where B is a ${\mathcal {B}}a$-measurable hull of A yields the operator $\Phi ^*$ satisfying conditions $1^*$–$4^*$. Finally, taking into account that the pair $\langle {\mathcal {B}}a,{\mathcal {I}}\rangle$ has a non-separation property we get the following theorem

Theorem 36

The family

$$\begin{aligned} {\mathcal {T}}^*=\{A\subset {\mathbb {R}}: A\subset \Phi ^*(A)\} \end{aligned}$$

is a strong generalized topology on ${\mathbb {R}}$ and the space $\langle {\mathbb {R}}, {\mathcal {T}}^*\rangle$ satisfies all separation axioms.

It is worth noting that in the case of the operator $\Phi _d^*$ (see Remark 28) and the operator $\Phi ^*_{{\mathcal {I}}}$ (see Example 35), they generate generalized topological spaces satisfying all separation axioms. However, in the case of the classical density operator $\Phi _d$ and ${\mathcal {I}}$-density operator $\Phi _{{\mathcal {I}}}$ generating topological spaces $\langle {\mathbb {R}}, {\mathcal {T}}_d\rangle$ and $\langle {\mathbb {R}}, {\mathcal {T}}_{{\mathcal {I}}}\rangle$, we obtain a completely different situation. In the case of the classical density operator, we have that the density topological space $\langle {\mathbb {R}}, {\mathcal {T}}_d\rangle$ is completely regular but not normal and the space $\langle {\mathbb {R}}, {\mathcal {T}}_{{\mathcal {I}}}\rangle$ is not even regular.