Theta - ω - Mappings in Topological Spaces

WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif Theta    Mappings in Topological Spaces RAJA MOHAMMAD LATIF Department of Mathematics and Natural Sciences Prince Mohammad Bin Fahd University P.O. Box 1664 Al Khobar KINGDOM OF SAUDI ARABIA Abstract: - In 2017 S. Ghour and B. Irshedat defined the   closure operator as a new topological operator and introduced   open sets as a new class of sets and proved that this class of sets is strictly between the class of open sets and the class of   open sets. In this paper we introduce   continuous,   irresolute,   open,   closed, pre    open, pre    closed, contra   continuous and almost contra   continuous mappings and investigate properties and characterizations of these new types of mappings in topological spaces. Key-Words:-   open,   continuous,   irresolute,   closed, pre    open, pre    closed, contra   continuous, almost contra   continuous. Received: January 2, 2020. Revised: May 14, 2020. Accepted: May 20, 2020. Published: May 27, 2020. 1 Introduction The notions of   open subsets,   closed denoted by A Cl  A  . A subset A of X is subsets and   closure were introduced by Velicko 39 for the purpose of studying the called   closed if A  Cl  A  . Dontchev and Maki 10 , Lemma 3.9 have shown that if A and B important class of H-closed spaces in terms of arbitrary filterbases. Dickman and Porter 8,9 ,  X ,   , then Cl  A U B   Cl  A  U Cl  B  and Cl  A I B   Cl  A  I Cl  B  . Note also that the   closure are Joseph  20 and Jankovic 18,19 continued the work of Velicko. Recently Noiri and Jafari 33 and Jafari 17  have also obtained several new subsets of a space of a given set need not be a   closed set. But it is always closed. The complement of a   closed set is called a   open set. The and interesting results related to these sets. In what follows  X ,    or X  denotes   interior of set A in X , written Int  A  , topological spaces on which no separation axioms are assumed unless explicitly stated. We denote the interior and the closure of a subset A of X by Int  A and Cl  A  , respectively. A consists of those points x of A such that for some open set U containing x, Cl U   A. A point x  X is called a   adherent point of A 10 , if A I Cl  A   for every open set V equivalently, X  A is   closed. The collection of all   open sets in a topological space containing x. The set of all   adherent points of A is called the   closure of A and is  X ,   forms a topology   on X , coarser than  and     if and only if  X ,   is regular. E-ISSN: 2224-2880 set A is   open if and only if A  Int  A  , or 186 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif Several authors continued the study of   closure operator,   open sets and their related topological concepts. Recently some authors have studied several generalizations of   open sets. A set A is   open set in  X ,   2 Preliminaries Definition 2.1. if for each x  A, there is U  and a countable set C  X such that x U  C  A. The family 39 Let  X ,   x  Cl  A  is well known that   forms a topology on X if Cl U  I A   for any U  and finer than  .   open sets played a vital role in x U .  b  . A is   closed if Cl  A  A. general topology research. Al Ghour used   open sets to define   regularity as a c . denoted by   . x X  F, Theorem 2.2. there exist U  and V  such that x U and F  V with U I V   . The closure of A in the topological  X ,    is called the in  X ,   and is denoted by Definition 2.3. operator to define the   closure operator in a similar way to that used in the definition of the A point x  X is in   closure operator. of A  x  Cl  A   if Cl  A  I A   for any U  with x  U . A set A is called   closed if Cl  A   A. The complement of a   closed set is called a   open set. The family of all   open sets in  X ,   denoted by   forms a topology on  Let  X ,   be a topological 16 Let  X ,  be X which is strictly between   and  . In this paper we introduce   continuous,   irresolute,   open,   closed, pre    open, pre    closed, contra   continuous and space. Then the following statements are true.  a  .   is a topology on X .  b  .     and     in general. Theorem 2.5. Let  X ,   be a topological space and let A  X . Then Cl  A   Cl  A and Cl  A   Cl  A  in general. almost contra   continuous and investigate properties and characterizations of these new types of mappings. Definition 2.6. 1 Let  X ,   be a topological space and let A  X . E-ISSN: 2224-2880 a topological space and let A  X .  a  . A point x in X is a condensation point of A if for each U  with x U , the set U I A is uncountable.  b  . A set A is   closed if it contains all its condensation points.  c  . A set A is   open if the complement of A is   closed. The family of all   open sets in a topological space  X ,   is denoted by   . For a subset A of a topological space  X ,   , it is known that A   if and only if for each x  A, there exists U  such that x  U and U  A is countable. Theorem 2.4. 3 Let  X ,   be a topological Cl  A  . In 2017 Al Ghour used the   closure   closure 39 space. Then  a  .   forms a topology on X .  b  .     and     in general. space   closure of A A is   open if the complement of A is   closed.  d  . The family of all   open sets in  X ,   is generalization of regularity as follows. A topological space  X ,   is   regular if for  X ,   and a topological space and let A  X .  a  . A point x in X is in the   closure of A of all   open sets in  X ,   is denoted by   . It each closed set F in be 187 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif  a  . A point x in X is in the   closure of A  x  Cl  A  if Cl U  I A   for any U  Theorem 2.12. x  A, there exists U  x U  Cl U   A. each A is called   closed if Cl  A   A.  c  . A set Let  X ,   be a topological space and A  X . Then A   if and only if for  with x  U .  b  . A set 1 such that Corollary 2.13. Every open   closed set in a topological space  X ,   is   open. Corollary 2.14. Every countable open set in a topological space  X ,   is   open. The following example shows that   open are strictly between   open sets and open sets. is called   open if the complement of A is   closed.  d  . The family of all   open sets in  X ,   A is denoted by    or O  X   O  X ,    .  e  . The family of all   closed sets in  X ,   is denoted by  C  X    C  X ,   . Theorem 2.7. 1 Let  X ,   be a topological Example 2.15. 1  a  . Cl  A   Cl  A  Cl  A .  b  . If A is   closed, then A is   closed,  c  . If A is   closed, then A is closed. Theorem 2.8. 1 Let  X ,   be a topological  Let   X ,   . Then the Kernel of A, denoted by Ker  A  , is the intersection of all open supersets of space A. Lemma 2.17. Let space.  a  . If A  B  X , then Cl  A   Cl  B  . topological space properties hold:  For each subsets A, B  X , b. Cl  A U B   Cl  A  U Cl  B  .  c  . For each subset A  X , Cl  A is closed   i .  Theorem 2.10.  ii  . A  , Cl  A   Cl  A   Cl  A  .  iii  . Let  X ,   be a topological space. Then  a  .  and X are   closed sets.   closed sets is  b  . Finite union of   closed.  c  . Arbitrary intersection of   closed sets is   closed. Theorem 2.11. 1 the following x  Ker  A  if and only if A I F   for every A  Ker  A and if A is open in  X ,   , then If A  B, then Ker  A   Ker  B  . 3   Continuous Mappings The purpose of this section is to investigate properties and characterizations of   continuous functions. Definition 3.1. A function f :  X ,    Y ,   is Let  X ,   be a topological said to be   continuous space. Then   is a topology on X . E-ISSN: 2224-2880  X ,   , then A  Ker  A  . A   , Cl  A   Cl  A  . 1 A and B be subsets of a closed set F in  X ,   containing x.  in  X , .  d  . For each  e  . For each ¥ Definition 2.16. Let A be a subset of a topological  X ,   be a topological  and Then     , ΅ , ¥  and     , ΅  . space. Then        . 1 ΅ , ¤ , ¤ c, denote, respectively the set of real numbers, the set of rational numbers, the set of irrational numbers and the set of natural numbers. Consider  X ,   where    , ΅ ,¥ ,¤ c ,¥ U¤  . space and let A  X . Then Theorem 2.9. Let every V  . 188 1 if f V    for Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Theorem 3.2. Let f :  X ,    Y ,   function. Then the following are equivalent: Raja Mohammad Latif be  1 f is   continuous;  2  2   1 : Let V  Y closed. The inverse image of each closed set in Y is a for every 1   5  : f Cl U    Cl  f U   , for every U  X ; Bd   f 1 V    f 1  Bd V   , for such that every f 1  Int V    Int  f 1 V   , for x  U and  6    8  : Let Bd  f 1 V    every  Proof . 1   2  : Let F  Y be closed. Since f is   continuous, f 1 Y  F   X  f 1  F  is   open. Therefore, f 1  F  is   closed in Hence V  Y. Then by hypothesis, f 1  Bd V    f 1 V   Int  f 1 V    f 1 V  Int V    f 1 V   f 1  Int V    X. f 1  Int V    Int  f 1 V   .  2    3 : Since Cl V  is closed for every V  Y , f 1 Cl V   is   closed. Therefore then f 1 Cl V    Cl  f 1  Cl V     Cl  f 1 V   .  f  x   f U   V . points. Therefore f 1 V    . V Y;  8    6  : Let V  Y . Then by hypothesis, f 1  Int V    Int  f 1 V   f 1 V   Int  f 1 V    f 1 V   f 1  Int V    f 1 V  Int V      3   4  : Let U  X and f U   V . Then Thus Cl  f 1 V    f 1 Cl V   . Cl U   Cl  f 1  f U     f 1 Cl  f U      Bd  f 1 V    f 1  Bd V   .  1   7  :  and f Cl U    Cl  f U   . It is   continuous obvious, since and f is  4 by f Cl U    Cl  f U   for each U  X . So be a closed set, and f  D U    Cl  f U   . f Cl U    Cl  f U    Cl  f  f 1 W     Cl W   W . Thus  7   1 : and E-ISSN: 2224-2880 f 1 V    . Let x U  f 1  f U    f 1 V  . It shows that f 1 V  is a   neighbourhood of each of its f  D U    Cl  f U   , for every U  X ;  4    2  : Let W  Y U  f 1 W  . Then be x  f 1 V  . Then f  x  V and there exists U   V Y;  is    open f :  X ,    Y ,   Let  5   1 : Let V  . We prove x  U and f U   V ; 8 V  is f U   f  f 1 V    V . Y containing f  x  , there exists U   such that 7 1   continuous. For any x  X and any open set V of Y containing f  x  , U  f 1 V    , and  5  For any point x  X and any open set V of  6 f 1 Y  V   X  f 1 V  Then in X . Cl  f 1 V    f 1 Cl V   , V Y;  4 be an open set. Then Y  V is    closed in X and hence f   closed set in X ;  3  Cl U   f 1  f Cl U    f 1 W   U . So U   is    closed. a 189 f 1 Let U  Y be an open set, V  Y  U V   W . Then by hypothesis Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif Remarks 3.4. 1 Every    continuous function is continuous but the converse may not be true. f  D W    Cl  f W   . Thus f  D f 1 V    Cl  f f 1 V    Cl V   V .      2 D  f V    f V  and f V  is    closed. Therefore, f is    continuous. 1 Then 1  8  : Let 1 1 V  Y . Then  3 Therefore  4  c each  f 1  B    f 1 Cl  B   for each B  Y . f Cl  A    Cl  f  A   for each A⊆X. Proof . Since A is    open in Y , there exists a    open set U  X such that A  Y I U . Thus A being the intersection of two    open sets in X , is    open in X . Hence, we have U I f 1  X  f 1 V     for every also for Lemma 3.5. Let A  Y  X , Y is    open in X and A is    open in Y . Then A is    open in X . V for every U   containing x. We is holds, then f is continuous. a point x of X . Then there exists an open set V  Y containing f  x  such that f U  is not a V  . f 1  Int  B    Int  f 1  B    b  Cl Pr oof . Suppose that f is not    continuous at follows g f :  X ,     Z ,  B  Y.    open sets containing f  x  . x  Cl  X  f is Let  X ,   and Y ,   be topological spaces. a  Theorem 3.3. #   c. is identical with the union of the    frontier of the inverse images of It f :  X ,    Y ,   function If f :  X ,    Y ,   is a function, and one of the following In the next Theorem, #   c. denotes the set of points x of X for which a function f :  X ,    Y ,   is not   continuous. x. a continuous, then    continuous.    continuous. 1 If is  containing is    continuous and a function g : Y ,     Z ,  8   1 : Let V  Y be an open set. Then f 1 V   f 1  Int V    Int  f 1 V   . Therefore, f 1 V  is    open in X . Hence f is U   f :  X ,    Y ,   function is    continuous, then g f :  X ,     Z ,  is    continuous. 1    open in X . Thus f  Int V    subset of a    continuous and a function g : Y ,     Z ,  f 1  Int V   is Int  f 1  Int V     Int  f 1 V   . f 1  Int V    Int  f 1 V   . If that have f :  X ,    Y ,   Let Theorem 3.6. be a x  f 1 V   Cl  f 1 V   . This means that x  Fr  f 1 V   . Now, let f be    continuous mapping and U i : i  I  be a cover of X such that at x  X and V  Y any open set containing f  x  . Then, x  f 1 V  is a    open set of    continuous. X. Thus. x  Int  f 1 V   and U i    for each i  I . Then prove that f is Proof . therefore Let  f U  V  1 x  Fr  f 1 V   for every open set V containing i be an open set, then is    open in U i for each i  I . Since U i is    open in X for each i  I . So by f  x. Lemma 3.5, E-ISSN: 2224-2880 V Y 190  f U  V  1 i is    open in X for Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18  f U  each i  I . But, f 1 V   U 1 i Raja Mohammad Latif V  : i  I  , Sufficiency. Let B   , let A  f 1  B  . We show that A is    open in X . For this let x  A. It implies that f  x   B. Then by hypothesis, there f 1 V    because   is a topology on X . This implies that f is    continuous. then  exists Ax    such that x  Ax and f  Ax   B. Ax  f 1  f  Ax    f 1  B   A. Thus A  U Ax : x  A . It follows that A is    open Then 4   Irresolute Mappings in X . Hence f is    irresolute. In this section, the functions to be considered are those for which inverses of    open sets are    open. We investigate some properties and characterizations of such functions. Definition 4.1.  X ,  Let Definition 4.4. Let  X ,  be a topological space. Let x  X and N  X . We say that N is a    neighbourhood of x if there exists a    open set M of X such that x  M  N. Y ,   be :  X ,    Y ,   and topological spaces. A function f is called    irresolute if the inverse image of each    open set of Y is a    open set in X . Theorem 4.5. Prove that a function f :  X ,    Y ,   is    irresolute if and only if for each x in X , the inverse image of every f  x , is a    neighbourhood of    neighbourhood of x. Theorem 4.2. Let f :  X ,    Y ,   be a function between topological spaces. Then the following are equivalent: 1 f is    irresolute. Proof . Necessity. Let x  X and let B be a    neighbourhood of f  x  . Then there exists  2 U    such that f  x  U  B. This implies that the inverse image of each    closed set in Y is a    closed set in X ;  3 Cl  f V Y;  4 1 V   f 1 x  f 1 U   f 1  B  . Since f is    irresolute, so f 1 U    . Hence f 1  B  is a Cl V   for every    neighbourhood of x. Sufficiency. Let B   . Put A  f 1  B  . Let f Cl U    Cl  f U   for every U  X ;  5 B Y. f 1  Int  B    Int  f 1  B   Theorem 4.3. Prove that x  A. Then B being    open set, is a    neighbourhood of f  x  . for every a f  x   B. But then, by hypothesis, A  f 1  B  is a    neighbourhood of x. Hence by definition, there So function exists Ax   such that x  Ax  A. Thus f :  X ,    Y ,   is    irresolute if and only if for each point p in X and each    open set B in A  U Ax : x  A . It follows that A is a    open set in X . Therefore f is    irresolute. such that p  A, f  A   B. Theorem 4.6. Y with f  p   B, there is a    open set A in X f  p   B. Let that    irresolute, A f  B. f 1 E-ISSN: 2224-2880 a function of f  x  , there is a    neighbourhood V of x is such that f V   U . A is    open in X . Also  B   A as f  p   B. f  A   f  f 1  B    B. p f  Since that f :  X ,    Y ,   is    irresolute if and only if for each x in X . and each    neighbourhood U Proof . Necessity. Let p  X and B   such 1 Prove Thus we have 191 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif    neighborhood V of f  a0  Proof . Necessity. Let x  X and let U be a    neighbourhood of f  x  . Then there exists    neighborhood U of a0 contains at least one O f  x    such that f  x   O f  x   U . It follows x f that 1 O f  x    f   1 U  . By element f  a0  V , and f 1 O f  x     . Let V  f 1 U  . Then it follows is a    neighbourhood f V   f  f 1 U    U . of Then f  x   B. Thus B is a it follows    irresolute. Let every    neighborhood V of f  a0  contains an be element consequently, f  a0   D  f  A  . necessity of the condition. f  a0  ; We Definition 5.1. f  a0  , Let be  X ,   and Y ,   function f :  X ,    Y ,   topological spaces. A is called    open if for every open set G in X , This proves f  G  is a    open set in Y . Sufficiency. Assume that f is not    irresolute Then by Theorem 4.6, there exists a0  X and a E-ISSN: 2224-2880 have The purpose of this section is to investigate some characterizations of    open mappings. every    neighborhood of f  a0  contains an from from 5   Open Mappings f  a   f  a0  . Thus different different Sufficiency. Follows from Theorem 4.7. exists, therefore, at least one element a U I A such that f  a   f  A  and f  a   f V  . Since f  A f  A f  a0   D  f  A  . therefore f  D  A    D  f  A   . From a0  D  A  , it follows that U I A  ; there of of consequently    neighbourhood of f  a0  . Since f is    irresolute, so by Theorem 4.6, there exists a    neighbourhood U of a0 such that f U   V . element a and, since f is one to one, f  a   f  a0  . Thus Assume that f  a0   f  A  and let V denote a f  a0   f  A  , we have be V a U I A such that a  a0 ; then f  a   f  A  a0  D  A  . A  X , and and But a0  D  A ; hence there exists an element f  D  A    f  A  U D  f  A   , for all A  X . f :  X ,    Y ,       neighborhood U of a0 such that f U   V . function f :  X ,    Y ,   is    irresolute if and only if Pr oof . Necessity. Let also f  a0  . Since f is    neighborhood of    irresolute, so by Theorem 4.6, there exists a is    open in X . Therefore, f is    irresolute. a f  a0   A;  A  X , a0  D  A  Ox   . Thus O  UOx : x  O . It follows that O that therefore Proof . Necessity. Let f be    irresolute. Let Ox   such that x  Ox  Vx . Hence x  Ox  O, Prove Then a0  A since f  D  A    D  f  A   , for all A  X . that x Vx  f 1  f Vx    f 1  B   O. Since Vx is a    neighbourhood of x, so there exists an Theorem 4.7. Put Theorem 4.8. Let f :  X ,    Y ,   be a one-toone function. Then f is    irresolute. if and only if there exists a    neighbourhood Vx of x such that Thus f  a  V . which which is a contradiction to the given condition. The condition of the Theorem is therefore sufficient and the theorem is proved.    neighbourhood of f  x  . So by hypothesis, f Vx   B. for f  a0   D  f  A  since V I V   f  a0   . So f  a0   f  D  A     f  A  U D  f  A      , x and Sufficiency. Let B    . Put O  f 1  B  . Let x  O. a U A  a  X : f  a   V  . hypothesis, that V such that every 192 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif Sufficiency. Let U  . Then by hypothesis, f  Int U    Int  f U   . Since Int U   U as U is open. Also Int  f U    f U  . Hence Theorem 5.2. Prove that a mapping f :  X ,    Y ,   is    open if and only if for each x  X , and   U   such that x U , there exists a    open set W  Y containing f  x  such f U   Int  f U   . Thus f U  is    open open in Y . So f is    open. that W  f U  . Proof . Follows immediately from Definition 5.1. Remark 5.6. The equality may not hold in the preceding Theorem. f :  X ,    Y ,   be Theorem 5.3. Let    open. If W  Y and F  X is a closed set Theorem 5.7. H  Y  f Y  F  . f 1 W   F , we f  Int  f 1  B    is    open in Y . Also we have f  Int  f 1  B     f  f 1  B    B. Hence, we have f  Int  f 1  B     Int  B  . Therefore, we obtain Int  f 1  B    f 1  Int  B   . Since f is    open, then H is    closed and f 1  H   X  f 1  f  X  F    X   X  F   F. Theorem 5.4. Let f :  X ,    Y ,   be a    open. function and let B  Y . Then    f 1 Cl Int Cl  B    Cl  f 1  B   .   Proof . Cl  f  B   1 Sufficiency. Let A  X . Then f  A   Y . Hence by hypothesis, we obtain 1 1 Int  A   Int  f  f  A     f  Int  f  A    . is closed in X containing f 1  B  . By Theorem 5.3, there exists a B  H Y set such that    closed f  H   Cl  f 1  B   . f 1 Cl Int Cl  B   1  f Cl Int Cl  H   1   Theorem 5.5. Therefore, we Prove 1 that a Thus f  Int  A    Int  f  A   , for all A  X . Hence, by Theorem 5.5, f is    open. obtain       f  H   Cl  f   1 function Pr oof . Necessity. Let B  Y . Since Int  f 1  B   f is    open, X and is open in Since f 1 Y  F   Y  W  . have a Int  f 1  B    f 1  Int  B   , for all B  Y . H  Y containing W such that f 1  H   F . Let that f :  X ,    Y ,   is    open if and only if containing f 1 W  , then there exists a    closed. Proof . Prove Theorem 5.8. Let f :  X ,    Y ,   be a mapping. Then a necessary and sufficient condition for to be is that f    open  B   . f 1 Cl  B    Cl  f 1  B   for every subset B of function Y. f :  X ,    Y ,   is    open if and only if f  Int  A   Int  f  A  , for all A  X . Proof . Necessity. Assume f is    open Let Pr oof . Necessity. Let A  X . Let x  Int  A . f  x   Cl  B  . Let U   such that x U . Since B Y. Then there exists U x   such that x  U x  A. So f  x   f U x   f  A  . f U x     . Hence and by x  f 1 Cl  B   . Then f is    open, then f U  is a    open set in Y. B I f U   . Then Therefore, hypothesis, f  x   Int  f  A   . Thus U I f 1  B   . f  Int  A   Int  f  A  . E-ISSN: 2224-2880 Let Hence x  Cl  f 1  B   . conclude that f 1 Cl  B    Cl  f 1  B   . 193 Volume 19, 2020 We WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif Y  B   Y . By hypothesis, f 1 Cl Y  B    Cl  f 1 Y  B   . X  Cl  f 1 Y  B    X  f 1 Cl Y  B   . Thus X  Cl  X  f 1  B    f 1 Y  Cl Y  B   . By f is    closed, then G is a    open set and Sufficiency. Let B  Y . Then f 1  G   X  f 1  f  X  E    X   X  E   E.  Theorem 6.4. Suppose that f :  X ,    Y ,   is a mapping. Then    closed   Int Cl  f  A     f Cl  A   for every subset A of X . applying a well-known result, it implies that Now form Int  f 1  B    f 1  Int  B   . Theorem 5.7, it follows that f is    open. Pr oof . Suppose f is a    closed mapping and A is an arbitrary subset of X . Then f Cl  A  is Y. in Then    closed 6   Closed Mappings  In this section we introduce    closed functions and study certain properties and characterizations of this type of functions.  Definition 6.1. A mapping f :  X ,    Y ,   is called    closed if the image of each closed set in X is a    closed set in Y . Theorem 6.2. Prove that a  Int Cl f  Cl  A    f Cl  A  . But also   Int Cl  f  A    Int Cl f  Cl  A   .   Hence Int Cl  f  A     f Cl  A   .  Theorem 6.5. Let f :  X ,    Y ,      closed function, and B, C  Y . f 1  B  , then there neighborhood V of 1 1 f  B   f V   U . Cl  f  A   f Cl  A  for each A  X . Proof . Necessity. Let f be    closed and let  2 A  X . Then f  A  f Cl  A  and f Cl  A   is a    closed set in Y . Thus Cl  f  A   f Cl  A  . a    open such that exists B If f is also onto, then if f 1  B  and f 1  C  have disjoint open neighborhoods, so have B and C. Proof. 1 V  Y  V  f U Sufficiency. Suppose that Cl  f  A   f Cl  A  , for each A  X . Let c Let c V  Y  f  X U .  . Since Then f is    closed, so V is a    open set. Since f 1  B   U , we have A  X be a closed set. Then V c  f U c   f  f 1  B c    B c . Hence, B  V , and thus V is a    open neighborhood of B. Cl  f  A    f Cl  A    f  A . This shows that f  A is a    closed set. Hence f is    closed. Further U c  f 1  f U c    f 1 V c    f 1 V   . This proves that f 1 V   U . c Theorem 6.3. Let f :  X ,    Y ,   be    closed. If V  Y and E  X is an open set 1 containing f V  , then there exists a    open  2 set G  Y containing V such that f 1  G   E. neighborhoods M and N , then by Let a Proof . 1 If U is an open neighborhood of mapping f :  X ,    Y ,   is    closed if and only if Proof . be If f 1  B  and f 1  C  have disjoint open 1 , we have    open neighborhoods U and V of B and C G  Y  f  X  E  . Since respectively such 1 1 f  B   f U   Int  M  and f 1 V   E, we have f  X  E   Y  V . Since f 1  C   f 1 V   Int  N  . Since M and that N are disjoint, so are Int  M  and Int  N  , hence E-ISSN: 2224-2880 194 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif so f 1 U  and f 1 V  are disjoint as well. It c  a  : follows that U and V are disjoint too as f is onto. Then Let A be an arbitrary closed set in X . XA is open in  f   X  A 1 1    ocontinuous, Theorem 6.6. Prove that a surjective mapping f :  X ,    Y ,   is    closed if and only if for each subset B of Y and each open set U in X containing f 1  B  , there exists a    open set V Y. But f 1 X . Since is is    open in  f   X  A  f  X  A  Y  f  A . 1 1 Thus f  A is    closed in Y . This shows that f is    closed. in Y containing B such that f 1 V   U . Remark 6.8. A bijection f :  X ,    Y ,   may be open and closed but neither    open nor    closed. Proof . Necessity. This follows from 1 of Theorem 6.5. Sufficiency. Suppose F is an arbitrary closed set in X . Let y be an arbitrary point in Y  f  F  . Then f 1  y   X  f 1  f  F     X  F  and  X  F  is open in X . Hence by hypothesis, there exists a    open set V y containing y such that f 1 Vy    X  F  . This implies y Vy  Y  f  F   . Thus Y  f  F   UVy : y  Y  f  F  . we So 7 Pre    Open Mappings The purpose of this section is to introduce and discuss certain properties and characterizations of pre     open functions. that obtain Y  f F  Definition 7.1. Let  X ,   and Y ,   be topological spaces. Then a function f :  X ,    Y ,   is said to be pre     open if and only if for each A   , f  A     . being a union of    open sets, is    open Thus its complement f  F  is    closed. This shows that f is    closed. Theorem 6.7. Let f :  X ,    Y ,   be a bijection. Then the following are equivalent: Theorem 7.2. f :  X ,    Y ,   Let and g : Y ,     Z ,   be any two pre     open functions. Then the composition function g f :  X ,     Z ,   is a pre     open function. a  f is    closed.  b f is    open. Pr oof . Let U   . Then f U     . Since f is c f 1 is    ocontinuous. pre     open But then g  f U     as g is Proof .  a    b  :  pre     open. Hence, g f is pre     open. Let U  . Then X  U is Theorem 7.3. closed in X . By  a  , f  X  U  is    closed in Y. But    open. So f U    f  U  there exists V    a mapping such that f  x  V and V  f U  . Proof . Routine. is    open in Theorem 7.4. Y . Hence f 1 is    ocontinuous. E-ISSN: 2224-2880 that for each x  X and for any U   such that x U , Let U  X . be an open set. Since f is 1 1 Prove f :  X ,    Y ,   is pre     open if and only if f  X  U   f  X   f U   Y  f U  . Thus f U  is    open in Y . This shows that f is    open.  b  c :  Prove that a mapping f :  X ,    Y ,   is pre     open if and only if for each x  X and for any    neighbourhood U 195 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif be an identity function defined as f  x   x, for of x in X , there exists a    neighbourhood V of f  x  in Y such that V  f U  . x X. Let A  ¤ c.   f  Int  A    Int  f  A    ¤ c . Proof . Necessity. Let x  X and let U be a    neighbourhood of x. Then there exists W    such x W  U . that Theorem 7.7. Then Hence is a    neighbourhood of f  x  and V  f U  . Proof . Int  f Sufficiency. Let U   . Let x  U . Then U is a a function f  Int  A    Int  f  A   , for all A  X .  Then there exists U x   such that x  U x  A. So f U x     . Hence by hypothesis, that a mapping B Y. Necessity. Let Cl  B   . Then f  x   Cl  B  . Proof . is    open. Also Int  f U    f U  . Hence f U   Int  f U   . Thus f U  is U   x f as U 1 such f U        open in Y . So f is pre     open. that x U . By 1  B   . So x  Cl  f  B   , f 1 Cl  B    Cl  f 1  B   . 1 Example 7.6. Let X  Y  R. suppose X be with Let Let hypothesis, f  x   f U  . and f U  I B  . Hence U I f We remark that the equality does not hold in Theorem 7.5 as the following example shows. Thus Therefore, we obtain Y  B   Y . By f 1 Cl Y  B    Cl  f 1 Y  B   .  f 1 Y  B    X  f 1 Cl Y  B   . Sufficiency. Let B  Y . Then Then     , ΅ , ¥  . Let Y be with discrete topology hypothesis, D   A : A  X   P  X  . Let f  Id : X  Y E-ISSN: 2224-2880 Prove f 1 Cl  B    Cl  f 1  B   , for every subset B of Y . f  Int U    Int  f U   . Since Int U   U .  f :  X ,    Y ,   is pre     open if and only if Sufficiency. Let U   . Then by hypothesis, c  Hence, Theorem 7.8. f  x   Int  f  A  . Thus    , ΅ , ¥ ,¤ c , ¥ U¤ is Hence, by Theorem 7.5, f is pre     open. f  Int  A    Int  f  A   . topology f   Proof . Necessity. Let A  X . Let x  Int  A  . and  and Sufficiency. Let A  X . Then f  A   Y . Hence by hypothesis, we obtain 1 1 Int  A   Int  f  f  A    f  Int  f  A   . This implies that  1 f  Int  A    f  f Int  f  A     Int  f  A  .   Thus f  Int  A    Int  f  A   , for all A  X . f :  X ,    Y ,   is pre     open if and only if f  x   f U x   f  A  is    open in X Since f  Int  f 1  B     Int  B  . Therefore Int  f 1  B    f 1  Int  B   .  B. points. Therefore f U  is    open. Hence f is pre     open. that  B  B Y. Let Necessity. 1     neighbourhood of each of its Prove function Y . Also we have f  Int f 1  B    f  f 1  B   that f  x  V f  x   f U  . It follows at once that Theorem 7.5. a pre     open, f  Int f 1  B   is    open in    neighbourhood of x. So by hypothesis, there exists a    neighbourhood V f  x  of f  x  such f U  is a that Int  f 1  B    f 1  Int  B   , for all B  Y .  V  f W  Prove f :  X ,    Y ,   is pre     open if and only if f  x   f W   f U  . But f W    as f is pre     open Then each So X  Cl 196    Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif f  A is    closed for each    closed subset A of X . So X  Cl  X  f 1  B    f 1 Y  Cl Y  B   . By a well-known result, it follows that Now by Int  f 1  B    f 1  Int  B   . set Theorem 8.2. The composition of two pre     closed mappings is a pre     closed mapping. Theorem 7.7, it follows that f is pre     open. Theorem 7.9. f :  X ,    Y ,   Let g : Y ,     Z ,   and Proof . The straight forward proof is omitted. be two mappings such that g f :  X ,     Z ,   is    irresolute. Then Theorem 8.3. is a pre     open injection, then f is    irresolute. If f is a pre     open surjection, then g is    irresolute. Necessity. Suppose f is a pre     closed mapping and A is an arbitrary subset of X . Then f Cl  A   is    closed in Y . Since f  A   f Cl  A  , we obtain  Cl  f  A    f Cl  A  .  : Sufficiency. Suppose F is an arbitrary    closed X . By hypothesis, we obtain set in f  F   Cl  f  F    f Cl  F    f  F  . f 1  g 1  g U     f 1 U  . Consequently f 1 U  is    open in X . This proves that    irresolute.  2 Let V    . Then  g f  V    f is f  F   Cl  f  F   . Thus f  F     closed in Y . It follows that f pre     closed. 1  since Theorem 8.4. Let f :  X ,    Y ,   pre     closed function, and B,C  Y .  be a If U is a    open neighborhood of f 1  B  , then there exists a    open neighborhood V of  B such that f 1  B   f 1 V   U .  2 If f is also onto, then if f 1  B  and f 1  C  have disjoint    open neighborhoods, so have B and C. 8 Pre     Closed Mappings Proof . In this last section, we introduce and explore several properties and characterizations of pre     closed functions. 1 Let V c  Y  V  f U c  . pre     closed, Definition 8.1. A function f :  X ,    Y ,   is said to be pre     closed if and only if the image E-ISSN: 2224-2880 is 1 1 1 f  g f  V     f   g f   V       1 1 1 1 1      f  f g  V    f f g V    g V  . Hence g is    irresolute.  is Hence g f is    irresolute. Also f is pre     open 1    open f  g f  V   is    open in Y . is surjective, we note that Since f  mapping Proof . Proof . 1 Let U    . Then g U     since g is pre     open. Also g f is    irresolute. 1 Therefore, we have  g f   g U     . Since g have a if Cl  f  A    f Cl  A   for every subset A of X .  2 is an injection, so we 1 1 1  g f   g U    f g   g U   that f :  X ,    Y ,   is pre     closed if and only 1 If g  Prove f 1  B  U , so V  Y  f  X U . Since V Then f is is    open. Since we have V c  f U c   f  f 1  B c    B c . Hence, B  V , and thus V is a    open neighborhood of B. 197 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif Further U c  f 1  f U c    f 1 V c    f 1 V   . This proves that f 1 V   U .  3 f 1 is    irresolute. c Proof . 1   2  : Let U   . Then X  U is   2 If f 1  B and f 1 C     closed    closed have disjoint    open neighborhoods M and N , then by 1 , we have    open neighborhoods U and V of B and C respectively such that and f 1  B   f 1 U   Int  M  C   f  2    3 : Let  is 1 A  X . Then by Theorem 4.2, it follows that f is    irresolute. set V in Y containing B such that f 1 V   U . from  3  1 : 1 of Let A be an arbitrary    closed set in X . Then X  A is    open in X . Since f 1 is Theorem 8.4.    irresolute. Sufficiency. Suppose F is an arbitrary    closed set in X . Let y be an arbitrary point in But Y  f F . y  Vy  Y  f  F   .  This implies  Hence Theorem 8.6. Let f :  X ,    Y ,   be bijection. Then the following are equivalent: 1 f is pre     closed.  2 f is pre     open. E-ISSN: 2224-2880  f   X  A  f  X  A  Y  f  A . 1 1 said to be contra    continuous if Y  f  F  , being a union of    open sets is Thus its complement f  F     open.    closed. This shows that f is    closed. is    open in Y . We introduce the definition of contra    continuous functions in topological spaces and study some of their properties in this section. Definition 9.1. A function f :  X ,    Y ,   is that Thus Y  f  F   U Vy y  Y  f  F  . 1 1 9 Contra    Continuous Mappings there exists a    open set V y containing y such f 1 Vy    X  F  .  f   X  A Thus f  A is    closed in Y . This shows that f is pre     closed. Then and f 1  y   X  f 1  f  F     X  F   X  F  is    open in X. Hence by hypothesis, that Thus 1 1 Thus Cl  f 1   A     f 1  Cl  A   , for all   in X containing f 1  B  , there exists a    open follows But Y. f 1 Cl  f  A     Cl  f 1  f  A    . It implies that Cl  f  A    f Cl  A  . f :  X ,    Y ,   is pre     closed if and only if for each subset B of Y and each    open set U This is A  X . Since f is pre     open, by Theorem 7.8, so Theorem 8.5. Prove that a surjective mapping Proof . Necessity. in f  X U  pre     open.  It follows that U and V are disjoint too as f onto. 1 , f U  is    open in Y . This shows that f is 1  X . By f  X  U   f  X   f U   Y  f U  . V   Int  N  . Since M and N are disjoint, so are Int  M  and Int  N  , and hence so f 1 U  and f 1 V  are disjoint as well. f 1 in  X ,  f 1 V  is is    closed in Y ,   . a Observe that if Observe that if X is a countable set, then every function f :  X ,    Y ,   is contra    continuous. for each open set V of f :  X ,    Y ,   be a function. Then the following are equivalent. 1 f is contra    continuous. Theorem 9.2. Let 198 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18  2 f 1  F  is    open in  X ,  Raja Mohammad Latif    closed in X . This shows that f is contra    continuous. Proposition 9.3. Let f :  X ,    Y ,   be contra for every closed subset F of Y ,   .  3 For each x  X and each closed set F in Y ,   containing f  x  ,there exists a    open. set U in  X ,   containing x such that f U   F .  4  f Cl  A  Ker  f  A for ever subset A of  X ,   .  5  Cl  f 1  B   f 1  Ker  B  for ever subset B of Y ,   . Proof . 1   2  : Let F be any closed set of    continuous. If one of the following conditions holds, then f is    continuous. 1 Y ,   is regular, Int  f 1  Cl V     f 1 V  for each open set V in Y ,   .  2  Proof . 1 Let x  X and V be an open set of Y ,   containing f  x  . Since Y ,   is regular,  there exists an open set W in Y ,   containing Y . Then Y  F is open. Hence by hypothesis f 1  Y  F  is Thus    closed. f  x  such that Cl W   V . Since f is contra    continuous, so by Theorem 9.2, there exists a f 1 Y  F   Cl  f 1 Y  F   . We can obtain X  f 1  F   X  Int  f 1  F   . Therefore, we    open set U in that x such f U   Cl W  ; hence f U   V . Therefore is    continuous. f 1  F   Int  f 1  F   . Thus f 1  F  is have  X ,   containing    open in X .  2 f Let V be an open set of Y ,   . Since f is x  X and F be a closed set of contra    continuous and Cl V  is closed, by Y containing f  x  . By  2  , x  Int  f 1  F   . Hence there exists U    X  containing x such Theorem 9.2, f 1 Cl V   is    open set in  2    3 : Let  X ,  f  3   4  : Let A be any subset of X . Let x  Cl  A  and F be a closed set of Y containing f  x  . Then by  3 there exists U  O  X  containing x such that f U   F ; hence x U  f  F  . Since x  Cl  A  , so U I A   subset  f Cl  f and we is called the f :  X ,    Y ,   be a every x  X . If g is contra    continuous, then f is contra    continuous. Proof . Let U be an open set in Y ,   , then X  U is an open set in  X  Y ,     . Since g is contra    continuous, g 1  X  U   f 1 U  is    closed in  X ,   . This shows that f is contra    continuous. obtain Thus Cl  f 1 V    f 1  Ker V    f 1 V  . Hence Cl  f 1 V    f 1 V  . f 1 V  is E-ISSN: 2224-2880 and function of f , defined by g  x    x, f  x   for  4 ,  B    Ker  f  f  B    Ker  B  2.15 So, function and g :  X ,     X  Y ,     the graph 1 Lemma  x, f  x   : x  X   X  Y Theorem 9.4. Let and hence Cl  f 1  B    f 1  Ker  B   .  5   1 : Let V be any open set of Y . Then by  5 1 graph of f and is denoted by G  f  . that Then by Lemma 2.15, we have f  x   Ker  f  A   and hence we obtain f Cl  A    Ker  f  A   . 1 1 consequently f 1 V  is    open in  X ,   . So f is a    continuous function. Recall that for a function f :  X ,    Y ,   , the   4    5  : Let B be any subset of Y . By it implies Cl V    Int  f  Cl V     f V  . we obtain f 1 V   Int  f 1  Cl V    that x U  f 1  F  . Then, x U and f U   F . and hence it follows   f U I A   f U  I f  A   F I f  A  . and hence by (2), 1 199 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif 1 g f is    continuous, if f is contra    continuous and g is contra  continuous. Definition 9.5. A subset A of a topological space  X ,   is said to be    dense in X if Cl  A   X . Definition 9.6. A topological space  X ,   is said  2 g f is contra    continuous, if f is contra    continuous and g is continuous.  (  3 g f is contra    continuous, if f is    irresolute and g is contra    continuous. to be a Urysohn space if for any two distinct points x, y  X , there exist open subsets U and V of  X ,   such that x U , y V and Cl U  I Cl V    . Theorem 9.7. Let f ,g :  X ,    Y ,   be two contra    continuous functions. If Y ,   is Urysohn, the following properties hold: 1 The set E   x  X : f  x   g  x  is Theorem 9.9. g Cl W   are Then g f :  X ,     Z ,      continuous if and only if    continuous. containing x. Let U  f 1 Cl V   1 g 1  F   f  f 1  g 1  F    is    open in Y ,   and we obtain that g is surjective, contra    continuous.. is contra    continuous. Let V be a closed set in  Z ,   . Then g 1 V  is    open in Y ,   . Since f is For  X ,  and converse, suppose g f 1  g 1 V     g f  1 V  is    open in  X ,   and so g f is a contra    continuous. Definition 9.10. A space topological  X ,   is said to be Strongly S  closed if every closed cover of  X ,   containing x. Now, f  A  I g  A   f U I G  I g U I G   f U  I g  G   Cl V  I Cl W    . This implies that A I E   , where A is    open in  X ,   . Hence x  Cl  E  . So E    closed in  X ,   .  2  Let E   x  X : f  x   g  x  . Since f and g are contra    continuous and Y ,   is X has a finite cover. Definition 9.11. A space topological  X ,   is said  to be    compact if every    open cover of X has a finite cover. Definition 9.12. A subset A of a space  X ,   is Urysohn, by the previous part, E is    closed in  X ,   . By assumption, we have f  g on A, said to be    compact relative to X if for any cover V :  V of A by    open sets of X, there exists a finite subset V0 of V such that A  UV :  V0  . Theorem 9.13. Let f :  X ,    Y ,   be contra    continuous surjection. 1 If A is    compact relative to  X ,   , then f  A is strongly S  closed in Y ,   . where A is    dense in  X ,   . Since A  E, A is    dense and E is    closed in  X ,   , so X  Cl  A   Cl  E   E. Hence f  g on  Theorem 9.8. Let f :  X ,    Y ,   and g : Y ,     Z ,   be functions, then the following properties hold: E-ISSN: 2224-2880 the    irresolute,    open set in  X , . is contra g is contra Then f 1  g 1  F     g f   F  is    open in Since f is pre    open and  X , . G  g 1 Cl W   and set A  U I G. Then A is  a Proof . Suppose g f :  X ,     Z ,   is contra    continuous. Let F be a closed set in  Z ,   . f 1 Cl V   and    open sets in be    irresolute and pre     open function and g : Y ,     Z ,   be any function. By assumption on the space Y ,   , there exist open sets V and W in Y ,   such that f  x   V , g  x  W and Cl V  I Cl W    . Since f and 1 f :  X ,    Y ,   surjective    closed in  X ,   .  2  f  g on  X ,   whenever f  g on a    dense set A  X . Proof . 1 Let x  X  E. Then f  x   g  x  . g are contra    continuous, let 200 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18  2  If  X ,   is strongly Raja Mohammad Latif S  closed, then Y ,   be a topological space. Then the    frontier of a subset A of X , denoted by closed sets of the subspace f  A . For  V, Fr  A , by is defined as Fr  A   Cl  A    Cl  X  A   there exists a closed set A of Y ,   such that V  A I f  A  . For each X,  Definition 9.15. Let is compact. Proof . 1 Let V :  V be any cover of f  A  Cl  A     Int  A   . x  A, there exists  x V such that f  x   A . x Now by hypothesis f is contra    continuous Theorem 9.16. The set of all points x of X at and hence by Theorem 9.2, there exists a    open set U x in  X ,   such that x  U and which cover of A U x : x  A is a by    open sets of  X ,   , there exists a subset of A0 A  UU x : x  A0  . such A Y containing f  x  . Proof.  f  A   U f U x  : x  A0   U A x : x  A0 . Thus  f  A   U V x : x  A0  f  A and hence is contra    continuous, f We have X  U f  X , . is every  f V  :  V is for Y ,    X  ,    :    Theorem 9.14. Let not contra U  O  X , x  , which Thus implies that x  Cl  f 1 Y  F   1  Cl  X  f 1  F   . Again, since x  f  F  , we get x  Cl  f 1  F   and so it follows that some finite subset V0 of V. Since f is surjective, Y  UV :  V0  . This shows that compact. be at a point U  f 1 Y  F    .  V  :  V0  f x  X . Then by Theorem 9.2, there exists a closed set F of Y containing f  x  such that f U   Y  F    for a    closed cover of the strongly S  closed space 1 Let Necessity:    continuous strongly S  closed .  2  Let V :  V be any open cover of Y . Since 1 contra not that Therefore,  is    continuous is identical with the union of   frontier of the inverse images of closed sets of f U x   A x . Since the family finite f :  X ,    Y ,   x  Fr  f 1  F   . is  Sufficiency: Suppose that x  Fr  f 1  F   be any  for some closed set F of Y containing f  x  and f is family of topological spaces. If a function f : X   X  is contra    continuous, then contra    continuous at x. Then there exists    f : X  X  is contra U   O  X , x  such that f U   F . V    continuous. for is the projection of  X  each   , where   x U  f 1  F  V Therefore and hence it follows that   onto X  . x  Int  f 1  F    X  Fr  f 1  F   . But this Proof. For a fixed   , let V be any open subset is a contradiction. So f is not contra of X  . Since   is continuous,   1 V  is open   continuous at x. in  X  Since Definition 9.17. A function f :  X ,    Y ,   V f 1   1  f is contra    continuous, V     f  V  Therefore,  f 1 is    closed in X . is called almost weakly    continuous, if, for each is contra    continuous, for x  X and for each open set V of Y containing each   . E-ISSN: 2224-2880 201 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif f  x  , there exists U   O  X , x  such that    ,Y ,a ,b be a topology for Y . Let f U   Cl V  . f :  X ,    Y ,   f  A   a and f  X  A   b . Suppose Theorem 9.18. that a function Then constant and contra    continuous such that Y ,   is T0  space. This is a contradiction. Hence  X ,   must be    connected. Definition 9.22. A topological space  X ,   is said to be    T2 if for each two distinct points x, y  X , there exist    open sets U and V in  X ,   such that x U , y V and U I V   . Definition 9.23. A topological space  X ,   is said to be weakly Hausdorff if each element of X is an intersection of regular closed sets. Definition 9.24. A topological space  X ,   is said to be ultra Hausdorff if every two distinct points of X can be separated by disjoint clopen sets. Definition 9.25. A topological space  X ,   is said to be ultra normal  resp.    normal  if each pair of non-empty disjoint closed sets can be separated by disjoint clopen resp .  open sets.    Theorem 9.26. Let f :  X ,    Y ,   be a contra    continuous injection, then the following f :  X ,    Y ,   is contra    continuous. Then f is almost weakly    continuous. Proof. For any open set V of Y , Cl V  is closed Y . Since in is contra f    continuous, f 1 Cl V   is    open set in X . We take U  f 1 Cl V   , then f U   Cl V  . Hence f is almost weakly    continuous.  X ,   is said to be Definition 9.19. A space    connected provided that X is not the union of two disjoint nonempty    open sets. Proposition 9.20. Let f :  X ,    Y ,   be surjective and contra    continuous. If  X ,   is    connected, then Y ,   is connected. Proof. Assume that Y ,   is not connected. Then, there exist nonempty open sets V1 , V2 of such that V1 I V2   and have f 1 V1  I f Y ,   V1 UV2  Y . Hence we 1 V2    and V1  U f V2   X . Since f is surjective, f 1 V1  and f 1 V2  are nonempty sets. Since f 1 be a function such that 1 is not properties hold: 1  X ,   is    T1 if Y ,   is weakly Hausdorff.  2   X ,   is    T2 if Y ,   is a Urysohn space or ultra Hausdorff.  3  X ,   is    normal if Y ,   is ultra normal and f is closed. Proof . 1 Suppose that Y ,   is weakly Hausdorff. For any distinct points x and y in  X ,   , there exist regular closed sets A, B in Y ,   such that f  x   A, f  y   A, f  x   B and f  y   B. Since f is contra    continuous, f 1  A  and f 1  B  are    open sets in  X ,   such that x  f 1  A , y  f 1  A  , x  f 1  B  and y  f 1  B  . This shows that  X ,   is f is contra    continuous and V1 , V2 are open sets. Hence f 1 V1  and f 1 V2  are    open sets in  X ,   . Therefore,  X ,   is not    connected. Theorem 9.21. If every contra    continuous function from a space  X ,   into any T0  space Y ,   is constant, then  X ,   is    connected. Pr oof . Suppose that  X ,   is not    connected and every contra    continuous function from  X ,   into any T0  space Y ,   is constant. Since  X ,   is not    connected, there exists a proper nonempty    open subset A of  X ,   . Let Y  a,b and E-ISSN: 2224-2880 f    T1 . 202 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18  2  Let x1 and x2 be any distinct points in Then, since f is injective, Raja Mohammad Latif f is contra    continuous at x1 and x2 , so there exists    open sets G and H in  X ,   X. f  x1   f  x2  . Moreover, since Y ,   is ultra-Hausdorff, there containing x1 and x2 , respectively, such that f  G   Cl U  and f  H   Cl V  . Hence we obtain G I H   . Therefore,  X ,   is    T2 . exist clopen sets V1 , V2 such that f  x1  V1 , f  x2  V2 and V1 I V2   . Since f is contra    continuous. So there exists U i  O  X ,  containing xi such that f U i   Vi for i  1,2. Clearly, we obtain U1 I U 2   . Thus  X ,   is Definition 9.28. .A function f :  X ,    Y ,   is called almost contra    continuous if    closed for every regular open set V of Y .    T2 . In case Y ,   is Urysohn space, there here exist open sets U 1 , U 2 such that f  x1  U1 , f  x2  U 2 and Cl U1  I Cl U 2    . Let G  f 1 Cl U1   and H  f 1 Cl U 2   . Then f :  X ,    Y ,   be a Theorem 9.29. Let function. Then the following statements are equivalent: a  f x1  G, x2  H and G I H   . Since f is contra    continuous. Therefore G and H are b    open sets in  X ,   . Thus  X ,   is    T2 .  3 Let F1 and F2 be disjoint closed subsets of is almost contra    continuous f 1  F  is    open in X for every regular closed set F of Y . Y ,   . Since f is closed and injective, f  F1  and f  F2  are disjoint closed subsets of Y ,   . Since Y ,   is ultra normal, f  F1  and f  F2  c for each x  X and each regular open set F of Y containing f  x  , there exists U  O  X  such that x U and f U   F . are separated by disjoint clopen sets V1 and V2 , respectively. Since f is contra    continuous, Fi  f 1 Vi  and f 1 Vi  is    open in  X ,   for i  1,2 and f 1 V1  I f 1 V2    . Thus  X ,   is    normal. Theorem 9.27. Let  X ,   be a topological space. If for each pair of distinct points x1 and x2 in X there exists a function f of  X ,   into a Urysohn space Y ,   such that f  x1   f  x2  and f is contra    continuous at d for each x  X of Y and each regular open set V non-containing f  x, there exists a    closed set K of X non-containing x such that f 1 V   K . Proof .  a    b  : Let F be any regular closed set of Y . Then Y  F  is regular open and therefore f 1 Y  F   X  f 1  F    C  X  . Hence, f 1  F    O  X  . The converse part is obvious. x1 and x2 , then  X ,   is    T2 .  b  c : Proof . Let x and y be any two distinct points of X . Then by the hypothesis, there exist a Urysohn space Y ,   and a function containing x  f 1  F  . f :  X ,    Y ,   which satisfies the condition of Let F be any regular closed set of Y f  x  . Then f 1  F    O  X  and Taking U  f 1  F  we get f U   F . the theorem. Let yi  f  xi  for i  1,2. Then c   b : y1  y2 . Since Y is Urysohn, there exist open sets U and V containing y1 and y2 , and respectively, such that Cl U  I Cl V    . Since E-ISSN: 2224-2880 f 1 V  is 203 Let F be any regular closed set of Y x  f 1  F  . Then, there exists Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif U x   O  X , x  such that f U x   F and so Corollary 9.32. If f :  X ,    Y ,   is a contra U x  f 1  F  .    continuous Also, we have f 1  F    x f 1 F  U x . Hence f 1  F    O  X  . Hausdorff, then X is Bc  T1. c  d : Theorem 9.33. Let Let V be any regular open set of Y non- of Y containing f(x). Hence by (c), there exists U   O  X , x  such that f U   Y  V  . Hence, we obtain U  f Y  V   X  f V  f 1 V    X  U  . X  U 1 Now, since U   O  X  , 1 1    continuous surjection, f U  and f V  X f containing f  x  . Then Cl  H  is a regular closed f is almost said weakly U   f V  . This shows that X is not to be countably f  X,   is an almost contra    continuous surjection. Then the following statements hold: f  y   U and a  being almost If X is    compact, then Y is S  closed.  b contra    continuous, f 1 U  and f 1 V  are If X is    Lindelof , 1   open subsets of X such that x  f U  , S  Lindelof . y  f 1 U  and y  f 1 V  , x  f 1 V  . This  c  If shows that X is    T1. countably S  closed. E-ISSN: 2224-2880 every f :  X ,    Y ,   be Theorem 9.36. Let Proof . Since Y is weakly Hausdorff, for distinct points x, y of Y, there exist regular closed sets U f  y  V , f  x  V . Now, is said to be    Lindelof if every    open cover of X has a countable subcover. f :  X ,    Y ,   be an f  x  U , X,     compact if Definition 9.35. A topological space almost contra    continuous injection and Y is weakly Hausdorff. Then X is   T1. and V such that and of countable cover of X by    open sets has a finite subcover. that   continuous. Theorem 9.31. Let X sets 1 Definition 9.34. A topological space set of Y containing f  x  . Then by Theorem 9.29, So 1    connected. But this is a contradiction. Hence Y is connected. Proof . For x  X , let H be any open set of Y f  G   Cl  H  .    open are contra    continuous. Then f is almost weakly    continuous. such an U and V are clopen sets in Y, they are regular open sets of Y. Again, since f is almost contra Theorem 9.30. Let f :  X ,    Y ,   be almost G  O  X , x  be connected. Then there exist disjoint non-empty open sets U and V of Y such that Y  U  V . Since and so is    closed set of X not containing x. exists f :  X ,    Y ,   Proof . If possible, suppose that Y is not The converse part is obvious. there is weakly almost contra    continuous surjection and X be    connected. Then Y is connected. containing f(x). Then  Y  V  is regular closed set 1 Y injection and 204 then Y is X is countably    compact, then Y is Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif Proof.  a  : Let V :   I be any regular closed cover of Y. f Since    continuous, is almost  f V  :   I  1 then    open cover of     closed cover of X . Again, since X    closed compact, there exists a finite subset I 0 contra of I such that X    f 1 V  :   I 0  and hence is a X . Again, since X Y  V :   I 0  . Therefore, Y is nearly compact. is   compact, there exist a finite subset I 0 of I such that X    f 1 V  :   I 0  and The proofs of hence  b  and  c  are being similar to Sets and functions in topological spaces are developed and used in many engineering problems, information systems and computational topology. By researching generalizations of closed sets, some new separation axioms and compact spaces have founded and are turned to be useful in the study of digital topology. In this paper we have introduced   continuous,   irresolute,   open,   closed, pre    open, pre    closed, contra   continuous and almost contra    Definition 9.37. A topological space  X ,   is said to be    closed compact if every   closed cover of X has a finite subcover. Definition 9.38. A topological space  X ,   is said to be countably    closed if every countable cover of X by    closed sets has a finite subcover. mappings and have investigated properties and characterizations of these new types of mappings in topological spaces. We have studied new types of functions using    open sets and these functions will have many Definition 9.39. A topological space  X ,   is said to be    closed Lindelof if every   closed cover of X has a countable subcover. f :  X ,    Y ,   possibilities of applications in computer graphics and digital topology. be an almost contra   continuous surjection. Then the following statements hold: a  being similar to 10 Conclusion  a  : omitted. Theorem 9.40. Let  b  and  c  are  a  : omitted. Y  V :   I 0  . Therefore, Y is S  closed. The proofs of is Acknowledgment If X is    closed compact, then Y is nearly The author is highly and gratefully indebted to the compact. Prince Mohammad Bin Fahd University, Al Khobar,  b Saudi Arabia, for providing all necessary research If X is    closed Lindelof , then Y is facilities during the preparation of this research nearly Lindeloff . paper.  c  If X is countably    closed compact, then Y is nearly countable compact. References: Proof .  a  : Let V :   I  be any regular open Y. Since    continuous, then cover of E-ISSN: 2224-2880 f is almost  f V  :   I  1  [1] Samer Al-Ghour, Bayan Irshedat, The Topology of   open Sets, Filomat (Faculty of Sciencees and Mathematics, University of Nis, Serbia) 31:16 (2017), 5369 – 5377. contra is a 205 Volume 19, 2020 WSEAS TRANSACTIONS on MATHEMATICS DOI: 10.37394/23206.2020.19.18 Raja Mohammad Latif [18] D.S. Jankovic, On some separation axioms and   closure, Mat. Vesnik 32(4) (1980), 439449. [19] D.S. Jankovic,   regular spaces, Internat. J. Math. & Math. Sci. 8(1986), 515-619. [20] J.E. 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