Almost continuity and its applications on weak structures

Abstract

In this paper, we introduce and study the concept of almost continuity in weak structures (Császár in Acta Math Hung 131(1–2):193–195, 2011) and discuss some of its characteristic properties. Finally, we give some applications of this new kind of continuity.

1 Introduction and preliminaries

Császár [1] defined generalized topology and studied some of its concepts like generalized open sets and continuity. Later on, Maki et al. [6] introduced minimal structures and investigated some of its concepts. Lately, Császár [2] presented the weak structures (A family $\mathcal {W}\subset P(X)$ is called a weak structure on X (briefly, WS) iff $\emptyset \in \mathcal {W}$). A non-empty set X with a weak structure $\mathcal {W}$ is called simply a space $(X,\mathcal {W})$. The members of $\mathcal {W}$ are $\mathcal {W}$-open subsets and their complements are $\mathcal {W}$-closed subsets. Moreover, Császár [2] presented the operations $c_{_\mathcal {W}} (A)$ and $i_{_\mathcal {W}} (A)$ in WS as the intersection of all $\mathcal {W}$-closed set containing A and the union of all $\mathcal {W}$-open subsets of A. Also, the properties of $c_{_\mathcal {W}} (A)$ and $i_{_\mathcal {W}}$ are introduced and discussed. For more details about weak structures, the readers should refer [4, 5, 7,8,9].

Theorem 1.1

[2] For any space $(X,\mathcal {W})$ and $A, B \subseteq X$, we have:

• (1) $A \subseteq c_{_\mathcal {W}}(A)$ and $A \supseteq i_{_\mathcal {W}}(A)$;

• (2) If $A\subseteq B$, then $c_{_\mathcal {W}}(A)\subset c_{_\mathcal {W}}(B)$ and $i_{_\mathcal {W}}(A)\subset i_{_\mathcal {W}}(B)$;

• (3) If A is $\mathcal {W}$-closed, then $A=c_{_\mathcal {W}}(A)$, and if A is $\mathcal {W}$-open, then $A=i_{_\mathcal {W}}(A)$;

• (4) $c_{_\mathcal {W}}(c_{_\mathcal {W}}(A))=c_{_\mathcal {W}}(A)$ and $i_{_\mathcal {W}}(i_{_\mathcal {W}}(A))=i_{_\mathcal {W}}(A)$;

• (5) $c_{_\mathcal {W}}(X-A)=X-i_{_\mathcal {W}} (A)$ and $i_{_\mathcal {W}}(X-A)=X-c_{_\mathcal {W}} (A)$;

• (6) $i_{_\mathcal {W}}(c_{_\mathcal {W}} (i_{_\mathcal {W}} (c_{_\mathcal {W}}(A))))=i_{_\mathcal {W}}(c_{_\mathcal {W}} (A))$ and $c_{_\mathcal {W}} (i_{_\mathcal {W}}(c_{_\mathcal {W}} (i_{_\mathcal {W}} (A))))=c_{_\mathcal {W}} (i_{_\mathcal {W}}(A))$;

• (7) $x \in c_{_\mathcal {W}}(A)$ iff $V \cap A\ne \emptyset$ for every $\mathcal {W}$-open subset V containing x;

• (8) $x \in i_{_\mathcal {W}}(A)$ iff there exists a $\mathcal {W}$-open subset V such that $x \in V \subset A$.

Theorem 1.2

[3] For a space $(X,\mathcal {W})$ and $U, V \subseteq X$, we have:

• (1) $i_{_\mathcal {W}}(U \bigcap V) \subseteq i_{_\mathcal {W}}(U)\bigcap i_{_\mathcal {W}}(V)$;

• (2) $c_{_\mathcal {W}}(U) \cup c_{_\mathcal {W}}(V) \subseteq c_{_\mathcal {W}}(U\cup V).$

Definition 1.1

Let $(X,\mathcal {W})$ be a space and $A \subseteq X$. Then

• (1) $A \in \alpha (\mathcal {W})$ [2] if $A \subseteq i_{_\mathcal {W}}(c_{_\mathcal {W}}(i_{_\mathcal {W}} (A)))$;

• (2) $A \in \pi (\mathcal {W})$ [2] if $A \subseteq i_{_\mathcal {W}}(c_{_\mathcal {W}}( (A)))$;

• (3) $A \in \sigma (\mathcal {W})$ [2] if $A \subseteq c_{_\mathcal {W}}((i_{_\mathcal {W}} (A))$;

• (4) $A \in \beta (\mathcal {W})$ [2] if $A \subseteq c_{_\mathcal {W}}(i_{_\mathcal {W}}(c_{_\mathcal {W}}(A)))$;

• (5) $A \in \rho (\mathcal {W})$ [2] if $A \subseteq i_{_\mathcal {W}}(c_{_\mathcal {W}}(A))\bigcup c_{_\mathcal {W}}(i_{_\mathcal {W}} (A))$;

• (6) $A \in r(\mathcal {W})$ [3] if $A=i_{_\mathcal {W}}(c_{_\mathcal {W}} (A))$;

• (7) $A \in rc(\mathcal {W})$ [3] if $A=c_{_\mathcal {W}} (i_{_\mathcal {W}}((A))$.

2 Almost $\mathcal {W}$-continuity

Lemma 2.1

For any space $(X,\mathcal {W})$ and $\mathcal {W}$-open set V, if $U \bigcap V=\emptyset$, then $c_{_\mathcal {W}} (U) \bigcap V=\emptyset$ for each subset U of X.

Proof

Let V be an $\mathcal {W}$-open set and $U \subseteq X$. Suppose $U \bigcap V=\emptyset$ and $c_{_\mathcal {W}} (U) \bigcap V\ne \emptyset$, then there exists $x \in X$ such that $x \in c_{_\mathcal {W}} (U)$ and $x \in V$. Since $x \in c_{_\mathcal {W}} (U)$ and V is an $\mathcal {W}$-open set containing x, then $U \bigcap V\ne \emptyset$. This is a contradiction. Therefore if $U \bigcap V=\emptyset$, then $c_{_\mathcal {W}} (U) \bigcap V=\emptyset$ for each subset U of X. $\square$

Definition 2.1

For any space $(X,\mathcal {W})$ and $S \subseteq X$, a point $x \in X$ is said to be:

• (1) $\mathcal {W}_ \theta$-cluster point of S if $c_{_\mathcal {W}}(V) \bigcap S \ne \emptyset$ for every $\mathcal {W}$-open set V containing x.

• (2) $\mathcal {W}_ \delta$-cluster point of S if $i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)) \bigcap S \ne \emptyset$ for every $\mathcal {W}$-open set V containing x.

The set of all $\mathcal {W}_\theta$ (resp. $\mathcal {W}_\delta )$-cluster points of S is called the $\mathcal {W}_\theta$ (resp. $\mathcal {W}_\delta )$-closure of S and is denoted by $c_{_\mathcal {W}}^{\theta }(S)$ (resp. $c_{_\mathcal {W}}^{\delta }(S))$. If $S=c_{_\mathcal {W}}^{\theta }(S)$ (resp. $S=c_{\mathcal {W}}^\delta (S))$, then A is called $\mathcal {W}_\theta$ (resp. $\mathcal {W}_ \delta )$-closed. The complement of an $\mathcal {W}_\theta$ (resp. $\mathcal {W}_\delta )$-closed set is called $\mathcal {W}_\theta$ (resp. $\mathcal {W}_\delta )$-open.

The union of all $\mathcal {W}_\theta$ (resp. $\mathcal {W}_\delta )$-open sets contained in S is called the $\mathcal {W}_\theta$ (resp. $\mathcal {W}_\delta )$-interior of S and is denoted by $i_{_\mathcal {W}}^{\theta }(S)$ (resp. $i_{_\mathcal {W}}^{\delta } (S))$. The class of $\mathcal {W}_\theta$ (resp. $\mathcal {W}_ \delta )$-open sets in $\mathcal {W}$ is denoted by $\theta (\mathcal {W})$ (resp. $\delta (\mathcal {W}))$ and the class of $\mathcal {W}_\theta$ (resp. $\mathcal {W}_ \delta )$-closed sets in $\mathcal {W}$ is denoted by $\theta c (\mathcal {W})$ (resp. $\delta c(\mathcal {W}))$.

Remark 2.1

One may notice that if $(X,\mathcal {W})$ a space and $V \subseteq X$, then

$$\begin{aligned} c_{_\mathcal {W}} (V)\subseteq c_{_\mathcal {W}}^{\delta }(V)\subseteq c_{_\mathcal {W}}^{\theta }(V). \end{aligned}$$

Lemma 2.2

Let $(X,\mathcal {W})$ be a space. Then

$$\begin{aligned} c_{_\mathcal {W}} (V)=c_{_\mathcal {W}}^{\delta }(V)=c_{_\mathcal {W}}^{\theta }(V) \end{aligned}$$

for each $\mathcal {W}$-open set V in X.

Proof

We aim to prove that $c_{_\mathcal {W}}^{\theta }(N)\subseteq c_{_\mathcal {W}} (N)$. Let N be an $\mathcal {W}$-open set in X and let $x \notin c_{_\mathcal {W}} (N)$. Then there exists $M\in \mathcal {W}$ such that $x\in M$ and $M\bigcap N=\emptyset$. By Lemma 2.1, we have $c_{_\mathcal {W}} (M) \bigcap N =\emptyset$ and hence $x \notin c_{_\mathcal {W}}^{\theta } (N)$. Thus $c_{_\mathcal {W}}^{\theta }(N)\subseteq c_{_\mathcal {W}} (N)$. Since $c_{_\mathcal {W}}(N)\subseteq c_{_\mathcal {W}}^{\theta }(N)$ for each subset N in X, then $c_{_\mathcal {W}}(N)=c_{_\mathcal {W}}^{\theta }(N)$. $\square$

Definition 2.2

A function $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ from a space $(X,\mathcal {W}_X)$ to a space $(Y,\mathcal {W}_Y)$ is called almost $\mathcal {W}$-continuous at $x \in X$ if for every $\mathcal {W}_Y$-open set N containing f(x), there is a $\mathcal {W}_X$-open set M including x such that $f(M)\subseteq i_{_\mathcal {W}}(c_{_\mathcal {W}}(N))$. A map f is called almost $\mathcal {W}$-continuous if it is almost $\mathcal {W}$-continuous at each $x \in X$.

Theorem 2.1

For a function $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$. The following statements are equivalent:

• (1) f is almost $\mathcal {W}$-continuous;

• (2) $f^{-1}(V)\subseteq i_{_\mathcal {W}} (f^{-1}(V))$ for each $V \in r(\mathcal {W}_Y);$

• (3) $c_{_\mathcal {W}} (f^{-1}(F))\subseteq f^{-1}(F)$ for each $F \in rc(\mathcal {W}_Y);$

• (4) $f^{-1}(V) \subseteq i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))))$ for each $\mathcal {W}$-open set V in Y;

• (5) $c_{_\mathcal {W}} (f^{-1}(c_{_\mathcal {W}} (i_{_\mathcal {W}}(F)))) \subseteq f^{-1}(F)$ for each $\mathcal {W}$-closed set F in Y;

• (6) $f^{-1}(V) \subseteq i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))))$ for each $V \in \pi (\mathcal {W}_Y)$;

• (7) $c_{_\mathcal {W}} (f^{-1}(c_{_\mathcal {W}} (i_{_\mathcal {W}}(F)))) \subseteq f^{-1}(F)$ for each $F \in \pi c(\mathcal {W}_Y)$;

• (8) $f^{-1}(V) \subseteq i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))))$ for each $V \in \alpha (\mathcal {W}_Y)$;

• (9) $c_{_\mathcal {W}} (f^{-1}(c_{_\mathcal {W}} (i_{_\mathcal {W}}(F)))) \subseteq f^{-1}(F)$ for each $F \in \alpha c(\mathcal {W}_Y)$;

• (10) For each a point $x \in X$ and a $V \in \pi (\mathcal {W}_Y)$ containing f(x), there exists an $\mathcal {W}_X$-open set U containing x such that $f(U)\subseteq i_{_\mathcal {W}}(c_{_\mathcal {W}}(V))$;

• (11) For each a point $x \in X$ and a $V \in \alpha (\mathcal {W}_Y)$ containing f(x), there exists an $\mathcal {W}_X$-open set U containing x such that $f(U)\subseteq i_{_\mathcal {W}}(c_{_\mathcal {W}}(V))$;

• (12) $c_{_\mathcal {W}} (f^{-1}(V))\subseteq f^{-1}(c_{_\mathcal {W}} (V))$ for each $V \in \beta (\mathcal {W}_Y)$;

• (13) $f^{-1}(i_{_\mathcal {W}} (F))\subseteq i_{_\mathcal {W}} (f^{-1}(F))$ for each $F \in \beta c(\mathcal {W}_Y)$;

• (14) $c_{_\mathcal {W}} (f^{-1}(V))\subseteq f^{-1}(c_{_\mathcal {W}} (V))$ for each $V \in \sigma (\mathcal {W}_Y)$;

• (15) $f^{-1}(i_{_\mathcal {W}} (F))\subseteq i_{_\mathcal {W}} (f^{-1}(F))$ for each $F \in \sigma c (\mathcal {W}_Y)$;

• (16) $c_{_\mathcal {W}} (f^{-1}(V))\subseteq f^{-1}(c_{_\mathcal {W}} (V))$ for each $V \in \pi (\mathcal {W}_Y)$;

• (17) $f^{-1}(i_{_\mathcal {W}} (V))\subseteq i_{_\mathcal {W}} (f^{-1}(V))$ for each $V \in \pi c (\mathcal {W}_Y)$;

• (18) $c_{_\mathcal {W}} (f^{-1}(V))\subseteq f^{-1}(c_{_\mathcal {W}} (V))$ for each $V \in \alpha (\mathcal {W}_Y)$;

• (19) $f^{-1}(i_{_\mathcal {W}} (V))\subseteq i_{_\mathcal {W}} (f^{-1}(V))$ for each $V \in \alpha c(\mathcal {W}_Y)$.

Proof

$(1)\Rightarrow (2)$: Suppose that $V \in r(\mathcal {W}_Y)$ and $x \in f^{-1}(V)$. Then $V=i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))$ and $f(x) \in V=i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))$, then there exists $U\in \mathcal {W}_Y$ containing f(x) such that $f(x) \in U \subseteq c_{_\mathcal {W}} (V)$. By (1), there exists an $\mathcal {W}_X$-open set W in X containing x such that $f(W)\subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}} (U))$. Thus

$$\begin{aligned} f(x) \in f(W) \subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}} (U)) \subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)) \end{aligned}$$

and hence $x \in W \subseteq f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))$. Therefore

$$\begin{aligned} x \in W \subseteq i_{_\mathcal {W}} (W) \subseteq i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))))=i_{_\mathcal {W}} (f^{-1}(V)). \end{aligned}$$

$(2)\Rightarrow (1)$: Suppose that V be an $\mathcal {W}_Y$-open set such that $f(x)\in V$. Then $x \in f^{-1}(V)\subseteq f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))$. Since $i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))\in r(\mathcal {W}_Y)$. By (2),

$$\begin{aligned} x \in i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))), \end{aligned}$$

then there is $x\in U\in \mathcal {W}_X$ provided that $x \in U \subseteq f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))$. Thus there exists an $\mathcal {W}_X$-open set U containing x and $f(U) \subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))$. Therefore f is almost $\mathcal {W}$-continuous.

$(2)\Leftrightarrow (3)$: Obvious.

$(1)\Rightarrow (4)$: Let V be an $\mathcal {W}_Y$-open set and $x \in f^{-1}(V)$. Then V is an $\mathcal {W}_Y$-open set containing f(x). From (1), there is a $\mathcal {W}_X$-open set U containing x such that $f(U)\subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))$. Thus $x \in U \subseteq f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V)))$. Since U is $\mathcal {W}_X$-open, then $x \in i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))))$. Therefore

$$\begin{aligned} f^{-1}(V) \subseteq i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))). \end{aligned}$$

$(4)\Rightarrow (1)$: Let V be an $\mathcal {W}_Y$-open set containing f(x). Then

$$\begin{aligned} x \in f^{-1}(V)\subseteq i_{_\mathcal {W}}(f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))) \end{aligned}$$

and then there is a $\mathcal {W}_X$-open set U containing x such that

$$\begin{aligned} x \in U \subseteq f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))). \end{aligned}$$

Thus $x \in U$ and $f(U) \subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))$. Therefore f is almost $\mathcal {W}$-continuous.

$(4)\Leftrightarrow (5)$: It is clear.

$(2)\Rightarrow (6)$: Let $V \in \pi (\mathcal {W}_Y)$. Then $V \subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))$ and hence $f^{-1}(V)\subseteq f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))$. Since $i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)) \in r(\mathcal {W}_Y)$, then by (2), we get $f^{-1}(V) \subseteq i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))))$.

$(6)\Leftrightarrow (7)$: It is clear.

$(6)\Rightarrow (4)$: It follows from $\mathcal {W}\subseteq \pi (\mathcal {W})$.

$(7)\Rightarrow (9)\Leftrightarrow (8)$: It follows from $\alpha (\mathcal {W}) \subseteq \pi (\mathcal {W})$.

$(8)\Rightarrow (4)$: It follows from $\mathcal {W}\subseteq \alpha (\mathcal {W})$.

$(1)\Rightarrow (10)$: Let $x \in X$ and $V \in \pi (\mathcal {W}_Y)$ containing f(x). Then $f(x) \in i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))$ and hence there exists $U \in \mathcal {W}_Y$ such that $f(x) \in U \subseteq c_{_\mathcal {W}} (V)$. By (1), there exists a $\mathcal {W}_X$-open set W containing x such that $f(W)\subseteq i_{_\mathcal {W}}(c_{_\mathcal {W}}(U))$. Hence $f(W)\subseteq i_{_\mathcal {W}}(c_{_\mathcal {W}}(U))\subseteq i_{_\mathcal {W}}(c_{_\mathcal {W}}(V))$.

$(10)\Rightarrow (11)$: It follows from $\alpha (\mathcal {W}) \subseteq \pi (\mathcal {W})$.

$(11)\Rightarrow (1)$: It follows from $\mathcal {W}\subseteq \alpha (\mathcal {W})$.

$(3)\Rightarrow (12)$: Let $V \in \beta (\mathcal {W}_Y)$. Then $c_{_\mathcal {W}}(V)=c_{_\mathcal {W}}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))$ and hence $c_{_\mathcal {W}}(V) \in rc(\mathcal {W}_Y)$. By (3), we get $c_{_\mathcal {W}}(f^{-1}(V)) \subseteq c_{_\mathcal {W}}(f^{-1}(c_{_\mathcal {W}}(V)))\subseteq f^{-1}(c_{_\mathcal {W}}(V))$.

$(12)\Rightarrow (3)$: Let $H \in rc(\mathcal {W}_Y)$. Then $H \in \beta (\mathcal {W}_Y)$ and hence by (12), we get $c_{_\mathcal {W}}(f^{-1}(H) \subseteq f^{-1}(c_{_\mathcal {W}}(H))=f^{-1}(c_{_\mathcal {W}} (i_{_\mathcal {W}}(H)))= f^{-1}(H)$.

$(12)\Leftrightarrow (13)$: Obvious.

$(12)\Rightarrow (14)\Leftrightarrow (15)$: It follows from $\beta (\mathcal {W}) \subseteq \sigma (\mathcal {W})$.

$(14)\Rightarrow (3)$: It is similar to that of $(12)\Rightarrow (3)$.

$(12)\Rightarrow (16)\Leftrightarrow (17) \Rightarrow (18)\Leftrightarrow (19)$: It follows from $\beta (\mathcal {W}) \subseteq \pi (\mathcal {W})$ and $\pi (\mathcal {W}) \subseteq \alpha (\mathcal {W})$.

$(19)\Rightarrow (3)$: It is similar to that of $(12)\Rightarrow (3)$. $\square$

Theorem 2.2

For a function $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$. Consider the following statements:

• (1) f is almost $\mathcal {W}$-continuous;

• (2) $f(c_{_\mathcal {W}} (V))\subseteq c_{_\mathcal {W}}^{\delta } (f(V))$ for each subset V of X;

• (3) $c_{_\mathcal {W}} (f^{-1}(U))\subseteq f^{-1}(c_{_\mathcal {W}}^{\delta } (U))$ for each subset U of Y;

• (4) $f^{-1}(i_{\mathcal {W}_\delta }(U))\subseteq i_{\mathcal {W}}(f^{-1}( (U))$ for each subset U of Y;

• (5) $c_{_\mathcal {W}} (f^{-1}(F))\subseteq f^{-1}(F)$ for each $F \in \delta c(\mathcal {W}_Y)$;

• (6) $f^{-1}(V)\subseteq i_{_\mathcal {W}} (f^{-1}(V))$ for each $V \in \delta (\mathcal {W}_Y)$.

Then $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4)\Rightarrow (5)\Rightarrow (6)$.

Proof

$(1)\Rightarrow (2)$: Let $x \in c_{_\mathcal {W}} (V)$ and U be an $\mathcal {W}$-open set of Y containing f(x). By (1), there exists an $\mathcal {W}$-open set W containing x such that $f(W) \subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}} (U))$ and hence $W \bigcap V\ne \emptyset$. Thus $f(W) \bigcap f(V)\ne \emptyset$ which implies $i_{_\mathcal {W}} (c_{_\mathcal {W}} (U)) \bigcap f(V)\ne \emptyset$. Then $f(x) \in c_{\mathcal {W}_\delta } (f(V))$ and hence $x \in f^{-1}(c_{\mathcal {W}_\delta } f(V))$ which implies $c_{_\mathcal {W}}(V)\subseteq f^{-1}(c_{_\mathcal {W}}^{\delta } f(V))$. Therefore $f(c_{_\mathcal {W}}(V))\subseteq c_{_\mathcal {W}}^{\delta } f(V))$.

$(2)\Rightarrow (3)\Leftrightarrow (4)$: It is clear.

$(3)\Rightarrow (5)$: Let $H \in \delta c(\mathcal {W}_Y)$. Then $H = c_{_\mathcal {W}}^{\delta } (H)$. By (3), we get $c_{_\mathcal {W}} (f^{-1}(H)\subseteq f^{-1}(c_{_\mathcal {W}}^{\delta } (H))=f^{-1}(H)$.

$(5) \Leftrightarrow (6)$: It is clear.

Lemma 2.3

Let $(X,\mathcal {W})$ be a space and $i_{_\mathcal {W}} (V)$ be $\mathcal {W}$-open for each $V \in rc(\mathcal {W})$. Then $rc(\mathcal {W}) \subseteq \delta c(\mathcal {W})$.

Proof

Let $V \in rc(\mathcal {W})$ and let $x \notin V$. Then $x \notin c_{_\mathcal {W}}(i_{_\mathcal {W}} (V))$ and hence there exists an $\mathcal {W}$-open set U containing x such that $U \bigcap i_{_\mathcal {W}} (V)=\emptyset$. Since $V \in rc(\mathcal {W})$, then $i_{_\mathcal {W}} (V)$ is $\mathcal {W}$-open. Then by Lemma 2.1, we have $c_{_\mathcal {W}} (U) \bigcap i_{_\mathcal {W}} (V)=\emptyset$ and hence $i_{_\mathcal {W}} (c_{_\mathcal {W}} (U)) \bigcap i_{_\mathcal {W}} (V)=\emptyset$. Since U is $\mathcal {W}$-open, then $c_{_\mathcal {W}} (U) \in rc(\mathcal {W})$ and hence $i_{_\mathcal {W}} (c_{_\mathcal {W}} (U))$ is $\mathcal {W}$-open. By Lemma 2.1, we have $i_{_\mathcal {W}} (c_{_\mathcal {W}} ((U)) \bigcap c_{_\mathcal {W}} (i_{_\mathcal {W}} (V))=\emptyset$. Thus $i_{_\mathcal {W}} (c_{_\mathcal {W}} (U)) \bigcap V=\emptyset$ and hence $x \notin c_{_\mathcal {W}}^{\delta }(V)$. Therefore $c_{_\mathcal {W}}^{\delta }(V) \subseteq V$ and hence $V \in \delta c(\mathcal {W})$. $\square$

Theorem 2.3

Let $(X,\mathcal {W})$ be a space and $V \in rc(\mathcal {W})$. If $i_{_\mathcal {W}} (V)$ is $\mathcal {W}$-open set, it leads to the equality of the statements in Theorem 2.2.

Proof

It is clear from Lemma 2.3 and Theorem 2.1(3). $\square$

Theorem 2.4

Let $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ be an almost $\mathcal {W}$-continuous function and let V be an $\mathcal {W}$-open set of Y. If $x \in c_{_\mathcal {W}} (f^{-1}(V))-(f^{-1}(V))$, then $f(x) \in c_{_\mathcal {W}} (V)$.

Proof

Let $x \in X$ and V be an $\mathcal {W}$-open set of Y such that $x \in c_{_\mathcal {W}} (f^{-1}(V))-f^{-1}(V)$ and $f(x) \notin c_{_\mathcal {W}} (V)$. Then there exists an $\mathcal {W}$-open set U containing f(x) such that $U \bigcap V=\emptyset$. Since f is almost $\mathcal {W}$-continuous, then there exists an $\mathcal {W}$-open set W containing x such that $f(W) \subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}} (U))$. Since $U \bigcap V=\emptyset$, then by Lemma 2.1, we have $c_{_\mathcal {W}} (U) \bigcap V=\emptyset$ and hence $i_{_\mathcal {W}} (c_{_\mathcal {W}} (U)) \bigcap V=\emptyset$. Thus $f(W) \bigcap V=\emptyset$. Since $x \in c_{_\mathcal {W}} (f^{-1}(V))$ and W is an $\mathcal {W}$-open set containing x, then $W \bigcap f^{-1}((V)\ne \emptyset$ and hence $f(W) \bigcap V \ne \emptyset$. This is a contradiction. Therefore $f(x) \in c_{_\mathcal {W}} (V)$. $\square$

Definition 2.3

For any WS $\mathcal {W}$ on X and $A\subset X$. A point $x \in X$ is called $\mathcal {W}$-boundary point of A iff $x \in c_{_\mathcal {W}}(A) \bigcap c_{_\mathcal {W}}(X-A)$. The family of all $\mathcal {W}$-boundary points of A is denoted by $Bd_\mathcal {W}(A)$.

Theorem 2.5

For any space $(X,\mathcal {W})$ and $A \subseteq X$, we have:

• (1) $Bd_\mathcal {W}(A)=Bd_\mathcal {W}(X-A)$;

• (2) $Bd_\mathcal {W}(A)=c_{_\mathcal {W}}(A) - i_{_\mathcal {W}}(A)$;

• (3) $A \cap Bd_\mathcal {W}(A)=\emptyset$ if $A\in \mathcal {W}$;

• (4) $Bd_\mathcal {W}(A)\subset A$ if $X-A\in \mathcal {W}$.

Proof

It follows from Theorem 1.1 and Definition 2.3. $\square$

Remark 2.2

It is clear that the converse of (3) and (4) in the above theorem are not correct in general as illustrated by the next example.

Example 2.1

Let $X=\{x, y, z\}$ and $\mathcal {W}=\{\emptyset , \{x\}, \{y\}, \{z\} \}$. It is clear that:

• (1) $A=\{x, z\}$ achieves $A \cap Bd_\mathcal {W}(A)=\emptyset$, but $A\not \in \mathcal {W}$;

• (2) $A=\{z\}$ achieves $Bd_\mathcal {W}(A)\subset A$, but A is not $\mathcal {W}$-closed.

Theorem 2.6

Let $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ be a function and let $A=\{x \in X:f$ is not almost $\mathcal {W}$-continuous at $x \}$. Then $A=Bd_\mathcal {W}(f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))))$ for each $\mathcal {W}$-open set V containing f(x).

Proof

Let $x \in A$. Then f is not almost $\mathcal {W}$-continuous at x and hence there exists an $\mathcal {W}$-open set V of Y containing f(x) such that $U \bigcap (X-f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V)))\ne \emptyset$ for each $\mathcal {W}$-open set U of X containing x and hence $x\in c_{_\mathcal {W}} (X-f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))))$. Since $f(x) \in V$, then $x \in f^{-1}(V) \subseteq f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V)))$ and thus $x\in c_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))))$. Thus we get $x \in Bd_\mathcal {W}(f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V))))$. Therefore

$$\begin{aligned} A\subseteq Bd_\mathcal {W}(f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))) \end{aligned}$$

(1)

Let $x \notin A$ and V be an $\mathcal {W}$-open set containing f(x). Then f is almost $\mathcal {W}$-continuous at x and V is an $\mathcal {W}$-open set containing f(x) and hence there exists an $\mathcal {W}$-open set U containing x such that $U \subseteq f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V)))$. Thus $x \in i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))))$ and hence $x \notin X-i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))))=c_{_\mathcal {W}} (X-f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))))$ which implies $x \notin Bd_\mathcal {W}(f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))$. Therefore

$$\begin{aligned} A\supseteq Bd_\mathcal {W}(f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))) \end{aligned}$$

(2)

From (1) and (2) we have $A= Bd_\mathcal {W}(f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}}(V)))).$ $\square$

Definition 2.4

A map $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ is said to be $r(\mathcal {W})$-continuous iff $f^{-1}(H)$ is an $\mathcal {W}$-open set in X for every $H \in r(\mathcal {W}_Y)$.

Remark 2.3

One may notice that each $r(\mathcal {W})$-continuous function is almost $\mathcal {W}$-continuous, but the converse need not be true in general as shown by the following example.

Example 2.2

Let $X=\{a, b, c\}$, $\mathcal {W}_X=\{\emptyset , \{a\}, \{b\}, \{c\} \}$, $Y=\{x, y\}$, $\mathcal {W}_Y=\{\emptyset , \{x\}, \{y\} \}$ and $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ be a map defined by $f(a)=f(b)=x$, $f(c)=y$. One may notice that:

• (1) $A=\{x\} \in r(\mathcal {W}_Y)$ and $f^{-1}(A)=\{a, b\}$ which is not an $\mathcal {W}$-open set in X.

• (2) For $\mathcal {W}_Y$-open set $\emptyset$, we have $c_{_\mathcal {W}} (\emptyset )=\emptyset$ and hence $i_{_\mathcal {W}} (c_{_\mathcal {W}} (\emptyset ))=\emptyset$ which implies $f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (\emptyset )))=f^{-1}(\emptyset )=\emptyset$. Thus $f^{-1}(\emptyset ) \subseteq i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (\emptyset ))))$. For $\mathcal {W}_Y$-open set $\{x\}$, we have $c_{_\mathcal {W}} (\{x\})=\{x\}$ and hence $i_{_\mathcal {W}} (c_{_\mathcal {W}} (\{x\}))=\{x\}$ which implies $f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (\{x\})))=f^{-1}(\{x\})=\{a,b\}$. Thus $i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (\{x\}))))= i_{_\mathcal {W}} (\{a,b\})=\{a,b\}$ and hence $f^{-1}(\{x\}) \subseteq i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (\{x\}))))$. For $\mathcal {W}_Y$-open set $\{y\}$, we have $c_{_\mathcal {W}} (\{y\})=\{y\}$ and hence $i_{_\mathcal {W}} (c_{_\mathcal {W}} (\{y\}))=\{y\}$ which implies $f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (\{y\})))=f^{-1}(\{y\})=\{c\}$. Thus $i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (\{y\}))))= i_{_\mathcal {W}} (\{c\})=\{c\}$ and hence $f^{-1}(\{y\}) \subseteq i_{_\mathcal {W}} (f^{-1}(i_{_\mathcal {W}} (c_{_\mathcal {W}} (\{y\}))))$. Then f satisfy (4) in Theorem 2.1 and hence f is almost $\mathcal {W}$-continuous. Hence f is almost $\mathcal {W}$-continuous but it is not $r(\mathcal {W})$-continuous.

Theorem 2.7

If a function $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ is almost $\mathcal {W}$-continuous and $c_{_\mathcal {W}}(V)$ is $\mathcal {W}$-closed for each $V \subseteq X$, then f is $r(\mathcal {W})$-continuous.

Proof

Let $H \in rc(\mathcal {W}_Y)$. Then by Theorem 2.1(3) we have $c_{_\mathcal {W}} (f^{-1}(H) \subseteq f^{-1}(H)$ and hence $c_{_\mathcal {W}} (f^{-1}(H) = f^{-1}(H)$. Since $c_{_\mathcal {W}}(V)$ is $\mathcal {W}$-closed for each $V \subseteq X$, then $c_{_\mathcal {W}} (f^{-1}(H)$ is $\mathcal {W}$-closed in X and hence $(f^{-1}(H)$ is $\mathcal {W}$-closed in X. Therefore f is $r(\mathcal {W})$-continuous. $\square$

Theorem 2.8

If a function $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ is $r(\mathcal {W})$-continuous, then for each $V \in r(\mathcal {W}_Y)$ such that $f(x)\in V$, there is a $\mathcal {W}_X$-open set U such that $x\in U$ and $f(U)\subseteq V$.

Proof

Let $V \in r(\mathcal {W}_Y)$ containing f(x). Then $f^{-1}(V)$ is an $\mathcal {W}_X$-open set containing x and hence there exists an $\mathcal {W}_X$-open set U such that $x \in U \subseteq f^{-1}(V)$. Thus there exists an $\mathcal {W}_X$-open set U containing x such that $f(U) \subseteq V$. $\square$

Remark 2.4

By the following example, we show that the converse of the above theorem need not be true in general.

Example 2.3

Let $X=\{a, b, c\}$, $\mathcal {W}_X=\{\emptyset , \{a\}, \{b\}, \{c\} \}$, $Y=\{x, y\}$, $\mathcal {W}_Y=\{\emptyset , \{x\}, \{y\} \}$ and $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ be a function defined by $f(a)=f(b)=x$, $f(z)=c$. One may notice that:

• (1) $A=\{x\} \in r(\mathcal {W}_Y)$ and $f^{-1}(A)=\{a, b\}$ which is not an $\mathcal {W}$-open set in X.

• (2) For $a \in X, f(a)=x \in \{x\}=V$ which is an $\mathcal {W}_Y$-open set, there exists an $\mathcal {W}_X$-open set $U=\{a\}$ containing a such that $f(U)=f(\{a\})=\{x\} \subseteq \{x\} =i_{_\mathcal {W}} (c_{_\mathcal {W}} \{x\}))=i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))$. For $b \in X, f(b)=x \in \{x\}=V$ which is an $\mathcal {W}_Y$-open set, there exists an $\mathcal {W}_X$-open set $U=\{b\}$ containing b such that $f(U)=f(\{b\})=\{x\} \subseteq \{x\} =i_{_\mathcal {W}} (c_{_\mathcal {W}} \{x\}))=i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))$. For $c \in X, f(c)=y \in \{y\}=V$ which is an $\mathcal {W}_Y$-open set, there exists an $\mathcal {W}_X$-open set $U=\{c\}$ containing c such that $f(U)=f(\{c\})=\{y\} \subseteq \{y\} =i_{_\mathcal {W}} (c_{_\mathcal {W}} \{y\}))=i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))$.

Theorem 2.9

Let $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ be a function. If for each $x \in X$ and each $V \in r(\mathcal {W}_Y)$ containing f(x), there exists an $\mathcal {W}_X$-open set U containing x such that $f(U)\subseteq V$, then f is almost $\mathcal {W}$-continuous.

Proof

Let $x \in X$ and V be an $\mathcal {W}_Y$-open set such that $f(x)\in V$. Then $i_{_\mathcal {W}} (c_{_\mathcal {W}} (V)) \in r(\mathcal {W})$ containing f(x) and hence there is $U\in \mathcal {W}_X$ such that $x\in U$ and $f(U)\subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))$. Therefore f is almost $\mathcal {W}$-continuous. $\square$

Theorem 2.10

Let $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ be a map. If $i_{_\mathcal {W}} (V)$ is an $\mathcal {W}$-open set for each $V \in r(\mathcal {W})$, then the converse of Theorem 2.9 is true.

Proof

Let $x \in X$ and $V \in r(\mathcal {W}_Y)$ containing f(x). Then $i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))= i_{_\mathcal {W}} (V)$ and hence $i_{_\mathcal {W}} (c_{_\mathcal {W}} (V)$ is an $\mathcal {W}_Y$ -open set containing f(x) and hence there exists an $\mathcal {W}_X$-open set U containing x such that $f(U)\subseteq i_{_\mathcal {W}} (c_{_\mathcal {W}} (V))=V$. $\square$

Remark 2.5

If we replaced a space $(Y,\mathcal {W}_Y)$ by a topological space $(Y, \tau )$ in Theorems 2.7, 2.8, and in $r(\mathcal {W})$-continuity definition, then the statements in these theorem are equivalents.

Definition 2.5

A family of sets $\xi =\{\lambda _{\alpha } :\alpha \in \Delta \}$ in a space $(X,\mathcal {W})$ is said to be a cover of X if $\bigcup \limits _{\lambda _{\alpha } \in \Delta } \lambda _{\alpha }=X$ and a subfamily of $\xi$ having a similar property is called a subcover of $\xi$.

Definition 2.6

A space $(X,\mathcal {W})$ is called:

• (1) $\mathcal {W}$-regular if for every $x \in X$ and $\mathcal {W}$-closed set U such that $x \notin U$, there exist $M,N\in \mathcal {W}$ such that $x \in M$, $U \subseteq N$ and $M \bigcap N=\emptyset$.

• (2) Almost $\mathcal {W}$-regular if for every $x \in X$ and $F \in rc(\mathcal {W})$ with $x \notin F$, there exist $M,N\in \mathcal {W}$ such that $x \in M$, $F \subseteq N$ and $M \bigcap N=\emptyset$.

• (3) $\mathcal {W}$-normal if for every two $\mathcal {W}$-closed sets U and V with $U \bigcap V=\emptyset$, there exist $M,V\in \mathcal {W}$ such that $U\subseteq M$, $V \subseteq N$ and $M \bigcap N=\emptyset$.

• (4) Almost $\mathcal {W}$-normal if for every $U, V \in rc(\mathcal {W}_Y)$ with $U \bigcap V=\emptyset$, there exist $M,N\in \mathcal {W}$ such that $U\subseteq M$, $V \subseteq N$ and $M \bigcap N=\emptyset$.

• (5) $\mathcal {W}$-compact if every $\mathcal {W}$-open cover of X has a finite subcover.

• (6) Nearly $\mathcal {W}$-compact if every cover $\xi =\{\lambda _{\alpha } :\alpha \in \Delta , \lambda _{\alpha } \in r(\mathcal {W})\}$ of X has a finite subcover.

Theorem 2.11

If $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ is an $r(\mathcal {W})$-continuous, $\mathcal {W}$-open function and $(X,\mathcal {W}_X)$ is $\mathcal {W}$-regular, then $(Y,\mathcal {W}_Y)$ is almost $\mathcal {W}$-regular.

Proof

Let f be $r(\mathcal {W})$-continuous and $\mathcal {W}$-open function and $F \in rc(\mathcal {W}_Y)$ with $x \notin F$. Then $f^{-1}(H)$ are $\mathcal {W}$-closed set in X with $f^{-1}(x) \notin f^{-1}(F)$. Since $(X,\mathcal {W}_X)$ is $\mathcal {W}$-regular, then there exist $M,N\in \mathcal {W}_X$ such that $f^{-1}(F)\subseteq M$, $f^{-1}(x) \in N$ and $M \bigcap N=\emptyset$ and hence $F\subseteq f(M)$, $x \in f(N)$ and $f(M) \bigcap f(N)=\emptyset$. Since f is $\mathcal {W}$-open, then f(M) and f(N) are $\mathcal {W}_Y$-open sets. Therefore $(Y,\mathcal {W}_Y)$ is almost $\mathcal {W}$-regular. $\square$

Theorem 2.12

If $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ is an $r(\mathcal {W})$-continuous, $\mathcal {W}$-open function and $(X,\mathcal {W}_X)$ is $\mathcal {W}$-normal, then $(Y,\mathcal {W}_Y)$ is almost $\mathcal {W}$-normal.

Proof

Let f be $r(\mathcal {W})$-continuous and $F_1, F_2 \in rc(\mathcal {W}_Y)$ such that $F_1 \bigcap F_2=\emptyset$, then $f^{-1}(F_1), f^{-1}(F_2)$ are $\mathcal {W}$-closed sets in X with $f^{-1}(F_1) \bigcap f^{-1}(F_2)=\emptyset$. Since $(X,\mathcal {W}_X)$ is $\mathcal {W}$-normal, then there exist two $\mathcal {W}_X$-open sets M and N such that $f^{-1}(F_1)\subseteq M$, $f^{-1}(F_2) \subseteq N$ and $M \bigcap N=\emptyset$ and then $F_1\subseteq f(M)$, $F_2 \subseteq f(N)$ and $f(M) \bigcap f(N)=\emptyset$. Since f is $\mathcal {W}$-open, then f(M) and f(N) are $\mathcal {W}_Y$-open sets. Therefore $(Y,\mathcal {W}_Y)$ is almost $\mathcal {W}$-normal. $\square$

Theorem 2.13

If $f :(X,\mathcal {W}_X) \rightarrow (Y,\mathcal {W}_Y)$ is an $r(\mathcal {W})$-continuous surjective function and $(X,\mathcal {W}_X)$ is $\mathcal {W}$-compact, then $(Y,\mathcal {W}_Y)$ is nearly $\mathcal {W}$-compact.

Proof

Let f be $r(\mathcal {W})$-continuous,$\mathcal {W}$-open and $\xi =\{\lambda _{\alpha } :\alpha \in \Delta , \lambda _{\alpha } \in r(\mathcal {W})\}$ be a cover of Y. Then $f^{-1}(\xi )=\{f^{-1}(\lambda _{\alpha }) :\alpha \in \Delta , \lambda _{\alpha } \in r(\mathcal {W})\}$ is an $\mathcal {W}$-open cover of X. Since $(X,\mathcal {W}_X)$ is $\mathcal {W}$-compact, then $\{f^{-1}(\lambda _{\alpha }) :\alpha =1,2,3,\ldots ,n\}$ is a finite subcover of $f^{-1}(\xi )$ and hence $X=\bigcup \nolimits _{\alpha =1}^n f^{-1}(\lambda _{\alpha })$. Thus $Y=\bigcup \nolimits _{\alpha =1}^n f(f^{-1}(\lambda _{\alpha }))=\bigcup \nolimits _{\alpha =1}^n \lambda _{\alpha }$. Therefore $(Y,\mathcal {W}_Y)$ is nearly $\mathcal {W}$-compact. $\square$