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Sequential space methods Kremsater, Terry Philip 1972

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. cl SEQUENTIAL SPACE METHODS by TERRY PHILIP KREMSATER B.SC, UNIVERSITY OF BRITISH COLUMBIA, 1969 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF t MASTER OF ARTS in the Department of MATHEMATICS We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA August 1972 In present ing th is thes is in p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make it f r e e l y a v a i l a b l e for reference and study. I fu r ther agree that permission for extensive copying o f th is t h e s i s for s c h o l a r l y purposes may be granted by the Head of my Department or by h is representa t ives . It i s understood that copying or p u b l i c a t i o n o f th is thes is f o r f i n a n c i a l gain s h a l l not be allowed without my wr i t ten permiss ion . Department of TT?^ 7%N^ ^^>^> The Un ive rs i t y of B r i t i s h Columbia Vancouver 8, Canada i i Abstract The class of sequential spaces and i t s successive smaller subclasses, the Frechet spaces and the first-countable spaces, have topologies which are completely specified by their convergent sequences. Because sequences have many advantages over nets, these topological spaces are of interest. Special attention i s paid to those properties of first-countable spaces which can or cannot be generalized to Frechet or sequential spaces. For example, countable compactness and sequential compactness are equivalent in the larger class of sequential spaces. On the other hand, a Frechet space with unique sequential limits need not be Hausdorff, and there i s a product of two Frechet spaces which i s not sequential. Some of the more d i f f i c u l t problems are connected with products. The topological product of an arbitrary sequential space and a T^ (.regular and T^) sequential space X i s sequential i f and only i f X is locally countably compact. There are also several results which demonstrate the non-productive nature of Frechet spaces. The sequential spaces and the Frechet spaces are precisely the quotients and continuous pseudo-open images, respectively, of either (ordered) metric spaces or (ordered) first-countable spaces. These characterizations follow from those of the generalized sequential spaces and the generalized Frechet spaces. The notions i i i of convergence subbasis and convergence basis play an important role here. Quotient spaces are characterized in terms of conver gence subbases, and continuous pseudo-open images in terms of convergence bases. The equivalence of hereditarily quotient map i and continuous pseudo-open maps implies the latter result. i v Table .of Contents Page Introduction . . . . . . . . . . . . 1 Notation . . . . . 6 .,Chap,.ter.„L: ._Sequential....Spaces 7 Chapter 2: Frechet Spaces 59 Chapter 3: Generalized Sequential Space Methods . . 81 Chapter 4: Generalized Sequential Spaces and their Properties in Ordered Topological Spaces . . . . 109 Bibliography 124 V L i s t of I l l u s t r a t i o n s Figure 1: The Graph of U(W : n e to} i n R (See Example 1.19) v i Acknowledgments The author is indebted to Dr. T. Cramer for suggesting the topic of this thesis and for his patience, encouragement, and invaluable assistance during the past year. Dr. J.V. Whittaker's ..careful reading and constructive criticisms of the f i n a l manuscript are also gratefully acknowledged. Finally, the author wishes to thank Miriam Swan for her conscientious typing work. Introduction A first-countable space i s a topological space whose open sets can be described by i t s convergent sequences alone. This i s so by either of two properties of first-countable spaces ([16], Theorem 2.8) : CI) A -set -is -open -if -and-on-l-y- i f -each -sequence-which converges to a point in the set i s , i t s e l f , eventually in the set. (2) A point l i e s in the closure of a set i f and only i f there i s a sequence in the set converging to the point. For more general spaces, i t is often assumed that sequences are inadequate and that nets or f i l t e r s must be used. There are, however, many topological spaces which do not satisfy the f i r s t "axiom "of couritabiTity and "yet 'sequences 'suffice' to determine open sets. The real line with the integers identified to one point i s an example of such a space. The topological spaces satisfying (1) are called sequential spaces and those satisfying (2), Frechet spaces. Each first-count-able space, and hence each metric space and each discrete space, i s both Frechet and sequential. Moreover, the real line with the integers identified is both a Frechet space and a sequential space. Consequently, since (2) implies (1) but (1) does not imply (2), the concepts of Frechet space and sequential space provide successive proper generalizations of first-countable space. In studying sequential spaces, one can r e s t r i c t oneself to sequential convergence. - 2 -Accordingly, since the language of sequences has many advantages over that of nets, i t i s of interest to know when a topological space i s sequential. A result due to Ponomarev characterizes first-countable T^-spaces as continuous open images of metric spaces. Analogously, S.P. Franklin {8J establishes that the sequential spaces are precisely the quotients of either metric spaces or first-countable spaces, and Arhangel' ski3C [2] asserts that "among Hausdorff spaces, Frechet spaces and only these, are continuous pseudo-open images of metric spaces". (The pseudo-open maps form a class between the open maps and the quotient maps.) In [22], P.R. Meyer extends Arhangel'skii's result by eliminating the Hausdorff hypothesis:--^ .pseudo-open images of either metric spaces or first-countable spaces. In order to obtain this result, he introduces the notions of convergence subbasis and convergence basis which, provide the foundation for studying topological spaces whose open sets are completely specified by any given class of nets. Meyer's general-ized sequential space methods are used to derive D.C. Kent's [18] characterizations of "spaces in which well ordered nets suffice." Recently, many mathematicians have researched sequential spaces and generalized sequential spaces. The purpose of this thesis i s to present the more important of their results in a unified theory. The author generalizes a few of these results and - ap-proves numerous statements asserted without proof in the original papers. Chapter 1 i s an investigation of sequential spaces, their properties, their characterization as quotients of metric spaces or first-countable spaces, and their relation to other topological properties. Their relation to the first-countable spaces i s of particular interest. It i s well-known that countable compactness and sequential compactness are equivalent in the class of f i r s t -countable spaces. Franklin asserts their equivalence in the larger class of sequential spaces. Franklin proves this result in [8] for Hausdorff spaces and in [10] for spaces with unique sequential limits. In this thesis, the author provides the proof of the same „r,esult ...f,o:r,,,ar.bi^  „,T.he,J,.author,valso',.shows that any countable product of countably compact sequential spaces i s countably compact. There are, however, many properties of f i r s t -countable spaces which cannot be generalized to sequential spaces. For example, the product of two sequential spaces need not be sequential. A result due to T.K. Boehme [3] shows that this situa-tion cannot occur in the presence of suitable compactness conditions. In addition, E. Michael [23] has proved that for any T^ sequential space X and sequential space Y, the topological product space X x Y i s sequential i f and only i f X i s locally countably compact. _ 4 -The second chapter i s concerned with. Frechet spaces, their properties, and their relation to sequential spaces. The character-ization of sequential spaces i s used to prove Arhangel'skii's characterization of Hausdorff Frechet spaces. In this section, the author proves Arhangel'skix's assertion that the continuous pseudo-open maps and the hereditarily quotient maps are equivalent. The author also provides the proof of a result due to P.W. Harley III I12J concerning the product of two Frechet spaces. Chapter 3 i s devoted to Meyer's generalized sequential space methods and his study of convergence subbases and m-sequential spaces. CAn m-sequential space i s a space for which m-nets (I.e., nets whose directed-set i s of cardinality <_ m) suffice to determine ^closed ,,.ssats..,.)^  given in terms of convergence subbasis. This result leads to a characterization of the m-sequential spaces. The author proves an analogous result for continuous pseudo-open images which, leads to Meyer's characterization of m-Frechet spaces. In the last chapter, the author employs many of the properties of convergence subbases to investigate weakly sequential spaces and m_ - sequential spaces (i.e., those spaces for which well-ordered nets and well-ordered m-nets, respectively, are sufficient to describe closed sets). These spaces are characterized i n terms of ordered topological spaces (i.e., those spaces which have the order - 5 -topology arising from a total order).. Finally, there i s a brief coda which demonstrates that the concepts of first-countable space, Frechet space, and sequential space are equivalent i n products of ordered spaces. N o t a t i o n For the most p a r t , the terminology and b a s i c n o t a t i o n used i n t h i s t h e s i s f o l l o w s K e l l e y I16J. The exceptions are l i s t e d below. C D X - A = {x e X : x i A} C2) For any t o p o l o g i c a l space X and subset A of X, i n t ^ ( A ) denotes the i n t e r i o r of A with, respect to X and c l ^ ( A ) i s the c l o s u r e of A w i t h respect to X. When no c o n f u s i o n seems p o s s i b l e these w i l l be abbreviated to i n t A and c l A. ( 3 ) R i s the set of r e a l numbers, Z i s the set of i n t e g e r s , Q. i s the set of r a t i o n a l s , and W = { 1 , 2 , 3 , . . . } i s the set of n a t u r a l numbers. (4) to i s the f i r s t i n f i n i t e o r d i n a l and ft i s the f i r s t uncountable o r d i n a l . ( 5 ) For any o r d i n a l a, a + 1 denotes the set of a l l o r d i n a l s which are l e s s than or equal to a; th a t i s , a + 1 i s the successor o r d i n a l of a. (6) A t o p o l o g i c a l space i s s a i d to be f i r s t - c o u n t a b l e , or a f i r s t - c o u n t a b l e space, i f and only i f i t s a t i s f i e s the f i r s t axiom of c o u n t a b i l i t y . S i m i l a r l y , a t o p o l o g i c a l space i s second-countable,, or a second-countable space, i f and only i f i t s a t i s f i e s the second axiom of c o u n t a b i l i t y . Chapter 1 Sequential Spaces Sequences have numerous advantages over nets. This i s so because many properties of sequences f a i l to generalize to •nets. For example, a converging sequence and i t s l i m i t i s compact, whereas t h i s i s not true f o r nets. Among Hausdorff spaces, each convergent sequence ( i . e . , the union of the sequence and i t s l i m i t ) s a t i s f i e s the second axiom of c o u n t a b i l i t y and i s therefore metrizable. These f a c t s together with other properties of sequences not a p p l i c a b l e to nets play a c r i t i c a l r o l e i n the i n v e s t i g a t i o n of sequential spaces. CJ.8J, 19] and J10]) survey of sequential spaces. There are, however, several important r e s u l t s due to Boehme [3] and Michael .123] r e l a t e d to t o p o l o g i c a l products. --1.1 D e f i n i t i o n Let-X be a t o p o l o g i c a l space. (1) A subset U of X i s s e q u e n t i a l l y open i f and only i f each sequence i n X converging to a point i n U i s eventually i n U. C2) A subset F of X i s s e q u e n t i a l l y closed i f and only i f no sequence i n F converges to a point not i n F. For any topological space, a subset A i s closed i f and only i f no net in A converges to a point not in A CI"! 6] 5 Theorem 2. Therefore closed sets are sequentially closed and open sets are sequentially open. The converses need not be true. 1.2 Example There are sequentially open sets which are not open and sequentially closed sets which are not closed. Proof Consider the ordinal topological space ft + 1 provided with the order topology. Let S be a sequence in ft + 1 which i s not eventually equal to ft. Then S is frequently in ft and hence there i s a subsequence S q of S in ft. But the supremum of S q i s less than ft, and therefore S q cannot converge to ft. This implies that S cannot converge to ft. Thus a sequence in ft + 1 converges to ft i f and only i f i t i s eventually equal to ft. Additionally, a sequence in ft can converge only to a member of ft. It follows that {ft} i s sequentially open and ft i s sequentially closed in ft + 1. But {ft} i s not open and ft i s not closed in ft + 1. 1.3 Proposition A subset of a topological space is sequentially open i f and only i f i t s complement i s sequentially closed. .Proof If U i s a sequentially open subset of a topological space X and S i s .a sequence in X-U converging to x, then x e X-U. This i s so because otherwise S i s eventually in U. Thus X-U i s sequentially closed. Conversely, suppose that F i s a sequentially closed subset of X and let S q be a sequence in X converging to y e X-F. Then S q i s not frequently in F since otherwise there i s a subsequence of S q in F converging to y t F. Hence S q i s event-ually in X-F, and therefore X-F i s sequentially open. 1.4 Proposition For any topological space X, the collection of a l l sequentially open subsets forms a topology for the set X. Proof Clearly, 0 and X are sequentially open. If {U : a e A} c L i s any family of sequentially open subsets of X and S i s a sequence in X which converges to x e \J^& - a £ A}, then x e U £ for some c e A. Consequently S i s eventually in U^ and therefore i n (J {U : a e A}. Hence [J {U : a E A} i s sequentially open. 3. Si Suppose now that U and V are sequentially open, and l e t {y n : n e ui} be a sequence in X converging to a point in UflV. Then {y^ : n e w} i s eventually in both U and V, and there exists n Q, n^ e OJ with {y : h > n } C U and {y : h > n. } C V. So y e U D V for a l l n > y n — o *^ n — 1 n. — sup {nQ> n^}. Thus UflV i s sequentially open, and the proof i s complete - 10 -1.5 Definition The set of a l l sequentially op.en subsets of a topological space is said to be the sequential closure topology. 1 ...6 Definition A topological space is sequential, or a sequential space, i f and only i f each sequentially open subset i s open. .(.In view of (1.3) -and ,(1.5), - i t -is .clear that a -topological - space i s sequential i f and only i f each, sequentially closed subset i s closed, or equivalently, i f and only i f i t s topology coincides with the sequential closure topology.). In first-countable spaces, a set i s open i f and only i f each sequence converging to a point in the set i s , i t s e l f , event-'ually iri' the' set "([16] /"Theorem 278) . '"ThereTore"•first-countable spaces, and hence metric spaces and discrete spaces, are sequential. On the other hand, by virtue of (1.2), the ordinal space n + 1 provided with the order topology i s not sequential. After a few preliminary results, several equivalent formulations for the notion of sequential space are given. 1.7 Definition Let X and Y be topological spaces; and l e t T be the topology on X. The space Y divides X i f and only i f no topology T a on X which i s s t r i c t l y larger than T leaves every T-continuous function from Y into X T-continuous. a 1.8 Proposition Let X and Y be topological spaces, let T be the topology on X, and l e t T^ = { B C X : f^Ci) is open in Y for each T-continuous function f : Y ——> X}. The space Y divides X i f and only i f T = T . a Proof -Since -inverse-set^functions-preserve -set^operations,-it i s clear that T i s a topology on X. Furthermore, T d T and every T-continuous function from Y into X i s T-continuous. If a Y divides X, then T CI T and so T = ' T . Conversely, suppose that T^ i s any topology for X which leaves every T-continuous function from Y into X'T --continuous. If g : Y -> X i s a T-continuous p function and B e T , then g ^CB) i s open in Y and hence B E T . p ct Thus T = T implies that Y divides X'. a 1.9 Lemma A mapping f of the ordinal space to + 1, provided with i t s order topology, into a topological space X i s continuous i f and only i f the sequence {f (n) : n e to} converges to f Cw) i n X. Proof If f : to + 1 > X i s continuous and U i s any neighbourhood of fCw), f ^(U) is a neighbourhood of to in to + 1. Then f .''"GJ) contains (m» to] = {n : m < n e to} for some m e to. Therefore {f(n) : n E to} i s eventually i n U and hence (f(n) : n E to} converges to f ( w ) . Conversely, suppose that V i s an open subset of X. If f (w) -t V then f'1 (V) =' {n e w' : f (n) e V} = U {{n} : n e w , f(n) e V}, which i s the union of open subsets of to + 1. If ..f(u)) e V, {f(n) : n e cj} i s eventually in V and consequently there exists p e w such that f(n) e V for each ri >^  p. Therefore f *(V) = X p . , J J (n ,.: -p >.,.n...e .to., ...f,(n.) ..e .V.},,.is .-open.„in,.co H-,1. .The lemma i s proved. 1.10 Definition A convergent sequence i s the union of the sequence and a l l of i t s limit points. (Let S be a convergent sequence in a topological space X , and let S q denote the range of S provided with the relative topology. The topology on S i s the largest topology in which the natural function f : S > S Q defined by f(x) = x i s open.) 1.11 Lemma Every convergent sequence in a Hausdorff space i s compact and metrizable. Proof Let S = {x n : n e w} \J {x} be a convergent sequence in a Hausdorff space, and suppose that U i s an open covering for S . Now x e U for some U e U. Furthermore, {x^ ; n e w} i s eventually in U and thus x e U whenever ri > m for some m e' to. For each n — n < m choose U e U such that x e U . Then {U} U {V : m > n e to} n n n n i s a f i n i t e subcovering of U for S, and so S i s compact. To see that S i s a metric space, let = ^-{x^ : Ic > n e to} for each k e to. The collection {V^ : n e to} i s clearly a countable neighbourhood basis at x. Because S i s compact Hausdorff and hence a regular T^-space, there exists open sets U and V satisfying x e V d c l V O'U and x^ £ U. Consequently {x.} = S-(cl V U {x : i ^ n e to, x £ c l V}), which i s open in S. x ^ n ' n r The family {V : n e to} (J {{x } : n e to} i s a countable open basis n n r for the topology on S. Therefore S i s a second-countable regular T^-space. In view of CI16], Theorem 4.17), S is metrizable. ,a0b s erv.e,*t hat ,..,ther e ^ defined by d C x , x ) = |l/m - l/n| and dCx ,x) 1/m. m n ' m 1.12 Theorem For any topological space X, properties C l ) and C2) are equivalent. If X is Hausdorff they are also equivalent to C3) and C 4 ) . C l ) X i s sequential. C2) to + 1, provided with i t s order topology, divides X. C3) Each subset of X which intersects every convergent sequence in a closed set i s closed. CA) Each subset of X which intersects every compact metric subspace of X in a closed set i s closed. Proof (1) < > (2) Suppose that U is a subset of X with f (U) open in u •+ 1 for each continuous function f : w + 1 > X. Let {x^ : n E w} be a sequence in X converging to x e U. Define g(w) = x and g(n) = for each new. Then (g(n) : n e OJ) converges to g(w) and i t follows from (1.9) that g : w + 1 > X is a continuous function. Thus g "''(U) is an open subset of w + 1 containing w, which implies that g "*"(U) contains (m,w] for some mew. So = g(n) e U for each n >_ m + 1, and hence {x^ : n e w} is eventually in U. There-fore U is a sequentially open subset of X. If X is sequential, U is open and consequently, by virtue of (1.8), w + 1 divides X. Assume now that U is a sequentially open subset of X, and "let"4!'-4: ~w , {f(n) : n e w} converges to f(w). If f(w) e U then f ^ (U) contains (k,w] for some kew. So f ^ "(U) = (k,w] (J {n : k _> n E W, f (n) E U} which is open in w + 1. If f (w) t U, f ''"(U) = {n E w : f (n) E U} is open in w + 1. Therefore f '''(U) is open in w + 1; then U is open i f w + 1 divides X. (1) < > (3) Suppose that F is a subset of X, and let S = {x^ : n e w} U {x} be a convergent sequence in X. Either Ff| S is finite or infinite. In the first case, F D S is obviously compact. In the second case, F contains a subsequence of {x : n e w}. Therefore, i f F is sequentially closed, x e F. Hence F H S i s compact because (F 0 S)-U i s f i n i t e for any open set U containing x. Now; let be a net in F fl S converging to y e X. Since F f) S i s compact, Sy has a cluster point in F f\ S. But, the Hausdorff hypothesis implies that y is the only cluster point of S^. Evidently, F f) S i s closed. Thus, i f a subset of X i s sequentially closed, i t intersects every convergent sequence in a closed set. Conversely, i f {x : n e a)} i s contained in F, {x : n e to} is also a sequence n n i n F f i S. If F D S is closed then x e Ff) S and hence x e F. -Consequently, a subset of X i s sequentially closed i f and only i f i t intersects every convergent sequence in a closed set. The ^equivalence Aof.<*X*l^^ (3) < > (t\) Suppose that F i s a subset of X intersecting every compact metric subspace of X in a closed set. According to (1.11), each convergent sequence i s a compact metric subspace of X. Therefore F intersects every convergent sequence in a closed set, and ~hence~(3)- implies that-F- i s closed-To-establish- the-converse, assume that E is a subset of X intersecting every convergent sequence i n a closed set. Let K be a compact metric subspace of X. Since X -is Hausdorff, K i s also closed. If S i s a sequence in E f] K converging to x, then x E K. Additionally, x e E because E A (S U{x}) i s closed. Thus ED K i s a sequentially closed subset of the closed -metric subspace K; consequently E D K i s closed, and the proof i s complete. The elementary properties of sequential spaces are summarized in the following theorem. 1.13 Theorem (1) A function f : X ——> Y of a sequential space X into a topological space Y i s continuous i f and only i f { f C x ^ ) i n s t o } converges to f ( x ) whenever { x ^ : n E to } converges to x . (2) Every quotient of a sequential space i s sequential. (3) The continuous open or closed image of a sequential space i s sequential. (4) The cartesian product of sequential spaces need not be sequential. However, i f the product i s sequential, so i s each of i t s coofH'inate spaces. (5) The disjoint topological sum of any family of sequential spaces i s sequential. (6) The inductive l imi t of any family of sequential spaces i s sequential. (7) A subspace of a sequential space need not be sequential. An open or closed subspace, however, is sequential. (8) Every loca l ly sequential space i s sequential. Proof (1) The necessity of the condition i s true f o r a r b i t r a r y t o p o l o g i c a l spaces. I f f i s continuous and U i s an open subset of Y containing f ( x ) , then f ^ (U) i s an open subset of X con t a i n -ing x. Moreover, {x n : n e to} Is ev e n t u a l l y i n f "^(U) and so f ( x n ) e U f o r a l l n s u f f i c i e n t l y l arge. Hence {fCx n) : n e to} converges to f ( x ) . Conversely, suppose that V Is an open subset of Y and l e t {y^ : n e to} be a sequence i n X converging to y e f ^  Ql) . By hypothesis, {f CyQ) : n e to} converges to f (y) and so eventually, fCy n) e V. But then {y n : n e to} i s eventually i n f ^"(V). I t follows that f ^ (V) i s a sequ e n t i a l l y open subset of X. Then, since X i s sequential, f "'"(V) i s open and hence f i s continuous. - •"•(2')'"*"Let 'f "T"'X—->""Y''b'e'*V'qu^'tieht map of a sequential space X onto a t o p o l o g i c a l space Y. Suppose that U i s a s e q u e n t i a l l y open subset of Y and that {x^ : n e to} i s a sequence i n X converging to x e f '''(U). Then, since f i s continuous, {f(x^) : n e to} converges to f(x) e U. Consequently {x n : n e to} i s eventually i n f ^"(U), which implies that f ^ (U) i s a sequ e n t i a l l y open subset of the sequential space X. Therefore f ^ (U) i s open and hence, by d e f i n i t i o n of the quotient topology, U i s open. (3) By (116], Theorem 3.8), i f f i s a continuous open or closed map of a to p o l o g i c a l space X onto a space Y, then Y i s the quotient space r e l a t i v e to f and X. It follows from part (2) that i f X i s a sequential space then the Image f(X) = Y i s sequen t i a l . • - 18 -(4) An example of a non-sequential product of sequential spaces w i l l be given i n (1.19). To prove the second part of (4), l e t X be the cartesian product of any family {X& : a e A} of -topological spaces. For each c e A l e t P : X > X^ denote the projection map of X onto i t s coordinate space X^. From the definition of the product topology on X, P^ i s continuous. Further-more, according to (.[16], Theorem 3.2), the projection of a product space onto each of i t s coordinate spaces i s open. Hence P^ i s a continuous open surjection. Thus, i f X i s sequential, part (3) implies that X^ i s sequential. (5) Let X be the disjoint topological sum of any family {X : a e A} of sequential spaces. If U is not open in X, there 3. exists c E A such that U/1 i s not open and hence not sequentially open in X . Consequently, there i s a point x e Dfl X and a C C' sequence in X^-U converging to x with respect to X c and therefore with respect to X. Then U i s not sequentially open and the contrapositive of "each sequentially open subset of X being open" i s established. (6) Assuming that (A, <) i s a directed set, l e t {X , (j> ^  : a, b e A; a < b} denote the family {X : a e A} of sequential spaces together with the set of continuous maps - 19 -^ab * ^a > "^ b s a t^ s^y^ nS t a e condition : i f a < b < c then 4 = (J) o (f) By definition, the inductive limit of {X : a e A} cLC DC 3 p 3. i s the quotient space X/^ where X is the disjoint topological sum of {'X : a e A} and R is the equivalence relation : two elements a x a e X^, x^ £ X^ in X are equivalent i f and only i f there exists c e A,such that a < c, b < c and 4 (x ) = 4, (x, ). It follows ' Tac a be b from parts (5) and (2) that X/R i s sequential. (7) The non-hereditary nature of sequential spaces w i l l be demonstrated in (1.15) and (1.17). To prove the second part of (7), assume f i r s t that Y i s an open subspace of a sequential space X and let U be a sequentially open subset of Y. If "S = - { x n J ' n e to} i s a sequence I n *X converging to x eVcz Y then, since Y i s open, S is eventually in Y. There exists m e <o such that x e Y for each n > m. Moreover, {x , : n e to} is a n — m+n sequence in Y converging to x e U. Then, since U is sequentially open in Y, i x m + n : n e u} i s eventually in U. This surely implies that S i s eventually in U. Hence U i s sequentially open and there-fore open in X. .Assume now that Y is a closed subspace of the sequential space X and let F be a sequentially closed subset of Y. Suppose t h a t S i s a sequence in F converging to y with respect to X. - 20. -Because Y is closed in X, y e Y and consequently S converges to y in Y. Since F i s sequentially closed i n Y, y e F. Thus F i s sequentially closed in X, and so F = Fft Y i s closed in Y. (8) Let U be a sequentially open subset of a loca l ly sequential space X. If G i s any sequential neighbourhood of x e U, int G i s sequential by part (7). Let V = (int G)n U. It is clear that V is sequentially open and hence open in int G. But then V i s open in X. By hypothesis, there exists a col lect ion {G : x e U} of sequential neighbourhoods satisfying x e G . X X For each x e U, = (.Int G^) H U i s open in X. Therefore U = (J {V x : x e U} i s open, and (8) i s established. As previously stated, first-countable spaces are sequential. The following shows that not a l l sequential spaces are first-countable. 1.14 Example There i s a sequential space which i s not f i r s t -countable. Proof Let.X be the real l ine R with the integers Z identified to the point 0. From (1.13.2), X i s sequential. Suppose that {U^ : i e to} i s a countable neighbourhood basis at 0 in X. Since each IK i s obviously open in R, there exists a col lect ion {V : .n e u} of open intervals satisfying n e V 0 U • For each - 21 -new. choose an open interval I with n e I CL V . Then n n n {x E R : x < 0} U ( U {I : n E to}) i s an open neighbourhood of .0 in X which does not contain U for any n E to. Hence X cannot be n first-countable. ,1.15 -Example A jSubspace of .a .sequential-space need not be sequential. Proof Let X be the real numbers provided with the topology generated by i t s usual topology and a l l sets of the form {0}U U where U is a usual open neighbourhood of the sequence {——r- : n e to}. The topology of the real line i s altered only —n-rx at 0. For each open subset G of X, {0}UG i s open i f and only i f {——j- : n E to} is eventually in G. Accordingly, each sequence in n-rx X converging to 0 is either eventually in {0} or eventually in every neighbourhood of {^J" : n E to}, and hence is either eventually -equal-to 0 or a subsequence of : n e w^ • Define a subspace Y = {(x, 0) : 0 4 x £ R} U {(^r» 1) n E to} U {(0, 1)} of the plane. The space Y is the disjoint topological sum of the punctured real line .{(x, 0) : 0 £ x E R} and the convergent sequence { ( — - , 1) : n E w l U U O , !)}• Since both {(x, 0) : 0 ^ x e R} and {(— -^p 1) : n e CJ} [) {(0, 1)} are first-countable, Y i s sequential. The relative topology for Y is generated by sets of the form {G^jip 1)}, (CO, 1)} U i-C^p -1) : 111 .1 n e ^ > a n d CU-{0}) x {0} where m e w and U i s a usual open subset of the real line*!?. Let I 5 : Y > X be .the surjectlon defined by "P(x, y) - x for each (x, y) e "Y. To establish that 'X i s sequential, i t suffices ..to prove that P i s a quotient map. It i s f i r s t shown that P i s continuous. Let U be an open subset of X. If 0 i U, U is open in R and cons equently P (U) = (U x {0}) \J {(^pD : n e co,—r=- e U} i s open in Y. I f O e U , U = {0} (J G where G is an open subset of the real l i n e such that t' ^  • n e to} is eventually in G. Assume 1 • e*g^ wii e K e v e r« n-^>~ i n :-- e~ w-; '«Afben n+1 P _ 1CU) = CG-{0} x {0}) {J (HO, 1)} U 1) : m < n £ o>}) (J { (~jip 1) : m > n e w, ~ j p e U}, which is open in Y. Hence P is continuous. Now let V be a subset of X such that P ^(V) is open in Y. If 0 i V then P - 1CV) = ' {(-^p 1) : n E u), ~ - e V} U (V x {0}), which i s open i n Y i f and only i f V i s open in R. If 0 £ V, CO, 1) £ P '''CO and so there exists k E w such that (~jip 1) £ P ^CV) for each ri 21 k. Then P _ 1CV) = C{C0,1)} U i C~-, -1) : k ^ n E U)}) U'{C^p D : k > n E CO, E V} (J CCV-{0}) x {0}). Since P^CV) i s open in Y, V-{0} i s an n+1 open subset of the real line containing -—^ for each h >_ k, and consequently V = {0} U (V-{0}) is open in X . Hence P is a quotient map, and therefore X i s sequential. Consider the subspace X-v^j- : n E to}. Because each-sequence in X-{ ^ ' n E to} converging to 0 must be eventually equal to 0, {0} is sequentially open. But then, since {0} i s not open, .Xrr.{^j- : n e to} Is a non-sequential subspace of the sequential space X . 1.16 Example (1) The continuous image of a sequential space need not be sequential. (2) The open and closed image of a sequential space need not be sequential. Proof (1) Let (X, T) and (X, T ) be topological spaces with the discrete topology and a non-sequential topology on X respectively. The identity map 1 : (X, T) > (X, T ) i s a continuous surjection X ot of the sequential space (X, T) onto the non-sequential space (X, T ). In particular, the continuous image of the identity map of ft + 1, provided with i t s discrete topology, onto i t s e l f , provided with i t s order topology, i s not sequential. (2) Let R be the real line and l e t X be the topological space of (1.15). The identity mapping of the first-countable space R-{^j^- : n e to} onto the non-sequential space X-v^j- : n e to} i s an open and closed surjection. - 24 - . The topological space M of the next example i s important for later reference. 1.17 Example There i s a countable, T^ (normal and T^) sequential .space with a non-sequential subspace. Proof Let M = (W x W) [J W C/{0} with each Cm, n) e W x "W an isolated point, where W denotes the set of natural numbers. For a basis of neighbourhoods at n e W, take a l l sets of the form ..{n} U { (n, m) : m >_ q} where q e W. Define a subset U to be a neighbourhood of 0 i f and only i f 0 e U and U is a neighbourhood of a l l but f i n i t e l y many natural numbers. Clearly, M i s countable _,and.Hausdorf f. To establish that M i s normal, let G be an open - subset of M containing the closed subset A. If 0 t A, choose {m e hi : n e W f) A} such that V = {n} U {(n, m) : m > m } i s n n — n contained in G. Since 0 i A and A i s closed, W f) A i s f i n i t e . Therefore, V = [ U {V n : n e W f) A}] (J [ (W x W) D A] i s open and —A-G--V-G c l V c G. Suppose now that 0 e A. Then, choosing {m e W : n e W D G} such that U = {n} U {(n, m) : m > m } i s n n ' — n contained in G, U = { U (0 i n c N n G}J U [ (N x W) C\ A] {J {0} i s open and A c U c c l U CG. Hence M i s normal. - 25 -To prove that M i s sequential, l e t U be a sequentially open subset of M. For each x E (hi X hi) D U, x e int U since {x} .is open. For each, x e hi f\ U, {(x, m+1) : m e to} i s a sequence in M converging to x. Then, since U i s sequentially open, there exists m e to such that V = {x} U {(x, m) : m > m } i s contained x x — x in U.. ..But V ...is .a .neighbourhood .of ..x,,...and,therefore .x ,e .int ..U.. X If 0 e U then.W-U i s f i n i t e because otherwise W-U contains a sequence converging to 0. Consequently {0} U ( U {Vx : x e hi C\ U}) •is a neighbourhood of 0 contained in U. Hence U i s open, and so :M i s sequential. ...Since 0 e cl^ihl x hi), {0} i s not open in K-hl. If --I^) ,: .1 ,.e .toj .is any,sje;q,uenc.e,in.^ x ^either., there .is some . n e hi such that n^ =-.n for infinitely-many i or there i s no such -~n. -In the f i r s t case, {(n^, m/) : i e to} has a cluster point in the set {n} \J {(n, m) : m e hi}. Indeed, either there exists m e hi such that nu = m for i n f i n i t e l y many i or there i s no such m. It follows that the subsequence {(n, nr ) : i e to} has a cluster point at either (n, m) or n. Then, since M i s Hausdorff, {(n_^, m_.) : i e to} cannot converge to 0 i n the f i r s t case. In the second case, •-{(n_^ , m^ ) : i E to} has a subsequence in which each point has a distinct f i r s t coordinate. Without loss of generality i t can be - 26 -assumed that the sequence ((n^, nu) : i e to} has distinct f i r s t coordinates. Choose a sequence {k_^  e W : i E to} such that k. > m.. For each i E to let V. = {n.} (J {(n., m) : m > k.}, and 1 1 1 1 l — i for each n e W-{n^  • : i E to} let U n = {n} U {(n, m) : m E W}. Then ijj iY.± : i. e .(•£). J / i.USV^ : ,n e W-{n : i E to.}.}.) (/_{.0} i s a neighbourhood of 0 disjoint from {(iu, nu) : i e to}. Accordingly, a sequence in M-W converges to 0 i f and only i f i t i s eventually equal to 0. Therefore {0} i s sequentially open i n the subspace M-W, and hence M-N is not sequential. The topological space M-W i s a countable Hausdorff space which i s not sequential. The following shows that such a space must f a i l to be locally compact. 1.18 Proposition Every countable, locally compact Hausdorff space is first-countable (and hence sequential). Proof Let X be a countable, locally compact Hausdorff space. Then X is regular. Let x e X. By hypothesis, there exists a compact neighbourhood K of x. The subspace K i s regular and compact. From the regularity condition, there i s a collection U = {Uy : x ^ y E X'} of neighbourhoods of x satisfying y I c l K. Clearly {x} = C\ U. The family B of a l l f i n i t e intersections of ' 1 - 27 -members ,of U is a neighbourhood basis at x. This i s so because otherwise there exists an open neighbourhood V of x such that no member of 8 i s contained in V. But then, the intersection of the closed subset K-V with any f i n i t e intersection of {el : y e X-{x}} is non-empty and yet (K-V) f) ( c l U y : y e X - { X } } = 0; this contradicts K"being compact. To complete the proof, i t i s only necessary to establish that B i s countable. Let = {n : i >_ n e W}. There i s a one-to-one correspondence between the set of functions A i U = {f : A^ > U; f is a function) and the set of a l l f i n i t e intersections of i elements of (J. Consequently the cardinality A i of 8 Is less than or equal to the cardinality of U {U : I e W}. A- i ""The card'inali'ty 'of 'U ^-^8^)^~••^-4^ , "andJ-henc'e"the 'cardinalit-y of B ±s <_ hfg' H0~ He ~ Ha ([16], Theorem 179, page 279). Since any countable product of first-countable spaces i s first-countable, i t is natural to ask i f there i s an analogous result for sequential spaces. This question i s answered negatively by the succeeding example. Indeed, the product of two sequential spaces need not be sequential. The construction used in this example i s slig h t l y different than that derived by Franklin ([8], Example 1.11). Using this construction, i t i s also possible to prove that the square of a sequential space need not be sequential. 1..19 Example There i s a product of two sequential .spaces which i s not sequent ia l . Proof Let OJ be the ra t iona l s 0^  wi th the integers i d e n t i f i e d , and l e t X = Q_ x Q.' • The space X i s the product of two sequential spaces but contains a sequent ia l ly open set w which i s not open. To describe W, l e t { x n : n E CO} be a sequence of i r r a t i o n a l numbers less than one converging monotoriically downward to 0. For.each n E co, l e t be the i n t e r i o r of the plane rhombus determined by the points ( - x » n ) , (0, n + ^)» ( x n > n ) a n ^ n - -j) ; l e t J n be the i n t e r i o r of the t r i ang le determined by the points (x , n ) , (1, n + 4) and (1, n - ; and l e t K be the r e f l e c t i o n of J i n the y - a x i s . Then n J W = H U J UK U {(x, y) E R 2 : | x | > x } 1/ { (x, y) E R 2 : y < 0} n n n n 1 1 o ' J J i s an open subset of the plane. Thinking of U {W^  : n E co} as a subset of the plane with the hor izon ta l integer l i n e s i d e n t i f i e d , l e t W = X n ( U (Wn : n E co}) . (See Figure 1) Let P 1 : X > Q_ and : X > OJ be the canonical pro jec t ions . For any open neighbourhoods U and U ' of 0 i n Q. and OJ respec t ive ly , P ^ ( U ) D P2^(U') i s not contained i n W because there ex i s t s m E CO such that P ^ ( U ) C\ P2^(U' D {x : m - y <x < m + y} i s not contained i n W . Therefore (0, 0) £ i n t W, and hence W i s m not open i n X . Figure 1: The Graph of U (W : n e co} in R n (See Example 1.19) - 29 -To establish that W i s sequentially open, let {y^ : n e to} be a sequence in X converging to y e W. If I^Cy) ^ 0> convergence in X is simply convergence in Q xQ and,since (Q. x Q) H ( U {W^ : n e to}) i s open in Q_ x {y^ : n e 03} i s eventually in W for this case. Assuming that ^2^^ = 0, i f P^(y) 4 0 then W can be replaced by a scaled down version of i t s e l f , in W, with y at the symmetric position. Therefore i t can be assumed without loss of generality that y = (0, 0) . Now {y n : n e to} > (0, 0) implies that ^2^1? '' net0* ° i n . But then, i f P i s the quotient map of Q. onto. Q.' and K is the set of integers k such that {P ^  0 ^ ^ n ^ : n e w^ ^veciuent^y ^n U-{k} for each neighbourhood U cf k, { p 2^ y n ^ : n e w ' * S c v e n t u a H y in {0} U V where V i s any neighbourhood of K. Furthermore, K i s f i n i t e because otherwise ^ 2^T? : n e <°} n a s a subsequence not converging to 0. To verify that K i s f i n i t e , let V be a neighbour-hood of K and suppose that K = {k : n e to} where k < k i f and v * n n m only i f n < m. There i s a sequence {l_^ : i e to} of open intervals satisfying (1) I± C V, (2) Z D I± = {1n±}t and (3) I ± n Ij f 0 i f and only i f i = j . For each i e to there exists n^ e to such that y n > e 1^. Next, l e t {U^ : n e to} be a sequence of open sets satisfying k. e U. <Z I-{yn.}, and let G be a neighbourhood of Z-K disjoint from U {I. : i e co}. It follows that G U ( U {U : n e co}) J i n is a neighbourhood of 0 in OJ disjoint from the sequence {P 0(y n -) : i E co}. Therefore, {P 0(y n>) : i £ to} is a subsequence of {P 9(y ) : n E co} not converging to 0, and hence K must be f i n i t e . •Let-q ;=-sup-fk : -k-E-'K-}. —Since {P^Cy^) -: -n "E CO-} -converges"to 0 -in Q_ and ^2^r? : n £ * S e v e n t u a l ± y i n ^} f V for any neighbour-hood V of K, {y^ : n e co} i s eventually in E = [X fl ( U {Wn :. n <_ q})] U (0. x 10}). But E is contained in W. Thus W i s sequentially open, and this completes the proof. Defining OJ and as above and thinking of U {Wn : n E co} as a subset of the plane with the horizontal integer lines identified and the v e r t i c a l Integer lines identified, i t is not d i f f i c u l t to see from (1.19) that (OJ x OJ) n ( U {W : n E co}) i s a sequentially open subset of OJ x OJ which i s not open. Hence OJ x OJ is not sequential, and therefore the square of a sequential space need not be sequential. After a few preliminary results i t w i l l be shown that the situation described in (1.19) cannot occur in the presence of suitable compactness and separation conditions. F i r s t , i t i s convenient to prove that countable compactness and sequential compactness are equivalent in sequential spaces. As i s well-known ([28], Proposition 9.8), these concepts are equivalent in the class of first-countable spaces. Since sequentially compact spaces are always countably compact, the following establishes their equivalence in the larger class of sequential spaces. The proof i s provided by the author. 1.20 Theorem Every countably compact sequential space i s sequentially compact. Proof Let X be sequential and countably compact, and suppose that S = {x^ : n e to} i s a sequence in X with no convergent subsequence. Let A = U {c l x : n e to}. If S = {y : n e to} n o n " t iTy , '"^either c l x for some m e to, or no such m exists . In the f i r s t case, m y e c l x C A. In the second case, there exists a subsequence J m {y : k e to} of S with y„ e c l x . . But then {x^ : k e to} \ ° n k ^ k n k i s a subsequence of S converging to y. The second case, therefore, cannot occur and so y e A. From this i t follows that A i s sequentially closed and hence closed.. Since X i s countably compact, A i s countably compact and consequently S has a cluster point x e A. Now x E c l x^ for only f in i t e ly many n e to because otherwise S would have a convergent sequence. Let k e to be such that x ft c l x^ whenever n >^  k. But then, applying the same argument at above, - 32 - . \J {cl : 11 >_ k} i s sequentially closed and yet does not contain i t s accumulation point x. A result due to Novak 127] demonstrates that the product of two countably compact spaces need not be countably compact. The following shows that one of the spaces being sequential i s enough. 1.21 Corollary The product of two countably compact spaces, one of which i s sequential, i s countably compact. Proof Let {(xn» y ) : n e to} be a sequence in the topological product space X x Y of a countably compact space-X and a countably 'compact saequenti-ai«spaee--»Yv-^By'-virtu e^of-s?2Q').j-Y--d?s 'sequentially compact. Accordingly, the sequence {y^ : n e to} has a subsequence {y : k e to} which converges to some point y c Y. Since X i s countably n k compact, {x : k e u} has a cluster point x e X. Then (x, y) i s a nk cluster point of the sequence t(xn» y ) : n e to}. 1.22 Corollary Let X be the topological product of any countable family {X^ : n e to} of sequential spaces. Then X i s countably compact i f and only i f each X i s countably compact. - 33 -Proof The necessity of the condition i s obvious since the continuous image of a countably compact space i s countably compact ([7], Theorem 11.3.6). To establish that X i s sequentially compact and hence countably compact, l e t {x^ : n E to} be a sequence i n X. For each l e w let P^ be the projection map of X onto X_^ . If each X^ is countably compact then, by (1.20), each X^ is sequentially compact. Hence there exists a sequence {k^  : i e to} of functions mapping .to into to such that {x^ : n e to} o i s a subsequence of {x : n E <JJ} and {P (x, , J : n e u} i s a n o k (n) o convergent subsequence of i P G ( x n ) : n e to}, and for 1 < i e ii , ~{x, , \ : n E to} Is a subsequence of {x, , ^  : n c to} and ..ac^ Cn) ^ - ki--T^'' -{P;(x, , »•) : n e to} i s a-convergent subsequence of {P.(x. , 5: n E to}, i % ^ ( n ) ' 1 ^i.^y The sequence {x, , ^  : n E to} i s the desired convergent subsequence n of {x : n e to}, n 1.23 Corollary Let X be an uncountable set, and let 2 denote the set {0, 1} provided with the discfete topology. Then the product X space 2 i s not sequential. Proof Suppose that 2 i s sequential. Since any product of compact X topological spaces i s compact, 2 i s (countably) compact and hence, by (1.20), sequentially compact. Let f : X > 2 W be a surjection. X •Define a sequence {x : n e to} i n 2 by x (a) = [f (a) ] (n) for each n n a e X. Let {x : k e to} be a subsequence of {x : n E to}. Now k there exists y E 2 U such -that y( n2j c) -=.0 and- y( n2i c+i) = f o r each k E to. Since f i s surjective, f(3) = y for some 3 e X. X Therefore, " i f "P "is the "canonical "pro jection'map-of 2 -onto the 3 3-th coordinate space, then {P„(x ) : k E to} cannot converge 3 n k since 2 i s discrete; clearly P.(x ) = x n (3) = [f(3)] (n^) = y^y)' k k X Thus {x : k e to} does not converge, and hence 2 cannot be sequentially compact. The contradiction shows that 2 must not be •sequential. 1.24 Theorem Let X and Y be sequential spaces, and assume that each point of X has a neighbourhood basis consisting of sequentially compact sets.-- Then the topological product space X x Y i s sequential. Proof Let G be a sequentially open subset of X x Y . To prove that G i s open, suppose that (u, v) E G and let G = {x : (x, .v) e G}. Clearly u E G^. If {s^ : n E to} i s a sequence in X converging to s E G^ then {(s n, v) : n E to} i s a sequence in X x Y converging to (s, v) E G. Since G i s sequentially open, {(s n, v) : n E to} i s - 35 -eventually i n ,G, and consequently {s^ : n E co} i s eventually ,in G^. Hence G^ i s sequentially open and therefore open in X. By hypothesis, there exists a sequentially compact neighbourhood U of u with U x {v} contained in G. Let V be the largest subset of Y such that U x V C G; that i s , V = {z : U x {z} C G}. If V i s not open, there is a sequence {y n : n E co} i n Y-V converging to y £ V. But then, for each n E co there exists x £ U with n (x^, y ) i G. Since U i s sequentially compact, {X r : n £ co} has a subsequence {x : k e co} which converges to some point x e U. It follows that {.(x , y ) : k E CO} converges to (x, y) E G and n k n k hence that (x , y ) E G for a l l k sufficiently large. The k n k contradiction shows that V must be open. Then,' since (u, v) e.U *• V C G, (u, v) E int G and so G i s open. 1.-25-Corollary -(1) - The -product of two sequential-spaces, -one of which i s regular and either locally countably compact or loc a l l y sequentially compact, i s sequential. (2) The product of two sequential spaces, one of which i s l o c a l l y compact and either Hausdorff or regular, i s sequential. • - 36 -Proof (1) Let X and Y be sequential spaces, and assume that X i s regular and locally countably compact. Each point x e X has a countably compact neighbourhood K. Let {U : a e A} be a neighbour-a hood basis at x such that each U i s a subset of K. Since X Is a regular, for each a e A there exists an open set V a with x e V a c l V a. U . Then each c l V is countably compact and Q, Q. 3. . cl {cl V : a e A} i s a basis of countably compact closed neighbour-a hoods of x. By virtue of (1.13.7) and (1.20), c l V is sequential a and hence sequentially compact. Accordingly, each x e X has a neighbourhood basis {cl V : a e A} consisting of sequentially a «Gompact,v,s.ubsefcs,..,,.«^The«.iprieGeding.,theorem.wimplies.uthat sthe...product space X x Y i s sequential. The second part of (1) is now clear since sequentially compact spaces are countably compact. (2) This follows from (1) because every locally compact Hausdorff space i s regular and every compact space i s loca l l y countably compact. As seen in Example 1.19, the product space Q * Q.' i s not sequential. Although both coordinate spaces Q and are regular, neither topological space is locally countably compact. It i s clear that 0J f a i l s to be locally countably compact at 0. The space Q i s not locally countably compact because regular l o c a l l y countably compact spaces are Baire spaces and Q is not a Baire space ([7], pp. 249-250). 1.2,6 Corollary If X and Y are sequential Hausdorff spaces then the product spaces ;X * Y and (X x Y) , provided with the usual s product topology and the sequential closure topology respectively, have the same compact sets. Proof Since the sequential closure topology is larger than the product topology, i t i s only necessary to show that each compact subset of X x y is compact in (X x Y) . Let K be compact in X x Y, s and l e t K = {x : (x, y) e K} and K = {y : (x, y) e K} be the x y -projections of K into X and Y respectively. The subspaces and K are compact Hausdorff and hence closed. Thus K and K are also y r x y -•^sequenfeial^spaeesw'^ x i s sequential. Consequently, the topology induced on K by K x K i s the same as that induced on K by either the usual product x y -topology or the sequential closure topology. Therefore U D K i s open in K whenever U i s open in (X x y) , and hence K is a compact s subset of (X x y) . s The foregoing corollary i s of interest in studying k-spaces. — ( A topological space X i s a k-space i f and only i f a subset A of X i s closed whenever A/1 K i s closed in K for every compact subset K of X.) - 38 -1.27 Proposition Every sequential space is a k-space. Proof Suppose that A i s a subset of a sequential space X with A H K closed i n K for every compact subset.K of X. Let S be a sequence in A converging to x. Since S U {x} is compact, A D (S U {x}) i s closed in S U {x}. This implies that x e A and hence "that-A -is •sequentially -closed. There are, however, k-spaces which are not sequential. For example, the ordinal space U + 1 provided with the order topology i s a k-space which i s not sequential. The space ft + 1 is a k-space because i t i s compact ([7], pp. 222, 162) and the locally compact spaces are k-spaces ([7], 11.9.3). In view of (1.26) and (1.27), i t is not d i f f i c u l t to see that the product of two"k-spaces need not be a"k-space. The non-sequential space Q_ x OJ i s , in fact, a product of two Hausdorff sequential spaces which i s not a k-space. This i s so because there exists a non-closed sequentially closed subset A of Q. x £)* such that A f\ K i s closed in K for every compact subset K of ca * a')s. The next two theorems are important results concerning the product of quotient maps and the product of sequential spaces respectively. For each cardinal number m, let denote the discrete space of cardinality m, l e t be the quotient space obtained from D x [0, 1] by identifying a l l points in D x {0}, - 39 -l e t g : -D x [0, 11 -—> Y be the quotient map, and let y denote fom m ' m ^ r ' "'o the point x {0} in Y^. For any topological space X, let 1^ be the identity function on X. (For any two functions f : X > Z and g : Y > Z q , define ( f x g)(x, y) = ( f ( x ) , g(y)) for each (x, y) e X x y.) 1.28 Theorem The following properties of a regular space X are equivalent. (1) X i s loc a l l y countably compact. (2) 1 x g is a quotient map for every quotient map g x with sequential domain. (3) h = 1„ x g i s a quotient map, where m is the X m smallest cardinal such that each x e X has a neighbourhood basis of cardinality <_ m. Proof (1) > (2) Let g : Y > Z be a quotient map with sequential domain Y , and l e t f denote the product map 1^ x g. Clearly f i s continuous; i f A x B i s a basic open subset of X x z then f '''(A x B ) = A x g ' ' ' (B), which i s open since g is continuous. Let G be a subset of X x Z with f *(G) open in X x Y . Suppose that (u, v) e G and l e t r e g ^(v). There i s a basic open set U * V i n X x Y such that (u, r) e U x v C f *(G). Since X i s loc a l l y countably compact, there exists a countably compact neighbourhood X. of u. - 40 -Then, since X i s regular, there i s an open subset of X satisfying u e c l I ^ C U D The set K = c l U 1 i s a countably compact neighbourhood of u contained in U. Let E = {z e Z : K x {z} C G}. Since (u, r) e K x V C f - 1 ( G ) , K x {r} <Z f^CG) and G ^ > f (K x {r}) = K x {g(r)} = K x {v}, which implies that v e E. It remains to prove that E i s open. But since g i s a quotient map, i t suffices to show that g ^(E) i s open. If a e Y and K x {a} d f - 1(G) then f(K x {a})dG, which implies that K x g(a) CZ G and hence that g(a) E E. Thus g - 1(E) = {y e Y : K x {y}<Cf _ 1(G)}. Suppose that g - 1(E) i s not open. Then, since Y i s sequential, there i s a sequence {y^ : n e to} i n Y-g ^"(E) converging to some y E g "*"(E). S O K x {y^} CZf"f "'"(G) for each n E to. Hence there exists a sequence {x : n e to} in K with each (x , y ) £ f ^(G). Because K i s n . n n countably compact, {X r : n E to} has a cluster point x E K. Then the sequence {(* n, y ) : n e to} has a cluster point (x, y) e f "'"(G). Since f "'"(G) i s open, {(xn» Y n) : n E to} i s frequently in f "*"(G) contradicting (xn» y^) £ f ^ "(G) for a l l n e to. Thus g ^"(E) must be open, and (1) implies (2). (2) > (3) Both D and [0, 1] are sequential spaces. m Since D i s discrete, D i s locally compact Hausdorff and i t follows m m from (1.25) that D '* [0» 1] is sequential. Thus g m i s a quotient map with sequential domain x [0, 1] and so (2) implies (3). _ 41 -(3) > (1) Suppose that X i s not locally countably compact at-some point X q. Let {U^ : a e D^ } be a neighbourhood basis at x . For each a e D , c l U i s not countably compact o m a a and thus has a countable family {F n : n e W} of distinct non-empty ,,xlos.ed,,suhse.ts...sa,tiafying^the..finite ...intersection...property .whose cl Si intersection i s empty. Let E n = D {F^ : n >_ k e W} for each " a a a n e W. It i s clear that f] {E : n e A/} = 0, E -> E ,, , and n n n+1' a each E i s closed and non-empty. Thus, for each a e D there n m a exists a countable well-ordered family {E^ : n e W} of distinct non-empty closed subsets of c l U satisfying the f i n i t e inter-a ' section property whose intersection i s empty. To establish that h i s not a quotient map, for each a e D m define S a C X x (D m x [0, 1]) by S& = U {E* x {( a, ^)} : n e W}, and define S C X x y b y S = U {h(S ) : a e D }. It suffices to m . a m show that h - 1(S) is closed i n X x (D x [0, 1]) and that S i s not m closed in X x y . Note that (x o, y Q) t S since (x, (a, 0)) I for any x e X. If U x V i s a neighbourhood of (x o > y ) in X x y^ then, since X i s regular, c l WQCZ U for some c e D^. Thus choosing (c, i ) e g^(y)n <<c, : n E"W}, 0 * h(E^ x {( c, | ) } ) C (U * V ) n S; the point (c, exists because V i s a neighbourhood of y Q and - 42 -. v ^ r : n e co} converges to 0 in [0, 1]. It follows that (x , y ) e cl„ „ (S) and hence that S i s not-closed in X x Y . N o' Jo X*Y m m Since D i s discrete, i t remains to prove that m h ~ 1 ( S ) n (X * ({a} x [0, 1])) i s closed in X x ({a} x [0, 1]) for each a e D . But since (X x {(a, 0 ) } ) D S » 0, h - 1(S) n (X x ({a} x [0,1])) HI 3. = S • Assume that 'the point (x, (a, a)') i s contained i n the closure of a a Sfl with respect to X x ({a} x [0, 1]). Since each i s closed, i t a a i s clear that x e E-. Then, since {E : n e W} i s well-ordered and 1 n , -a a f) {E n : n e W} = 0, there i s a smallest set E^ containing x. It follows that a j> -^ because otherwise G x ({a} x (a - e, a + e)n[°, 1]), where 0 < e < ?~ . >. and G is a neighbourhood of x disjoint from a Ep +^, i s a neighbourhood of (x, (a, a)) disjoint from S g. Moreover, since [0, 1] is Hausdorff, a E : n <_ p}. Therefore a -1 (x,(a, a)) £ E x {(a, a)} C S . Thus h (S) is closed in ot a X x (D x [0, 1]), and the proof i s complete, m 1.29 Theorem The following properties of a T^ sequential space X are equivalent. (1) X is locally countably compact. (2) X x Y i s sequential for each sequential space Y . (3) X x y i s a k-space, where m i s the smallest m cardinal such that each point of X .has a neighbourhood basis of cardinality £ m* Proof The proof that (1) implies (2) i s given in (1.25), and • i t i s ^ obvious -that *(2-) implies (3) .«since Y^ ,is •sequential »and every sequential space is a k-space. By virtue of (1.28), to establish that (3) implies (1) i t suffices to prove that h = l v x g is a A m quotient map. F i r s t , i t i s convenient to prove a preliminary lemma. Let K be a compact subset of X * Y and let P be the f i r s t coordinate m projection map of X x Y m onto X. If (X x {y^}) r\ K = 0, h _ 1(K) = K *ahd-"'K*-'±s-^Iso^-xbmpaet^ih*X'x-(.D^x«['0,-"'"x '{y^)nYc'!f''^} for any a e Dffi the set E = (K-(X x {y Q})) U (P(K) x { ( a , 0)}) i s a compact subset of X x (D x [0, 1]) because any open neighbourhood of (a, 0) in x [0, 1] is contained in some open neighbourhood of •y in Y . In addition, h(E) = K. It follows that every compact o m subset of X x y is the image under h of a compact subset of m X x (D x [0, 1]). Suppose now that B is a subset of X x Y with m m h _ 1(B) closed in X x (D x [0, 1]). Since X x y i s a Hausdorff m m k-space, to prove that B is closed i t i s sufficient to show that B A K i s compact in K for every compact subspace K of X x y - 44 -•But K •= h(C) for some compact subset C of X x (D x [0, 1]). Thus m E = h "^(B) n 'C i s a compact, subset of C. Clearly h(E) i s contained i n B A I . If .b e B n K then b e K and so there exists some a e C such that h(a) = b. Furthermore, b e B implies that a e h "''(B) and hence that a e h - 1(B)n C = E. Therefore b = h(a) e h(E) and consequently h(E) = B o K. Then, since E i s compact in C and h is continuous, B/~) K i s a compact subset of h(C) = K. Thus B i s closed and h i s a quotient map. A characterization of the sequential spaces follows from the next theorem. 1.30 Theorem Every sequential space i s a quotient of a disjoint topological sum of convergent sequences. Proof Let X be a sequential space. For each x e X and for each sequence {Xr : n e to} in X converging to x, let SCx^, x) = {x n : n e to} U {x} be a Hausdorff topological space in which each x n is isolated and the sequence {x^ : n e to} converges to x. Although the elements of S(x n, x) need not be distinct i n X, they are taken to be distinct i n S(x n, x) . Thus s ( x n > x) is homeomorphic to to + 1 provided with the order topology. Clearly S(x n > x) is a convergent sequence in S(x n > x). Let W be the disjoint topological sum of a l l possible S(x , x). Since for ,.each x e X the convergent sequence {x : x^ = x, n e co} U {x} is a .summand >of W, the natural function f : W > X defined by f(x) = x i s a surjection. In addition, f i s continuous because i t 'is'continuous on each summand. To complete the'proof, i t -remains to establish that f i s a quotient map. Let U be a subset of X with f "\(11) opjen in W. If {.y : n e co} i s a sequence in X converging to y e U, y e f ^ (U)n ^ ^n» Y^ which is open in S(y n > y). Then {y n : n e co} as a subset of S(y n > y) i s eventually in f ^"(U), and hence • {y : n e co} as a subset of X i s eventually in U. Consequently U i s sequentially open and therefore open in X. 1.31 Corollary A Hausdorff space i s sequential i f and only i f i t i s a quotient of the disjoint topological sum of i t s convergent sequences. Proof The necessity of the condition i s clear from (1.30). It i s only necessary to remark that i f X i s Hausdorff then W i s precisely the disjoint topological sum of a l l the convergent sequences in X. Conversely, (1.11) implies that each summand of W i s a metric space and hence a sequential space. Then X i s sequential by (1.13.5) and (1.13.2). - 46 -1.32 Corollary Every sequential space is the quotient of a zero-dimensional, locally compact, complete metric space. Proof It suffices to show that W i s a zero-dimensional, loc a l l y compact, complete metric space. Suppose that U i s a neighbourhood basis at y e W. Each D e 0 i s both open and closed in W because U r\ S(x , x) i s bo.th open and .closed in S(x , x). Hence W i s n r n zero-dimensional. According to (1.11), each S(x n, x) i s a compact metric space. Obviously W is locally compact. If d . v i s a S vx_ »x / metric on S(x , x), then n f d n S ( x n , x ) ( u ' V ) l f U> V £ S ( x n » X ) 1 otherwise i s a metric on W. Lastly, W i s complete by virtue of ([7], Corollary 14.2.4). 1.33 Corollary The following are equivalent. (1) X i s sequential. (2) X i s the quotient of a metric space. (3) X i s the quotient of a first-countable space. Proof In view of the preceding corollary, (1) implies (2). Clearly (2) implies (3) because metric spaces are first-countable. Since first-countable spaces are sequential, (3) implies (1) by (1.13.2). - 47 -1.34 Example There is a separable sequential space which is not the quotient of a separable metric space. Proof Let H be the real numbers provided with the half-open interval topology. Sets of the form {x : a <^  x < b} = [a, b) are a basis for this topology. Since for each x E H the collection {[x, n^j_) 5 n e co} i s a countable neighbourhood basis at x, H is first-countable. Then H x H i s first-countable and hence sequential. The space H x H i s also separable because {(x, y) : x, y rational} i s a countable dense subset. Suppose that H x H is the quotient of a separable metric space X, and let P : X > H x H be the quotient map. According to ([7], .9.5.6), every separable metric space i s •iMndelof•^.TliuS'^ H^--x-••>H. -"Indeed., i f {U : a e A} Is an open covering of H x H, {P ^ (U ) : a e A} i s a a an open covering of X and hence i t has a countable subcovering {P ^(U ) : n E co}; then {U : a e A} has a countable subcovering a a • n ' {U : n E co} of H x H. a n Consider the subspace K = {(x, -x) : x irrational} of H x H. For any E > 0 each (x, y) e H x H such that x + y _> 0, ([x, x+e) x [y, y+e)) r\ K £ 0 i f and only i f (x, y) E K. In addition, for each (x, y) E H x H such that x + y < 0, ([x, x+6) x [y, - x - 6 ) ) ^ K = 0 whenever 0 < 6 < -x -y. Thus K is a closed subspace of the Lindelof space H x H. Since K i s discrete and uncountable, ,K i s not Lindelof. However, i f {V - : a e A} is an open covering of K, Si {V : a e A} U {(H x H)-K} i s an open covering of H x H. Then, a since H*x H is Lindelof, there i s a countable subcovering {Va : n e co} U {(H x H)-K} of H x H. It follows that K i s n covered by {V : n e co} and hence that K i s Lindelof. The a n contradiction shows that H x H must not be the quotient of a separable metric space. It has been shown that the notion of sequential space i s neither hereditary nor productive. The following i s a characterization of those subspaces and those products of sequential spaces which are themselves sequential. 1.35 Proposition For X sequential, let 4^ denote the quotient map of X* onto X, where X* i s the disjoint topological sum of convergent sequences in X as derived in (1.30). (1) A subspace Y of a sequential space X i s sequential i f and only i f 4 v i , - l , v v i s a quotient map. 'X (2) The product of two sequential spaces X and Y i s sequential i f and only i f <j>Y .x 4 i s a quotient map. - 49 -Proof ( 1 ) Let Y = c C^Y) and cf> = <j>v | ' . Let g : Y* > Y X X • A 1 ^ be the function defined by g(x) = x for each x e Y*. Then g i s a quotient map i f and only i f Y i s sequential. It suffices to show that (f> i s a quotient map i f and only i f g i s a quotient map. Let U be a subset of Y. If 4>~1(U) i s open in Y 1 then <j)-1(U)/^\ S is open in 'S for each summand S of Y^. From the definition of the relative topology on Y i t i s clear that each convergent sequence in Y i s a convergent sequence in X. Therefore each summand of Y* i s a summand of Y^, and consequently g ^ (U) = <f> ^ (U)/n Y* i s open in Y*. Conversely, suppose that g ^ (U) i s open in Y*. Then 4 "*"(U)/*"} S = g ^ (U)n S i s open in S for each summand S of Y^ which i s also a summand of Y*. Let S( x n> x) be a summand of X*; then = S(x n, x) D Y i s a summand of Y^. The topological space is either f i n i t e or i n f i n i t e . In the f i r s t case, each point of is isolated and thus C\ <J> ^ (U) must be open in S^. Assume now that the second case occurs. If x t U then S^n I "*"(U) = {x n : n e w} n <{> ^(U) which i s certainly open in S^. If x e U, each sequence in converging to x e O <i> "^ (U) i s a subsequence {x^ : k e to} of {x : n E to}. But since \ n * ^(U) H S(x , x) i s open in S(x n , x), {x : k e to} i s eventually °k k n k in D <f> ''"(U). It follows that S 1 f) <j> "*"(U) i s a sequentially open subset of the first-countable space S^. Therefore <j> (U) i s open in Y^, and the proof of (1) i s complete. (2) Since X* and Y* are first-countable, the topological product space X* x Y* i s first-countable and hence sequential. .Then i f h '= - . < ( > . x A i s a quotient map, X x y i s sequential. To establish X Y the converse, assume that .X x y is sequential and let G be a subset of X x y with h~1(G) open in X* x y*. Suppose ( (x »' y ) : n e co} i s a sequence in X x Y converging to (x, y) e G. Then (x, y) e h "^(G) and there exists a basic open subset U x V of X* x y* such that (x, y) e U x V C h - 1(G). Accordingly, U n S(x n > x) is open in S(x , x) and V fl S(y , y) i s open in S(y , y). This implies that n n n the sequences {x^ : n e co} and {y n : n e co} are eventually i n U and V respectively, and hence that (x n > y ) e h 1.(G) for a l l n sufficiently large. Therefore (x , y ) = (<j> (x ), ^ ( y ^ ) ) = h(x , y ) e G for n xi A H x n n i l a l l n sufficiently large, and so G i s a sequentially open subset of the sequential space X x y. Thus G i s open and h i s a quotient map. A first-countable space with unique sequential limits i s Hausdorff since otherwise i t is possible to find a sequence converging to two distinct points. That i s , i f x and y are distinct points which cannot be separated by disjoint open sets and {U n : n e co} and {V : n E co} are countable neighbourhood bases of x and y - 51 -respectively, then the sequence {x : n e to} satisfying x n e U ^ H V R converges to both x and y. The succeeding examples show that this result cannot be generalized to sequential spaces. The construction used in the f i r s t example is based on Sorgenfrey's [29] well-known result concerning the product of normal spaces. It w i l l be shown that the square of the normal space H i s not normal. 1.36 Example There i s a sequential space with unique sequential limits which i s not Hausdorff. Proof Let H be the real numbers provided with the half-open *inter-vai i n^ sequential. If A = {(x, y) : x + y = 1} i s the antidiagonal of H x H, let A and A. be those points of A with rational and q l irrat i o n a l coordinates respectively. For any e > 0 and each (x, y) e H x H such that x + y _> 1, ([x, x+e) x [y, y+e)) f) A^ ^ 0 i f and only i f (x, y) e A^; similarly, ([x, x+e) x [y, y+e))D A^ ^ 0 i f and only i f (x, y) e A ^ And for each (x, y) e H x H such that x + y < 1, ([x, x+6) x [ y > l-x-5))O A = 0 whenever 0 < 6 < l - x - y . Therefore A and A. are disjoint closed subsets of' H x H. To prove that and A^ cannot be separated by d i s j o i n t open sets , l e t U be an open neighbourhood of A ^ . For each i r r a t i o n a l x , l e t f (x) = sup{e > 0 : [x, x+e) X [1-X , l -x+e)c~U}. Then f i s a function on the set of i r r a t i o n a l numbers and f i s never zero. The set of i r r a t i o n a l s i s the countable union of sets of the form {x : f (x) > —} where n e W. In tthe r e a l l i n e R, the — n i r r a t i o n a l s are of the second category ( [7] , pp. 249-251) and consequently there ex i s t s m e U such that {x : f (x) i l ~} i s not nowhere dense i n R. Hence there i s a r a t i o n a l number r which i s an accumulation point of {x : f(x) i l ~} • For any neighbourhood V of the point ( r , 1-r) , there ex i s t s p e R such that P < ~ and [ r , r+p) x [1-r , 1-r+p) d V . But there i s an i r r a t i o n a l 1 1 number s e ( r -p , r+p) such that [s , s + —) x [1-s, 1-s -\—) C U. r > n m m Then <[r, r+p) x [1-r , 1-r+p)) fl ( U , s + ~) x [ l - s , 1-s + ^)) j 0, and hence U H V ^ 0 for every neighbourhood V of ( r , 1-r) . There-fore ( r , 1-r) e c l U , and so A ^ / T c l U 0 for every neighbourhood U of A ± . The set E = (A q x A ) U (A ± x A ± ) {J {(x, x) : x e (H x H) ( A ^ U A^)} i s an equivalence r e l a t i o n i n the f i r s t -countab le space H x H. The quotient space X = (H x H) /„ i s sequential and T, but hi I not Hausdorff. Let 4 : H x H > X be the quotient map, and l e t q = * (A ) and i = <|>CA.). Then q and i are the only pair of distinct points of X which cannot be separated by open sets, and consequently i f some sequence in X converges to two distinct points, ..they must be q and I. Suppose that S = "{x : n e w} converges to q. Since X i s T^, i t can be assumed that x^ ^  i for a l l n e co. Again since X i s T^, i f frequently x^ = q then S cannot converge to i . However, i f X r i s eventually different from q, there must be some q^ e A^ and a subsequence S q of S converging to q^ in H x H. But then there i s a neighbourhood of A^ disjoint from S q , and thus S cannot converge to i . Hence X has ^ unique,..sequential ..limits... 1.37 Example There i s a countable, compact, sequential space with unique sequential limits which i s not Hausdorff. Proof Let M be the Hausdorff sequential space of (1.17). Let p be some point not i n M and let = M 0 {p} with M open in and where the basic neighbourhoods of p are of the form {p} U ((W x W)-F) with F the union of the ranges of a f i n i t e number of convergent sequences i n M. Since M is Hausdorff and a convergent sequence in M cannot also converge to p, has unique sequential limits. However, Mn i s not Hausdorff because - 54 -0 and p have no disjoint neighbourhoods. It i s also clear that i s countable and compact. To verify that i s compact, let U and V be any open neighbourhoods of 0 and p respectively. Then W-(U U V) is ^finite .and M^-(U U V) is the union of the ranges of a f i n i t e number of convergent sequences in W x hi. But each "sequence i n *M^ -(U XI V) converges "to some "meniber Of "W-"(U"U V) . Hence for any f i n i t e collection {U^ : n e W-(U U V)} of open sets satisfying n e U , M1-(U U V V [ (J{U :n e W-(U U V)}]) i s f i n i t e , n 1 n To see that i s sequential, let V be a sequentially open subset of M^ . If p ft V, V i s sequentially open in the sequential space M. Then, since M i s open in M^ , V i s also open in M^ . Assume now that p £ V. Clearly, V i s a neighbourhood of each point in V-{p} because V-{p} is sequentially open in M. Since any subset {(m, n) : m e A C W, n e B C W} contains a sequence converging to p whenever A is i n f i n i t e , M^ -V must contain points of W x W having only f i n i t e l y many first-coordinates. Thus V contains a basic neighbourhood of p and so p E int V. Therefore V i s an open subset of M1 and consequently M1 i s sequential. - 5 5 -1.38 Proposition (1) A sequential space with unique sequential limits i s T^. (2) If X i s a topological space with unique sequential limits and X x X is sequential, then the diagonal A = {(x,x) : x e of X x X i s closed (and hence X i s Hausdorff). Proof (1) For each.member .y of a sequential space Y, the singleton {y} is sequentially closed and hence closed in Y. (2) Every sequence i n A i s of the form t( x n> x n ) : n e -Since X has unique sequential limits, {( x n> x n ) : n e to} converges to (x, x) i f and only i f {x^ : n e to} converges to x. Therefore A i s sequentially closed and hence closed in X x X. It follows frqmjpart (2) that the product spaces X x x and x M^ , where X = (H x H)^ £ and are the non-Hausdorff sequential spaces of (1.36) and (1.37) respectively,are not sequential. If these products were sequential, X and would be Hausdorff. After a preliminary result i t w i l l be shown that a sequential space with unique sequential limits in which each point has a neighbourhood basis consisting of countably compact sets i s Hausdorff. The topological spaces X = (H x H)^ £ and do not satisfy this compactness condition. For any neighbourhood U of q - 56 -i n X , l e t { x n : n e to} be a sequence i n U such tha t x E ( n , n+e ) x (1-n, 1-n+e ) CZ U where 0 < E < -i-. The sequence n n n n 2 ^ {x^ : n E 10} has no c l u s t e r p o i n t i n the neighbourhood U . T h e r e -f o r e U i s not c o u n t a b l y compact, and consequent ly X i s not even l o c a l l y c o u n t a b l y compact. Suppose now t h a t V i s any n e i g h b o u r -hood of p i n M ^ . Then, s i n c e {n+1 : n E to} i s a sequence i n M c o n v e r g i n g to 0, V-(W U {0}) i s a l s o a neighbourhood of p . I t i s c l e a r t h a t t h e r e e x i s t s m e W such t h a t {(m, y ) : n e to} i s a sequence i n V - ( M (J {0}) w i t h y = y, i f and o n l y i f n = k . But {(m, -y ) : n e to} has no c l u s t e r p o i n t i n V-(W U {0}). So V-(WU{0}) i s not c o u n t a b l y compact, and does not have a c o u n t a b l y compact 1 neighbourhood b a s i s a t p . 1.39 P r o p o s i t i o n A s e q u e n t i a l space has unique s e q u e n t i a l l i m i t s i f and o n l y i f each c o u n t a b l y compact subset i s c l o s e d (and hence s e q u e n t i a l ) . P r o o f L e t X be a s e q u e n t i a l space . Suppose X has unique s e q u e n t i a l l i m i t s and K i s a c o u n t a b l y compact subset of X . L e t S = { x n : n e to} be a sequence i n K c o n v e r g i n g to x . S ince X has unique s e q u e n t i a l l i m i t s , {x} U range S i s s e q u e n t i a l l y c l o s e d and hence c l o s e d i n X . Then i f y i s a c l u s t e r p o i n t of S, e i t h e r y = x or f r e q u e n t l y x = y . - 57 -Again since X has unique sequential limits, y = x . Thus x i s the only cluster point of S, and consequently x e K.. Therefore K i s a .sequentially closed subset of the sequential space X. Assume now that S i s a sequence in X converging to two distinct points x and y. Since {x} U range S i s compact, {x} U range . S i s closed and hence ..contains y. . This implies that S i s frequently equal to y. But then {y} i s a non-closed compact subset of X. ,1.40 Corollary A sequential space has unique sequential limits i f and only i f each sequentially compact subset is closed. Proof Since every sequentially compact set i s countably compact, the necessity of the condition follows from (1.39). Conversely, .suppose that S = {x^ : n E .to} i s a sequence in a sequential space X converging to points x and y. Then {x} U range S i s sequentially compact because each sequence in the set has a subsequence which i s either a subsequence of S or eventually equal to x m for some .m .e-.io. Therefore...{x} U range .S_is .closed.,. and_consequently either y = x or frequently x^ = y. The latter case cannot occur since otherwise {y} i s a non-closed sequentially compact subset of X. 1.41 Proposition Let X be a sequential space with unique sequential limits. If each point has a neighbourhood basis consisting of countably compact sets, then X i s Hausdorff. - 58 -Proof Each countably compact subset of X i s closed (by 1.39), sequential (by 1.13.7), and hence sequentially compact (by 1.20). Thus each point of X has a neighbourhood basis consisting of sequentially compact sets, and so (1.24) implies that X x X i s sequential. Then, according to (1.38.2), X i s Hausdorff. 1.42 Corollary If X i s a sequential space with unique sequential limits and each point has a neighbourhood basis consisting of compact or sequentially compact sets, then X i s Hausdorff. Chapter 2 Frechet Spaces The Frechet spaces form an important subclass of the sequential spaces -which-conta-ins the first-countable spaces-. The study of Frechet spaces i s closely related to that of both first-countable spaces and sequential spaces. For example, every subspace of a Frechet space i s Frechet and the quotient of a Frechet space need not be Frechet. On the other hand, there i s a Frechet space with unique sequential limits which i s not Hausdorff and the product of two Frechet spaces need not be ~Ffe%he"t. "This chapter emiilates'Tranklin's"([8], *[9] and [ 10]) investigation of Frechet spaces. There are, however, several results concerned with Arhangel*skii 1s [2] study of pseudo-open maps and a result due to Harley [12] connected with the product of Frechet spaces. 2.1 Definition A topological space X i s Frechet, or a Frechet space, i f and only i f the closure of any subset A of X i s the set of limits of sequences in A. - 60 -In first-countable spaces, a point x i s an accumulation point of a set A i f and only i f there i s a sequence in A-{x} which converges to x ([16], Theorem 2.8). Therefore first-count-able spaces, and hence metric -spaces and discrete spaces, are Frechet. 2 . 2 Proposition Every Frechet space i s sequential. However, there are sequential spaces which are not Frechet. Proof By definit ion of sequentially closed and Frechet space, i t i s obvious that every sequentially closed subset of a Frechet space i s closed. Thus every Frechet space i s sequential. On the «other>*hand,-, (the^iitopologi respectively are examples of sequential spaces which are not Frechet. In both spaces, 0 E cl(W X W) but no sequence in W x W converges to 0. 2 . 3 Theorem ( 1 ) Every subspace of a Frechet space i s Frechet. ( 2 ) The disjoint topological sum of any family of Frechet spaces i s Frechet. ( 3 ) Every loca l ly Frechet space i s Frechet. (4) If A i s any subset of a Frechet space X then Y = ^/^» the topological space X with the points in A identif ied, i s Frechet. - 61 -Proof (1) Let Y be a subspace of a Frechet space X and l e t A be a subset of Y. Then c l (A) is the set of limits in X of sequences in A. Hence c l (A) = c l v ( A ) n Y i s the set of limits Y X in Y of sequences i n A. (2) Let.X be the disjoint topological sum of the family {X : a e ,A} of Frechet ..spaces. Let B be a subset of X and l e t B' a be the set of limits of sequences in B. For each c e A, B'n i s the set of limits in X of sequences in B(1 X ; because X i s c c c Frechet, B' n X i s closed in X . Then B' i s closed and (2) i s c c proved. / ' (3) Let B' be the set of limit points of sequences In a subset B «of-a locally -Frechet'<space-X. -';For-•each'-'x e X-B' there i s a neighbourhood G of x which is Frechet. By part (1), int G i s Frechet. Let V = (int G)O (X-B'). The intersection (int G ) O B' i s the set of limits of sequences in (int G) H B. Then (int G ) n B' i s closed in the subspace int G, and consequently V i s an open subset of int G. It follows that V Is open i n X. By hypothesis, there exists a collection {G^ : x e X-B1} of Frechet subspaces of X such that each G x is a neighbourhood of x. Each \r = (int G x>H (X-B') is open in X. Therefore X-B' = U {Vx : x e X-B*} i s open i n X and so B1 i s a closed subset of X. Hence X i s Frechet. (4) If g : X > Y is the quotient map, let i = g(A). Let B' be the set of limit points of sequences in a subset B of Y. If i I cl Y(B) then, since X i s Frechet, clyCB) = C 1 X ( B ) = B ' « If i e c l (B), either k e c l (B) for some k e A or no such k exists. Y A In the second case, there i s a collection {U : x e A} of open x subsets of X satisfying x e U x and U^H B = .0. But then U {U x : x e A} i s a neighbourhood of i disjoint from B. The contradiction shows that the f i r s t case must occur. Thus k e c l v ( B ) and consequently there exists a sequence in B converging to k. Hence i e B' and the proof i s complete. 2.4 Example (1) There i s a Frechet space which i s not f i r s t -countable. (2) The product of two Frechet spaces need not be Frechet. Proof (1) The real line with the integers identified i s Frechet (by 2.3.4) but does not satisfy the f i r s t axiom of countability (by 1.14). (2) It follows from (2.3.4) that the topological space £)', the rationals with the integers identified, i s Frechet. Then Q x Q.' i s the product of two Frechet spaces which, by (1.19), i s not sequential and hence not Frechet. Similarly, the square x Q_' i s not a Frechet space. ',2,;5 .-Example (1) The open and closed image of a Frechet space need not be Frechet. (2) . The quotient of a Frechet space need not be Frechet, Proof (1) Since every Frechet space i s sequential and every f i r s t -countable space i s Frechet, the proof of (1) i s the same as that of (1.16.2). (2) Let X be the topological space of (1.15) and let A = X-({ : n e <JJ} U {0}). For each n e w there i s a sequence n 1 { X j , : i e w} in A converging to n +^- By the theorem on iterated limits ([16], Theorem 2.4), 0 e c l A. However, every sequence in X converging to 0 i s eventually equal to 0 or a subsequence of : n e to}-. Hence X i s not Frechet. But X is a quotient of the first-countable space Y. 2.6 Definiton A surjective function f : X > Y of the topological space X onto the topological space Y i s pseudo-open i f and only i f for any y e Y and any open neighbourhood U of f ^(y), y e int f(U). 2.7 Definition A map f : X > Y satisfying some property C i s said to be hereditarily C i f and only i f for each subspace Y^ of f(X) and X 1 = f " 1 ^ ) the induced map f | x : > Y 1 satisfies property C. - 64 -2.8 Proposition (1) Every open or closed surjection i s pseudo-open . (2) Every continuous pseudo-open map i s a quotient map. (3) Every pseudo-open map i s hereditarily pseudo-open. Proof Let f : X > Y be a surjective mapping of the topological space X .onto .the top.oiogic.al space "Y. (1) Let y e Y and let U be an open neighbourhood of f _ 1 ( y ) . If f is an open map, y e f ( f _ 1 ( y ) ) C f(U) = int f(U). Thus open surjections are pseudo-open. Suppose now that f i s a closed map. Then, since X-U is closed and f ^(y)c^ X-U, y i s not contained in the closed set f(X-U). Therefore Y-f(X-U) is open and consequently, since f(X-U) ~D f(X)-f(U), y e Y-f(X-U)C Y-(f(X)-f(U)) = f(U) implies that y e int f(U). (2) Let V be a subset of Y such that f - : L(V) i s open i n X. For each y e V, f ^(V) i s an open neighbourhood of f ^(y). Then, since f i s pseudo-open, y e int f ( f "'"(V)) = int V and hence V i s open. (3) The function f induces a map h : X^ > Y^ where Y^ i s a subspace of Y, X^ = f "''(Y^), and h = f |^ . Suppose that f i s pseudo-open. Let y e Y^ and l e t be an open neighbourhood of h _ 1(y) i n X^. Then = U PI X 1 where U i s open i n X and f " " 1 ^ ) = h - 1 ( y ) C ^ C U . Accordingly, y e i n t y f ( U ) . But . y e ( i n t y f(U ) )n Y 1 = ( i n t y f(U ) ) n Y 1 = i n t ^ ( f ( U ) n = int hCU^). Hence h Is pseudo-open. ,2.9 Example (1) There i s a pseudo-open map which i s not open. (2) There i s a quotient map which i s not hereditarily quotient. Proof (1) The quotient map from the real line onto the real line with the integers identified i s a pseudo-open map which i s .not open. (2) Let X and Y be as in (1.15), and let P be the quotient map of Y onto X. If X^ = X-{^£j- : n e to}, then Y± = P~ 1(X 1) = {(0, 1)} U {(x, 0) : 0 t x e RM^jip 0) : n e to} and P induces the map P^ = P|Y • But P^ is not a quotient map sines P^({0}) = {(0, 1)} is open in Y^ and yet {0} i s not open in The next result was asserted without proof by Arhangel'skix [2]. The proof i s provided by the author. 2.10 Proposition A function i s continuous pseudo-open i f and only i f i t i s hereditarily quotient. Proof Let f : X > Y be a continuous surjection, and l e t f. = f L where Y. is a subspace of Y and X = f^CY-). If U. i s open i n Y^, = UO Y^ for some open subset U of Y. Then, since f i s continuous, f^CO^) " f ^ C U O Y 1) = f - 1 ( U f l Y ^ = f "^(U)n X- i s open in X-. Thus continuous maps are hereditarily - 66 -continuous. It follows from (2.8) that i f f is continuous pseudo-open then f^ is a continuous pseudo-open map and therefore a quotient map. Conversely, assume that f is hereditarily quotient. Suppose y e Y and V is an open neighbourhood of f "*"(y). Let Y 2 = (Y-f(V))U {y>, X2 = f _ 1 (Y 2 ) •= [X-f~ 1(f(V))].U f _ 1 ( y ) , and "f 2 "=;f-|x . Then f~*(y) '=-f ~*(y)•;*«• V'n;X2, • which :±s open- in X2» Since f 2 is a quotient map, {y} is open in Y2 and thus {y} = G H Y2 for some open subset G of Y. This implies that G is contained in f(V) and hence that y e int f(V). 2.11 Proposition Every continuous pseudo-open image of a Frechet Proof Let f : X ——>'-Y.be a continuous pseudo-open function of a Frechet space X onto a topological space Y. Let B be a subset of Y and suppose that y £ c l B. If f _ 1(y)n c l f _ 1(B) =0, U = X-cl f ^ (B) is an open neighbourhood of f "'"(y). Then, since f is pseudo-open, y E int f(U)ci f(U) = f(X-cl f ' 1 ^ ) ) C f (X-f - 1(B)) d Y-B contradicting y e c l B. Hence there is some x E f ^ (y)O c i f ^(B) and, since X is Frechet, there exists a sequence {x^ : n E OJ} in f "^(B) converging to x. Thus {t(* n) : n E to} is contained in B and, since f is continuous, {f(x n) : n E w} converges to f(x) = y. There-fore Y is a Frechet space. - 6 7 -2.12 Corollary Cl) The continuous open or closed image of a Frechet space i s Frechet. C2) If a product space i s Frechet, so i s each of i t s coordinate spaces. Proof CD follows from C2.8.1) and C2.H), and Cl) implies C2). The following i s a slight generalization of the necessity condition of Franklin's Proposition 2.3 C[8]). The Hausdorff hypothesis is replaced by "unique sequential limits". 2.13 Proposition If f : X > Y is a quotient map of a topological space X onto a Frechet space Y having unique sequential limits, then f i s pseudo-open. Proof Let y e Y and suppose that U i s an open neighbourhood of f - 1 C y ) . Assume that y I int fCU). Then y e Y-int fCU) = c l CY-fCU)), and consequently there exists a sequence S in Y-fCU) converging to y. Because Y has unique sequential limits, c l Crange S) = {y} U range S. If F = f - 1 C S ) then, since f is continuous, c l F = c l C f - 1 C S))C f ~ 1 C c l S) = f'Hs U (y>) = f~*(S) U f _ 1 ( y ) = F U f _ 1 ( y ) . But f - 1 C y ) C U and U n F = 0. This implies that f - 1 C y ) n c l F = 0 and therefore that F i s closed. Hence f - 1CY-S) = f - 1CY)-f _ 1CS) = X-F i s open. Then, since f i s a quotient map, Y-S i s an open neighbourhood of y, contradicting the supposition that S converges to y. - 68 -2.14 Theorem A Hausdorff space i s Frechet i f and only i f i t i s • a continuous pseudo-open image of the disjoint topological sum of i t s convergent sequences. Proof Each Frechet Hausdorff space is sequential Hausdorff and hence, .by (1.31), a quotient of the disjoint topological sum of i t s convergent ..sequences. Then, .by ,(2.13), the quotient map must be pseudo-open. Conversely, for any Hausdorff space X each convergent sequence in X i s a metric space and hence a Frechet space. It follows from (2.3.2) and (2.11) that X i s Frechet. 2.15 Corollary Among Hausdorff spaces, the following statements are equivalent. (1) X i s a Frechet space. (2) X i s the continuous pseudo-open image of a metric space. (3) X is the continuous pseudo-open image of a f i r s t -countable space. Proof By virtue of (1.32) and (2.14), X i s the continuous pseudo-open image of a zero-dimensional, locally compact, complete metric space. Since metric spaces are first-countable and first-countable spaces are Frechet, (2) implies (3) and (3) implies (1) by (2.11). - 69 -As previously stated, first-countable T^-spaces are precisely the continuous open images of metric spaces. In view of (1.30) and (2.14), Franklin posed and answered negatively the question of whether every first-countable (Hausdorff) space i s the continuous open image of a disjoint topological sum of convergent sequences. Any such sum Is a Baire space as are continuous open images of Baire spaces ( [5 ] , p.767). J.de Groot's Corollary i s applicable here because every convergent sequence in a Hausdorff space i s metrizable. However, many spaces are first-countable Hausdorff but not Baire spaces. The rationals Q_ is an example of such a topological space. An unanswered question of Alexandroff asks whether or «>not there "is -a >«*f ir-st-countable'ecompact- -Hausdorff - space vwith cardinality > c. The corresponding question for Frechet spaces i s t r i v i a l l y answered by the following. 2.16 Proposition The one point compactification of any discrete space i s a Frechet space. Proof Let X* = X U {<*>} be the one point compactif ication of the discrete space X. For any subset A of X*, 0 0 e(cl A)-A i f and only i f A contains i n f i n i t e l y many points. Moreover, any sequence {x^ : n E io} i n A satisfying = x m i f and only i f n = m converges to 0 0. Therefore, i f °° e e l A, A contains a sequence converging to 0 0. - 70 -If x E (cl A)n X then, since X i s discrete, x e A. Hence X* i s Frechet The topological spaces X = (H x H ) / E a n d M^., of (1.36) and (1.37) respectively, are sequential spaces with unique sequential limits which are not Hausdorff. Although i s not Frechet, the next result shows that X i s . 2.17 Example There are Frechet spaces with unique sequential limits which are not Hausdorff. Proof Let H and X = (H x H ) ^ be as in (1.36). Recall that <f> : H x H — > X i s a quotient map of the first-countable space ^x^H^onto^t'he^non-HauBdor-f'f^sequent'ial^'sp'ace"X-'which *has unique sequential limits. To establish that X i s Frechet, i t suffices to prove that <j> i s a pseudo-open map. Let x e X and suppose that U i s an open neighbourhood of <J> ^ (x). If x E X-{q, i} then <j> ''"(x) = x and for any neighbourhood V of x such that V fl A = 0, x e U H V = *(U n V) C <J)(U). If x = q and G i s an open neighbour-hood of A^ disjoint from A ^ q e <J> (U H G) = int <j>(un G) C int <J>(U) . Similarly, i f x = i then i E <f>(U). Thus <j> i s continuous pseudo-open and X i s Frechet. - 71 -Let Y be a Frechet space with, unique sequential limits. It follows from (1.41) that i f each point of Y has a neighbourhood basis consisting of countably compact sets, then Y i s Hausdorff. The succeeding example shows that simple compactness i s not enough to ensure that Y i s Hausdorff. 2.18 Example There i s a countable, compact, Frechet space with unique sequential limits which i s not Hausdorff. Proof Let Y = (W x hi)U {p, q> with p f q and {p, q} H (hi x -hi) j 0. Let each ( i , j) e W x W be an isolated point. For a basis of neighbourhoods of p take a l l sets of the form {p} U ( U {(i,j) : i , j e hi; i . >_ k}) where k e M, and for q take a l l sets of the form i^U (\J{±, j) : i , j e W; j >_ j ' } ) where each j e hi. The topological space Y i s compact because i f U and V are open neighbourhoods of p and q respectively, then Y-(U U V) is f i n i t e . It i s also clear that Y i s not Hausdorff since p and q cannot be separated by disjoint open sets. Then, since Y-{p, q} i s discrete, i f some sequence converges to two distinct points, they must be p and q. However,a sequence {(i ^ , j ) : n E co} in W x W can converge to p only when { i ^ : n e w} is unbounded, and to q only when { i ^ : n e co} i s bounded. Therefore Y has unique sequential limits. It remains to prove that Y i s Frechet. Let A be a subset of Y. Each point i n Y-{q} has a countable - 72 -neighbourhood basis. Thus for each y e Y-{q}, y E c l A i f and. only i f there exists a sequence in A converging to y. If for each i E W, A n ({i} x hi) i s f i n i t e , q £ c l A. If for some i i t i s inf ini te , there i s a sequence in A converging to q. Hence Y i s a Frechet space. The topological product spaces X x X and Y x Y, where X and Y are the non-Hausdorff Frechet spaces of (2.17) and (2.18) respectively, are not Frechet. In fact, these products are not even sequential. If these product spaces were sequential then, by (1.38.2), X and Y would be Hausdorff. Similarly, the spaces Q x and £' x OJ of (2.4.2) are products of Frechet spaces which are not sequential. Example 2.19 shows that this need not always be the case. This example also demonstrates that the term "sequential" cannot be replaced by "Frechet" in (1.24) and (1.25). 2.19 Example There i s a product of two Hausdorff Frechet spaces which i s sequential but not Frechet. In addition, one of the spaces i s normal, compact, and first-countable. Proof Let X be the real l ine with the integers ident if ied, and let I = [0, 1] be the closed unit in terval . Both X and I are Hausdorff. Furthermore, X i s Frechet and I i s a normal, compact first-countable space. It follows from (1.25.2) that X x I i s - 73 -sequential. To see that X x I i s not Frechet, define A C X x I by A = U {A : n e W} where A = {(n - -) : k E W}. Then J n n k n (0, 0) E c l A since (0, 0) e c l {fr-1, -) : n E W} and n {(n - 1 , : n e W} CL A. But no sequence i n A converges to (0, 0) because no sequence contained i n A converges i n R x I to (k, 0) for any k e Z. 2.20 Example The product of two continuous pseudo-open maps may be a quotient map without being continuous pseudo-open. Proof Let X and I be the Hausdorff Frechet spaces of (2 .19) . By (2.13), the quotient maps <|>x : X* > X and cf> : I * > I are continuous pseudo-open. Since X x I i s sequential, (1.35.2) implies that tb x <j) • X* x I * > X x I i s a quotient map. X i However, <f> x A cannot be continuous pseudo-open because X* x I * X X. i s f i r s t - c o u n t a b l e but X x I i s not Frechet. The next two r e s u l t s , which further i l l u s t r a t e the non-productive nature of Frechet spaces, generalize Harley's Theorem ( [12]) . The author provides the proof. 2.21 Lemma The product of two Frechet spaces, one of which i s d i s c r e t e , i s Frechet. - 74 -Proof Let X and Y be Frechet spaces, and assume that Y i s discrete. Let B be a subset of the topological product space X x Y , and suppose that (u, v) e c l B. Since Y i s discrete, U x {v} is a neighbourhood of (u, v) for any neighbourhood U of u. Therefore v e {y e Y : (x,y) e B}. Then u e c l {x e X : (x,v) e B} and there is a sequence '{u^  : n e to } in {x : *(x,v) e *B} converging to u. The sequence {(u n > v) : n e t o } i s contained in B and converges to (u, v). 2.22 Theorem Let X be a Frechet space. Let A be a subset of X satisfying the property : there i s a sequence {U^ : n e to } ''"o'f open subsets of X such that"""(T)""U^C."*un+1 >' (2) A i s •contained in U = U {U : n e t o } , and (3) A U U i s not open. Let X / A n n r /A denote the quotient space obtained from X by identifying the points in A. Then i f X is ^ and Y i s Hausdorff, X^A x y i s Frechet i f and .only i f Y is discrete. If A i s closed, the T^ hypothesis may be replaced by regularity. Proof If .Y i s discrete then, since i s Frechet (by 2.3.4), the topological product space X ^ x Y i s Frechet (by 2.21). To establish the converse, let i = g(A) where g : X > i s the quotient map, and suppose that y is not an isolated point - 75 -of Y. Then y e cl(Y-{y}) and consequently, since Y is Frechet (by 2.3.2), there i s a sequence {y^ : n E co} in Y-{y} converging to y. Since Y i s Hausdorff, i t can be assumed without loss of generality that the y n are distinct. Let W = -U {U-(UnU A) x {yn} : n E co}, Then ( i , y) E c l W. Indeed, this Is so because AU being not open implies that ( i , y^) e c l (U-(U n U A) x {y n}). Since X^A x Y i s Frechet, there exists a sequence {(r^, s^) : k E to} in W converging to ( i , y) . The Hausdorff hypothesis on Y implies that {s^ : k E to} i s a subsequence of {y : n E co}. Let y = s, for each k E co. n Jn n, k k Thus {(r, , y ) : k E to} i s a sequence in W converging to ( i , y) k n k with each r, e U-(Un U A). Since X i s T, (or X is regular and A k k ± i s closed), for each k E to there i s an open neighbourhood of A k such that r, t G, . Let U = U D ( H {G. : j < k}) for each n, k k n n j — k e co; and let m^  be the largest member of to satisfying r ^ i U^m^_i-j« It follows that U (U : k E to} is an open neighbourhood of A ™k disjoint from the sequence {r^ : k E to}. But then {r^ : k E to} cannot converge to i which contradicts {(r, , y ) : k E to} converging K. II. k to ( i , y). Therefore {y} must be open, and the proof i s complete. - 76 -Several examples of sequential spaces which are not Frechet have already been given. After another such example, a characterization of those sequential spaces which are also Frechet follows. The non-Frechet sequential space M of (1.17) is used to give a characterization in the Hausdorff case. 2.33 Example There is a compact sequential Hausdorff space which i s not Frechet. Proof Let F be an i n f i n i t e maximal pairwise almost disjoint family of i n f i n i t e subsets of the natural numbers hi (Two sets U and V are almost disjoint i f and only i f UH V is f i n i t e ) . To establish the existence of F, let G be the collection of a l l i n f i n i t e pairwise almost disjoint families of i n f i n i t e subsets of W. The set G is partially ordered by set inclusion. Note that G ^ 0. Indeed, for each real number r there i s a sequence {x^ : n E co} of rational numbers converging to r. Then i f f : Q > W i s a bijection between the rationals and the natural numbers, {{f(x ) : n e co} : r e R} e G. Now, let {E : a £ A} Vi 3. be a chain in G and let E = U {E : a E A}. For any pair E, cl F £ E there is some c e A such that E, F e E which implies that c E O F i s f i n i t e . Therefore E E G and consequently E is an upper bound of the chain {E : a E A}. Then, by Zorn's lemma, - 77 -there i s a maximal element F of the set G. Let i() = F U N with points of W isolated and neighbour-hoods of F E F those subsets of ty containing F and a l l but f i n i t e l y many points of F. Clearly ty is Hausdorff. Furthermore, ty i s local l y compact because F U {F} i s a compact neighbourhood of F in It ...follows-.from ..(.[,16.].,. JTheorem 5.21) -that •= 4-..U .{•«*>•}., ...t-he-one -point compactification of ty, i s a compact Hausdorff space. By definition of the topology on ty*t i t is clear that °° e c l W. However, i f {x n : n e to} i s a sequence of distinct points in W then, since F is maximal, {x : n E to} f l F i s i n f i n i t e for some F e F and hence n {x^ : n E to} converges to F. Therefore no sequence in W converges ,„to .^and—so l.^ *.,.is.„not,.iEr.echet. It remains to show that ty* is a sequential space. Suppose that V i s a sequentially open subset of ty*, and let x E V. If x e W, x E int V because {x} is open. If x e F then, since any sequence of distinct points in x converges to x, x-V is f i n i t e and hence {x} U {n : n e x O V} is a neighbourhood of x contained in V. Assume now that x = 00. Since F U {F} is a neighbourhood of F E F and ty*-\] is compact for each open neighbourhood U of °°, any sequence of distinct points of F converges to °°. Accordingly, V contains a l l but f i n i t e l y many members of F. Let F-V = {F i : i <_ m} where m E W. If {U^ : i _< m} i s any f i n i t e collection of open sets satisfying F . d U. then, since F i s maximal, U-[V.U ( (J {U : i £m})] i s f i n i t e - 78 -and consequently IJJ*-V is compact. Hence TJJ* being Hausdorff implies that ifj*-V is a closed compact subset of i f i . Therefore V is open whenever » e V. Thus every sequentially open subset of i s a neighbourhood of each.of i t s points, and so i s sequential. . Since no sequence in hi converges to •», the singleton {<»} i s sequentially open but not open in the subspace i p*-F. Hence ^*-F is a non-sequential subspace of ty*. The next result proves that such a subspace must always exist in sequential spaces which are not Frechet. 2.24 Proposition A sequential space i s Frechet i f and only i f i t is' hereditarily sequential. Proof If a sequential space i s Frechet, every subspace i s Frechet and hence sequential. Conversely, i f a topological space X i s hereditarily sequential, (1.35.1) implies that 6^ i s a hereditarily quotient map with Frechet domain. Then -$ is continuous pseudo-open A and therefore X i s Frechet. 2.25 Theorem A Hausdorff sequential space i s Frechet i f and only i f i t contains no subspace which, with the sequential closure topology, i s homeomorphic to the topological space M of (1.17). Proof Let Y be a subspace of a sequential space X, and l e t h : M > Y g be a homeomorphism of M onto the subspace Y provided with the sequential closure topology. If Y ^  Y , Y i s a non-sequential subspace of X and hence, by (2.24), X i s not Frechet. If Y = Y then, since M-W is a non-sequential subspace of s M, Y-h(W) = h(M)-h(W) = h(M-W) i s a non-sequential subspace of X. Again by (2.24), X i s not Frechet. Conversely, suppose that X i s not Frechet. Then there exists a subset B of X such that c l B ^  B' where B' i s the set of limits of sequences in B. Since X i s sequential, there i s a sequence S = {x_^  : i e co} in B' converging to some point x e(cl BVB'. The sequence S i s not frequently in B because otherwise i t has a subsequence in B converging to x t B'. Consequently, i t can be assumed that the x. are distinct and S C B'-B. Then, since l x^ e B'-B, there exists a sequence {x^ : j e co} In B converging to x^. The x ^ (for i , j e to) may be taken a l l distinct. This i s so because X i s Hausdorff and S converges to x; that i s , there exists a family {U. : i e to} of pairwise disjoint open sets satisfying - 80 -e U\. Since x.-^ .B', no sequence consisting of the x^ converges to x. Thus the topological space Y where the subspace s Y = {x} U (x. : i e W} U {x. . : i , j e hi} is homeomorphic to M. 1 xj It i s easy to verify that the non-Frechet Hausdorff sequential space ty* of (2.23) satisfies (2.25). If {E^ : i c co} i s any collection of pairwise disjoint subsets of hi then, since "F i s maximal, for each i e co there exists F. e F such that F.H E. ' x x x i s i n f i n i t e . Thus {F^H E. : i e w} i s a pairwise disjoint sequence in F converging to °°. Let in±^ 3 E co} be any sequence i n E^ such that the n i j (j E w) are distinct. Clearly, each i^^j : j e co} converges to F^ D E^ and the n ^ ( i , j e co) are a l l distinct. Then, since no sequence in hi converges to «>,• the subspace {<»} U {F^O E_^  : i e hi} U {n^j : i , j e hi} provided with the sequential closure topology i s homeomorphic to M. Chapter 3 Generalized Sequential Space Methods When a topology is specified by i t s open sets, the importance of basis and subbasis i s well-known. In the same way, .the ...concepts ...of ,,cony.er,genc.e „b,asis.„and ..convergence...subbasis .are prominent in the study of topological spaces whose topologies are determined by their convergence classes. For example, one can study convergence subbases and convergence bases consisting of convergent sequences in lieu of studying sequential spaces and Fre'chet spaces respectively. The notion of convergence subbasis i s also useful in the investigation of generalized sequential spaces. In this chapter, --•the'"topological "''spates ^  sets are examined; these spaces are called m-sequential (m-Frechet) spaces. It w i l l be shown that any topological space can, for sufficiently large m, be so described. 3.1 Definition Let X be a set and let C be a class of pairs (S, x) where S i s a net and x i s a point in X. The class C i s a p-convergence class on X i f and only i f i t satisfies : If ( { x n : n e D},x)eC and E i s a cofinal subset of D, then ({x : n e E}, x) e C. A n p-convergence class in which a l l of the nets are sequences i s called a sequential p-convergence class. - 82 -Observe that each convergence class Q16] , p..74) i s a p-convergence class but the converse need not be true. The convergence associated with a p-convergence class C on a set X can be studied topologically by means of the largest topology on X in which the C-nets (i.e., a l l of the nets in C) are convergent. 3.2 Theorem Let C be a p-convergence class on a set C. For any subset A of X, l e t t - c l A be the smallest set containing A and closed with respect to the formation of limits of C-nets. Then t - c l i s a closure operator and hence defines a topology T(t) for X ([16], Theorem 1.8). This i s the largest topology on X in which the C-nets converge. Proof It is f i r s t shown that t - c l i s a closure operator. Since a net i s a function on a directed set, and the set i s non-empty by definition, t - c l <f> i s empty. By definition of t - c l , A C t - c l A for each subset A of X. Then t - c l A C t - c l ( t - c l A). Again by definition of t - c l , t - c l ( t - c l A) i s the smallest set containing t - c l A and closed with respect to the formation of limits of C-nets. But t - c l A i s closed with respect to the formation of limits of C-nets and so t - c l ( t - c l A) CZ t - c l A. Hence t - c l A = t - c l ( t - c l A). It remains to prove that t-cl(A U B) = ( t - c l A) (J ( t - c l B). Clearly A U B C ( t - c l A) U ( t - c l B) C t - c l (A U B) . To establish the opposite inclusion, l e t S = {x : n e D} be a net in ( t - c l A) U ( t - c l B) with n (S, x) e C. Let D A = {n E D : X r e t - c l A) and D g = {n E D : X R E t - c l - 83 -Since D.U D_ = D, either D. or D„ is cofinal in D. It can be A B A B assumed without loss of generality that is cofinal in D. Then S.. = {x : n e D.} is a subnet of S in t - c l A and (S,, x) e C. 1 n A 1 Hence x e t - c l A C" ( t - c l A) U ( t - c l B) and consequently ( t - c l A) U ( t - c l B) i s closed with respect to the formation of limits of C-nets. Let (S, x) e C. If S does not converge to x in (X, T(t)), there i s an open neighbourhood U of x such that S i s not eventually in U. Then S i s frequently in X-U and there i s a subnet S^  in X-U with (S^, x) e C. But since U i s open, X-U = t - c l (X-U) and hence x e X-U. The contradiction shows that S must converge to x with respect to T(t), and hence that each C-net converges in (X, T(t)). Suppose now that T^ i s a topology on X in which the C-nets converge. I f V e T^ then for each net pair (S, x) e C such that S C X-V, x e X-V. Thus X-V = t-cl(X-V) which implies that V e T(t). / 3.3 Corollary Let X be the topological space provided with the topology T(t) derived from a p-convergence C. Then X i s T^ i f and only i f C satisfies : If S = {x : n e D} i s a net i n X such that n x = x for each n e D and y f x, then (S, y) £ C. - 84 -Consequently, i f C i s a sequential p-convergence class and X has unique sequential limits, X i s T^. Proof If X is and y ^  x, S cannot converge to y and so (3.2) implies that ( S , y) £C. Conversely, i f S q is a net in {x} and z ^  x then, since (S Q, Z) £ C, z £ t-cl{x}. Hence {x} is closed and there-fore X is T 1. 3.4 Proposition Let X be the topological space provided with the topology T(t) derived from a sequential p-convergence class C. Let C(T(t)) denote the class of convergent sequences in X . Then ' T = C(T'(t)') i f C s a t i s f i e s : (1) If S = {x : n e to} i s a sequence such that x = x n n for each n e to, then ( S , x) e C. (2) If S is a sequence and ( S , x) £ C then there i s a subsequence of S, no subsequence of which together with x i s a member of C. (3) If ( S , x) e C and (S, y) e C then x = y. Proof It i s clear that C<lC(T(t)). To prove the opposite inclusion, suppose that ( S , x) £ C. By (2), there i s a subsequence S q = {y^ : n e to} of S, no subsequence of which together with x i s a member of C. It can be assumed without loss of generality that y 4 x for each n e to. - 85 -Either there exists ( S ^ , z) e C such that is a subsequence of S q or no such sequence pair exists. In the f i r s t case, any C-net in S^U {z} is either a subsequence of or frequently equal to some point in S^. Then conditions (2) and (3) imply that U {z} is a closed subset of X disjoint from x. In the second case, S o is a closed subset of X disjoint from x. In both cases, S has a subsequence not converging to x. Hence ( S , x) £ C(T(t)). Another closure operator is defined in the following proposition. 3.5 Proposition Let C be a p-convergence class on a set X, and for each subset A of X let c-cl A be the union of A and the limits of those C-nets contained in A. Then i f C is a convergence class, c-cl i s a closure operator, and ( S , x) e C i f and only i f S converges to x with respect to the topology associated with c - c l . Proof This is given in ([16], Theorem 2.9). In the presence of a convergence class, c-cl is precisely the closure operator described in that theorem. 3.6 Corollary If C^ and C^ are convergence classes on a set X and T and T„ are the associated topologies, then C C C. i f and a 3 a 3 only i f Ta^> T . - 86 -Proof If (S, x) e C^, S i s eventually i n each neighbourhood of x in (X. T ). Thus T.CT T implies that S i s eventually in each a 0 a neighbourhood of x i n (X, T.) and hence that (S, x) e C . Conversely, p p suppose that C C C„ and let U e T„. If (S, x) e C and S CI X-U, then (S, x) E C c and U e T Q implies that x e X-U. Therefore X-U ""p P i s closed in (X, T ) and so U e T . a a 3.7 Proposition Let C be a p-convergence class on a set X, and let T(t) be the topology for X associated with the t-closure operator. Then T(t) Is the topology with the smallest convergence class containing C. Proof Let C(T(t)) denote the convergence class for (X, T(t)). According to (3.2), T(t) i s the largest topology on X in which the C-nets converge. Then, i f T i s any topology on X whose convergence class C(T) contains C, T O T ( t ) . The preceding result implies that C(T(t))CC(T).-If C i s a p-convergence class on a set X, then c - c l need not be idempotent and hence not a closure operator. Let T(c) denote the topology associated with c-cl whenever c - c l ( c - c l A) = c-cl A for each subset A of X. Clearly, c - c l A i s a subset of t - c l A, and i t can be a proper subset. Observe that a topological space i s Frechet i f and only i f for each subset A of X, c l A = c-cl A with respect to the convergent sequences in X; similarly, X i s sequential i f and only i f c l A = t - c l A. Therefore, since not a l l sequential spaces are Frechet, c - c l and t - c l need not coincide. In particular, consider the topological space M of (1.17). As previously observed, no sequence in hi x W converges to 0. Hence c-cl(W x hi) = M-{0} whereas t-cT(W x hi') = "M. The t-closure operator, however, can be constructed inductively by iteration of c - c l . Define A° = A and for each ct 6 successor ordinal a l e t A = c-cl A where a = B + 1 for some ordinal 8. ( 3+1 denotes the ordinal successor of 3). If a i s a limit ordinal define A a = U {A^ : 3 < a}. It i s clear that A a C t - c l A for each ordinal a, and i f A a = t - c l A then A^ = t - c l A whenever 3 _> a. For any subset A of X the cardinality of the number of iterations of c-cl to obtain t - c l A i s <_ 2 m where m i s the cardinality of the set X. Then, since the ordinals are well-ordered, for each x e t - c l A there Is a smallest ordinal n such that x belongs to the n-th iterate of c - c l on A; that i s , x e A whenever a _> n and x i A whenever a < n. 3.8 Definition Let C be a p-convergence class on a set X and l e t A be a subset of X. (1) A point x e t - c l A is said to be of Baire order n (write ord x = n) with respect to C and A i f and only i f n i s the smallest ordinal such that x is a member of the n-th iterate of c-cl on A. (2) The Baire order of a set A Cord A) i s defined as sup {ord x : x e t - c l A}. 3.9 Proposition Let C be a p-convergence class for a set X. Then TCt) = TCc) i f and only i f every subset of X has Baire order <_ 1. Proof The ^ topology T(t) coincides-with-'T-(c-) — i f -and -only i f t - c l A = c-cl A for each subset A of X, which occurs i f and only i f every subset of X has Baire order <_ 1. 3.10 Definition Let C be a p-convergence class on a topological space X. Cl) C is a convergence subbasis for X, or for the topology on X, i f and only i f the topology on X Is the topology with the smallest convergence class containing C. C2) C i s a convergence basis for X, or for the topology on X, i f and only i f C i s a convergence subbasis for X and every subset of X has Baire order _< 1. 3.11 Proposition Let C be a p-convergence class on a topological space X. Cl) C is a convergence subbasis for X i f and only i f X has topology TCt). (2) C i s a convergence basis for X i f and only i f X has topology TCc). - 89 -Proof By virtue of (3.7) and (3.10), (1) i s clear. Then (1) together with (3.9) implies (2). 3.12 Definition Let m be an.infinite cardinal number. An m-net i s a net whose directed set i s of cardinality <_ m. 3.13 Definition Let X be a topological space, and let m be an i n f i n i t e cardinal number. (1) X i s m-sequential, or an m-sequential space, i f and only i f i t has a convergence subbasis in which a l l of the nets are m-nets. (2) X i s m-Frechet, or an m-Fre'chet space, i f and only ..-if. -it-^as^a 1^ 3.14 Proposition (1) Every m-Frechet space i s m-sequential. (2) If a topological space i s m-sequential then i t is m^-sequential whenever m^ >_ m. Similarly, an m-Frechet space is m^-Frechet i f > m. Proof The proof of (1) is obvious because, by definition, every •convergence basis i s a convergence subbasis. Since every m-net is an m.-net for mn >_ m, (2) i s also clear. - 90 -The next two results give several equivalent formulations of the definitions of m-sequential space and m-Frechet space respectively. 3.15 Proposition The following statements about an arbitrary -topological space X-are-equivalent. (1) X is m-sequential. (2) X has topology T(t) with respect to some p-convergence class consisting of m-nets in X. (3) A subset F of X i s closed i f and only i f no m-net in F converges to a point not in F. (4) A subset U of X is open i f and only i f each m-net "in X converging to a point lri""U i s eventually in U. (5) The class C of a l l pairs (S, x) where S is an m-net in X converging to the point x i s a convergence subbasis for X. Proof If X i s m-sequential then, by definition, X has a convergence subbasis in which a l l of the nets are m-nets. According to (3.11), X has topology T(t) with respect to and therefore (1) implies (2). If F i s a subset of X with no m-net in F converging to a point not in F, no C^-net in F converges to a point not in F; consequently F = t - c l F = c l F and (2) implies (3). Suppose that U Is a subset of X such that each m-net i n X converging to a point in U i s eventually i n U. Let S be an m-net in X-U converging to a point x . Then x e X-U - 91 -because otherwise S i s eventually in U. Thus (3) implies that X-U i s closed and hence that U is open. To establish that (4) implies (5), l e t T be any topology for X in which the C-nets converge. Then i f V e T, every m-net in X converging to a point in V i s eventually in V, and so V i s open by (4). Accordingly X has the largest topology i n -which-the .C-nets-converge. .It .follows ..from (3.6) that (4) implies (.5). By definition, (5) obviously implies (1). 3.16 Proposition For any arbitrary topological space X, the following are equivalent. (1) X i s m-Frechet. (2) The class C of a l l pairs (S, x) where S i s an m-net "in X converging to x i s a cohvergerice"'basisJ ¥ or X. (3) The closure of any subset A of X is the set of limits of m-nets in A. (4) X has topology T(c) with respect to some p-convergence class consisting of m-nets in X. Proof If X i s m-Frechet, X i s m-sequential and hence the class C i s a convergence subbasis for X. Moreover, every subset of X has Baire order <^  1. Therefore C i s a convergence basis and (1) implies (2). It follows from (2) that X has the topology T(c) associated with C. Consequently (2) implies (4). In addition, (2) i s equivalent to (3) because for each subset A of X, x e c l A i f and only i f there exists an m-net in A converging to x. The proof that (4) implies (1) i s clear form (3.11). The following corollary together with (3.15) and (3.16) shows: (1) A topological space is sequential i f and only i f i t has a convergence subbasis in which a l l of the nets are sequences. (2) A topological space i s Frechet i f and only i f i t 'has a convergence "'basis i n which a l l "of "the "nets are sequences. Furthermore, this result implies thkt.'eyiery Frechet space i s m-Frechet and that every sequential space i s m-sequential. 3.17 Corollary (1) A topological space i s sequential i f and only i f i t is /^-sequential. (2) A topological space i s Frechet i f and only i f i t i s U -Frechet. Proof Since every sequence i s an h{o~net, the sequential spaces and the Frechet spaces obviously satisfy (3.15.3) and (3.16.3) respectively. To prove the converses, i t i s f i r s t shown that every •non-trivial /^-rnet has.a cofinal sequence. Let {x n : n e D} be an }-jo-net and l e t g :. to '> D be a bisection. Because D i s a directed set, for each k E to there exists n^ E to such that g C 1 1 ^ ) ^_ g C O for every i < k . Then {x , N : k E to} i s a subnet of {x : n E D}. - g ( n k ) n Suppose that F i s a sequentially closed subset of a topological space X. If S i s an hj-net i n F converging to some point x, x e F since otherwise S has a cofinal sequence which i s eventually in X-F. Then i f X i s ^ - s e q u e n t i a l , F is closed by (3.15). Thus (1) i s proved. To complete the proof of (2), suppose that A i s a subset of an |y -Frechet space X and let x e c l A. By virtue of (3.16), there i s an /^-net in A converging to x. Then, since every non-t r i v i a l • rj -net has a cofinal sequence, there exists a sequence in ,A...converging,,to x.. In view of (3.15) and (3.16), i t is easy to see that many of the properties of sequential spaces and Frechet spaces can be generalized to m-sequential spaces and m-Frechet spaces respectively, by simply replacing "sequences" with "m-nets". This i s so whenever those properties of sequences used, can be generalized to m-nets. Nevertheless, for greater generality i t i s convenient to state "result's ih^terWs**'^ . -Let C be the class of convergent m-net pairs in a topological space X. According to (3.15), C is a convergence subbasis for X i f and only i f X i s m-sequential. However, i t is possible to have a convergence subbasis which is a proper subset of C. Although using a smaller convergence subbasis may increase Baire order, there i s an upper bound. 3.18 Proposition If X i s an m-sequential space with any convergence subbasis, then no element of X has Baire order equal to the least ordinal of cardinality m . (m denotes the cardinal successor of m) - 94 -Proof Let to denote the least ordinal of cardinality m~*\ Then a i s of the form 8 + 1 for some ordinal 3, where to is the least p ordinal of cardinality m. Thus to^ i s regular and hence not the supremum of any set B of s t r i c t l y smaller ordinals i f the cardinality of B i s _< m. Assume that A is a subset of X and let x e c l A. Since X is m-sequential, there exists an m-net {x n : n E D} such that x e c l A and ord x < ord x. Then ord x = sup {ord x : n e D} n n ^ n and consequently ord x < ">a-3.19 Theorem Let C be a convergence subbasis for a topological space X, l e t Y be a subset of X, and l e t V be the trace of C on Y. (i.e., V = {((x n : n e D}, x) e C : X r e Y for each n e D, x e Y}.) Then P i s a convergence subbasis for a topology on Y which is larger than the relative topology. This induced topology coincides with the relative topology on Y i f Y i s closed or open in X. The two topologies coincide for a l l subsets of X i f and only i f C i s a convergence basis. Proof The space X has the topology T(t) associated with C, and the trace V i s clearly a p-convergence class on Y. For each subset A of Y define u-cl A to be the smallest set containing A and closed with respect to the formation of limits'of P-nets. By (3.2), u-cl i s a closure operator on Y and hence defines a topology T(u) for Y. It - 95 -follows from (3.11.1) that V i s a convergence subbasis for T(u). Furthermore, T(u) i s larger than the relative topology on Y because the P-nets converge in the relative topology and T(u) i s the largest topology on Y in which the P-nets converge. To establish that these topologies on Y coincide when Y i s closed or open, or when C i s a convergence basis, i t suffices to show that the two topologies have the same closed sets. Let F be a subset of Y. By definition of the u-closure operator, u-cl F CZ ( t - c l F) p| Y. To prove.the opposite inclusion, assume that x e ( t - c l F) H Y with ord x = X with respect to C and F, and proceed by transfinite induction on X• If X = 0, ( t - c l F) H Y = F° H Y m F fVY = F C u-cl F. If X = 1 then "(t-cT F) Pi Y =~r 1 n' Y = ( c - c T T ) r T T C ti-cl F. Thus "the proof i s complete for the case in which C is a convergence basis. For X: > 1 consider separately the cases Y i s closed and Y i s open. F i r s t , suppose that Y i s closed. Then t - c l F C. t - c l Y = Y. By the induction hypothesis, there exists a net pair ( { x n : n e ^W, e C with x n E ( t - c l F) H Y and ord X r < X for each n E D. Consequently each x e u-cl F and ({x : n e D}, x) e V, which implies that n n x e u-cl F. Assume now that Y i s open. By the induction hypothesis, there i s a net pair ({y : n e D}, x) e C with y E t - c l F and r Jn 'n ord y < X for each n E D. Since Y i s open and x e Y, the net •'n {y : n E D) i s eventually i n Y. Thus E = {n £ D : x e Y} i s a - 96 -cofinal subset of D and so ({y n '• n e E}, x) e V. Therefore y^ e u-cl F for each n e E implies that x e u-cl F. It remains to show that i f the two topologies are the same then C i s a convergence basis. If C is not a convergence basis for X, there i s at least one Baire order 2 situation. That i s , there exists a subset B of X with x e t - c l B and ord x = 2. Let Y = B U {x}. Then x e ( t - c l B) H Y but, since ord x = 2, (c-cl B) n Y = B and no net in B converges to x e Y. Thus x £ u-cl B and the topologies are different. 3.20 Corollary Every open or closed subspace of an m-sequential space i s m-sequential. A topological space i s m-Frechet i f and only i f i t i s hereditarily m-Frechet i f and only i f i t i s hereditarily m-sequential. Proof Let Y be an open or closed subspace of an m-sequential space X with a convergence subbasis C consisting of m-nets. According to (3.19), the trace of C on Y i s a convergence subbasis for the relative topology on Y. Hence Y i s m-sequential. It i s obvious that every hereditarily m-Frechet space i s m-Frechet. Conversely, i f X i s an m-Frechet space with a convergence basis C consisting of m-nets then, by (3.19), for every subspace Y of X the trace of C on Y i s a convergence subbasis for the relative topology on Y. Thus Y i s m-sequential and consequently every m-Frechet space is hereditarily m-sequential. In addition, any subspace of an m-Frechet space is hereditarily m-sequential. It remains to show that every hereditarily m-sequential space i s m-Frechet. Assume that X i s a hereditarily m-sequential space with a convergence subbasis C in which a l l of the nets are m-nets. If Y i s a 'subspace of -X, -then -Y -is- m-sequenti-a-L- -and - therefore -has a convergence subbasis V consisting of m-nets. The trace of C on Y surely coincides with V. The preceding theorem implies that C i s a convergence basis for X and hence that X i s m-Frechet. 3.21 Proposition If X i s the disjoint topological sum of any family {X : a E A} of topological spaces where each X has a a a convergence subbasis C , then C =U{C : a e A} i s a convergence Si SL subbasis for X. If each C is a convergence basis, so i s C. Si Proof Let (S, x) E C and suppose that U i s an open neighbourhood of x in X. Then (S, x) e C for some a E A and therefore, since cL U f l X i s open in X , x E U H X for a l l n sufficiently large, a a n a . Thus the convergence class on X contains C. Now let T denote the usual topology on X and l e t T^ be any topology for X whose convergence class CCT )^ contains C. If V £ T, V D X £ i s not open in X for some c E A. From this, i t follows that there exists a - 98 -C-net in X£-V converging to a point y e V Pi X c > Then, since (Sn , y) e C and C C C ( I ), V £ T . Consequently, T i s the largest X ex ct topology on X in which the C-nets converge, or equivalently by (3 .6) , T is the topology with the smallest convergence class containing C. Hence C i s a convergence subbasis for X. - •Assume -now -that - each -C^.is- a -convergence-.basis.. To complete the proof, i t suffices to show that every subset of X has Baire order <^  1 with respect to the convergence subbasis C. Let F be a subset of X. Then (c-cl F) f) X = c- c l (F) = c l (F) = a A a A a (cl F) D X . Therefore (c-cl F) fi X i s closed in X for each a a a a e A, and hence c-cl F i s closed in X. 3.22 Corollary The disjoint topological sum of any family of m-sequential spaces i s m-sequential. The disjoint topological sum of any family of m-Frechet spaces i s m-Frechet. 3.23 Definition Let C be a p-convergence class on a set X. For any function with domain X let fC denote the set of a l l net pairs ({f(x n) : n e D}, f(x)) for ({X r : n e D>, X) e C. - 99 -3.24 Theorem Let f : X > Y be a function of a topological space X into a topological space Y, and l e t C be a convergence subbasis for X. (1) The function f i s continuous i f and only i f fC i s contained in the convergence class of Y. (2) If f is surjective, fC i s a convergence subbasis for Y i f and only i f Y is a quotient space. Proof (1) Let ({f(x ) : n e D}, f(x)) e f C. Since C i s a n convergence subbasis for X, each C-net belongs to the convergence class of X. Accordingly, {x^ : n e D} converges to x in X. Then, since f i s continuous, ( f ( x n ) : n e D} converges to f(x) in Y. Conversely, "let A be a closed subset of Y and suppose that ({x^ : n e D}, x) e C with x^ e f ''"(A) for every n e D. Clearly each f ( x n ) £ A. Because A i s closed and by hypothesis {f(x n) : n e D} converges to f(x), f(x) e A. Thus x e f '''(A) and f \ A ) = t - c l f ^"(A). Then, since C i s a convergence subbasis for X, f "'"(A) is closed. (2) By definition, fC i s a convergence subbasis for Y i f and only i f the topology on Y is the topology with the smallest convergence class containing fC. According to (3.6) and part (1), fC i s a convergence subbasis for Y i f and only i f the topology on Y i s the largest topology such that f i s continuous. - 100 -3.25 Corollary Every quotient of an m-sequential space i s m-sequential. Proof Let f : X > Y be a quotient map of an m-sequential space X onto a topological space Y. The space X has a convergence subbasis C i n which a l l of the nets are m-nets. It i s obvious that each net "pair 'in -f C -is 'an-m-net-'pair. Then,-since "(3.24) implies that fC i s a convergence subbasis for Y, Y i s m-sequential. Example 2.5 shows that the quotient of a Frechet space need not be Frechet. Consequently, i f C i s a convergence basis for a topological space X and f i s a quotient map with domain X, i t i s only possible to conclude that fC i s a convergence subbasis for the quotient space. However, fC i s a convergence basis when-ever f is continuous pseudo-open. 3.26 Theorem Let f : X > Y be a surjection of the topological space X onto the topological space Y, and let C be a convergence basis for X. Then f is continuous pseudo-open i f and only i f fC i s a convergence basis for Y. Proof Let Y^ be a subspace of Y and let V be the trace of C on f "^(Y^). By virtue of (3.19), V i s a convergence subbasis for the relative topology on f ^"(Y^). If f i s continuous pseudo-open, then f i s hereditarily quotient and so fV i s a convergence subbasis for the relative topology on Y 1. Moreover, fV coincides with the trace - 101 -of fC on and hence (3.19) implies that fC i s a convergence basis for Y . . • Assume now that fC i s a convergence basis for Y. By (3.24.1), f i s continuous. Let x E Y and let U be an open neighbour-hood of f "*"(y) in X. If f i s not pseudo-open, y E c l (Y- f (U)) and hence there i s a net pair ({f(x n) : n e D}, f(x)) E f C such that f(x) = y and each f ( x n ) e Y-f(U). Consequently x n £ U for every n E D. Then, since x E U, the net {X r : n E D) does not converge to x in X. Therefore ({x^ : n E D}, X) £ C and the theorem i s proved by contradiction. >3...27~>,Cor,ollar.y...aEy space i s m-Frechet. Proof This follows from (3.26) i n the same way that (3.25) followed from (3.24). 3.28 Definition Let m be an inf in i te cardinal number. A topological space i s m-first-countable, or an m-first-countable space, i f and only i f each point has a neighbourhood basis of cardinali ty <_ m. (Note that Ho -first-countable and first-countable are equivalent concepts.) - 102 -3.29 Proposition If X i s a topological space and m i s an in f in i t e cardinal, then each of the following implies the next. (1) X i s m-first-countable. (2) X is m-Frechet. (3) X i s m-sequential. (4) For any subset A of X, each point in c l A i s in c l B for some subset B of A with the cardinality of B <^  m. (5) X i s 2m-Frechet. Proof (1) > (2) Let F be a subset of X and suppose that x e c l F. By hypothesis, x has a neighbourhood basis {U : a e A} a. with the cardinality of A <_m. Defining a < c i f and only i f U C U , A i s a directed set with order <. Because x e c l F, there C 3. exists x e U n F for each a e A. Then {x : a e A} i s an m-net in a a a F converging to x, and hence X i s m-Frechet. (2) > (3) This is clear since every convergence basis i s a convergence subbasis. (3) > (4) If X i s m-sequential, X has a convergence subbasis C in which a l l of the nets are m-nets. For x e c l A = t - c l A, the existence of a subset B satisfying (4) i s established by trans-f in i te induction on the Baire order X of x with respect to A and C. If X = 0 then x e A and x e c l {x}. If X = 1, x e c - c l A and there i s an m-net pair (S, x) e C such that S i s an m-net in A converging to x. Clearly x e c l S and the cardinality of S i s <_m. By the - 103 -induction hypothesis, there exists an m-net pair ({x^ : n E D}, X) E C with x^ E c l A and ord x^ < X for each n E D. Consequently each x E c l B for some subset B of A with the cardinality of B < m. n n n J . n — . But then x E c l ( U {B : n e D}), L/{B : n E D} C A, and the n n 2 cardinality of . N {.B : n E J)} i s < .m = m. n — (4) > (5) Let A be a subset of X and suppose that x E c l A. By hypothesis, x E c l B for some subset B of A with the cardinality of B <^  m. If {U\ : i e 1} i s a neighbourhood basis at x, U^H B 0 for every i e I. Since B has cardinality <_ m, there are at most 2 m distinct sets U^H B. Define an equivalence relation • • -"on '-I "by atLd~eiit;±fying . - Let D'b e-the index set I under this equivalence relation. The cardinality of D is £ 2 m. Order D by specifying a < c i f and only i f U 3> U , and cl C for each n e D choose x E U fl B. Then {x : n E D} is a 2 m-net n n n in B converging to x. For the case m = M e , there are examples which show that a l l of the conditions in the foregoing proposition are distinct. As previously observed, the real line with the integers identified i s a Frechet space which i s not first-countable and the space M of (1.17) i s a sequential space which is not Freshet. The countable space 0, x of (1.19) clearly satisfies (4) but i t i s not sequential. - 104 -Finally, the ordinal space Q + 1 with the order topology i s a H 2 ° -Frechet space which does not satisfy (4). The next result i s a characterization of m-sequential spaces which generalizes the characterization of sequential spaces. An interesting corollary to this theorem i s a characterization of •*m-Fr^chet-'&paees -whieh-leads^t'0^n-^x-tj«&sien'-"e€'>-t'he^eha>rac-ter~'ization of Hausdorff Frechet spaces given in (2.15). The Hausdorff hypothesis i s eliminated. 3.30 Theorem A topological space i s m-sequential i f and only i f i t i s a quotient of an m-first-countable space. Proof By virtue of (3.29) and (3.25), a quotient of an m-first-countable space i s m-sequential. Conversely, let X be an m-sequential space with a convergence subbasis in which a l l of the nets are m-nets. For each ({x : n E D}, x) e C, l e t S(x , x) = {x : n e D} U {x} be n n n a topological space in which the x^ are taken to be distinct and x •/ x for every n e D, and which has the convergence basis generated by the C-net pair ({* n '• n e D}, x). Each x n is isolated, and x has a neighbourhood basis indexed by the directed set D whose cardinality is <_ m. Thus each S( x n> x) is m-first-countable. The disjoint topological sum W of a l l such S(x n, x) i s therefore m-first-countable and has, by (3.21), a convergence basis E formed by taking the union - 105 -of the convergence bases for the s(x n> x) . The theorem how follows from (3.24). The surjection f : W > X defined by f(x) = x is a quotient map since fE = C. 3.31 Corollary A topological space i s m-Frechet i f and only i f i t i s a continuous pseudo-open image of an m-first-countable space. Proof The continuous pseudo-open image of an m-first-countable space i s m-Frechet by (3.29) and (3.27). The converse coincides with that of (3.30) with the exception that C i s a convergence basis and the fact that fE = C together with (3.26) implies f i s continuous pseudo-open. 3.32 "Corollary For any topological space "X, 'the "following statements are equivalent. (1) X is Frechet. (2) X i s a continuous pseudo-open image of a f i r s t -countable space. (3) X is a continuous pseudo-open image of a metric space. Proof According to (3.31) and (3.17.2), (1) i s equivalent to (2). Clearly (3) implies (1). To establish the opposite implication, let X be a Frechet space with a convergence basis C in which a l l of the nets are sequences. Then W has a convergence basis E consisting - 106 -of sequences, and fE = C implies that f i s a continuous pseudo-open map of W onto X. Each summand . S ,(x , x) of W is a convergent sequence in the Hausdorff space S(x n, x). Hence, by (1.32), W i s metrizable. In general, the product of m-sequential or m-Frechet •spaces -need not -he -m-sequential. -SeveraT<examples"f or -the-case m = Ha have already been given. However, the product of two m-sequential spaces, one of which is such that each point has a neighbourhood basis consisting of m-sequentially compact sets, i s m-sequential. (A topological space i s m-sequentially compact i f and only i f every m-net has a convergent m-subnet.) The proof of this result i s analogous to that of (1.24). The following i s a generalization of (1.23). 3.33 Proposition Let X be the product of any family {X : a e A} a. of non-trivial topological spaces (each space has at least one non-empty proper open set). If the cardinality of A i s > m, then X i s not m-sequential. In particular, no uncountable product of non-t r i v i a l spaces is sequential. Proof By hypothesis, each coordinate space X contains two points, £1 denoted by 0 and 1, and a neighbourhood of 1 not containing 0. Let e be the function i n X whose a-th value i s 1 for each a e A, and - 107 -let E be the subset of X consisting of a l l characteristic functions of f i n i t e subsets of A. Clearly e e c l E. It suffices to prove that no iteration of m-nets in E can converge to e. For convenience, define the cozero set of a function f, denoted by coz f, to be the set {a e A : f(a) ^  0}. The functions in E have f i n i t e cozero sets. .Suppose that Xf^ :„n.e..D.}. is,„.an m-net of functions .converging to f with the cardinality of each coz f < m. Since coz f CZ U{coz f : n e D}, J n — n 2 the cardinality of coz f i s < m = m. Thus by forming iterated limits of m-nets in E i t i s only possible to obtain functions whose cozero sets have cardinality j< m. Consequently, i f the cardinality of A i s > m, then the cardinality of coz e i s > m and hence no m-net in (cl E)-{e} converges to e. 3.34 Proposition If X is the product of any family {X& : a e A} of non-trivial topological spaces, then each point of X has a neighbourhood basis of cardinality less than or equal to the maximum r of the cardinality p of A and q = sup {X(X ) : a e A}. (Write cl X(Y) = m i f and only i f Y is m-first-countable.) Proof Let x e X and assume that {U. : i e I } i s a neighbourhood i a basis for the a-th coordinate of x with the cardinality of I <_X(X ). a a For each a e A let P denote the canonical projection map of X onto X . 1 - 108 -Then the set B of a l l f i n i t e intersections of elements in - l a B = ^{{P (Un) : i e I } : a e A} i s a neighbourhood basis 2 for x. Since the cardinality of B i s £ pq <_ r = r, B has cardinality <_ • r = r. (See 1.18.) Chapter 4 Generalized Sequential Spaces and their Properties in Ordered Topological Spaces The properties of convergence subbasis and convergence bases are applied, in this chapter, to the investigation of topological spaces whose open sets are specified by well-ordered nets. 4.1 Definition A well-ordered net i s a net whose directed set i s well-ordered. (A well-ordered m-net i s a net whose directed set Is well-ordered and of cardinality _< m.) 4.2 Definition (1) A topological space i s weakly sequential, or a weakly sequential space, i f and only i f i t has a convergence subbasis in which a l l of the nets are well-ordered. (2) A topological space i s weakly Frechet, or a weakly Frechet space, i f and only i f i t has a convergence basis consisting of well-ordered nets. 4.3 Definition Let X be a topological space, and le t m be an in f in i t e cardinal. (1) X i s m-sequential, or an m-sequential space, i f and only i f i t has a convergence subbasis in which a l l of the nets - 110 -are well-ordered m-nets. (2) X is m-Frechet, or' an m-Frechet space, i f and only i f i t has a convergence basis consisting of well-ordered m-nets. Since sequences are well-ordered H 0-nets, these generalized sequential spaces and generalized Frechet spaces clearly contain the sequential and Frechet spaces respectively. In particular, a topological space i s (sequential, Frechet) i f and only i f i t i s ( ^ -sequential, H0 -Frechet). Observe that a space i s m-sequential i f and only i f i t i s both weakly sequential and m-sequential. Similarly, a topological space i s m-Frechet i f and only i f i t i s both weakly Frechet and m-Frechet. The aim of the f i r s t part of this chapter is to characterize the generalized sequential spaces of (4.2) and (4.3). Their characterizations lead to new characterizations of the Frechet spaces and the sequential spaces in terms of orderable spaces. To avoid tedious repetition, the elementary properties of these generalized .sequential .spaces will.not be formally stated.. .The. preceding chapter's survey of convergence subbases greatly f a c i l i t a t e s their study. It i s easy to see that the investigation of these spaces i s analogous to that of the m-sequential spaces. - I l l -4.4 Definition (1) A topological space is weakly first-countable, or a weakly first-countable space, i f and only i f each of i t s points has a well-ordered neighbourhood basis. (A collection {F & : a e A} of sets i s called well-ordered whenever A i s well-ordered and F C~ F i f and only i f a > c in A.) a c 3 (2) .A .topological -space ,is„m-:f.ir.s.t-coun.tahle,, .or..,an m-first-countable space, i f and only i f each of i t s points has a well-ordered neighbourhood basis of cardinality <_ m. (Note that first-countable, H0 -first-countable, and \\0-iirst-countable are equivalent concepts.) 4.5 Proposition (1) Every weakly first-countable space i s weakly "*"" (2) Every m-first-countable space i s m-Frechet and hence m-sequential. Proof Let F be a subset of a weakly first-countable space X and suppose that x e c l F. By definition, x has a well-ordered neighbour-hood basis {U : a e A}. Then, since x e c l F, there exists a well-a ordered net {x : a e A} which satisfies x e U f) F and therefore a a a converges to x. If X is m-first-countable, the cardinality of A i s < m. - 112 -4.6 Example For any uncountable cardinal m, there i s an m-first-countable space which is not weakly sequential and hence not m-sequential. Proof Let D be the family of a l l f i n i t e subsets of a set whose cardinality i s m, and order D by Z2> . Then D i s a directed set of cardinality m. Let S = {x : n e D} XJ {x} be a topological space in which the x are distinct and x £ x for every n e D, and which n n J ' has a convergence basis generated by the net pair ({x^ : n e D}, x). Each x^ i s isolated and x has a neighbourhood basis indexed by D. Consequently S i s m-first-countable. However, S is not weakly sequential because x E cl(S-{x}) and yet no well-ordered net in S-{x} converges to x. To verify that this i s so, suppose that {x„,„ v : k e K} i s a well-ordered subnet of {x : n e D}. Choose N(k) n a countable collection {n. : i E to} of distinct elements in D. i For each i E to there exists k e K such that, i f k j> k^ then N(k )25n^ . From the description of D, i t is obvious that there i s no supremum of {k^ : i e to} in K. But then {k^ : i E to} i s a cofinal subset of K and hence {N(k_^ ) : i E to} is a cofinal subset of D. This i s impossible since U {N(k_^ ) : i E to} is only a countable subset of the given set of cardinality m. - 113 -4.7 Definition An ordered topological space is a space which, has the order topology arising from a total order on the set. A topological space is orderable i f and only i f some total order can be imposed on the set relative to which the given topology coincides with the order topology. 4.8 Proposition Let A be a subset of an ordered topological space X. If an m-net in A converges to a point x e X-A, then there i s a s t r i c t l y monotone well-ordered m-net in A converging to x. Proof It i s f i r s t shown that every totally ordered set has a cofinal well-ordered subset. Let F = {F^ : i e 1} be the family of a l l well-ordered subsets of a totally ordered set Y. P a r t i a l l y order F by defining F. < F, whenever F. = F. or F. i s an i n i t i a l & i j i j l segment of F.. Note that F. < F. implies that F.C F,, . Let C J 1 J i j be a chain in F and suppose that B i s a subset of UC. There exists C e C such that C fi B ^ 0, and C H B has a least element b since C i s well-ordered. The total order < on C implies that b i s the least element of B and hence that U C e F. By Zorn's lemma, F has a maximal well-ordered element Y^. Then Y^ i s also a cofinal subset of Y because otherwise there exists y e Y~Y^ with no element of Y greater than y; from this, Y1 \J {y} e F - 114 -contradicting the maximality of Y^. Let A q be the intersection of A with the range of the m-net in the hypothesis. The cardinality of A ^ is clearly <^ m. Let A . = {y e A : y. < x} and A „ = {y E A : y > x}. Since X has 1 o ^ 2 ^ o the order topology, x e c l A^ , for i = 1 or i = 2; assume the former. The set A ^ i s directed by the total order inherited from X, and thus the identity map on A ^ i s a s t r i c t l y monotone m-net converging to x. Moreover, A ^ has a cofinal well-ordered subset A ^ and the identity map on A ^ i s the desired net. 4.9 Theorem , ^.^^^ topological space X are equivalent. (1) X i s weakly sequential. (2) X is the quotient of a weakly first-countable orderable space. •- (3) X is the quotient of a weakly first-countable space. (4) X i s the quotient of an orderable space. Proof Clearly (2) implies both (3) and (4) . In addition, (3) implies (1) by (4.5-1) and (3.24). It remains to show that (1) > ( 2) and (4) > (1). - 115 -To establish the latter implication, let f : y > X be the quotient map of an ordered space Y onto a topological space X. Suppose U is a subset of X such that any well-ordered net converging to a point in U is eventually in U. It suffices to prove that f "^ (U) i s open in Y. Let y e f "^(U). If y £ int f "*"(U), there is a net {y : n e D} which is disjoint from f "''(U) and .n converges to y. According to (4.8), i t can be assumed that D i s well-ordered. But then {f(y ) : n e D} i s a well-ordered net in n X-U converging to f(y) e U. Hence (4) implies (1) by contradiction. Assume now that X i s a weakly sequential space with a convergence subbasis C in which a l l of the nets are well-ordered nets. For each ({x : n e D}, x) e C, let S(x , x) = {x : n e D'} U {x} n n n be a topological space in which the x n are taken to be distinct and x n ^ x for every n e D', and which has the order topology arising from the total order defined as follows. Let a be the least element o of D and let D' = (w x {a o»(J(Z * (D-{aQ})) with w and Z ordered i n the usual way. Totally order D' by specifying ( i , n) < ( j , m) whenever n < m or i < j and n = m. Each element of D' has an immediate successor, and each element other than (0, a Q) has an immediate predecessor. (Such an order i s called a discrete order.) Now l e t X,. N = x for each ( i , n) e D', and totally order S(x , x) (i,n) n ' J n' by defining x < x i f n < m in D' and x < x for each n e D'. J b n m n - 116 -In the ordered space S(x , x), each x i s isolated and x has a . n n well-ordered neighbourhood basis indexed by a set order isomorphic to D. Then SCx^, x) is weakly first-countable and has a convergence basis generated by the net pair ({x^ : n e D'}, x). The disjoint ...topological,,sum W .of a l l such S.(x , x) i s also weakly first-countable The natural mapping of W onto X defined by x > x is a quotient map because the net pairs (i^n '• n e D'}, x) form a convergence subbasis for X. To demonstrate that W i s orderable, let {S :• a e A} denote a the set of a l l S(x , x) and define a discrete order on Z x A in the n "same'-'way ••*as*tD1. -"'(In^th-^ -order with least element a Q.) Using the existence of a one-to-one correspondence between A and Z x A, this discrete order can be imposed.on A. Let W be totally ordered by specifying x < y whenever x < y in S where x, y e S , or a < b in A where x e S and y E S, . J a a a J b Because of the discrete orderings and the fact that each has a greatest element and a least element, the order topology on W coin-cides with i t s usual disjoint topological sum topology. Thus W i s orderable, and the theorem i s proved. - 117 -4.10 Corollary For any topological space X and any i n f i n i t e cardinal m, the f i r s t three statements are equivalent. If m = H0 they are also equivalent to (4). (1) X is m-sequential. (2) X i s the quotient of an orderable m-first-countable space. (3) X i s the quotient of an m-first-countable space. (4) X i s the quotient of an orderable metric space. Proof This i s analogous to (4.9). It i s only necessary to remark that each S(x , x) i s m-first-countable and hence so is W. n — 4.11 ,P,roppsit ion The ..fallowing are .equivalent. (1) X is weakly Frechet. (2) X i s the continuous pseudo-open image of an orderable weakly first-countable space. (3) X i s the continuous pseudo-open image of a weakly first-countable space. (4) X i s the continuous pseudo-open image of an orderable space. Proof Clearly (2) implies both (3) and (4), and (3) implies (1) by (4.5.1) and (3.26). The fact that (1) > (2) follows from (4.9) i n the same way that (3.31) followed from (3.30). - 118 -T o p r o v e t h a t (A) > (1), l e t f : Y > X b e a c o n t i n u o u s p s e u d o - o p e n f u n c t i o n o f a n o r d e r e d s p a c e Y o n t o a t o p o l o g i c a l s p a c e X . L e t A b e a s u b s e t o f X a n d s u p p o s e t h a t x e c l A. T h e n s o m e y z f ^ ( x ) D c l f "'"(A). T h i s i s s o b e c a u s e o t h e r w i s e t h e r e i s a n o p e n n e i g h b o u r h o o d U o f f ^ ( x ) d i s j o i n t f r o m f ^(A); f r o m t h i s , ATI ' f ( U ) = '0 c o n t r a d i c t i n g x e " ( c l "A)'"fY""±n't f ' ( t l ) . S i n c e Y i s a n o r d e r e d t o p o l o g i c a l s p a c e , t h e r e e x i s t s a w e l l - o r d e r e d n e t { y ^ : n e D} i n f "'"(A) c o n v e r g i n g t o y . T h e n { f ( y ^ ) : n e D} i s a w e l l - o r d e r e d n e t i n A c o n v e r g i n g t o f ( y ) = x . 4 . 1 2 C o r o l l a r y F o r a n y t o p o l o g i c a l s p a c e X a n d a n y i n f i n i t e c a r d i n a l m , t h e f i r s t t h r e e s t a t e m e n t s a r e e q u i v a l e n t . I f t h e y a r e a l s o e q u i v a l e n t t o ( 4 ) . ( 1 ) X i s m - F r e c h e t . ( 2 ) X i s t h e c o n t i n u o u s p s e u d o - o p e n i m a g e o f a n o r d e r a b l e m - f i r s t - c o u n t a b l e s p a c e . ( 3 ) X i s t h e c o n t i n u o u s p s e u d o - o p e n i m a g e o f a n m - f i r s t - c o u n t a b l e s p a c e . ( 4 ) X i s t h e c o n t i n u o u s p s e u d o - o p e n i m a g e o f a n o r d e r a b l e m e t r i c s p a c e . T h e f i n a l r e s u l t s a r e c o n c e r n e d w i t h t h e s e q u e n t i a l p r o p e r t i e s o f o r d e r e d t o p o l o g i c a l s p a c e s a n d t h e r e l a t i o n b e t w e e n t h e n o t i o n s o f f i r s t - c o u n t a b l e s p a c e , F r e c h e t s p a c e , a n d s e q u e n t i a l s p a c e i n p r o d u c t s o f t h e s e s p a c e s . I t i s n o w k n o w n t h a t F r e ' c h e t - 119 -spaces and sequential spaces are successive proper generalizations of first-countable spaces, and.that the product of two Frechet spaces need not be sequential. For topological spaces which are products of ordered spaces, the situation i s quite different. 4.13 Theorem If X i s an ordered topological space, the following are equivalent. The f i r s t three statements are equivalent whenever X i s a product of ordered spaces. (1) X i s m-first-countable. (2) X i s m-Frechet. (3) X i s m-sequential. (4) X i s m-Frechet. .^4^-)M^XMi&am-3S.e.queatial. Proof From (3.29), (1) > (2) > (3). Obviously (4) > (5) — > (3) and (4) > (2). Furthermore, (2) > (4) by (4.8). It remains to show that (2) > (1) and (3) > (2). Assume f i r s t that X i s an ordered topological space. (2) > (1). Let x e X. If x is an isolated point then x has a neighbourhood basis consisting of the singleton {x}. Suppose that x i s not isolated but has either an immediate predecessor or an immediate successor. In the former case, x e c l {y e X : y > x}. By hypothesis, there exists an m-net {x n : n E D} in {y : y > x} converging to x. Let A = { y : x _ < y < x } . The collection - 120 -{A^ : n e D} i s a neighbourhood basis at x and the cardinality of D i s _< m. Similarly, x has a neighbourhood basis of cardinality <_ m in the latter case. Suppose now that x i s not isolated and has neither an immediate successor nor an immediate predecessor. By hypothesis, there i s an m-net {x^ : n e D} converging to x with x < x for each n. There i s also an m-net {y : n e E} converging n n e ° to x with each y > x. The open sets {z : x < z < y } where (n, m) e D x E form a neighbourhood basis for x and the cardinality of D x E i s < = i . (3) > (2). Suppose that the nets S = {x" : i E E } r r n i n converge to x n and S = {x n : n e D} converges to x, with the cardinalities of E and D < m. It suffices to construct an m-net n — in the union of the ranges of the nets S n converging to x. According to (4.8), i t can be assumed that a l l of the given nets are s t r i c t l y monotone and directed by ordinal numbers. Either.S i s increasing or decreasing, and the nets S n are either frequently increasing or frequently decreasing with respect to the directed set D; that i s , the nets S^ are frequently (increasing, decreasing) i f and only i f for each p e D there exists q ^  p such that S^ i s (increasing, decreasing). There are four cases to consider. F i r s t , assume that the nets S are increasing and the net S is decreasing. Since x > x - 121 -for each n, there i s i(n) e E with x > x V / N > x. Then {x7, N : n e ' n n i(n) — i(n) is the desired m-net; i t converges to x because i t is bounded above by the net S which converges downward to x. For the second case, suppose that a l l of the nets are s t r i c t l y increasing. Let D' denote the set of a l l isolated ordinals in D. Clearly D' i s a cofinal subset of D; for each n e D the successor ordinal n + 1 is isolated. Then x , < x for each n e D', and there exists i(n) e E satisfying n-1 n ' n ° x , < x n, It follows that { x V / . : n e D'} i s an m-net converging n-1 i(n) i(n) e " to x. The remaining two. cases are similar to the f i r s t and second cases. This completes the proof for the case in which X consists of one ordered space. • -: >a -e-'-A<} a. of ordered topological spaces. It i s only necessary to prove that (3) implies (1). By virtue of (3.33), the cardinality of A i s <_ m. Then, since each X & i s m-sequential and hence m-first-countable, (3.34) implies that X i s m-first-countable. 4.14 Corollary If X i s the product of any family {X : a e A} of Si non-trivial ordered m-sequential (or equivalently m-Fre'chet) spaces, then X i s m-sequential (or equivalently m-Frechet) i f and only i f the cardinality of A i s < i . - 122 -Proof If X i s m-sequential then, by (3.33), the cardinality of A i s / m and hence <_m. To establish the converse, observe that each X is m-first-countable by the preceding theorem. Then i f the a cardinality of A i s < m, in view of (3.34), X is m-first-countable. "4.15 "Corollary "An ordered topological space i s weakly Frechet i f and only i f i t is weakly sequential. Proof This i s the same as (3) < > (2) of (4.13). In this case, the cardinalities of D and E n are not important. 4.16 Example There i s an ordered topological space which i s not ~ >'w *'ahy*uncouhtabie 'cardlnai -m, an m-first-countable ordered space need not be m-first-countable. Proof Let X = ( a + 1) + t o * where a i s the i n i t i a l ordinal of cardinality m and t o * has the reverse order to that of t o . By definition, X has the order : x < y i f a) x, y E a + 1 and x < y in cc + 1, or b) x, y e t o * and x < y in t o * , or c) x e a + 1 and y e t o * . - 123 -L e t X have the order topology a r i s i n g from t h i s t o t a l o r d e r . I t i s c l e a r tha t X i s m - f i r s t - c o u n t a b l e . To e s t a b l i s h tha t X i s not weakly f i r s t - c o u n t a b l e and hence not m - f i r s t - c o u n t a b l e , assume the o p p o s i t e and l e t {U : a e A} be a w e l l - o r d e r e d neighbourhood b a s i s a a t the p o i n t a . O b v i o u s l y a has n e i t h e r an immediate predecessor nor an immediate s u c c e s s o r . For each a e A , l e t x be t h e l e a s t a element of (a + 1 ) D U , and l e t y be the g r e a t e s t element of cl H to* n U . The range of {y : a E A} i s s u r e l y c o u n t a b l e a n d , s i n c e a a {U : a E A} i s w e l l - o r d e r e d , t h e r e are l e s s than m elements i n the a range of {x & : a E A} a s s o c i a t e d w i t h each element i n the range of {y : a £ A } . The supremum of {x 5 a E A} i s t h e r e f o r e l e s s than a , 3- cl and so {x : a E A} cannot converge to a . > - 124 -B i b l i o g r a p h y Arens, R. "Note on Convergence i n Topology", Mathematics  Magazine, 23 (1950), 229-234. A r h a n g e l ' s k i i , A. "Some Types of Fact o r Mappings, and the R e l a t i o n s between Classes .of . T o p o l o g i c a l Spaces,,11  Soviet Math. Do k l . , 4 (1963), 1726-1729. Boehme, T.K. "Linea r s-spaces", Symposium on Convergence S t r u c t u r e s , U n i v e r s i t y of Oklahoma, 1965, unpublished. C u l l e n , H.F. "Unique S e q u e n t i a l L i m i t s , " B o l l . Un.. Mat. I t a l . , Ser. 3-20 (1965), 123-124. de Groot, J . "Subcompactness and the B a i r e Category Theorem", Nederl. Akad. Wetensch. P r o c , Ser. A 66 (1963), 761-767. Dudley, R.M. "On S e q u e n t i a l Convergence," Trans. Amer. Math. S o c , 112 (1964), 483-507. Dugundji, J . Topology. Boston : A l l y n and Bacon, Inc., 1966. F r a n k l i n , S.P. "Spaces i n which Sequences S u f f i c e , " Fundamenta  Mathematicae, 57 (1965), 107-115. . "On Unique S e q u e n t i a l L i m i t s , " Nieuw A r c h i e f voor Wiskunde, 14 (1966), 12-14. - 125 -[10] . "Spaces in which Sequences Suffice II," Fundamenta Mathematicae, 61 (1967), 51-56. . [11] Gillman J. and M. Jerison. Rings of Continuous Functions. Princeton : D. Van Nostrand Company, Inc., 1960. [12] Harley III, P.W. "Products of Frechet Spaces," Notices Amer. Math. Soc, 19 (1972), 378. [13] Hu, S.T. Elements of General Topology. San Francisco : Holden-Day, Inc., 1964. [14] Jones, F.B. "On the Fi r s t Countability Axiom for Locally Compact Hausdorff Spaces," Colloquium  Mathematicum, 7 (1959), 33-34. [15] Kamke, E. Theory of Sets. New York : Dover Publications, Inc., 1950. [16] Kelley, J. L. General Topology. Princeton : D. Van Nostrand Company, Inc., 1955. [17] Kent, D.C. "Decisive Convergence Spaces, Frechet Spaces, and Sequential Spaces," Rocky Mountain Journal of  Mathematics, 1 (1971), 367-374. [18] . "Spaces in which Well Ordered Nets Suffice," Proceedings of the Washington State University on General Topology, (1970), pp. 87-101. - 126 -{19] Kisyffski, J.I. "Convergence du Type L," Colloquium Mathematicum, 7 (1959), 205-211. (20] Meyer, P.R. "The Baire Order Problem for Compact Spaces," Duke Mathematical Journal, 33 (1966), 33-39. 121] . "Sequential Properties of Ordered Topological .Spaces-," -Compositio Mathematicae.,-21 •,(-1-96.9.)-, 102-106. [22] • . "Sequential Space Methods in General Topological Spaces," Colloquium Mathematicum, 22 (1971), 223-228. [23] Michael, E. "Local Compactness and Cartesian Products of Quotient Maps and k-spaces," Ann. Inst. Fourier ....,i.,«,(GEen.ah,i.e:.x, .jLB«M6a)^£flj«2a&. [24] . " H 0 -spaces", J. Math. Mech., 15 (1966), 983-1002. [25] Mrowka, S. "Compactness and Product Spaces", Colloquium Mathematicum, 7 (1959), 19-22. [26] Nagata, J.I. Modern General Topology. Amsterdam : North Holland Pub. Co., 1968. 9 [27] Novak, J. "On the' Cartesian Product of two Compact Spaces," Fundamenta Mathematicae, 40 (1953),106-112. [28] Royden, H.L. Real Analysis, 2nd. ed. New York : The Macmillan Co., 1968. - 127 -[29] Sorgenfrey, R.H. "On the Topological Product of Paracompact Spaces", B u l l . Amer. Math. S o c , 53 (1947), 631-632. [30J Venkataraman, M. "Directed Sets i n Topology," Mathematics Student, 30 (1962), 99-100. [31] Whitehead, J.H.C. "Note on a Theorem due to Borsuk", B u l l . Amer. Math. S o c , 54 (1948), 1125-1132. 

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