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\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. Neubrunn, {\it Quasi-continuity}, Real Anal. Exch.
{\bf 14} (1988-89), 259--306.
\end {thebibliography}
\end {document}
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