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Retrospective Theses and Dissertations 1971 On bitopological spaces Marcus John Saegrove Iowa State University Follow this and additional works at: http://lib.dr.iastate.edu/rtd Part of the Mathematics Commons Recommended Citation Saegrove, Marcus John, "On bitopological spaces " (1971). Retrospective Theses and Dissertations. Paper 4914. This Dissertation is brought to you for free and open access by Digital Repository @ Iowa State University. It has been accepted for inclusion in Retrospective Theses and Dissertations by an authorized administrator of Digital Repository @ Iowa State University. For more information, please contact [email protected]. 71-26,888 SAEGROVE, Marcus John, 1944ON BITOPOLOGICAL SPACES. Iowa State University, Ph.D., 1971 Mathematics University Microfilms, A XEROX Company, Ann Arbor, Michigan THIS DISSERTATION HAS BEOi MICROFILMED EXACTLY AS RECEIVED On bitopological spaces by Marcus John Saegrove A Dissertation Submitted to the Graduate Faculty in Partial Fulfillment of The Requirements for the Degree of DOCTOR OF PHILOSOPHY Major Subject: Mathematics Approved: Signature was redacted for privacy. In Charge oàMajor Work Signature was redacted for privacy. Signature was redacted for privacy. Iowa State University Of Science and Technology Ames, Iowa 1971 PLEASE NOTE: Some pages have indistinct print. Filmed as received. UNIVERSITY MICROFILMS. ii TABLE OF CONTENTS Page I. II. III. IV. V. INTRODUCTION 1 SEPARATION PROPERTIES 4 AN INTERNAL CHARACTERIZATION OF PAIRWISE COMPLETE REGULARITY l8 PRODUCTS AND BICOMPACTIFICATIONS 26 GENERALIZED QUASI-METRICS AND QUASIUNIFORM SPACES 58 VI. LITERATURE CITED 49 VII. ACKNOWLEDGEMENTS 5% 1 I. INTRODUCTION A bitopological space topologies, P and Q, (X,P,Q) on it. is a set X with two Bitopological spaces arise in a natural way by considering the topologies induced by sets of the form form B metrics on B^^^ = fy I p(x,y) < e] = (y i g(x,y) < e], X and where p q(x,y) = p(y,x). and sets of the and q are quasi- Quasi-uniform spaces, which are generalizations of quasi-metric spaces, also induce bitopological spaces. Kelly [8j was one of the first to study bitopological spaces. Later work in the area has been done by Fletcher [3] et al., Kim L9j, Lane [10], Patty [13], Pervin [14], Reilly [16] and others. Reilly [I6] discusses separation properties of bitopo logical spaces at some length in his Ph.D. dissertation. In Chapter II of this thesis, we identify a new separation property which we call weak pairwise T^ and we attempt to organize the separation properties into what appears to be two natural types, weak and strong. We also generalize to bitopological spaces some results concerning regularity given by Davis [1]. We include in Chapter II a number of defi nitions and results previously given. definitions has been changed slightly. continuous function into (A,R,L)" The wording of some The words "pair (which will be defined in Chapter II) have replaced the words "P-lower semi- 2 continous Q-upper semi continuous function" since the two are equivalent. A "P-lower semi-continuous Q-upper semi- continuous function" is a function from a bitopological space (X,P,Q) into A (the real line) with the usual topology such that inverse images of open right rays are P-open and inverse images of open left rays are Frink [4] Q-open. and Steiner [I7] have given internal charac terizations of complete regularity. In Chapter III we gener alize this result to bitopological spaces using a generali zation of Steiner's method of proof. In Chapter IV we define products of bitopological spaces and obtain a bicompactification of a weak pairwise T 1 soace. To the author's knowledge products in bitopo- logical spaces have not previously been defined. Fletcher [3] and Kim [9] have given definitions of pairwise compactness in bitopological spaces, but under their definitions products of pairwise compact spaces are not pairwise compact, nor are real valued pair continuous functions bounded. We give a different definition of bicompactness which has these de sirable properties. Kalisch [6] has defined a generalized metric space and has proven that any generalized metric space is a uniform space and that every uniform space is isomorphic to a gener alized metric space. In Chapter V we define a generalized 5 quasi-metric space and show that a generalized quasi-metric space is a quasi-uniform space and that every separated quasi-uniform space is quasi-uniformly isomorphic to a generalized quasi-metric space. Since it has previously been shown that topological spaces are quasi-uniformizable, we have essentially shown (if we disregard separation) that topological spaces are generalized quasi-metric spaces. The following figure indicates some of the inter relationships that exist among these structures. metric spaces quasi-metric spaces 1 uniformizablegeneralized spaces metric spaces topological spaces generalized quasi-metric spaces quasiuniformizable spaces 1 bitopological spaces 4 II. SEPARATION PROPERTIES In this chapter we consider various properties of bitopological spaces. Some effort was made toward complete ness and the systematic presentation of the usual degrees of separation. We begin with a rather weak form of sepa ration due to Reilly [16,Definition 1.1], who calls it pairwise T^. We prefer to call this property weak pairwise Definition 2.1: pairwise T^ The bitopological space (X,P,Q) T^. is weak iff for each pair of distinct points, there is a set which is either P-open or Q-open containing one but not the other. Definition 2.2: pair (x,y) (X,P,Q) is pairwise of distinct points of P-open set containing open set containing x y but not but not T^ X, y iff for each there is either a or there exists a Q- x. Definition 2.2 was given by Fletcher, Hoyle, and Patty [3,Definition 1]. It is easy to see that topology implies weak pairwise T^ topology implies pairwise Paiirwise TQ set T^. and T^ T^ T^ in either in either does not imply in either topology as can be seen by considering the X = {a,b,c} with 0 = {0jX,{b,c},[a]}, neither P nor Q P = {0,X,{a,b},{c}} where pairwise is T^. Pairwise T^ T^ and is clear, but implies weak 5 pairwise but the following example shows the converse is not true. Example 2.1: and Consider R = {(x, o o ) I X € where and R] both topologies are But if a < b, tains a tains b, so L R (RjR,L) Definition 2.3: { ( - o o ,x) 1 (&,R,L) every and every L = R is the real line x € is weak pairwise open set containing open set containing is not pairwise (X P,Q) b a iff for x,y, set such that either y X U and a and Q-open set y€V, x;^V V or also con T^. each pair of distinct points U T^. also con is weak pairwise J Since R}. there exists a x€V, y/v P-open x € U, and y € U, / U. Reilly [l6,Definition 2.1] defined the following property as pairwise Definition 2.4: T^. (X,P,o) is pairwise pair of distinct points x,y, U such that and a Q-open set V iff for each there exists a x € U, y P-open set U and y € V, X X V. Clearly weak pairwise and pairwise implies weak pairwise implies pairwise is equivalent to Obviously, pairwise T^. Also pairwise in each topology [l6,Theorem 2.2]. implies weak pairwise but 6 Example 2.1 shows that weak pairwise to pairwise ing a b for if also contains must also contain a b. a < b, and any where P weak pairwise R-open set containing is not equiva as any bitopological space is discrete and T^, L-open set contain Weak pairwise lent to weak pairwise (X,P,Q) any is not equivalent Q is indiscrete is but not weak pairwise Reilly [16] states that Kim called the following property pairwise Definition 2.5: T -, . (X,P,0) is weak pairwise Tg (or weak pairwise Hausdorff) iff for each pair of distinct points X and set or V X y in X, there is a disjoint from U P-open set such that either U and a x € U, y € V € V, y € U. Definition 2.6: (x, P,Q) is pairwise (or pairwise Hausdorff) iff for each pair of distinct points of X, 0-open there is a disjoint from U P-open set such that U x € U and a and x Q-open set and y V y € V. This definition was originally given by Weston [I8]. Pairwise Tg Tg implies pairwise implies weak pairwise implies weak paiirwise Tg, are not equivalent, for if T^. and weak pairwise Clearly pairwise but Example 2.1 shows the two a < b, any L-open set contain 7 ing b also contains also must contain pairwise pairwise a b. and any R-open set containing Reilly [l6,Example 3-2] shows that is not equivalent to pairwise Tg Tg and that a space is not simply a pair of spaces [16,Example topological The following example shows weak pairwise is not equivalent to weak pairwise Example 2.2: Consider (O) where topology and Q R - {x} contains y but not x x is the cofinite R. but not . (R,P,0) however, since any T^. For y, P-open set which contains Example 2.1 shows weak pairwise T^. and weak pairwise Example 2.2 is pairwise Tg T-j^ Example 2.3: Let points T^, 1,2. y. does not imply (as it is T^. in both The following does not imply weak pairwise T^. X = {1,2,3], P = {0,X,{1,2},{3]} and Q = {0JX,[1], {2,3}}pairwise x does not imply pairwise topologies), but is not weak pairwise example shows pairwise so is not weak pair- must contain points of any interval which contains pairwise x p y, and contains is weak pairwise Tg, P is the usual topology on ^x - ^ ^ ^ ^ ^ J X + • ^^ ^ ^ ^ wise a It is easy to check that but not weak pairwise T^, (X,P,Q) is for consider the 8 Davis [1] has given for topological spaces regularity axioms such that i = 0,1, or and 2. together are equivalent to We extend this to bitopological spaces. Definition 2.7: given a a Q-closed set P-open set (Q)R Q (X,P,Q) U is A P and (Q)R lent to pairwise T^. (X,P,Q) x ^ A. open set such that and x ^ U U : y€A y We define iff it is T^. For each y € A, (Q)R . P. R^ R^. is equiva Let A there exists a y c U . x ^ TJ . Thus y y thus we have (P)R^ Since pairwise we prove pairwise P-closed and U y A c U. and pairwise be pairwise implies pairwise iff there exists with respect to Q Pairwise Let and is pairwise Proposition 2.1: Proof: ^ U x ^ A, Q in the same way. Definition 2.8: (X,P,Q) Q with respect to and a point such that with respect to with respect to (PjR^ (P)R be Q- U U =) A ySA ^ follows in a similar manner. Conversely, suppose pairwise containing T^, x x, y € X, x ^ y. suppose first that we have a but not set, so there exists a y. Then Q-open set x ^ X - U, V Since we have P-open set a such that U P-closed x jc V and 9 V Z 5 X - U . Y € V, X ^ V and we have the desired result in this case. Now suppose we have a y x. but not P-open set y^ U y ^X - V U such that y ^ U (X,P^q) and X - V C U . x€U, such that U fl V = 0 and if x ^ {y} , then there is a Q-open set V such that iff for P-open set V {y} containing Q-closed so there is a is pairwise ^ { y t h e n there is a X set which is V and we have the desired result in this case also. Definition 2.9: if Q-open set and x 6 V U and a and P-open set U D V = 0 and x,y € X, x € U Q-open {y c U U and a and c V. Proposition 2.2: Pairwise equivalent to pairwise Proof: Pairwise and pairwise Tg. implies pairwise [y}^ = [ y = [y] is in pairwise and since spaces, pairwise R^ is obvious. Conversely, if wise R^ open set {y}^ c V, x ^ y, then condition, there is a V such that that is x ^ {y}^ P-open set U fl V = 0 and x € U so by the pairU and a Q- and y € V. The next two definitions are due to Kelly [8]. Definition 2.10: P Let (X,P,Q) is regular with respect to be a bitopological space. 0 if for each point x € X 10 and each P-closed set P-open set U and and X Ç U and a A Definition 2.11: Rg) if P A such that Q-open set V x ^ A, there is a such that U fl V = 0 V. c (X^P^q) is oairwise recrular (or pairwise is regular with respect to lar with respect to Q and Q is regu P. Reilly [l6,Proposition 4-3] gives the following equivalent formulations of Definition 2.10. Proposition 2.3: If (x,P,Q) is a bitopological space, the following are equivalent: a) P is regular with respect to b) For each point ing X, x Ç X there is a and 0- P-open set P-open set U' U contain such that X € U' c ÏT^ c u. c) For each point that X X ^ A, € U Proof: and are P-closed set P-open set U A such such that n A = 0. (X,P,Q) is pairwise Rg, then it is R^. Let x, y € X. Q-open sets U n V = 0 If and there is a and Proposition 2.4: pairwise x € X and P-open and U x c V If and V and Q-open sets x;^ [y there are P-open respectively such that {y}^ c U. U and If V x [y]^, there respectively such 11 that U n V = 0 and Proposition 2.5: pairwise Proof: X x € U If and (X,P,q) Let A open sets be and R^, Similarly,, (X,P,0) is (o)R^ we have pairwise R^. Definition 2.12: (X,P,Q) R^ x ^ A. there exist y G A, P-open and fl A c U U and y€A ^ (p )R. with respect to with respect to is weak pairwise (X,P,0) is pairwise Proposition 2.6: Pairwise Rg T^. and pairwise (X,P,Q) Let be pairwise X = y. Since singleton points are spaces, U respectively and and V P; 1% Q. hence iff it is T^. iff it is pairT^. and x, y € X P-closed and Thus there exist which are P-open and and where Q-closed x ^ {y}^- x = fx}^ c v = 0 is equivalent to pairwise Proof: in pairwise Q- Then {pairwise regular) and pairwise disjoint sets For (pairwise regular) and weak pairwise Definition 2.13: Rg then it is respectively such that € V and {y} c u . ^ ^ X JC U U , and (x, P,Q) is y€A y wise is pairwise Q-closed and X pairwise c V. R^. X {yso by pairwise and {y} Q-open y = {y}^ c U. 12 Definition 2.14: [I5] (Y,S,T) from A function f from (X,P,Q) into is pair continuous iff the induced functions (X,P) into (Y,S) and (X,Q) into f (Y, T) are continuous. Remark: Lane [10] proves a number of results for semi-continuous P-lower Q-upper semi-continuous functions which we restate in terms of pair continuous functions into (R,R,L): i) If f and (R,R,L), ii) If f + g is also. and g > 0 are pair continuous functions (ft,R,l), f and (S,R,l), iv) If f then are pair continuous functions into then f > 0 If into iii) g g then so is f • g. are pair continuous functions into then min(f,g) and (-f) Definition 2.15: is pair continuous into R and A space regular iff for each L f(x) = 1 (X,P,Q) P-closed set B and (ft,L,R). is pairwise completely A f(A) = {0}, and each point (ft,R,L), have been interchanged.) there is a pair continuous function closed set are also. is a pair continuous function into (Note that such that max(f,g) y ^ B, and each point x ^ A, f : (X,P, Q) — ([0,1],R, L) and for each Q- there exists a pair 15 continuous function g(y) = 0 and as pairwise g : (X,P,Q) g(B) = {l}. ([0,1],R,L) such that We also refer to this property R,. Definition 2.16: (X ?,q) J is weak pairwise T ^ (or weak 5 pairwise Tychonoff) iff it is pairwise completely regular and weak pairwise Definition 2.17: T^. (X,P,Q) is pairwise T ^ iff it is pair- 5 wise completely regular and pairwise T^. It is easily seen that weak pairwise completely regu lar implies weak pairwise regular, and weak pairwise implies weak pairwise T^. Also pairwise completely regular 4 implies pairwise regular and pairwise wise T, implies pair- . Definition 2.18: P-closed set there is a U 4 T -, A (x,P,q) and Q-closed set P-open set disjoint from V property as pairwise Definition 2.19: weak pairwise is pairwise normal iff for each V B disjoint, from containing containing A. B and a A, Q-open set We also refer to this . (x,P,Q) is weak pairwise and pairwise normal. iff it is 14 Definition 2.20: (X,P,Q) wise is pairwise iff it is pair- and pairwise normal. Reilly [16,Proposition 5-2] proves the following theorem on equivalences of pairwise normality. Proposition 2.7: The following are equivalent a) (X,P,Q) b) is pairwise normal. For each P-closed set containing that c) A, disjoint from d) For each disjoint from A and a such that 4 and pairwise fl Definition 2.21: space and A and T,, such and Q-closed set A and U Q-closed set P-open set Q-open set B U V B con containing A = 0. implies weak pairwise implies pairwise T ,. 4 [11] Let (X,P,Q) be a bitopological B pletely separated from subsets of B f(A) = {0} and X. Then with respect to exists a pair continuous function such that A there is a Obviously, weak pairwise T -, U there is a Q-open set —P such that U fl B = 0. A B 0-open set V A, P-closed set taining Q-open set c V. P-closed set containing and there is a A c u c For each A A Q is P-com- if there f : (X,P,Q) — ([0,1],R,L) f(B) = {1}. A is Q-completely 15 separated from B with respect to pair continuous function that f(B) = {0} and Notice that if with respect to from A into ([0, 1],R, L), f : (X,P,Q) -* ([0,1],R,L) is then with respect to ([0,1],L,R). then B P. is Also if 1 - f f and each B Q-closed subset [8] If (x,P,Q) P A of Q from each B of X X is from each point B is Definition 2.22: (X,P,Q) [11] into Q If from f (R,R, L), P-zero set with respect to Q 0-zero set with respect to P. with respect to P X Q-closed, then rated with respect to spect to is pairwise normal and are disjoint subsets of closed and from is pair- X - B. Theorem 2.1: and (X,P,Q) P-closed subset Q-completely separated with respect to of is pair continuous is pair continuous into It is clear that a space X - A B Q-completely separated P-completely separated with respect to point in such P-completely separated from wise completely regular iff each is if there exists a f(A) = [ij. A Q, P Q A is A is P- P-completely sepa B. is a pair continuous function then and fx 1 f(x) <0} is a {x I f(x) >0] is a We usually call P-zero sets and Q-zero sets. such that A P-zero sets Q-zero sets with re 16 clearly are P-zero sets are Q-closed. Lane [10] proves that any {x I g(x) =0} also of the form ous into P-closed and (R, R, L) and where g Q-zero sets P-zero set is is pair continu g > 0. The following three propositions are due to Lane [10]. Proposition 2.8: respect to and a a Q If from A B, Q-zero set is P-completely separated with then there exist a such that Q-neighborhood of A and B^ If A is a then A is P-zero set A^ f! B^ = 0 is a and A^ is P-neighborhood of B. Proposition 2.Q: Q-zero set B, respect to Q from Proposition 2.10: regular iff the sets and the P-zero set disjoint from the P-completely separated with B. The space (X,P,Q) is pairwise completely P-zero sets form a base for the Q-zero sets form a base for the P-closed Q-closed sets. 17 Figure 2.1 indicates some implications proven in Chapter II concerning separation in bitopological spaces. PR WPT PT, Î PT, WPT, I WPT, PTg PT^ Î PT PT^ Figure > WPT^ 2.1. Implications concerning separation The letters weak pairwise T^ PTo and WPTo respectively. refer to ^ pairwise T^ o and 18 III. AN INTERNAL CHARACTERIZATION OF PAIRWISE COMPLETE REGULARITY In this chapter we give an internal characterization of pairwise complete regularity^ that is, a characterization which does not depend on an outside space, namely, the real numbers, as in the original definition and the character ization given by Lane. This generalizes a result of E. F. Steiner [l?]. Definition ^.1: 5 a family of sets. Then for each exists and A c S" C € Q A c X - C a family of (X,P,Q) be a bitopological space and P-closed sets and (3,Çj Definition 3.2: Q Let Q a family of Q-closed is called a pairwise-normal pair iff and BEG D € 5 and and such that such that A n B = 0, there X - c n x - D = 0 B c X - D. Let 5 be a family of Q-closed sets. P-closed sets and Then is called a pairwise-separating pair iff i) and ii) hold; i) If F A € Q is and P-closed, B € J? x / F, such that then there exist x Ç A, F c B and A n B = 0. ii) If F A c 2F is and A n B = 0. Q-closed and B f G x ^ F, such that then there exist x € A, F c B and 19 Theorem 3.1: A hitopological space (X,P,Q) is pairwise completely regular iff it possesses a pairwise normal pairwise separating pair. Proof: We prove that the family and the family Q. of all 5 A is P A € 3, B € Q P-zero sets Q-zero sets form a pairwise normal pairwise separating pair Let of all and (3,Q). A N B = 0. By Proposition 2.9 completely separated with respect to Q from A and B. Thus there is a pair continuous function f :(X P Q) - ([O 1],R,l) J f = 1 J on B. Then D = -^x I f (x) _< f f = 0 C = ^x| f(x) > € Q € 5 X - C nx- D =0 if such that J and on and are the required sets such that A c X - C is pair continuous, then and B c X - D, f - is also. since Thus is a pairwise normal pair. Now assume F is P-closed and space is pairwise completely regular, separated with respect to Q Proposition 2.8 there exist a zero set such that A^ H from x ^ F. P is Since the P x € X - F. P-zero set = 0, A^^ Thus i) of Definition $.2 is satisfied. A^ F completely Thus by and a and Q- x € B^. Similarly we could show ii) is satisfied. The proof of the converse uses a generalization of the 20 method used to prove Urysohn's lemma. (X,P,Q) has a pairwise normal pairwise sepa Assume rating pair Since (5,Q). (S*, q) and € Ç such that F^ € 3 2 F c X - G, o X he a x € F^ (3,%) since and F Q-closed set and is pairwise separating, there are F^ n G^ = G^ € Ç Let and F c G^ x ^ F. F^ € 5 and is pairwise normal there exist such that X - G^ H X - F^ = # 2 and 2 and 2 G, c X - F,. Thus 1 _i 2 2 € F^ c X - G^ c F^ c X - Gj^. Now F^ H G^ = 0, so again 2 2 by pairwise normality of in 3 and and 2 (3, Q) there are F^ and G^ 4 4 respectively, such that X-F^nx-G^=0 Q, F^ c x - G^ fI F^, G^ and 4 such that G^ c: x - F^. 2 è I F, c X - G^, ? and ¥ 4 Similarly we get sets ' I G, c: x - F_. Thus we now have € F_ c X - G, c F, c X - G, c F, c X - G, c F, o 11 11 1 1 4 4 2 2 4 4 By continuing this process we get collections X and ^ Q between F O and 1) D such that for f : X -» [0,1] f(x) = 1 for by x € G^. X - G,. 1 ^^ is the set of dyadic rationals i, j € D d X - G . C F . c x - G . c F . c x - G ,. 11 ] ] 1 function and 0 (where C and i < j, Now define a f(x) = inf{t € D | x € X - G^} We show that f is a pair 21 (R,R,L), continuous function into on {x} and 1 on We first show X € f~ ^ ( - o o ,a); 0 F. f : (X,Q) then dyadic rational which is obviously is continuous. f(x) = t < a. t' € D Let Thus there exists a such that f(x) = t < t' < a. U (X - G ) so f"^(-oo,a) c U (X - G ). t<a t<a t€D t€D Now if X € U (X - G, ), then x € X - G . j t < a . t<a o t€D f(x) = inf{t € D 1 X € X - G^} = t^ < a, so x 6 f ^(-oo,a) X € X - G. . c and U (X - G.) c f ^(-oo^a). t<a t€D f ^(-00,a) = U (X - G. ), t<a t6D and hence is Q-open. Thus which is a union of Before beginning the proof that Q-open sets f : (x,P) — (R,R) is continuous, we prove the following useful fact: f(x) < a If X assume a iff € X - G^ for all for all t > a, t € D then such that f(x) < a; f(x) = inf{t €D|x €X- G^} > a. t^ 6 D such that contradiction. a X € X - G^ Conversely, if such that i,i $ D, i < i t.o > a and f(x) < a x for Then there is X - G^ , o a and there exists x ^ ^ X - G. t , then o f(x) = inf{t € D I X € X - G^} > t^; for by construction, if t_o € D f(x) > t^ > a t > a. iff and X - G^ c X - Gj. But this is a 22 contradiction since f(x) < a. From the above we have Obviously Pi (X - G. ) c fl (X - G )^. t>a ^ t>a ^ t€D t€D the other way, let exists f ^(-M,a] = s G D c (X - G, ^ r € D such that ® f : (XjP) -* (&,R) Hence the is 0 on X F Then there •p Pi (X - G. ; t>a t€D Now is P Q P. Thus is continuous also. we have defined is pair continuous and 1 F being respect to f and separated from to c X - G^. ^ r > a. H (X - G c X - G t>a ^ t€D for all r > a, so H (X - G c H (X - G ) and we t>a r>a tcD r6D have n (X - G = 0 (X - G ) = f-l(-œ,a]. This means t>a ^ t>a ^ t€D t€D that CP To show containment be such that r > s > a. 0 (X - G. ). t>a t€D on F, so {x} with respect to Q is completely separated from In a similar fashion, each Thus Definition "3-3: (X,P,Q) completely which is equivalent completely separated with respect to it excludes. P {x} with P-closed set Q from points is pairwise completely regular. [1?] A family of sets is called an inter section ring if it contains all finite unions and countable intersections of its members. 23 Definition 3.4: Q a family of elements of Let 3 be a family of Q-closed sets. 3 elements of A sequence is called a nest in iff there exists a sequence Q such that Definition 3.5: P-closed sets and 3 [X - of with respect to of complements of X - c cx- G^ c F^. A family 5 of generated with respect to Q iff each member of intersection of a Theorem 3.2: 3 is If P-nest in 3 and G P-closed sets in P-nest 3 is the 3. Q are intersection rings where P-nest generated with respect to Q-nest generated with respect to P, Q then and (3,Ç) Q is is a pairwise normal pair. Proof: Let exist nests A € 3, B € Q {A^} B = n E . Let n and and [B^] {X - G } n and such that and sequences of complements. Choose a member Vtfhich forms U which forms V. and a member If X - Gn ^ n X - Am = 0 . X - F^ n X - B^ = 0. fX - F } ^ n Define V = U {X - F^ n X - A^}. U n V = 0. A D B = 0. Then there A = n A^ and be the associated U=UfX-G^ n x - B^} We first show that X - G^ H X - B^ of the union X - F_ m H X - A.m n > m, X - G^ n c A^; m' of the union hence If n — < m' , X - F m^ c B n' ^; Thus in either case, (X - G^ n X - B^) n (X - F^ n X - A^) = 0; hence hence 24 U n V = 0. Now we show U and V contain A and B respectively. Let some X n ° € A. Since such that implies x € An A fl B = 0 x B , "o for all Similarly A c U. Thus B = fl x € X - B . "o Thus Q. and V x € U and we have U is the complement of a is the complement of a member of n{G^UB^}ÇQ U since the desired result. Similarly complement of a member of is the set of which is If 5 V but 5. U B^ € Q is an intersection ring. Q and we have could be shown to be the 5. is the set of Q-zero sets, then 3 P-zero sets and Q is an intersection ring P-nest generated with respect to intersection ring which is to Q is the complement of a member of Proposition 3.1: xGA x € A.n +1 c X - G_n^. o o U = U { X - G ^ n x - B^3 = X - n [G^ U B^}, and so there is B c v. It remains to show that member of that is n, ' so that X € X - B n x -G . ^o "o Thus and Q and Q is an Q-nest generated with respect P. Proof: Let functions Fg € 3. f^ and F^ = {x I f ^ f x ) = 0 } fg > 0 [10,p.21]. f^ and Thus Then there exist pair continuous into (R, R,L) such that Fg = [x 1 f2(x) = 0 } and f^ > 0, Fj^ U Fg = {x 1 f^fxjfgfx) = 0}. 25 Since is pair continuous [10,p.g], {x I f^(x)f2(x) =0} is a P-zero set and U Fg € 3. Lane [10,Proposition 2.22] shows that under countable intersection. is F € ?. respect to Q tinuous function f > 0 F =-!xif(x)<—L n L \ / n; to show X - [F^} F = f; F^. such that Then clearly 1 •K 1 [F^] ^ îîrfTà; - Q-2ero set since (S.,L R) J LxO,p.$j. - f 3 with F = {x I f(x) = 0}. F = H F , n Q. Let so it remains Let Then clearlv X - G - c F -, c X c F„. If we can show n-RX n-rl n n Q-zero set, we have the desired result. r in There is a pair con is a nest with respect to = -.x 1 f(x) < 3 Q. We wish to find a nest such that is closed It then remains to show P-nest generated with respect to Let 3 ^ G n is a "hich is a is pair continuous into 26 IV. PRODUCTS AND BICOMPACTIFICATIONS Given a family of bitopological spaces, there is a verynatural way to define a product of these spaces. Definition 4.1: ffX ,P ,Q )} be a family of ^ a a a acA bitopological spaces. The product of this family. T is the bitopological space where irP TT (X ,P ) a€A for Let represents the usual product topology for and ttQ represents the usual product topology TT (X ,0 ). oGA G a The following three propositions are immediate conse quences of the definition and corresponding results for single topological spaces [5]. Proposition 4.1: (X ,P ,Q ) takes The proiection ^ TT : ir (X ,0 ) -* a a' a' is pair continuous and pair open (that is TP-open sets to P^-open sets and TT TrQ-open sets to Q^-open sets). Proposition 4.2: product A function TT (X ,P ,Q ) a€A " ct a € A, TT 0 f a f from (X,P,Q) into a is pair continuous iff for each is pair continuous. ^ Proposition 4.3: Let tinuous functions where ^a€A ^ family of pair con fn : (X \ ny,P ,Q *ry') -• (Y ^ rt,S ^ ,Tn ^). Then 27 TT f : (X ,P ,Q ) - TT (Y ,S .T ) a€A G o€A ° ° oGA ^ a a is pair continuous. Since the product of compact spaces is compact in usual topological spaces we would hope to get a similar result for bitopolocical spaces. Fletcher [5] and Kim [9] have given definitions of pairwise compactness. a bitopological space (X,P,Q) is called pairwise compact if every pairwise open cover of where a pairwise open cover P U Q such that C A P P with respect to C X has a finite subcover, is a cover by sets from contains a non-empty set and contains a non-empty set. of According to Fletcher C A O Kim defines the adjoint topology V € Q by p(v) = {p,x} U {U U V I U € P}, and he defines a bitopo logical space to be pairwise compact if for every non-empty element V of pact for every non-empty element Q U P(v) and of is compact Q(u) P. is com It is clear that the real line with the right ray topology and left ray topology (ft,RjL.) definitions. is pairwise compact under both of these However x (ft,R,l) is not pairwise compact under either of these definitions, for consider the cover of a x ft by [(0,oo) U [(-œ,n} X (-œ,1)]^^^ x U {(-oo,l) which is pairwise open. x Obviously it can not be reduced to a finite cover, so the product of pairwise compact spaces is not pairwise compact under Fletcher's definition. Now 7rL((0,oo) x (O^oo)), the adjoint 28 topology of TTL with respect to not compact either so (0, o o ) x (0, o o ) , (R, R,L) x (R,R,L) compact according to Kim's definition. clearly is is not pairwise Olie definition of bicompactness we shall give has the property that the product of bicompact spaces is bicompact. Our definition also has the desirable property that every pair continuous function from a bicompact space into (R, R,L) is bounded, which is not the case for either of the previously mentioned defi nitions, as one can see by considering the identity function on (R,R,L). Definition 4.2: A bitopological space (X,P,Q) compact iff every pair continuous function from into (R,R,L) Definition 4.3: is pseudo(X,P,Q) is bounded. A weak pair T -, space is pair real com- ^2 pact iff it is pair homeomorphic to the intersection of a 7rR-closed subset with a 7rL-closed subset of a product of (R,R,L). Remark: Definition 4.3 is a generalization of a character ization of real-compactness given by Engelking [2]. Definition 4.4; A bitopological space is bicompact iff it is pseudo-compact and real-compact. 29 Example 4.1: ([0,1],R,L) ness is obvious as is bicompact, for real compact [0,1] = (-co, 1] fl [0,oo). If it were not pseudo-compact, there would exist a pair continuous function f into right. (fô,R,l) which is unbounded, assume unbounded to the Then there exists a sequence that f(x^) > n for each point with respect to the usual topology and hence x^ n. ^^n^n€N This sequence must have a cluster will also be a cluster point with respect to the gy. f~^(-oo, f(x^) + e) elements of ^ contradiction. Let (X,P,Q) be bicompact. is pair homeomorphic to the intersection of a subset C _ ttR (R,R,L ) with a where topolo will then contain infinitely many f^n^n€N' Proposition 4.4: L x^ irL-closed subset D C _ TTL Then (X,P,Q) irR-closed of a product of is contained in a product of bounded intervals with the left ray and right ray topologies. Proof: Let ^ttR h be the pair homeomorphism between ^ TT ( R, R, L). bounded since (X,P,Q) For a € A, is bicompact. bound for the map Let oh a product of bounded Proposit ion 4.5: is bicompact, then and (X,Q) (X,P,Q) are compact. and is be a lower ir oh and b an a a n c TT ([a^,b^],R,L), a€A intervals. If upper bound. a^ X Then (X,P) 30 Proof: Let 0 be a net in homeomorphisin from Then ho0 X to X. Let h represent a pair C „ fl C _ c tt ([a ,b ],R, L). a€A is a net in C FL C _ c TT ([a ,b ],R,L). TTxv TTj-i ^/'TV Co O a cA. Since the latter set is compact in the usual topology, there is a cluster point (with respect to usual topology) ho 0 in TT ([a ,b J,R, L). Since both irR and subtopologies of the usual product topology on and are closed in the usual topology. u Li y € C „ n C - and y is a ttR TTL and a irL-cluster point for a P-cluster point and a (X,P) and (X,Q) TTL Thus are Thus ho 0 h ^(y) = x Q-cluster point for for R, TrR-cluster point for ho 0. y 0, is so are compact. Bicompactness is not equivalent to compactness in both topologies, however, as the following example shows. Example 4.2: of Consider the subspace of [-1,0) U (0,1] (R,R,L) with the induced topologies. consisting This space is compact in both topologies since any cover by left rays must contain the whole space as an element and simi larly for any cover by right rays. However, if this space were bicompact it would be pair homeomorphic, with pair homeormorphism h, TT ([a^,b^], R, L;. sequence say, to an intersection Consider the sequence {h(-H)}n5N i" SrR ^ Sri, = a€A fninCN" D C^^ c 51 a cluster point Hence y and the with respect to the usual topology. will be a cluster point with respect to the ttL consider in}n€N' y topologies also. x = h ^(y). x Thus y € fl so should be a cluster point for with respect to both topologies, but if is not an TTR L-cluster point and if x > 0, x < 0 it it is not an R-cluster point. We show in this chapter that given a weak pair space T -, there exists a bicompact space such that (X,P,Q) (S,P*,Q*) of Q*-dense in is pair homeomorphic to a subspace (X^,P*,Of) where S is 4 space P*-dense and (X*,P^,Q*). Theorem 4.1: A weak pair T -, (X, P,Q) homeomorphic to a subspace of a product of Proof: Consider the family P = is pair ([0,1],R,l). of all pair continuous functions f^ : (X,P,Q) — ([0,1],R,L). Define a Wf ; (X,P,Q) - ir ([0,1],R,L) by (f(x)) = f (x). function By Proposition 4.2 prove f""^ Let have or is pair continuous. is pair continuous and x,y € X U € P, V € Q x^V f such that such that either x€V, yj^V X € U, y U and and f x 7^ y. is one-one. Then there exist x € U, y / U y € U, x ^ U. y € V, x ^ V. It remains to and y € V, Suppose we Then there exists 52 a pair continuous f (X - U) =0. ^o and (f(y)) = 0. ^o To show and f € F such that f (x) = 1 and ^o ^o Thus f(x) / f(y) since (f(x)) = 1 ^o f ^ X = f~^(y) €U. xj^X-U so there exists a pair continuous map f € F such that ^1 ( X - u ) = 0 . The set A = {y € f(x) TT ([0,1],R,L) I y and y = f(x) € A. y' = f(x') for some f^ (x") =0, TrR-open. weak pair If a contradiction. 4.6: f (x) = 1 ^1 fl f(x) > x'€X. Similarly for Proposition U € P, y € f(u) is pair continuous let y' If is and irR-open in is any other point in y' ^ f(u), x' ^ U Thus V F 0, so A, so y' € f(u), f(u) is f ^ is pair continuous. The product of weak pair T^ spaces is T^. TT fx ,P ,Q ), x v where each acA a a. (X_,P ,Q^) is weak pair T,. If x ^ y, there exists ^a a a 1 / ^ some a € A such that x ^ y , so there exist ° *0 Gg/ Proof: u^o € Let x,y € -a and y e u , x j ^ V Go Go Go € 0,^ such that ^ or ê x^^ € s o suppose t h e latter. i% ^ Then and 53 jz € X IT (X ,P ,Q ) \ Z but not y and weak pair Proof: is TTQ-open and contains jz € TT (X ,P ,Q ) ! Z ^ a€A a a open and contains y Proposition 4.7: 1 € V but not x, is TTP- the desired result. 4 The product of weak pair T -, spaces is T . 5? We have just shown in Proposition product € U } a^j IT (X ,P jQ ) a(A = ° TP-closed subset and is weak pair x^ ^ F. 4.6 T,. 1 X - F is that the Let F be a TP-open and 0 € X - F. There exists a base element B 3 Xo € B c X - F. We wish to find a function which is X at X o B = n IT (U ) i=l °'i^ °'i X oa- and € U a. for each such that 0 on where for each i, X - B. U °^i is a base element are all P °^i open. ) = 1 so by weak pairwise and f.(X F. c 77 : 77-(X jP ,Q ) ~* ([0,1],R,L) 1 a^ ^a a a ^ Let B T , condition -2A 2 there exists a pair continuous function f^ f.(x^ i Since - U ) = 0. pair continuous and ç(x ) = 1 and if y 1 < j < n — — so for some j Then is pair continuous. ^ g(x) = min{f. o - (x) I i = 1,2,...,n}. "i U 1 Then y )e B, f.(y' ") = 0 ]^ ^ g then so is 54 g(y) = 0, hence g(X - B) = 0 . tinuous function X - B =3 F, g which is 1 Thus we have a pair con at x and 0 on the desired result. Theorem 4.2: The product of a family [ bicompact spaces is bicompact. Proof: pair From Proposition 4.7 we have the product being weak T . It remains to show that the product is pseudo- compact and pair real-compact. Since each each a is bicompact, we can find for a pair homeomorphism the intersection of a such that Consider the map h = pair continuous. onto 7r(C„ aSA from onto TrR-closed set closed set By Proposition 4.5, h^ ir h a€A h Since n C T ), n defined on with a c [a^,b^],R,L) TT (X ,P ,Q ). a€A ° ° is pair continuous and h FL- h ^ is obviously one-one and is h is it remains to show that 7r( C _ P I C ) is the intersection of a TrR-closed set a6A with a TL-closed set in order to show real compactness. T a €A Engelking [2,p.71] a cA a €A ir C _ is TrR-closed and v C ^ is aSA. a€A Trii-closed in tt ( rr (R,R,L)1. a6A X€Aa If TT (X ,P ,Q ) is not pseudo-compact, there exists o€a/ a' a' 55 a pair continuous function f on the product which is unbounded (assume to the right). fx } such that f(x ) > n n ntN ^ n induced sequence j^h Thus we can get a sequence for each n. Consider the " =,rla) = which is compact in the usual topology since it is the product of compact spaces. It has a cluster point respect to the usual topology, but point with respect to the gy, so y_ € TrC„ ° a€A ^o ^ fl t( X ,P ,Q ) ^ a a x^, hfx^) = y^ and ^^n^n€N topolo x^ is a ^ S) an obvious contradiction. is pseudo-compact and hence a TTL Thus there is an TrQ-cluster point for contains infinitely many with is also a cluster topology and the irC^. a€A s^-^h that TTP- and a Thus -R y^ y^ bicompact. The following corollary is immediate from the Theorem and Example 4.1. Corollary 4 .1: Proposition 7 r ([0,1 ], R, L) A 4.8: is bicompact. P-closed (or Q-closed) subset of a bicompact space is bicompact. Proof: Let (X,P,Q). ^ A be a (X,P,Q) ^ P-closed subset of a bicompact space is pair homeomorphic by RjL). Thus h(A) h is to TrR-closed, 36 so h(A) = (h(A) N N closed set with a the intersection of a /rL-closed set. If A TTR- were not pseudo- compact, there would be an unbounded pair continuous function f on A. Assume it to be unbounded to the right. Then there exists a sequence f(x^) > n for each n. Proposition 4.5 (X,p) P-cluster point x^. f~^(-oo, f(x^) + e) diction. Thus Since ^ (X,P,Q) is compact; hence Since A is Q^) such that is bicompact, by^^n^n€N P-closed, contains infinitely many (A, Proposition 4.Q: ^^n^n€N & x^ € A x^, so a contra is bicompact. The intersection of a P-closed subset with a 0-closed subset of a bicompact space is bicompact. Proof: Let Cp P-closed and is (X,P,Q) pair homeomorphic by be bicompact and is h A = Cp fl Q-closed. to Since h(Cp) is -R-closed and (h(Cp) P. C_^) n (h^Cg) n closed set with a (X,P,Q) is n h(A).h(CPNCG) = h(Cp)nh(CQ) = (h(Cp) n and since where n (h(Cg) n ^(^ ) Q is TrL-closedj is the intersection of a —L-closed set and A irR- is real compact. Pseudo-compactness follows by extending slightly the same argument used in the proof of Proposition Definition 4.4: A bitopological space called a bicompactification for 4.8. (X*,P*,Q*) {X,P,Q) provided will be 57 (X*,P*,0*) is bicompact and to a subspace dense in S of X* (X,P,0) where S is pair homeomorphic is P*-dense and Q*- X*. Theorem 4.3: If (X,P,Q) is a weak pair T, bitopologi- ^2 cal space, then there exists a bicompactification for (X,P,Q). Proof: Since (X,P,Q) Is weak pair homeomorphic to a subspace TT ([0,l],R,l). a€A A^^ n A^^, N A A T it is pair of a product obviously is —R-dense and -rL-dense in and by Proposition 4.9 and Theorem 4.2, is bicompact in ir ([0,1],R,L). Thus a desired a€A bicompactif ication would be (A^^ fl A^^,7rR,TrL). The question of whether every pair continuous function from a pair T ^ space into (ft,R,L) can be extended to a pair continuous function on the bicompactification is not known. 58 V. GENERALIZED QUASI-METRICS AND QUASI-UNIFORM SPACES The concept of a uniform space [7] is a generalization of a metric space. void family K i) A uniformity for a set of subsets of each member of 'V. X x x such that contains the diagonal ii) if U Ç. M, then U ^ iii) if U 6 ^, then Vov c U iv) if U and V v) if U € "U and and is a non- X for some are members of ^ , U c V c: X x X, V c then then A; ; U fl V € t/; V cî/. A quasi-uniform space is a non-void family "U sets of XXX satisfied. of sub such that i), iii), iv), and v) above are The symmetry condition ii) is deleted. A quasi- uniform space is a generalization of a quasi-metric space. Recall that a quasi-metric is a metric except that the usual symmetry condition is deleted. A uniformity ^ topology sets 6 U.) T( of X for so ^ X induces in a natural way a where T(^) x € U[x] c G. is the set of all sub x € 9, there exists By the symmetry condition, are the same, where ^ T(^) = T(^^). = {U ^ 1 U and When ^ is a quasi-uniformity,^ ^ need not be the same as %( ; T(t(). X such that for each U € K such that ^ and for thus Thus a quasi-uniform space natural way a bitopological space T(tt ^) (X,t() may differ from induces in a (X, T(t()_, T(tC^)). 59 In [6], G. K. Kalisch showed that spaces possessing a generalized metric are uniform spaces and uniform spaces are We generalize these results to generalized metric spaces. generalized quasi-metrics and quasi-uniform spaces. In what follows, we let I be a non-void partially ordered directed set, which means that for all i,j,k in i), ii) and iii) hold: i) i < j and j < i implies i = j ii) i < j and j < k i < k iii) i,j € I i < k We also let and < implies implies there exists and G = and where I, such that j < k. R is the set of real numbers is the usual ordering on in both ft k E I S2. Note that we use _< relying upon context to distinguish the usage intended. Definition 5.1: For exists a j € I such that exists a k f I, k < j, all i < k, i) ii) g^ <' g^ g^(i) > ggCj)^ such that iff if there then there g^fk) < g2(k) and for g^fi) < ggfi)• Definition 5.2: function from g^^g^ € G, Let S x S S be a non-void set and G such that for all 6(s,t) >'0 6(s,t) = 0 » s = t 6 a s,t^r € S, I. 40 iii) Then 6(s,t) <' 5(s,r) + ô(r,t). (S,5) will be called a generalized quasi-metric space. Now let ^ denote the collection of all sets of the form N(e ji^, ig,..• where e = {g € G lg(ij)l < e, j = 1,2,...,n}, is a positive real number and a finite subset of define I. For {ij^, i^,..., i^^} N(e;i^,ig,...,i^} € d., ig,...,1^) = {h € G 1 h <' g g € Nfeji^,ig,...,i^)}, is and let 8 for some denote the collection of all such sets. We can then form a bitopological space defining P and Q as follows; U^*(s) = {t 1 6(s,t) € N*} and For call P by s € S, N* € iS, let U^^(s) = {t I &(t,s) € N*}. It is easy to show that the collections {U^^(s)3 (S,P,0) {U^^(s)} and form neighborhood systems for topologies which we and Q Now let respectively [?]. = {(s,t) i t € U^^(s)} collection and consider the order to show this collection forms a base for a guasi-uniformity for S, it is suf ficient to show the following [10,p.65]: i) ii) 0 ^ ^^N*^N*€lS For ^N* ^ iii) A c N^, € S, there exists ^ ^N* for all N* € i® N* € IS such that 41 iv) For N* € fi, o for some N| € B. i) is obvious. For N|,N^ € G N* C= N* n N|. that satisfied. clearly there is Thus H c iii) is obvious. N* such and ii) is Before proving iv), we prove a lemma. Lemma 5.1: Proof: p <' q Suppose and r <' t k < j say, such that q(i) < p(i). If and r(i) < t(i) for all such that such that Then we have (p + r)(n) < (q + t)(n) hence i < k, n and i < n (p + r)(i) < (q + t)(i) for p + r <' q + t. netting r then there exists such that . ^ there exists an for all and for all Now to prove iv) above, we show = we are However, if there r(m) > t(m), r(n) < t(n) there is i < k, (p + r)(i) < (q + t)(i) an i < n; or p <' q, and for all m < k all Since (p + r)(k) < (q + t)(k). r(i) < t(i). Then q(j) > p(j) q(k) < p(k) exists an n < m so q(j) > p(j). finished since then i < k p + r <' q + t. (q + t)(j) > (p + r)(j). g ( j ) + t ( j ) > p(j) + r(j); t(i) > r(i), implies such that - \ * * * n' (s, t) € (s,r) 6 ^ . . \ ^ 4-2 and (r,t) € Ihus &(s,r) € N*(e/3; i^,...,i^) so there exist 6(s,r) < g^ and 6( r , t ) € N*(e/5i € N(e/3; i2_j • • • j ij^) and &(r,t) <' g^. G(s,t) <• &(s,r) + 6(r,t) < 9l + 92 « .. . , such that By Lemma $.1, g^ + g^; but as 1(91 + S2)(i]^)l = 191(1%) + ggC^k)! 5 < e/5 + s/5 < Ei therefore + 192(1%) I 5(s,t) € N* ( e ; i^^, ..., i^^) and (s,t) e Hence we have shown i) - iv) are satisfied; therefore is a base for a quasi-uniformity "K We now show that Let such that 3 € P P = T{^) and s 6 9- s € U^^(s) c 6. Ug,*[s] = [t I (s,t) € s € U^*[s] c 8; s F 8, Since so U € We have shown Q = T(1^ ^)• Then there exists But 6 6 T(% . U^^(s) € "U and Now if U € t( there exists Thus S x S. = ft i 6(s,t) e N*3 = U^*(s). then there exists Uj-^ c U. and on 0 € T(W) such that where Similarly N* € B Q = T(î/ and s € U[s] c 8. s c U^*(s) = U^^[s] c U[s] c 8 P = T(t(). Thus such that and 8 € P. . So far in this chapter we have taken a generalized quasi-metric space uniformity (s^i) and we have exhibited a quasi- such that the bitopological space induced by the generalized quasi-metric & is the same as the ; 45 bitopological space induced by the quasi-xiniformity X . Ihus we have proven the following: Theorem 5.1: Every generalized quasi-metric space is a quasi-uniform space. We now show the following: Theorem 5.2: Every quasi-uniform space is separated (nt( = A), (X,*U)^ vAiere is quasi-uniformly isomorphic to a generalized quasi-metric space. Before beginning the proof of this theorem, we prove a generalization of a theorem from Kelley. The next three definitions are due to Reilly [16]: Definition 5.2: spaces. Let (X,W) The function f : (X,t() continuous if for each that (x,y) € U Definition 5.5: spaces. f and If f ^ V 6 implies Let f:X - Y and (Y, V) be quasi-uniform (Y,y) is quasi-uniformly there exists U € (/ such (f(x),f(y)) € V. (X,^) and ,V) be quasi-uniform is a one-to-one function such that are both quasi-uniformly continuous, then f is a quasi-uniform isomorphism. Definition 5.4: form space (X,î^ A real valued function from the quasi-uni is quasi-uniformly upper semi-continuous (q.u.u.s.c) if for each e > 0 there is a U 6 Î/ such that 44 (x,y) 6 U implies f(y) < f(x) -f e. f is quasi-uniformly lower semi-continuous (q.u.l.s.c) if for each V Ç is a such that Theorem S.3: (x,y) 6 implies e > 0 there f(y) > f(x) - e. is quasi-uni- Each quasi-uniform space formly isomorphic to a suLspace of a product of quasipseudo-metric spaces. Proof: By Reilly [16, Theorem 2.13], ^ family P of all quasi-pseudo-metrics which are q.u.u. s.c .-1 xt/). (X X X, u~ on f ; X -• Z Xp Let be defined by Z = 7rf(X,p) I p € P} (f(x))p = x for all be assigned the quasi-uniformity and Z We need to show f and and x € X. Let generated by the product quasi-uniformity. one-one. is generated by the Obviously, f ^ f P is are quasi-uni formly continuous. Let W be a subbasic element of the product quasi- uniformity for Z, that is W = {(u, v) | (u^, y^) € the quasi-uniformity generated by implies U is such that f € since 1/ is induced by (x,y) € U implies But P. U €^ Thus (f(x),f(y)) € W, U €^ and is quasi-uniformly continuous. Now to show U p}. ^ f ^ is quasi-uniformly continuous, let be a subbasic element of^ , U = [(x,y) | p(x,y) < e} 45 for some p € P, e > 0. Choose V = {(u,v) e f (x) X f (x) I (u ,V ) e u } . P P PG îhen clearly (u,v) 6 V and implies (f~^(u),f~^(v)) E U quasi-uniformly continuous. isomorphism from X Hius f(x) onto f(x) isomorphic to a subspace f f ^ is is a quasi-uniform and X is quasi-uniformly of a product of quasi- pseudo-metric spaces. Proof of 'Theorem 5.2: By the previous theorem we can let the quasi-uniform space to a subspace (X,^ of a product Z be quasi-uniformly isomorphic P = TT P , where P (i€M ^ are ^ quasi-pseudo-metric spaces with quasi-pseudo-metries which generate *U . sets of M. For m € M , where component of distance in be the set of all finite sub x,y € Z, r x P |i Let p^ define &(x,y) = (r^) for = sup(dist(x ,y )) where x is the 1J.€M ^ ^ ^ in P^ and dist(x^,y^) is the ordinary between x and n y , and n (r_) ^ m' is an MQ element of G = R & : Z X Z -* G . makes It remains to show this definition of (Z,6) a generalized quasi-metric space which is quasi-uniformly isomorphic to Obviously X 0 < S(x,y) = y, 6(x,y) =0. If (x,y) U as X for all Y3 is separated. (xfU)• x^y € Z, there is and if U € %( such that ^ is generated by 46 so there is a finite number of n U =) n U ^ . As i=l Pp^Gi there is some So p , e. ^i ^ (x,y) / U, (x,y) X k, 1 ^ k < n, (x,y) > p such that such that n n U i=l (x,y) and there is at least one , i so U ^ . %^ m such that r r 0. Hence (r^) = ô(x,y) / 0. To verify that 5 is a m' ^ m ^ ' generalized quasi-metric, we have yet to show that the triangle inequality holds. Let ô(x,y) + 6(y,z) = (r^) + (r^) = (r^+ 5(x,z) = (r^). then If r;; > r^ + r^ o o o for some and m^ € sup (dist(x ,z )) > sup (dist(x ,y )) )j€m^ ^€m^ + sup (dist(y ,z )); nat.^ " hence there is a Uq ^ m^ such that Pu (x ,2 ) > sup (dist(x ,y )) + sup(dist(y ,z )). •"^o '"^o '^o LiGno ""^o o ii€mo '"^o o Hence " contradiction. (Z,6) Thus the triangle inequality holds and is a generalized quasi-metric space. (x/U) Now it remains to show that is quasi-uniformly isomorphic to the quasi-uniform space induced by the gener alized quasi-metric space uniformly isomorphic to (Z,5). Z We have quasi- with the product quasi-uniformi- ty, so it is sufficient to show Z with the product quasi- uniformity isomorphic to the quasi-uniform space induced by 47 (Z,6). We show the identity function is a quasi-uniform isomorphism. Clearly the identity function is one-one, so it is sufficient to show tinuous. Let induced on V Z i and i~^ be an element of the quasi-uniformity by V„„/ , , \ N*(e;kj_,...,k^J 6. Then V n m = U k. and consider ° j=l ] < e ,j = 1,2,...,n}. Z. since If (x,y) € U, ^ and i then formity. m^ = subsets of 6(x,y) € N*(e;k^,—,k^) r^ = sup {p^(x^,y|_^)} 3 P-€kj therefore (x,y) « V(e;lc^,.. is quasi-uniformly continuous. Now let by U = {(x,y) € Z x Z 1 p (x,y) < e, ^ ô(x,y) = (r^) where - Let Let which is an element of the product quasi-uniformity for c V contains a base element as defined before where N*(s;k^,= {(r^) 1 r^ ^ are quasi-uniformly con U Then be an element of the product quasi-uniU contains a base element I i = 1,2,...,n} m^; m = {k.}f '• 1/1=1 then m and let ^ ^ = 1.2.•••.'>}• m be the set of all is finite and can be represented Consider then . \ N*( e;k^,...,k^j element of the quasi-uniformity induced on (x.y) € Z by a base 6. If 5(x,y) f N*(e;ki,...,k,); 48 so ô(x,y) = (r^), Thus where sup {p (x ,y )} < e; ia€k^ m ^ r^ < e hence for i = 1,2, sup {p (x ,y )} < s; k^6m hence p (x ,y ) < e for i = 1,2,...,n, which means ^i ^^i ^i fx.y) € U„„/ \ c U. Thus i~^ is quasi-uniformi- ly continuous. Ihis completes the proof of Theorem 5-2- 49 VI. LITERATURE CITED Davis, A. S. Indexed systems of neighborhoods for general topological spaces. Amer. Math. Monthly. 68: 886-895. 1961. 2 Engelking, R. 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