% spellchecked: pdh11-10-98 % last edit: pdh11-10-98 jch 23/10/98 % gallies sent: % gallies corrected: % set in production style: jch 23/10/98 % Section Inroads editor Foran \documentclass{rae} \usepackage{amsmath,amsthm,amssymb} %\coverauthor{Zbigniew Grande} %\covertitle{On Continuity and Generalized Continuity with Respect to Two Topologies} \received{September 17, 1997} \MathReviews{Primary: 54C08, 54C30, 54C05} \keywords{bitopological continuity, quasicontinuity, cliquishness} \firstpagenumber{1} \markboth{Zbigniew Grande}{On Continuity and Generalized Continuity} \author{Zbigniew Grande\thanks {Supported by Bydgoszcz Pedagogical University grant 1997}, Institute of Mathematics, Pedagogical University, plac Weyssenhoffa 11, 85-072, Bydgoszcz, Poland\\ e-mail: {\tt grande@wsp.bydgoszcz.pl}} \title{ON CONTINUITY AND GENERALIZED CONTINUITY WITH RESPECT TO TWO TOPOLOGIES} %%%%%%Put Author's Definitions Below Here %%%%%%%%%%% \newcommand{\mathR}{\mathbb R} \newcommand{\consecutive} { \newtheorem {thm} {Theorem} \newtheorem {lem} {Lemma} \newtheorem {rem} {Remark} \newtheorem {example} {Example} } \def\osc{\mbox{\rm osc\,}} \def\cl{\mbox{\rm cl\,}} \def\Int{\mbox{\rm int\,}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \consecutive \maketitle \begin{abstract} Let $\tau _1$ and $\tau _2$ be topologies in $X$ and let $\tau = \tau _1 \cap \tau _2$. Some conditions concerning the topologies $\tau$, $\tau _1$ and $\tau _2$ and describing the relations between the $\tau$-continuity (quasicontinuity) [cliquishness] and the $\tau _i$-continuity (quasicontinuity) [cliquishness], $i = 1,2$, of functions defined on $X$ are considered. \end{abstract} \bigskip Let ${\cal R}$ denote the set of all reals and let $(X,\tau)$ be a topological space. A function $f:X \mapsto {\cal R}$ is called $\tau$-quasicontinuous ($\tau$-cliquish) at a point $x \in X$ (\cite{3} if for every positive real $\eta$ and for every set $U \in \tau$ containing $x$ there is a nonempty set $V \in \tau$ such that $V \subset U$ and $|f(v) - f(x)| < \eta$ for all points $v \in V$ ($\osc_Vf < \eta$, where $\osc_Vf$ denotes the diameter of the set $f(V)$). We will consider two topologies ${\cal T}_1$ and ${\cal T}_2$ in $X$. Let $${\cal T} = {\cal T}_1 \cap {\cal T}_2.$$ \section{${\cal T}$- and ${\cal T}_i$-continuity} Obviously, if a function $f:X \mapsto {\cal R}$ is ${\cal T}$-continuous (${\cal T}$-quasicontinuous)\break [${\cal T}$-cliquish] at a point $x \in X$ then it is also ${\cal T}_i$-continuous (${\cal T}_i$-quasicontinuous) [${\cal T}_i$-cliquish] at $x$ for $i = 1,2$. The converse implication need not be valid. \begin{example}\rm Let $$U = \bigcup_n\left[\frac{1}{4n + 1},\frac{1}{4n}\right] \cup \{ 0\}$$ and $$V = \bigcup_n\left[\frac{1}{4n + 3},\frac{1}{4n + 2}\right] \cup \{ 0\} .$$ Denote by ${\cal T}_1$ the topology of subsets of $X = {\cal R}$ generated by the family $$\{ (-r,r) \cap U;r > 0\}$$ and by ${\cal T }_2$ the topology of subsets of $X$ generated by the family $$\{ (-r,r) \cap V:r > 0\} .$$ For $n = 1,2,\ldots$, let $$I_n = \left[\frac{1}{4n + 2},\frac{1}{4n + 1}\right]$$ and let $f_n:I_n \mapsto [0,1]$, be a continuous function (with respect to the Euclidean topology ${\cal T}_e$) such that $$f_n(I_n) = [0,1] \;\; \wedge \;\;f_n\Bigl(\frac{1}{4n + 2}\Bigr) = f_n\Bigl(\frac{1}{4n + 1}\Bigr) = 0.$$ Put \begin{displaymath} f(x) = \left\{ \begin{array}{ccl} f_n(x) &\mbox{for} & x \in I_n,\;\; n = 1,2,\ldots \\ 0 & \mbox{for} & x \in {\cal R} \setminus \bigcup_nI_n \end{array} \right. \end{displaymath} and observe that the function $f$ is ${\cal T}_i$-continuous at $0$ for $i = 1,2$. Since $${\cal T} = {\cal T}_1 \cap {\cal T}_2 = \{ {\cal R},\emptyset \} ,$$ the function $f$ is not even ${\cal T}$-cliquish at $0$. \end{example} \begin{thm} Let $x \in X$ be a point. Suppose that the topologies ${\cal T}_1$ and ${\cal T}_2$ satisfy to the following condition $$\forall_{x \in A \in {\cal T}_1}\forall_{x \in B \in {\cal T}_2}\exists_{x \in E \in {\cal T}}E \subset A \cup B.\leqno(1)$$ If a function $f:X \mapsto {\cal R}$ is ${\cal T}_i$-continuous at the point $x$ for $i = 1,2$ then $f$ is also ${\cal T}$-continuous at $x$. \end{thm} \begin{proof}[{\sc Proof.}] Fix a positive real $\eta$. From ${\cal T}_i$-continuity of the function $f$ at $x$, $i = 1,2$, follows that there are sets $A_i \in {\cal T}_i$ such that $x \in A_i$ and for each point $t \in A_i$ the inequality $$|f(t) - f(x)| < \eta$$ holds. By the condition (1) there is a set $E \in {\cal T}$ with $$x \in E \subset A_1 \cup A_2.$$ Evidently, $$|f(t) - f(x)| < \eta$$ for all points $t \in E$ and the proof is completed.\end{proof} We will show the necessity of condition (1) for Theorem 1 to hold. \begin{rem} Let $x \in X$ be a point. If there are sets $A_i \in {\cal T}_i$, $i = 1,2$, such that $x \in A_1 \cap A_2$ and $$\forall_{x \in A \in {\cal T}}A \setminus (A_1 \cup A_2) \neq \emptyset ,$$ then there is a function $f:X \mapsto {\cal R}$ which is ${\cal T}_i$-continuous at $x$ for $i = 1,2$ and which is not ${\cal T}$-continuous at~$x$. \end{rem} \begin{proof}[{\sc Proof.}] Put \begin{displaymath} f(x)=\left\{ \begin{array}{ccl} 0 & \mbox{for} & x \in A_1 \cup A_2\\ 1 & \mbox{for} & x \in X \setminus (A_1 \cup A_2) \end{array} \right. \end{displaymath} and observe that $f$ is ${\cal T}_i$-continuous at $x$ for $i = 1,2$. Since $$\forall_{\emptyset \neq A \in {\cal T}}\osc_{A \cup \{ x\} }f = 1,$$ the function $f$ is not even ${\cal T}$-quasicontinuous at $x$. This completes the proof.\end{proof} \section{Quasicontinuity} \begin{example}\rm Let $X = {\cal R}$ and let the sets $U$, $V$ be the same as these from Example 1. Denote by ${\cal T}_1$ and ${\cal T}_2$ the topologies generated by family $$\{ U \cap (0,r);r > 0\}$$ and respectively by the family $$\{ V \cap (0,r);r > 0\} .$$ Let the function $f$ be the same as that from Example 1. Then the topologies ${\cal T}_1$ and ${\cal T}_2$ satisfy to the condition (1) from Theorem~1, the function $f$ is\break ${\cal T}_i$-quasicontinuous at the point~$0$ (even everywhere) for $i = 1,2$, but $f$ is not ${\cal T}$-quasicontinuos at $0$ (it is not even ${\cal T}$-cliquish at~$0$). \end{example} So, Theorem 1 is not true for quasicontinuity. For the case of quasicontinuity we recall the following notion:\newline A set $A \subset X$ is said $\tau$-semiopen if $$A \subset \cl_{\tau }(\Int_{\tau }(A)),$$ where $\cl_{\tau }$ and $\Int_{\tau }$ respectively denote the operations of the closure and the interior with respect to the topology~$\tau$. Let $S(\tau )$ be the family of all $\tau$-semiopen sets $A \subset X$. It is well known (\cite{3}) that a function $f:X \mapsto {\cal R}$ is $\tau$-quasicontinuous at a point $x$ if and only if for every positive real $\eta$ there is a $\tau$-semiopen set $A \subset X$ containing $x$ such that for each point $t \in A$ the inequality $$|f(t) - f(x)| < \eta$$ is true. \begin{thm} Let $x \in X$ be a point. Suppose that the topologies ${\cal T}_1$ and ${\cal T}_2$ satisfy to the following condition $$\forall_{x \in A \in S({\cal T}_1)}\forall_{x \in B \in S({\cal T}_2)} \exists_{\emptyset \neq E \in {\cal T}}E \subset A \cup B.\leqno(2)$$ If a function $f:X \mapsto {\cal R}$ is ${\cal T}_i$-quasicontinuous at $x$ for $i = 1,2$, then $f$ is also ${\cal T}$-quasicontinuous at~$x$. \end{thm} \begin{proof}[{\sc Proof.}] Fix a positive real $\eta$ and a set $A \in {\cal T}$ containing $x$. By the ${\cal T}_i$-quasicontinuity of the function $f$ at $x$ for $i = 1,2$, there are nonempty sets $$A_i \in S({\cal T}_i), \;\; i = 1,2,$$ containing $x$ and such that for all points $t \in A_i$, $i = 1,2$, the inequality $$|f(t) - f(x)| < \eta$$ holds. Observe that $$A \cap A_i \in S({\cal T}_i)$$ for $i = 1,2$. By the condition (2) there is a nonempty ${\cal T}$-open set $$B \subset A \cap (A_1 \cup A_2).$$ Evidently, $$\forall_{t \in B }|f(t) - f(x)| < \eta .$$ So, the proof is completed.\end{proof} \begin{rem} Let $x \in X$ be a point. Suppose that there are nonempty sets $A_i \in S({\cal T}_i)$, $i = 1,2$, containing $x$ and such that $$\forall_{\emptyset \neq A \in {\cal T}}A \setminus (A_1 \cup A_2) \neq \emptyset .$$ Then there is a function $f:X \mapsto {\cal R}$ which is ${\cal T}_i$-quasicontinuous at the point $x$ for $i = 1,2$ and which is not ${\cal T}$-quasicontinuous at the point~$x$. \end{rem} \begin{proof}[{\sc Proof.}] The construction of such a function is the same as the construction of the function $f$ in the proof of Remark~1. \end{proof} Observe that the case of quasicontinuity is different from the case of continuity. If a function $f$ is ${\cal T}$-continuous at a point $x$ then it is also ${\cal T}_i$-continuous at $x$ for $i = 1,2$. For quasicontinuity the situation is different. \begin{example}\rm Let ${\cal T}_1 = {\cal T}_e^+$ and respectively ${\cal T}_2 = {\cal T}_e^-$ be the topologies generated by the families $$\{ [x,x + r);x \in {\cal R}, \;\;r > 0\}$$ and $$\{ (x - r,x];x \in {\cal R}, \;\;r > 0\} .$$ Then $${\cal T} = {\cal T}_1 \cap {\cal T}_2 = {\cal T}_e$$ and the function \begin{displaymath} f(t)=\left\{ \begin{array}{ccl} 0 & \mbox{for} & t \geq 0\\ 1 & \mbox{for} & t < 0 \end{array} \right. \end{displaymath} is ${\cal T}$-quasicontinuous at $0$ and is not ${\cal T}_2$-quasicontinuous at $0$. Analogously, the function $$g(t) = f(-t), \;\; t \in {\cal R},$$ is ${\cal T}$-quasicontinuous at $0$ and is not ${\cal T}_1$-quasicontinuous at~$0$. \end{example} {\catcode\@=11 \global\tagsleft@true } \begin{thm} Let $x \in X$ be a point. Then the following conditions are equivalent: \begin{align*} &\vtop{\hsize9.5truecm \noindent every function $f:X \mapsto {\cal R}$ which is ${\cal T}$-quasicontinuous at $x$ is also ${\cal T}_i$-quasicontinuous at $x$ for $i = 1,2$;}\tag{$2'$}\\ &\vtop{\hsize9.5truecm \noindent every ${\cal T}$-semiopen set containing~$x$ contains also ${\cal T}_i$-semiopen sets $U_i$, $i = 1,2$, containing~$x$.}\tag{$2''$} \end{align*} \end{thm} {\catcode\@=11 \global\tagsleft@false } \begin{proof}[{\sc Proof.}] $(2') \Rightarrow (2'')$. If $(2'')$ does not hold then there are a ${\cal T}$-semiopen set $U$ containing $x$ and an index $i \leq 2$ such that for every ${\cal T}_i$-semiopen set $V$ containing $x$ the relation $$V \setminus U \neq \emptyset$$ holds. Let \begin{displaymath} f(t)=\left\{ \begin{array}{ccl} 0 & \mbox{for} & t \in U \\ 1 & \mbox{for} & t \in X \setminus U . \end{array} \right. \end{displaymath} Then $f$ is ${\cal T}$-quasicontinuous at $x$, but it is not ${\cal T}_i$-quasicontinuous at $x$. So, we obtain a contradiction with~$(2')$. $(2'') \Rightarrow (2')$. Suppose that a function $f:X \mapsto {\cal R}$ is ${\cal T}$-quasicontinuous at~$x$. Fix a positive real $\eta$ and an index $i \leq 2$. From the ${\cal T}$-quasicontinuity of $f$ at $x$ follows the existence of an ${\cal T}$-semiopen set $U$ containing $x$ and such that $$|f(t) - f(x)| < \eta$$ for each point $t \in U$. By $(2'')$ there is a ${\cal T}_i$-semiopen set $V \subset U$ containing $x$. Since $$|f(t) - f(x)| < \eta$$ for all points $t \in V$, the function $f$ is ${\cal T}_i$-quasicontinuous at $x$ and the proof is completed.\end{proof} \section{Cliquishness} \begin{rem} Assume all hypotheses of Remark~$2$. Moreover suppose that for each nonempty set $A \in {\cal T}$ there are sets $B$, $D$ which are dense (with respect to the topology ${\cal T}$) in $A$ and such that $$B \cap D = \emptyset \;\; \wedge \;\; B \cup D = A.$$ Then there is a function $f:X \mapsto {\cal R}$ which is ${\cal T}_i$-quasicontinuous at the point $x$ for $i = 1,2$ and which is not ${\cal T}$-cliquish at~$x$. \end{rem} \begin{proof}[{\sc Proof.}] If $$E = \Int_{{\cal T}}(X \setminus (A_1 \cup A_2)) \neq \emptyset$$ then let $$E = B \cup D,$$ where the sets $B$, $D$ are disjoint and dense (with respect to the topology ${\cal T}$) in $E$. If $E = \emptyset$ we put $$B = X \setminus (A_1 \cup A_2).$$ Define \begin{displaymath} f(x)=\left\{ \begin{array}{ccl} 1 & \mbox{for} & x \in B\\ 0 & \mbox{for} & x \in X \setminus B. \end{array} \right. \end{displaymath} Then $f$ is ${\cal T}_i$-quasicontinuous at $x$ for $i = 1,2$, but $f$ is not ${\cal T}$-cliquish at~$x$.\end{proof} \medskip The following example shows that all hypotheses of Remark 3 are essential. \begin{example}\rm Let \begin{align*} X & = \{ 0,1,2,3\} ,\\ {\cal T}_1 & = \{ \emptyset ,\{ 0,1\} ,\{ 3\} ,\{ 0,1,3\} ,X\}, \text{ and } {\cal T}_2 &= \{ \emptyset ,\{ 0,2\} ,\{ 3\} ,\{ 0,2,3\} ,X\}. \end{align*} Then $${\cal T} = \{ \emptyset ,\{ 3\} ,X\} .$$ Let $f:X \mapsto {\cal R}$ be a function which is ${\cal T}_i$-quasicontinuous at the point $0$ for $i = 1,2$. Then the function $f$ is also ${\cal T}_i$-continuous at $0$ for $i = 1,2$ and $$f(0) = f(1) = f(2).$$ If $0 \in A \in {\cal T}$, then $$A = X \;\; \wedge \;\; \{ 3\} \subset A$$ and $\osc_{\{ 3\} }f = 0$. So $f$ is ${\cal T}$-cliquish at the point $0$. \smallskip Moreover, \begin{align*} 0 \in U &= \{ 0,1\} \in {\cal T}_1,\\ 0 \in W &= \{ 0,2\} \in {\cal T}_2, \text{ and } \forall_{\emptyset \neq A \in {\cal T}}A &\setminus (U \cup W) \neq \emptyset .\end{align*}\end{example} \begin{thm} Let $x \in X$ be a point. Suppose that $$\vtop{\hsize10.5truecm\noindent there is a set  A \in {\cal T} containing x and such that for all nonempty sets A_i \in {\cal T}_i, i = 1,2, contained in A there exists a nonempty set D \in {\cal T} contained in A_1 \cup A_2 and for all nonempty disjoint sets B_i \in {\cal T}_i, i = 1,2, there are an index i \leq 2 and a nonempty set E \in {\cal T} with}\leqno(3)$$ $$E \subset B_1 \;\; \vee \;\; E \subset B_2.$$ If a function $f:X \mapsto {\cal R}$ is ${\cal T}_i$-cliquish at the point $x$ for $i = 1,2$, then $f$ is also ${\cal T}$-cliquish at~$x$. \end{thm} \begin{proof}[{\sc Proof.}] Fix a positive real $\eta$ and a set $U$ with $$x \in U \in {\cal T}.$$ Let a set $A$ satisfies to the condition (3). The ${\cal T}_i$-cliquishness of $f$ at $x$ implies the existence of a nonempty sets $B_i \subset U \cap A$, $i = 1,2$, with $$\osc_{B_i}f < \frac{\eta}{2} \;\; \wedge \;\; B_i \in {\cal T}_i,\;\; i = 1,2.$$ Observe that $$B_1 \cap B_2 \neq \emptyset \Rightarrow \osc_{B_1 \cup B_2}f < \eta .$$ By (3) there is a nonempty set $E \in {\cal T}$ such that: $$B_1 \cap B_2 = \emptyset \Rightarrow \exists_{i \leq 2}E \subset B_i$$ or $$B_1 \cap B_2 \neq \emptyset \Rightarrow E \subset B_1 \cup B_2.$$ Since in both cases $\osc_Ef < \eta$, the proof is completed.\end{proof} \begin{rem} Let $x \in X$ be a point. Suppose that the topologies ${\cal T}$, ${\cal T}_1$ and ${\cal T}_2$ are such that for each point $t \neq x$ there are sets $U_1 \in {\cal T}_1$, $U_2 \in {\cal T}_2$ and $U_3 \in {\cal T}$ for which $U_1 \cap U_3 = \emptyset$, $U_2 \cap U_3 = \emptyset$, $x \in U_3$ and $t \in U_1 \cap U_2$. Moreover we assume that there is a countable base of neighborhoods of $x$ for the topology ${\cal T}$ and that for which set $A \in {\cal T}$ containing $x \in X$ there are nonempty sets $A_i(A) \in {\cal T}_i$, $i = 1,2$, such that: $$A_1(A) \cap A_2(A) \neq \emptyset \Rightarrow \forall_{\emptyset \neq B \in {\cal T}}B \setminus (A_1(A) \cup A_2(A)) \neq \emptyset ,$$ $$A_1(A) \cap A_2(A) = \emptyset \Rightarrow \forall_{\emptyset \neq B \in {\cal T}}\forall_{i \leq 2}B \setminus A_i(A) \neq \emptyset ,$$ and $$G = X \setminus \bigcup \{ A_1(A) \cup A_2(A);x \in A \in {\cal T}\} \setminus \{ x\} = H \cup K,$$ where $H \cap K = \emptyset$ and $H$, $K$ are dense in $G$ with respect to the topology~${\cal T}$. Then there is a function $f:X \mapsto {\cal R}$ which is ${\cal T}_i$-cliquish at the point $x$ for $i = 1,2$ and which is not ${\cal T}$-cliquish at~$x$. \end{rem} \begin{proof}[{\sc Proof.}] At the beginning we assume that $\{ x\}$ is not in ${\cal T}_i$ for $i = 1,2$. Let $$W_1,\ldots ,W_n,\ldots$$ be an enumeration of all elements of some basis of neighborhoods of $x$ in the topology ${\cal T}$. Let $$A_i(W_1) \in {\cal T}_i,\;\;i = 1,2,$$ be nonempty sets satisfying our hypothesis for $A = W_1$. Since in this case there are points $t_i \in A_i(W_1)$, $i = 1,2$, we can find nonempty sets $V_{1,i} \in {\cal T}_i$, $i \leq 2$, and $V_{1,3} \in {\cal T}$ with $t_i \in V_{1,i}$, $i \leq 2$, $x \in V_{1,3}$, and $$V_{1,3} \cap (V_{1,1} \cup V_{1,2}) = \emptyset .$$ If $V_{1,1} \cap V_{1,2} = \emptyset$ we put $$f(t) = (-1)^i, \;\; t \in V_{1,i}, \;\; i \leq 2.$$ In the contrary case, where $V_{1,1} \cap V_{1,2} \neq \emptyset$, we put $$f(t) = 1, \;\; t \in V_{1,1} \cup V_{1,2}.$$ Analogously, for $n > 1$ we find nonempty sets $V_{n,i} \in {\cal T}_i$, $i \leq 2$, and $V_{n,3} \in {\cal T}$ such that: $$x \in V_{n,3} \subset X \setminus (V_{n,1} \cup V_{n,2}),$$ $$V_{n,1} \cup V_{n,2} \subset W_n \cap V_{1,3} \cap \ldots \cap V_{n-1,3},$$ $$V_{n,1} \cap V_{n,2} \neq \emptyset \Rightarrow \forall_{\emptyset \neq B \in {\cal T}}B \setminus (V_{n,1} \cup V_{n,2}) \neq \emptyset ,$$ and $$V_{n,1} \cap V_{n,2} = \emptyset \Rightarrow \forall_{\emptyset \neq B \in {\cal T}}\forall_{i \leq 2}B \setminus V_{n,i} \neq \emptyset .$$ If $V_{n,1} \cap V_{n,2} = \emptyset$ then we put $$f(t) = (-1)^in, \;\; t \in V_{n,i}, \;\; i \leq 2.$$ In the contrary case, where $V_{n,1} \cap V_{n,2} \neq \emptyset$ we put $$f(t) = n, \;\; t \in V_{n,1} \cup V_{n,2}.$$ Moreover, let \begin{align*} f(x) &= 0,\\ f(t) &= 0, \;\; t \in H, \text{ and } f(t) & = \frac{1}{2}\end{align*} at all other points of the space $X$. Observe that the function $f$ satisfies all requirements. \noindent Evidently, $$\neg (\{ x\} \in {\cal T}_1 \cap {\cal T}_2).$$ In the case, where $\{ x\} \in {\cal T}_i$ for $i = 1$ or $i = 2$, the construction of such a function $f$ is simpler, since starting from some index $n$ we find only one set $V_{k,i_0}$ instead a pair $(V_{k,1},V_{k,2})$. \end{proof} \begin{thm} Let $x \in X$ be a point. Suppose that \begin{itemize} \item[(a)] all sets $A$ with $$x \in A \in {\cal T}_1 \cup {\cal T}_2$$ contains some sets $B \in {\cal T}$ with $x \in B$. \end{itemize} Then every function $f:X \mapsto {\cal R}$ which is ${\cal T}$-cliquish at $x$ is also ${\cal T}_i$-cliquish at $x$ for $i = 1,2$. \end{thm} \begin{proof}[{\sc Proof.}] Fix an index $i \leq 2$, a positive real $\eta$ and a set $U \in {\cal T}_i$ with $x \in U$. By our hypothesis there is a set $B \in {\cal T}$ such that $$x \in B \subset U.$$ Since $f$ is ${\cal T}$-cliquish at $x$, there is a nonempty set $V \in {\cal T}$ such that $$V \subset B \;\; \wedge \;\; \osc_Vf < \eta .$$ So, $$V \subset U \;\; \wedge \;\; V \in {\cal T}_i \;\; \wedge \;\; \osc_Vf < \eta ,$$ and the proof is completed.\end{proof} \medskip Next example will show that the hypothesis (a) from the last theorem is not necessary. \begin{example}\rm Let \begin{align*} X & = \{ 0,1,2,3\} ,\\ {\cal T}_1 &= \{ \emptyset , X, \{ 1\} \} , \text{ and } {\cal T}_2 & = \{ \emptyset , X, \{ 2\} \} .\end{align*} Then $${\cal T} = \{ \emptyset , X \} ,$$ every function $f:X \mapsto {\cal R}$ which is ${\cal T}$-cliquish at $x$ is constant and the condition (a) is not satisfied. \end{example} The proof of the next theorem gives a characterization of the cliquishness. \begin{rem} Let $\tau$ be a topology of subsets of $X$ and let $x \in X$ be a point. A function $f:X \mapsto {\cal R}$ is $\tau$-cliquish at $x$ if and only if for each positive real $\eta$ there are $\tau$-open sets $A_s$, $s \in S$, where $S$ is a set of indexes, such that $x$ belongs to the $\tau$-closure $\cl_{\tau }(\bigcup_{s \in S}A_s)$ of the set $\bigcup_{s \in S}A_s$ and for every $s \in S$ the inequality $\osc_{A_s}f < \eta$ is true. \end{rem} \begin{proof}[{\sc Proof.}] Fix a positive real $\eta$. If $f$ is $\tau$-cliquish at $x$ then for every set $U \in {\tau }$ containing $x$ there is a nonempty $\tau$-open set $B(U) \subset U$ with $\osc_{B(U)}f < \eta$. If $$\{ U_s;s \in S\}$$ is a directed family of all $\tau$-open sets contained the point $x$, then the family $$\{ A_s = B(U_s);s \in S\}$$ satisfies all the required conditions. The proof of the second implication is evident.\end{proof} \begin{thm} Let $x \in X$ be a point. Suppose that the topologies ${\cal T}_1$ and ${\cal T}_2$ satisfy the following condition $$\vtop{\hsize10.5truecm \noindent for each {\cal T}-open set A such that x \in \cl_{{\cal T}}(A) and for each index i \leq 2 the point x \in \cl_{{\cal T}_i}(A).}\leqno{(3')}$$ If a function $f:X \mapsto {\cal R}$ is ${\cal T}$-cliquish at $x$ then it is also ${\cal T}_i$-cliquish at $x$ for $i = 1,2$. \end{thm} \begin{proof}[{\sc Proof.}] Fix a positive integer $i \leq 2$ and a positive real $\eta$. Since $f$ is ${\cal T}$-cliquish at $x$, by the last Remark there are nonempty sets $A_s \in {\cal T}$, where $S$ is a set of indexes, such that $$x \in \cl_{{\cal T}}(\bigcup_{s \in S}A_s)$$ and for every $s \in S$ the inequality $\osc_{A_s}f < \eta$ is true. From $(3')$ follows the existence of a ${\cal T}_i$-open set $D \subset \bigcup_{s \in S}A_s$ with $$x \in \cl_{{\cal T}_i}(D).$$ For $s \in S$ the sets $$D_s = D \cap A_s \in {\cal T}_i.$$ Let $$S' = \{ s \in S;D_s \neq \emptyset \} .$$ Since for $s \in S'$ the inequality $$\osc_{D_s}f < \eta$$ is true and since $$x \in \cl_{{\cal T}_i}(\bigcup_{s \in S'}D_{s}),$$ by the last Remark $f$ is ${\cal T}_i$-cliquish at $x$. So, the proof is completed.\end{proof} \begin{rem} Let $x \in X$ be a point. Suppose that there are an index $i \leq 2$, a set $V \in {\cal T}_i$, disjoint sets $Y,Z \subset V$ and a set $U \in {\cal T}$ such that $$U \cap V = \emptyset \;\; \wedge \;\; x \in \cl_{{\cal T}}(U) \cap V,$$ and $$x \in Y \;\; \wedge \;\; \cl_{{\cal T}_i}(Y) = \cl_{{\cal T}_i}(Z) = \cl_{{\cal T}_i}(V).$$ Then there is a function $f:X \mapsto {\cal R}$ which is ${\cal T}$-cliquish at $x$ and which is not ${\cal T}_i$-cliquish at~$x$. \end{rem} \begin{proof}[{\sc Proof.}] The function \begin{displaymath} f(t)=\left\{ \begin{array}{ccl} 1 & \mbox{if} & t \in Z\\ 0 & \mbox{if} & t \in X \setminus Z \end{array} \right. \end{displaymath} is ${\cal T}$-cliquish at $x$, but it is not ${\cal T}_i$-cliquish at $x$. So, the proof is completed.\end{proof} \medskip The following example shows the importance of all hypothesis of the last remark. \begin{example}\rm Let \begin{align*} X &= \{ 0,1,2\} ,\\ {\cal T}_1 & = \{ \emptyset , X, \{ 1\} \}, \text{ and } {\cal T}_2 &= 2^X.\end{align*} Then ${\cal T} = {\cal T}_1$ and for $$x = 0 \;\; \wedge \;\; A =\{ 1\} \;\; \wedge \;\; B = \{ 0\}$$ we have $$A \in {\cal T} \;\; \wedge \;\; B \in {\cal T}_2 \;\; \wedge \;\; A \cap B = \emptyset$$ and $$x \in B \;\; \wedge \;\; x \in \cl_{{\cal T}}(A).$$ Moreover, each function $f:X \mapsto {\cal R}$ is ${\cal T}_2$-continuous at the point $0$, because $\{ 0\} \in {\cal T}_2$. So, every function $f:X \mapsto {\cal R}$ which is ${\cal T}$-cliquish at $0$ is also ${\cal T}_2$-cliquish at~$0$. \end{example} \begin{thebibliography} {9} \bibitem {1} R. Engelking, {\it General Topology}, Warsaw PWN, 1976. \bibitem{2} J. L. Kelly, {\it General Topology}, New York 1955. \bibitem{3} T. 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