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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 i n the Department of MATHEMATICS  We accept t h i s thesis as conforming  to the  required standard  THE UNIVERSITY OF BRITISH COLUMBIA August 1972  In  presenting  this  an advanced degree the I  Library  further  for  agree  in  at  University  the  make  it  partial  freely  fulfilment  of  of  Columbia,  that permission for  this  representatives. thesis  for  It  financial  of  The U n i v e r s i t y  TT?^  of  7% ^ N  by the  gain  shall  ^^>^>  not  the  requirements  reference copying o f  Head o f  i s u n d e r s t o o d that  B r i t i s h Columbia  Vancouver 8, Canada  for  extensive  written permission.  Department  British  available  s c h o l a r l y purposes may be g r a n t e d  by h i s of  shall  thesis  I  agree  and this  be a l l o w e d  that  study. thesis  my Department  copying or  for  or  publication  w i t h o u t my  ii Abstract  The class of sequential spaces and i t s successive smaller subclasses, the Frechet spaces and the f i r s t - c o u n t a b l e spaces, have topologies which are completely s p e c i f i e d by t h e i r convergent sequences.  Because sequences have many advantages over nets, these  topological spaces are of i n t e r e s t .  Special attention i s paid to  those properties of f i r s t - c o u n t a b l e spaces which can or cannot be generalized to Frechet or sequential spaces.  For example, countable  compactness and sequential compactness are equivalent i n the larger class of sequential spaces.  On the other hand, a Frechet space  with unique sequential l i m i t s 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 a r b i t r a r y sequential space and a T^ (.regular and T^) sequential space X i s sequential i f and only i f X i s l o c a l l y countably compact. which demonstrate  There are also several r e s u l t s  the non-productive nature of Frechet spaces.  The sequential spaces and the Frechet spaces are p r e c i s e l y the quotients and continuous pseudo-open images, r e s p e c t i v e l y , of either (ordered) metric spaces or (ordered) f i r s t - c o u n t a b l e spaces. These characterizations follow from those of the generalized sequential spaces and the generalized Frechet spaces.  The notions  iii of convergence r o l e here.  subbasis and convergence basis play an important  Quotient spaces are characterized i n terms of conver  gence subbases, and continuous pseudo-open images i n terms of convergence bases.  The equivalence of h e r e d i t a r i l y quotient map i  and continuous pseudo-open maps implies the l a t t e r r e s u l t .  iv Table .of Contents Page Introduction  . . . . . . . . . . . .  1  . . . . .  6  Notation .,Chap,.ter.„L: ._Sequential....Spaces Chapter 2:  Frechet Spaces  Chapter 3:  Generalized Sequential Space Methods .  Chapter 4:  Bibliography  Generalized Sequential Spaces and t h e i r Properties i n Ordered Topological Spaces . . . .  7 59  .  81  109 124  V  L i s t of  Figure 1:  The Graph of  Illustrations  U(W  : n e to} i n R  (See Example 1.19)  vi  Acknowledgments The author i s indebted to Dr. T. Cramer f o r suggesting the topic of t h i s thesis and f o r h i s patience, encouragement, and invaluable assistance during the past year.  Dr. J.V. Whittaker's  ..careful reading and constructive c r i t i c i s m s of the f i n a l manuscript are also g r a t e f u l l y acknowledged.  F i n a l l y , the author wishes to  thank Miriam Swan f o r her conscientious typing work.  Introduction  A f i r s t - c o u n t a b l e 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 f i r s t - c o u n t a b l e spaces ([16], Theorem 2.8) : CI)  A -set -is -open -if -and-on-l-y- i f -each -sequence-which  converges to a point i n the set i s , i t s e l f , eventually i n the set. (2)  A point l i e s i n the closure of a set i f and  only  i f there i s a sequence i n the set converging to the point. For more general spaces, i t i s often assumed that sequences are inadequate and that nets or f i l t e r s must be used.  There are,  however, many t o p o l o g i c a l spaces which do not s a t i s f y the f i r s t "axiom "of couritabiTity and "yet 'sequences 'suffice' to determine open sets.  The r e a l l i n e with the integers i d e n t i f i e d to one point  i s an example of such a space. The topological spaces s a t i s f y i n g (1) are c a l l e d spaces and those s a t i s f y i n g (2), Frechet  spaces.  sequential  Each f i r s t - c o u n t -  able space, and hence each metric space and each d i s c r e t e space, i s both Frechet and sequential.  Moreover, the r e a l l i n e with the  integers i d e n t i f i e d i s 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 proper generalizations of f i r s t - c o u n t a b l e space.  In  successive  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 i n t e r e s t to know when a t o p o l o g i c a l space i s sequential. A r e s u l t due to Ponomarev characterizes f i r s t - c o u n t a b l e T^-spaces as continuous open images of metric spaces. S.P.  Analogously,  F r a n k l i n {8J establishes that the sequential spaces are  p r e c i s e l y the quotients of either metric spaces or f i r s t - c o u n t a b l e 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 c l a s s between the  open maps and the quotient maps.)  In [22], P.R. Meyer extends  Arhangel'skii's r e s u l t by eliminating the Hausdorff hypothesis:--^ .pseudo-open images of either metric spaces or f i r s t - c o u n t a b l e spaces.  In order to obtain t h i s r e s u l t , he introduces the notions  of convergence subbasis and convergence basis which, provide the foundation  for studying topological spaces whose open sets are  completely s p e c i f i e d by any given c l a s s of nets.  Meyer's general-  ized sequential space methods are used to derive D.C. Kent's [18] characterizations of "spaces i n which well ordered nets s u f f i c e . " Recently, many mathematicians have researched spaces and generalized sequential spaces. thesis i s to present u n i f i e d theory.  sequential  The purpose of t h i s  the more important of t h e i r r e s u l t s i n a  The author generalizes a few of these r e s u l t s and  - approves numerous statements asserted without proof i n the o r i g i n a l papers. Chapter 1 i s an i n v e s t i g a t i o n of sequential spaces, t h e i r properties, t h e i r characterization as quotients of metric spaces or  f i r s t - c o u n t a b l e spaces, and t h e i r r e l a t i o n to other t o p o l o g i c a l  properties.  Their r e l a t i o n to the f i r s t - c o u n t a b l e spaces i s of  particular interest.  I t i s well-known that countable compactness  and sequential compactness are equivalent i n the class of f i r s t countable spaces.  Franklin asserts t h e i r equivalence i n the larger  c l a s s of sequential spaces.  F r a n k l i n proves t h i s r e s u l t i n [8] f o r  Hausdorff spaces and i n [10] f o r spaces with unique sequential limits.  In t h i s t h e s i s , the author provides the proof of the same  „r,esult ...f,o:r,,,ar.bi^  „,T.he,,.author,valso',.shows J  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 r e s u l t due to T.K. Boehme [3] shows that t h i s s i t u a -  t i o n cannot occur i n the presence of suitable compactness conditions. In addition, E. Michael [23] has proved that f o r 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 l o c a l l y countably compact.  _ 4The second chapter i s concerned with. Frechet spaces, their properties, and t h e i r r e l a t i o n to sequential spaces.  The character-  i z a t i o n of sequential spaces i s used to prove Arhangel'skii's characterization of Hausdorff Frechet spaces.  In t h i s section, the  author proves Arhangel'skix's assertion that the continuous pseudoopen maps and the h e r e d i t a r i l y quotient maps are equivalent. The author also provides the proof of a r e s u l t due to P.W. Harley I I I I12J concerning the product of two Frechet spaces. Chapter 3 i s devoted to Meyer's generalized sequential space methods and h i s study of convergence subbases and m-sequential spaces.  CAn m-sequential space i s a space f o r which m-nets  (I.e.,  nets whose directed-set i s of c a r d i n a l i t y <_ m) s u f f i c e to determine ^closed ,,.ssats..,.)^ given i n terms of convergence subbasis.  This r e s u l t leads to a  characterization of the m-sequential spaces.  The author proves an  analogous r e s u l t for continuous pseudo-open images which, leads to Meyer's characterization of m-Frechet  spaces.  In the l a s t chapter, the author employs many of the properties of convergence subbases to investigate weakly sequential spaces and m_ - sequential spaces ( i . e . , those spaces f o r which w e l l ordered nets and well-ordered m-nets, respectively, are s u f f i c i e n t to describe closed s e t s ) .  These spaces are characterized i n terms of  ordered t o p o l o g i c a l spaces (i.e.,  those spaces which have the order  - 5 topology a r i s i n g from a t o t a l order)..  F i n a l l y , there i s a b r i e f  coda which demonstrates that the concepts of f i r s t - c o u n t a b l e space, Frechet space, and sequential space are equivalent i n products of ordered spaces.  Notation  For  t h e most p a r t , t h e t e r m i n o l o g y and b a s i c  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.  notation  The e x c e p t i o n s a r e  l i s t e d below. CD  X - A = {x e X : x i A}  C2)  F o r any t o p o l o g i c a l space X and s u b s e t A o f X,  i n t ^ ( A ) d e n o t e s t h e i n t e r i o r of A with, r e s p e c t t o X and c l ^ ( A ) i s the closure possible  o f A w i t h r e s p e c t t o X.  t h e s e w i l l be (3)  When no c o n f u s i o n seems  a b b r e v i a t e d t o i n t A and c l A.  R i s t h e s e t o f r e a l numbers, Z i s t h e s e t o f  i n t e g e r s , Q. i s t h e s e t o f r a t i o n a l s , and W = set of n a t u r a l (4)  {1,2,3,...}  i s the  numbers. to i s t h e f i r s t i n f i n i t e o r d i n a l and ft i s t h e f i r s t  uncountable o r d i n a l . (5)  F o r any o r d i n a l a, a + 1 d e n o t e s t h e s e t o f a l l  o r d i n a l s w h i c h a r e l e s s t h a n o r e q u a l t o a; t h a t i s , a + 1 i s t h e s u c c e s s o r o r d i n a l o f a. (6)  A t o p o l o g i c a l space i s s a i d t o be f i r s t - c o u n t a b l e ,  o r a f i r s t - c o u n t a b l e s p a c e , i f and o n l y i f i t s a t i s f i e s t h e f i r s t axiom o f 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-  c o u n t a b l e , , o r a s e c o n d - c o u n t a b l e s p a c e , i f and o n l y i f i t s a t i s f i e s t h e second a x i o m o f c o u n t a b i l i t y .  Chapter 1 S e q u e n t i a l Spaces  Sequences have numerous advantages over n e t s . so because many p r o p e r t i e s of sequences f a i l •nets.  to generalize to  F o r example, a c o n v e r g i n g sequence and i t s l i m i t  whereas t h i s i s n o t t r u e f o r n e t s . convergent  This i s  i s compact,  Among H a u s d o r f f spaces,  each  sequence ( i . e . , t h e u n i o n o f t h e sequence and i t s l i m i t )  s a t i s f i e s t h e second metrizable.  axiom o f c o u n t a b i l i t y and i s t h e r e f o r e  These f a c t s t o g e t h e r w i t h o t h e r p r o p e r t i e s o f sequences  not a p p l i c a b l e to nets p l a y 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 s e q u e n t i a l spaces.  CJ.8J, 19] and J10]) s u r v e y o f s e q u e n t i a l s p a c e s .  There a r e ,  however, s e v e r a l important r e s u l t s due t o Boehme [3] and M i c h a e l .123]  related  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 (1)  L e t - X be a t o p o l o g i c a l  space.  A subset U o f X i s s e q u e n t i a l l y open i f and o n l y i f  each sequence i n X c o n v e r g i n g t o a p o i n t i n U i s e v e n t u a l l y i n U. C2)  A subset F o f X i s s e q u e n t i a l l y c l o s e d i f and o n l y i f  no sequence i n F converges  t o a p o i n t n o t i n F.  For any topological space, a subset A i s closed i f and only i f no net i n A converges to a point not i n A CI"! 6] 5 Theorem 2. Therefore  closed sets are sequentially closed and open sets are  sequentially open.  1.2  Example  The converses need not be true.  There are sequentially open sets which are not open  and sequentially closed sets which are not Proof  closed.  Consider the o r d i n a l topological space ft + 1 provided  the order topology.  with  Let S be a sequence i n ft + 1 which i s not  eventually equal to ft. Then S i s frequently i n ft and hence there i s a subsequence S  q  of S i n ft. But the supremum of S  than ft, and therefore S  q  q  i s less  cannot converge to ft. This implies that  S cannot converge to ft. Thus a sequence i n ft + 1 converges to ft i f and only i f i t i s eventually equal to ft. A d d i t i o n a l l y , a sequence i n ft can converge only to a member of ft. It follows that {ft} i s sequentially open and ft i s sequentially closed i n ft + 1. But {ft} i s not open and ft i s not closed i n ft + 1.  1.3 Proposition  A subset of a t o p o l o g i c a l space i s 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 t o p o l o g i c a l space  X and S i s .a sequence  i n X-U converging to x, then x e X-U.  This i s so because otherwise S i s eventually i s sequentially closed.  Then S  Thus X-U  Conversely, suppose that F i s a sequentially  closed subset of X and l e t S y e X-F.  i n U.  be a sequence i n X converging to  q  i s not frequently i n F since otherwise there i s  q  a subsequence of S  q  i n F converging to y t F.  Hence S  q  i s event-  u a l l y i n X-F, and therefore X-F i s sequentially open.  1.4 Proposition  For any topological space X, the c o l l e c t i o n of a l l  sequentially open subsets forms a topology f o r the set X. Proof  C l e a r l y , 0 and X are sequentially open.  If {U  : a e A} cL  i s any family of sequentially open subsets of X and S i s a sequence i n X which converges to x e \J^ c e A. Consequently S i s eventually  a  &  (J {U  : a e A}.  3.  Hence [J {U  £ A}, then x e U f o r some i n U^ and therefore i n £  : a E A} i s sequentially open. Si  Suppose now  that U and V are sequentially open, and l e t {y  be a sequence i n X converging to a point i n UflV. i s eventually {y n y  n  Q  n^}.  : n e ui}  Then {y^ : n e w}  i n both U and V, and there e x i s t s n , n^ e OJ with Q  : h > n } C U and {y : h > n. } C V. — o *^n — 1  sup { >  n  So y e U n.  D V for a l l 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 i s said to be the sequential closure topology.  1 ...6 D e f i n i t i o n  A topological space i s 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 f i r s t - c o u n t a b l e spaces, a set i s open i f and only i f each sequence converging to a point i n 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 d i s c r e t e spaces, are sequential. On the other hand, by v i r t u e of (1.2), the o r d i n a l space provided with the order topology i s not  n  + 1  sequential.  A f t e r a few preliminary r e s u l t s , several equivalent formulations  1.7  f o r the notion of sequential space are given.  Definition  Let X and Y be topological spaces; and l e t T be  the topology on X. T  a  The space Y divides X i f and only i f no topology  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, l e t T be  the topology on X, and l e t T^ = { B C X each T-continuous function f : Y ——>  : f ^ C i ) i s open i n Y f o r 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 c l e a r 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. a Y divides X, then T CI T and so T = ' T .  If  Conversely, suppose that  T^ i s any topology f o r 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 i n Y and hence B E T . p  Thus T = T  1.9 Lemma  a  ct  implies that Y divides X'.  A mapping f of the o r d i n a l space to + 1, provided w i t h  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  I f f : to + 1  of fCw),  > X i s continuous and U i s any  f ^(U) i s a neighbourhood of to i n to + 1.  contains (m» to] = {n : m < n e to} f o r some m e to.  neighbourhood  Then f .''"GJ) 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' (V) 1  =' {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. I f ..f(u)) e V, {f(n) : n e cj} i s eventually i n V and consequently there exists p e w  such that f(n) e V f o r 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 D e f i n i t i o n  A convergent sequence i s the union of the (Let S be a convergent  sequence and a l l of i t s l i m i t points.  sequence i n a t o p o l o g i c a l space X , and l e t S of S provided with the r e l a t i v e topology.  q  denote the range  The topology on S i s  the largest topology i n which the natural function f : S  >S  Q  defined by f ( x ) = x i s open.)  1.11 Lemma  Every convergent sequence i n a Hausdorff space i s  compact and metrizable.  Proof  Let S = {x  n  : n e w} \J {x} be a convergent sequence i n a  Hausdorff space, and suppose that U i s an open covering f o r S . Now x e U f o r some U e U. i n U and thus x  n  n < m choose U  n  Furthermore,  {x^ ; n e w} i s eventually  e U whenever ri > m f o r some m e' to.  —  e U  such that x  n  e U .  n  For each  Then {U} U {V  n  : m > n e to}  i s a f i n i t e subcovering of U for S, and that S i s a metric space, l e t k e to.  The c o l l e c t i o n {V^  neighbourhood basis at x.  = ^-{x^  so S i s compact. : Ic > n e to}  To  see  for each  : n e to} i s c l e a r l y a countable Because S i s compact Hausdorff and  hence a regular T^-space, there e x i s t s open sets U and V satisfying x e V d  {x.} x  c l V O'U  and x^ £ U.  Consequently  = S - ( c l V U {x : i ^ n e to, x £ c l V}), which i s open i n S. ^ n ' n r  The family  {V  n  : n e to} (J {{x } : n e to} i s a countable open basis n r  f o r the topology on S. T^-space.  Therefore  S i s a second-countable regular  In view of CI16], Theorem 4.17), S i s metrizable.  , 0b s erv.e,*t hat ,..,ther e ^ a  defined by d C x , x ) = |l/m m n '  1.12  Theorem  1/m.  For any t o p o l o g i c a l space X, properties C l ) and  C2) are equivalent. to C3) and  - l / n | and dCx ,x) m  If X i s Hausdorff they are also equivalent  C4).  Cl)  X i s sequential.  C2)  to + 1, provided with i t s order topology, d i v i d e s X.  C3)  Each subset of X which i n t e r s e c t s every convergent  sequence i n a closed set i s closed. CA)  Each subset of X which i n t e r s e c t s every compact metric  subspace of X i n a closed set i s closed.  Proof  > (2) Suppose that U i s a subset of X with f (U)  (1) <  open i n 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 n e w .  Then (g(n) : n e OJ) converges to g(w)  and i t follows from (1.9) that g : w + 1 function.  > X i s a continuous  Thus g "''(U) i s 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} i s eventually in U. fore U i s a sequentially open subset of X.  There-  If X i s sequential, U i s  open and consequently, by virtue of (1.8), w + 1 divides X. Assume now that U i s a sequentially open subset of X, and "let" !'- : ~w 4  ,  4  {f(n) : n e w} converges to f(w). (k,w] for some k e w .  If f(w) e U then f ^(U) contains  So f ^"(U) = (k,w] (J {n : k _> n E W , f (n) E U}  which i s open i n w + 1. If f (w) t U, f ''"(U) = {n E w : f (n) E U} i s open i n w + 1. Therefore f '''(U) i s open in w + 1; then U i s open i f w + 1 divides X. (1) <  > (3) Suppose that F i s a subset of X, and l e t  S = {x^ : n e w} U {x} be a convergent sequence in X. Either F f | S i s f i n i t e or infinite.  In the f i r s t case, F D S i s obviously compact.  In the second case, F contains a subsequence of {x  : n e w}.  Therefore,  i f F i s 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. let  be a net i n F fl S converging to y e X.  Sy has a c l u s t e r point i n F f\ S.  i s closed.  Since F f) S i s compact,  But, the Hausdorff  implies that y i s the only c l u s t e r point of S^.  hypothesis  Evidently, F f) S  Thus, i f a subset of X i s sequentially closed, i t  i n t e r s e c t s every convergent sequence i n a closed set. i f {x  n  : n e a)} i s contained  i n F f i S.  Now;  i n F, {x  n  Conversely,  : n e to} i s also a sequence  I f F D S i s closed then x e F f ) S and hence x e F.  -Consequently, a  subset of X i s sequentially closed i f and  i t i n t e r s e c t s every convergent sequence i n a closed set.  only i f The  ^equivalence Aof.<*X*l^^ (3) <  > (t\)  Suppose that F i s a subset of X i n t e r s e c t i n g  every compact metric subspace of X i n a closed set.  According  to  (1.11), each convergent sequence i s a compact metric subspace of X. Therefore  F i n t e r s e c t s every convergent sequence i n a closed set, and  ~hence~(3)- implies that-F- i s c l o s e d - T o - e s t a b l i s h - the-converse, assume that E i s a subset of X i n t e r s e c t i n g every convergent sequence i n a closed set.  Let K be a compact metric subspace of X.  X - i s Hausdorff, K i s also closed. converging to x, then x E K. i s closed.  Thus ED  Since  I f S i s a sequence i n E f]  A d d i t i o n a l l y , x e E because E A  K (S  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.  U{x})  The elementary properties of sequential spaces are summarized i n 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 {fCx^)  to  ins  to}  converges to f ( x ) whenever  { x ^  : n  E to}  converges  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 i m i t of any family of sequential  spaces i s sequential. (7) sequential.  A subspace of a sequential space need not be An open or closed subspace, however, i s sequential.  (8) Every l o c a l l y sequential space i s sequential.  Proof  (1) The n e c e s s i t y o f t h e c o n d i t i o n i s t r u e f o r a r b i t r a r y  t o p o l o g i c a l spaces. of Y c o n t a i n i n g i n g x.  I f f i s c o n t i n u o u s and U i s an open subset  f ( x ) , then f ^(U) i s an open subset o f X c o n t a i n -  Moreover, { x : n e to} I s e v e n t u a l l y i n f "^(U) and so n  f ( x ) e U f o r a l l n sufficiently large. n  converges t o f ( x ) .  Hence {fCx ) n  : n e to}  C o n v e r s e l y , suppose t h a t V I s an open subset  of Y and l e t {y^ : n e to} be a sequence i n X c o n v e r g i n g t o y e f ^ Ql) . By h y p o t h e s i s , fCy ) n  e V.  {f Cy ) Q  : n e to} converges t o f (y) and so e v e n t u a l l y ,  But then { y : n e to} i s e v e n t u a l l y n  i n f ^"(V). I t  f o l l o w s t h a t f ^(V) i s a s e q u e n t i a l l y open subset o f X.  Then, s i n c e  X i s s e q u e n t i a l , f "'"(V) i s open and hence f i s c o n t i n u o u s . - •"•(2')'"*"Let 'f "T"'X—->""Y''b'e'*V'qu^'tieht map o f a s e q u e n t i a l space X onto a t o p o l o g i c a l space Y.  Suppose t h a t U i s a s e q u e n t i a l l y  open subset o f Y and t h a t {x^ : n e to} i s a sequence i n X c o n v e r g i n g t o x e f '''(U). t o f ( x ) e U.  Then, s i n c e f i s c o n t i n u o u s , { f ( x ^ )  : n e to} converges  Consequently { x : n e to} i s e v e n t u a l l y n  i n f ^"(U), which  i m p l i e s t h a t f ^(U) i s a s e q u e n t i a l l y open subset o f t h e s e q u e n t i a l space X. quotient  Therefore  f ^(U) i s open and hence, by d e f i n i t i o n o f t h e  t o p o l o g y , U i s open. (3)  By (116], Theorem 3 . 8 ) ,  i f f i s a c o n t i n u o u s open  o r c l o s e d map o f a t o p o l o g i c a l space X onto a space Y, then Y i s t h e quotient  space r e l a t i v e t o f and X.  I t f o l l o w s from p a r t  (2)  that  i f X i s a s e q u e n t i a l space then t h e Image f ( X ) = Y i s s e q u e n t i a l .  • (4)  - 18 -  An example of a non-sequential  spaces w i l l be given i n (1.19).  product of sequential  l e t X be the cartesian product of any family {X -topological spaces. projection map  For each c e A l e t P  of X onto i t s coordinate  &  : a e A} of  : X  > X^ denote the  space X^.  From the  d e f i n i t i o n of the product topology on X, P^ i s continuous. more, according  to  (.[16], Theorem 3.2),  Further-  the projection of a product  space onto each of i t s coordinate spaces i s open. continuous open s u r j e c t i o n .  (4),  To prove the second part of  Hence P^ i s a (3)  Thus, i f X i s sequential, part  implies that X^ i s sequential. (5) {X  Let X be the d i s j o i n t topological sum  : a e A} of sequential spaces.  of any  If U i s not open i n X,  family there  3.  e x i s t s c E A such that U/1 open i n X .  i s not open and hence not sequentially  Consequently, there i s a point x e D f l X  C  and  a  C'  sequence i n X^-U  converging to x with respect to X  with respect to X.  c  and  Then U i s not sequentially open and  therefore 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  4  = (J) DC  cLC  "^b  >  s a t  o (f) 3p  ^ ^y^ S s  n  -  condition : i f a < b < c then  t a e  By d e f i n i t i o n , the inductive l i m i t of {X  3.  : a e A}  i s the quotient space X/^ where X i s the d i s j o i n t t o p o l o g i c a l sum of {'X x  a  : a e A} and R i s the equivalence  a  r e l a t i o n : two elements  e X^, x^ £ X^ i n X are equivalent i f and only i f there e x i s t s  c e A,such that a < c, b < c and 4 (x ) = 4, (x, ). ' ac a be b  It follows  T  from parts (5) and (7)  (2) that X/  i s sequential.  R  The non-hereditary  be demonstrated i n (1.15) and  nature of sequential spaces w i l l  (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 l e t U be a sequentially open subset of Y. "S = - {  x  n  J' n e to} i s a sequence I n *X converging to x e V c z Y then,  since Y i s open, S i s eventually i n Y. that x  n  If  e Y for each n > m. —  Moreover, {x , : n e to} i s a m+n  sequence i n Y converging to x e U. open i n Y, i  x m  +  n  There exists m e <o such  Then, since U i s sequentially  : n e u} i s eventually i n U.  that S i s eventually i n U.  This surely implies  Hence U i s sequentially open and  there-  fore open i n X. .Assume now  that Y i s a closed subspace of the sequential  space X and l e t F be a sequentially closed subset of Y.  Suppose  t h a t S i s a sequence i n F converging to y with respect to X.  - 20. Because Y i s closed i n X , y e Y and consequently S converges to y i n Y.  Since F i s sequentially closed i n Y, y e F.  Thus F i s  sequentially closed i n X, and so F = F f t Y i s closed i n Y. (8)  Let U be a sequentially open subset of a l o c a l l y  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 i s clear that V i s sequentially open and hence open i n i n t G. But then V i s open i n X. By hypothesis, there exists a c o l l e c t i o n {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 i n X.  Therefore  U = (J {V : x e U} i s open, and (8) i s established. x  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 i n e 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 i n X . Since each IK i s obviously open i n R, there exists a collection {V  :.n e u} of open intervals satisfying n e V 0 U • For each  - 21 new.  choose an open i n t e r v a l I with n e I CL V . n n n  {x E R : x < 0} U ( U {I  Then  : n E to}) i s an open neighbourhood of .0  i n X which does not contain U  f o r 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 r e a l 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 i s a usual open neighbourhood of the sequence {——r- : n e to}. —n-rx at 0.  The topology of the r e a l l i n e i s altered only  For each open subset G of X, { 0 } U G  {——j- : n E to} i s eventually i n G. n-rx  i s open i f and only i f  Accordingly, each sequence i n  X converging to 0 i s either eventually i n {0} or eventually i n every neighbourhood of {^J" : n E to}, and hence i s e i t h e r eventually -equal-to 0 or a subsequence of  :  n  e  w  ^•  Define a subspace Y = {(x, 0) : 0 4 x £ R} U {(^r» U {(0, 1)} of the plane. of the punctured r e a l l i n e  1) n E to}  The space Y i s the d i s j o i n t t o p o l o g i c a l sum .{(x, 0) : 0 £ x E R}  { ( — - , 1) : n E w l U U O , !)}•  and the convergent sequence  Since both {(x, 0) : 0 ^ x e R}  and  {(—^-p 1) : n e  sequential. of the form  {(0, 1)} are f i r s t - c o u n t a b l e , Y i s  C J } [)  The r e l a t i v e topology f o r Y i s generated by sets  {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 r e a l  line*!?.  > X be .the surjectlon defined by  Let I  :Y  5  "P(x, y) - x f o r each (x, y) e "Y.  To e s t a b l i s h that 'X i s sequential,  i t s u f f i c e s ..to prove that P i s a quotient map. that P i s continuous.  Let U be an open subset of X.  U i s open i n R and cons equently P i s open i n Y.  1  • *g^ ii e  w  P CU) = _1  eKever  « -^>~  in:  •  n  i s continuous.  e  : m > n e w, ~ j p  e  1) : m < n £ o>})  U}, which i s open i n Y.  Hence P  Now l e t V be a subset of X such that P ^(V) i s open : n E u), ~ - e V} U (V x  which i s open i n Y i f and only i f V i s open i n R. and so there e x i s t s k E w such that (~jip 1)  n+1  Assume  w  -1  _1  : n e co,—r=- e U}  - - ~ - ; '«Afben  If 0 i V then P CV) = ' { ( - ^ p 1)  Then P CV) =  \J {(^pD  e to} i s eventually i n G.  CG-{0} x {0}) {J (HO, 1)} U  (J { (~jip 1)  i n Y.  n  (U) = (U x {0})  I f 0 i U,  U = {0} (J G where G i s an open subset of the  IfOeU,  r e a l l i n e such that t' ^  n+1  It i s f i r s t shown  C{C0,1)} U i C~-, -1) : k ^ n  E V} (J CCV-{0}) x {0}).  Since P^CV)  E  {0}),  I f 0 £ V, CO, 1)  £ P '''CO  £ P ^CV) f o r each ri 21 k. U)}) U'{C^p D  i s open i n Y, V-{0}  :k > n i s an  open subset of the r e a l l i n e containing -—^ f o r each h >_ k, and  E  CO,  consequently V = {0} U (V-{0}) i s open i n X . map,  Hence P i s a quotient  and therefore X i s sequential. Consider the subspace X - v ^ j - : n E to}.  ' n E to} converging to 0 must be eventually  sequence i n X-{ ^  equal to 0, {0} i s sequentially open. open,  .Xrr.{^j-  Because each-  But then, since {0} i s not  : 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  d i s c r e t e topology and a non-sequential topology on X r e s p e c t i v e l y . The i d e n t i t y map  1  : (X, T)  X  > (X, T ) i s a continuous s u r j e c t i o n ot  of the sequential space (X, T) onto the non-sequential space (X, T ).  In p a r t i c u l a r , the continuous image of the i d e n t i t y 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 r e a l l i n e and l e t X be the t o p o l o g i c a l  space of (1.15).  The i d e n t i t y mapping of the f i r s t - c o u n t a b l e space  R-{^j^- : n e to} onto the non-sequential space X - v ^ j - : n e to} i s an open and closed s u r j e c t i o n .  - 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 i s a neighbourhood of a l l but f i n i t e l y many natural numbers. _,and.Hausdorf f.  Clearly, M i s countable  To establish that M i s normal, l e t 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 i n 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 e W f) A}] (J [ (W x W) D A] i s open and n  —A-G--V-G c l V c G. Suppose now that 0 e A. Then, choosing {mn e W : n e W D G} such that U = {n} U {(n, m) : m > m n ' n } is —  contained i n 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 C G .  Hence M i s normal.  - 25 To prove that M i s sequential, l e t U be a sequentially open subset of M. .is open.  For each x E (hi X hi) D U, x e i n t U since {x}  For each, x e hi f\ U, {(x, m+1) : m e to} i s a sequence  i n M converging to x.  Then, since U i s sequentially open, there  e x i s t s m e to such that V = {x} U {(x, m) : m > m } i s contained x x — x i n 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 {V  •is a neighbourhood of 0 contained  i n U.  x  : x e hi C\ U})  Hence U i s open, and so  :M i s sequential. ...Since 0 e cl^ihl  x hi), {0} i s not open i n K-hl.  --I^) ,: .1 ,.e .toj .is any,sje q,uenc.e,in.^ x ;  If  ^either.,there .is some  . n e hi such that n^ =-.n f o r infinitely-many i or there i s no such -~n. the  -In the f i r s t case, {(n^, m/) : i e to} has a c l u s t e r point i n set {n} \J {(n, m) : m e hi}. Indeed, either there e x i s t s 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. I t follows that the subsequence {(n, nr ) : i e to} has a c l u s t e r point at e i t h e r (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 i n which each point has a d i s t i n c t f i r s t coordinate.  Without l o s s of generality i t can be  - 26 assumed that the sequence ( ( n ^ , nu) : i e to}  Choose a sequence {k_^ e W : i E to}  coordinates. k. > m.. 1 1  has d i s t i n c t  first  such that  For each i E to l e t V. = {n.} (J {(n., m) : m > k.}, and 1 1 l — i  for each n e W-{n^  • : i E to}  let U  n  = {n} U {(n, m)  Then ijj iY. : i. e .(•£). J / i.USV^ : ,n e W-{n ±  : i  E  : m E W}. to.}.}.)  i s a neighbourhood of 0 d i s j o i n t from { ( i u , nu) : i e Accordingly, a sequence i n M-W i t i s eventually equal to 0. i n the subspace M-W,  to}.  converges to 0 i f and only i f Therefore {0} i s sequentially open  and hence M-N  The topological space M-W space which i s not sequential.  (/_{.0}  i s not sequential. i s a countable Hausdorff  The following shows that such a  space must f a i l to be l o c a l l y compact.  1.18 Proposition  Every countable, l o c a l l y compact Hausdorff  space i s f i r s t - c o u n t a b l e (and hence sequential).  Proof  Let X be a countable, l o c a l l y compact Hausdorff space.  Then X i s regular.  Let x e X.  compact neighbourhood K of x. compact.  By hypothesis, there exists a The subspace K i s regular and  From the r e g u l a r i t y condition, there i s a c o l l e c t i o n  U = {Uy : x ^ y E C l e a r l y {x} = C\ U.  X'} of neighbourhoods of x s a t i s f y i n g y I c l The family B of a l l f i n i t e i n t e r s e c t i o n s of  K.  '  1  - 27 members ,of U i s a neighbourhood basis at x.  This i s so because  otherwise there e x i s t s an open neighbourhood V of x such that no member of 8 i s contained  i n V.  But then, the i n t e r s e c t i o n  of the closed subset K-V with any f i n i t e i n t e r s e c t i o n of {el  : y e X-{x}} i s non-empty and yet  t h i s contradicts K"being compact.  (K-V) f) ( c l U  y  :y e  X - { X } }  To complete the proof, i t i s only  necessary to e s t a b l i s h 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  U  i = {f : A^  > U; f i s a function) and the set of a l l f i n i t e  i n t e r s e c t i o n s of i elements of (J.  Consequently the c a r d i n a l i t y A  i  of 8 I s l e s s than or equal to the c a r d i n a l i t y of U {U A  ""The card'inali'ty 'of 'U ^-^ ^)^~••^-4^ 8  B ±s <_ hfg' H ~ 0  : I e W}.  i  -  He ~ H  a  , "and -henc'e"the 'cardinalit-y of J  ([16], Theorem 179, page 279).  Since any countable product of f i r s t - c o u n t a b l e spaces i s f i r s t - c o u n t a b l e , i t i s natural to ask i f there i s an analogous r e s u l t for sequential spaces. by the succeeding example.  This question i s answered negatively  Indeed, the product of two sequential  spaces need not be sequential.  The construction used i n t h i s  example i s s l i g h t l y d i f f e r e n t than that derived by F r a n k l i n ([8], Example 1.11).  =  Using t h i s construction, i t i s also possible to  prove that the square of a sequential space need not be sequential.  0;  1..19 Example  There i s a product of two sequential .spaces which  i s not s e q u e n t i a l . Proof  Let OJ be the r a t i o n a l s 0^ w i t h the integers i d e n t i f i e d ,  and l e t X = Q_  Q.' •  x  The space X i s the product of two s e q u e n t i a l  spaces but contains a s e q u e n t i a l l y 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  l e s s than one converging monotoriically downward to 0. n E co, l e t  be the i n t e r i o r of the plane rhombus determined by  the p o i n t s ( x » n ) , (0, n + ^ ) » ( > -  x  n  n  J  n  of J n  )  a n  ^  n - -j) ; l e t  be the i n t e r i o r of the t r i a n g l e determined by the p o i n t s  (x , n ) , (1, n + 4) and (1, n -  W  For.each  i n the y - a x i s .  n  J  = H U J  n  n  U  UK n  ; and l e t K be the r e f l e c t i o n  Then { ( x , y)  E  R  i s an open subset of the plane.  2  : | x | > x } 1/ { (x, y) o ' 1  1  J  E  R  2 J  : y < 0}  Thinking of U {W^ : n E co} as a  subset of the plane w i t h the h o r i z o n t a l integer l i n e s i d e n t i f i e d , l e t W = X n ( U (W  : n E co}) . (See Figure 1)  n  Let P projections.  1  : X  > Q_ and  : X  > OJ be the c a n o n i c a l  For any open neighbourhoods U and U ' of 0 i n Q. and  OJ r e s p e c t i v e l y , P ^ ( U ) D P2^(U') i s not contained i n W because there e x 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} i n R n  (See Example  1.19)  - 29 To e s t a b l i s h that W i s sequentially open, l e t {y^ : n e to} be a sequence i n X converging to y e W. i n X i s simply convergence i n Q xQ  If I ^ C y ) ^ 0> convergence  and,since (Q. x Q) H ( U {W^ : n e to})  {y^ : n e 03} i s eventually i n W for t h i s case.  i s open i n Q_ x  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 , i n W, with y at the symmetric p o s i t i o n . Therefore i t can be assumed without loss of generality that y = (0, 0 ) . in set  .  Now { y  n  of integers k such that {P ^  ^^n^  0  p  : n  e  y  w  ^ i ^y vec  ^  : n  e  w  n  i n {0} U V where V i s any neighbourhood of K. f i n i t e because otherwise ^2^T? converging to 0.  : n  e  v  only i f n < m. s a t i s f y i n g (1)  *  n  a  s  a  S  n  c v e n t u a  Hy  Furthermore, K i s  subsequence not  : n e to} where k  n  < k  n  i f and  m  There i s a sequence {l_^ : i e to} of open i n t e r v a l s  I  ±  C V, (2) Z D  and only i f i = j . e 1^.  <°}  ' *  ^  uent  To v e r i f y that K i s f i n i t e , l e t V be a neighbour-  hood of K and suppose that K = {k  n >  net0  But then, i f P i s the quotient map of Q. onto. Q.' and K i s the  U-{k} f o r each neighbourhood U c f k, { 2 ^ ^  y  '' *  > (0, 0) implies that ^2^1?  : n e to}  I  ±  =  {1n } ±  t  and (3)  I n ±  Ij  f0  if  For each i e to there e x i s t s n^ e to such that  Next, l e t {U^ : n e to} be a sequence of open sets  s a t i s f y i n g k. e U. <Z I-{y .}, and l e t G be a neighbourhood of Z-K n  °  d i s j o i n t from U {I. : i e co}. i  It follows that G U ( U {U : n e co}) n  J  i s a neighbourhood of 0 i n OJ d i s j o i n t from the sequence { P ( y ) : i E co}. 0  n -  of {P (y 9  Therefore,  {P (y >) : i £ to} i s a subsequence 0  n  ) : 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  *  £  hood V of K, {y^  Se  v  e  n  t  u  a  l  ±  y  i  n  ^} f V for any neighbour-  : n e co} i s eventually i n  E = [X f l ( U {W :. n <_ q})] U (0. n  x  10}).  But E i s contained  Thus W i s sequentially open, and t h i s completes the Defining OJ and  i n W.  proof.  as above and thinking of U {W : n E co} n  as a subset of the plane with the h o r i z o n t a l integer l i n e s  identified  and the v e r t i c a l Integer l i n e s i d e n t i f i e d , i t i s not d i f f i c u l t see from (1.19) that (OJ x OJ) n ( U {W open subset of OJ x OJ which i s not open.  to  : n E co}) i s a s e q u e n t i a l l y Hence OJ x OJ i s not  sequential, and therefore the square of a sequential space need not be sequential. After a few preliminary r e s u l t s i t w i l l be shown that the s i t u a t i o n described  i n (1.19) cannot occur i n the presence of  s u i t a b l e compactness and separation conditions.  First, i t is  convenient to prove that countable compactness and compactness are equivalent i n sequential spaces.  sequential As i s well-known  ([28], Proposition 9.8), these concepts are equivalent i n the class of first-countable spaces.  Since sequentially compact  spaces are always countably compact, the following establishes their equivalence i n 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 i n X with no convergent subsequence.  Let A = U { c l x : n e to}. n  If S = {y : n e to} o n  " t i T y , '"^either  c l x for some m e to, or no such m e x i s t s . m y e c l x C A. m J  {y \  In the f i r s t case,  In the second case, there exists a subsequence  : k e to} of S with y„ e c l x . . ° k ^k n  i s a subsequence of S converging to y.  But then {x^ k n  : k e to}  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 i n i t e l y many n e to because otherwise S would have a convergent sequence. whenever n >^ k.  Let k e to be such that x ft c l x^  But then, applying the same argument at above,  - 32 - . : 11 >_ k} i s sequentially closed and yet does not contain  \J { c l  i t s accumulation point x. A r e s u l t 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 C o r o l l a r y  The product of two countably compact spaces,  one of which i s sequential, i s countably compact. Proof  Let {(x » y ) : n e to} be a sequence i n the topological n  product space X x Y of a countably compact space-X and a countably 'compact aequenti-ai«spaee--»Yv-^By'-virtu e^of-s?2Q').j-Y--d?s 'sequentially s  compact.  : k e to} which converges to some point y c Y.  {y n  Accordingly, the sequence {y^ : n e to} has a subsequence Since X i s countably  k  compact, {x n  k  : k e u} has a c l u s t e r point x e X.  Then (x, y) i s a  c l u s t e r point of the sequence t(x » y ) : n e to}. n  1.22 C o r o l l a r y  Let X be the topological product of any countable  family {X^ : n e to} of sequential spaces. compact i f and only i f each X  Then X i s countably  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 e s t a b l i s h that X i s  sequentially compact and hence countably compact, l e t {x^ : n E to} be a sequence i n X. of X onto X_^.  For  each l e w  I f each X^ i s countably compact then, by (1.20),  each X^ i s sequentially compact. {k^  l e t P^ be the projection map  Hence there exists a sequence  : i e to} of functions mapping .to into to such that {x^ o  i s a subsequence of {x : n E <JJ} and {P (x, , J n o k (n) o convergent subsequence of i P ( G  x n  )  : n e to}  : n e u} i s a  : n e to}, and f o r 1 < i e ii,  ~{x, , \ : n E to} Is a subsequence of {x, , ^ : n c to} and ..ac^Cn) ^ - i--T^'' k  -{P;(x, , »•) : n e to} i s a-convergent subsequence of {P.(x. , 5: n E to}, i ^(n)' ^i.^y %  1  , ^ : n E to} i s the desired convergent subsequence  The sequence {x, n of {x : n e to}, n  1.23 Corollary  Let X be an uncountable set, and l e t 2 denote the  set {0, 1} provided with the d i s c f e t e topology.  Then the product  X space 2 i s not sequential. Proof  Suppose that 2  i s sequential.  Since any product of compact  X t o p o l o g i c a l spaces i s compact, 2  i s (countably) compact and hence,  by (1.20), sequentially compact.  Let f : X  > 2  W  be a s u r j e c t i o n .  •Define a sequence {x a e X.  Let {x  n  : n e to} i n 2  X  by x ( a ) = [f (a) ] (n) for each n : n E to}.  : k e to} be a subsequence of {x  Now  k there e x i s t s y E 2 each k E to.  U  such -that y( 2j ) -=.0  and- y( 2i +i)  n  n  c  =  f o r  c  = y f o r some 3 e X.  Since f i s s u r j e c t i v e , f(3)  X Therefore, " i f "P  "is the "canonical "pro jection'map-of 2  -onto the  3 3-th  ) : k E to} cannot converge  coordinate space, then {P„(x  3  n  k  since 2 i s d i s c r e t e ; c l e a r l y P.(x  )  =  k  x n  (3)  [f(3)] (n^) =  =  y^y)'  k X  Thus {x  : k e to} does not converge, and hence 2  sequentially compact.  cannot be  The contradiction shows that 2  must not  be  •sequential.  1.24  Let X and Y be sequential spaces, and assume that  Theorem  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 .  G i s open, suppose that (u, v) E G and C l e a r l y u E G^.  = {x : (x, .v) e G}.  If {s^ : n E to} i s a sequence i n X converging to  s E G^ then { ( s , v) n  (s, v) E G.  let G  To prove that  : n E to} i s a sequence i n X x Y converging to  Since G i s sequentially open, { ( s , v) n  : n E to} i s  - 35 eventually i n ,G, and consequently {s^ : n E co} G^.  i s eventually ,in  Hence G^ i s sequentially open and therefore open i n X.  hypothesis,  there exists a sequentially compact neighbourhood  U of u with U x {v} contained  i n G.  Let V be the largest subset  of Y such that U x V C G; that i s , V = {z : U x { } z  V i s not open, there i s a sequence {y to y £ V.  By  n  C  : n E co} i n Y-V  But then, for each n E co there exists x  G}.  If  converging  £ U with n  (x^, y ) i G.  Since U i s sequentially compact, {X  a subsequence {x  , y n  , y k  : n £ co} has  : k e co} which converges to some point x e U.  It follows that {.(x hence that (x  r  k  ) : k E CO} converges to (x, y) E G and n  k  ) E G for a l l k s u f f i c i e n t l y large. n  The  k  c o n t r a d i c t i o n shows that V must be open.  Then,' since (u, v) e.U *• V C  (u, v) E i n t 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 l o c a l l y countably compact or l o c 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.  G,  • - 36 Proof  (1) Let X and Y be sequential spaces, and assume that X  i s regular and l o c a l l y countably compact. countably compact neighbourhood K.  Each point x e X has a  Let {U : a e A} be a neighboura  hood basis at x such that each U i s a subset of K. a regular, f o r each a e A there e x i s t s an open set V  Since X Is with  a  x e V a c l V a. U . Then each c l V i s countably compact and Q, Q. 3. . cl {cl V : a e A} i s a basis of countably compact closed a hoods of x. By v i r t u e of (1.13.7) and (1.20), c l V  neighbour-  i s sequential  a and hence sequentially compact. neighbourhood basis { c l V  Accordingly, each x e X has a  : a e A} c o n s i s t i n g of sequentially  a «Gompact,v,s.ubsefcs,..,,.«^The«.iprieGeding.,theorem. implies. that sthe...product w  space X x Y i s sequential.  u  The second part of (1) i s now c l e a r  since sequentially compact spaces are countably compact. (2)  This follows from (1) because every l o c a l l y compact  Hausdorff space i s regular and every compact space i s l o c a l l y countably compact. As seen i n Example 1.19, the product space Q * Q.' i s not sequential.  Although both coordinate spaces Q and  are regular,  neither t o p o l o g i c a l space i s l o c a l l y countably compact.  It i s  c l e a r that 0J f a i l s to be l o c a l l y countably compact at 0.  The  space Q i s not l o c a l l y countably compact because regular  locally  countably compact spaces are Baire spaces and Q i s 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 r e s p e c t i v e l y , have the same compact sets. Proof  Since the sequential closure topology i s larger than the  product topology, i t i s only necessary to show that each compact subset of X x y i s compact i n (X x Y) . Let K be compact i n 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 r e s p e c t i v e l y . K are compact Hausdorff and hence closed. y r  The subspaces Thus K x  and K y  and are also  -•^sequenfeial^spaeesw'^ x K  x  i s sequential.  Consequently, the topology induced on K by  x K i s the same as that induced on K by either the usual product y -  topology or the sequential closure topology.  Therefore U D K i s  open i n K whenever U i s open i n (X x y) , and hence K i s a compact  s subset of (X x y) . s The foregoing c o r o l l a r y i s of i n t e r e s t i n studying k-spaces. — ( A t o p o l o g i c a l 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 i n K f o r every compact subset K of X.)  - 38 1.27 Proposition  Proof AH  Every sequential space i s a k-space.  Suppose that A i s a subset of a sequential space X with  K closed i n K for every compact subset.K of X.  sequence i n A converging to x.  Let S be a  Since S U {x} i s compact,  A D (S U {x}) i s closed i n 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 o r d i n a l space U + 1 provided with the order topology i s a k-space which i s not sequential.  The space ft + 1  i s a k-space because i t i s compact ( [ 7 ] , pp. 222, 162) and the l o c a l l y compact spaces are k-spaces ([7], 11.9.3). In view of (1.26) and (1.27), i t i s 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 , i n f a c t , a product of two Hausdorff sequential spaces which i s not a k-space.  This i s so because  there e x i s t s a non-closed sequentially closed subset A of Q. x £)* such that A f\ K i s closed i n K for every compact subset K of  ca * a') . s  The next two theorems are important r e s u l t s concerning the product of quotient maps and the product of sequential spaces respectively.  For each cardinal number m, l e t  d i s c r e t e space of c a r d i n a l i t y m, l e t obtained from D  denote the  be the quotient space  x [0, 1] by i d e n t i f y i n g a l l points i n D  x {0},  -  let g m fo  : -D x [ 0 , 11 -—> m '  the point  Y  x {0} i n Y^.  m  39  -  be the quotient map, ^ ' r  and l e t y "'o  denote  For any t o p o l o g i c a l space X, l e t 1^ be  the i d e n t i t y function on X.  (For any two  functions f : X  > Z  and g : Y  > Z , define ( f x g)(x, y) = ( f ( x ) , g(y)) for each  (x, y) e X x  y.)  q  1.28 Theorem are  The following properties of a regular space X  equivalent. (1)  X i s l o c a l l y countably compact.  (2)  1  x g i s a quotient map  for every quotient map  g  x with sequential domain. (3)  h = 1„ x g X  i s a quotient map,  where m i s the  m  smallest c a r d i n a l such that each x e X has a neighbourhood basis of c a r d i n a l i t y <_ 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. C l e a r l y 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 i s continuous.  Let G be a subset of X x Z with f *(G) open i n X x Y . (u, v) e G and l e t r e  g ^(v).  X x Y such that (u, r) e U x v C  Suppose that  There i s a basic open set U * V i n f *(G).  Since X i s l o c a l l y countably  compact, there e x i s t s a countably compact neighbourhood X.  of u.  - 40 Then, since X i s regular, there i s an open subset satisfying u e  c lI^C U D  The set K = c l U  countably compact neighbourhood of u contained E = {z e Z : K x { } C z  G}.  of X  Since  1  is a  i n U. Let  (u, r) e K x V C f ( G ) , - 1  K x { } <Z f^CG) and G ^> f (K x { }) = K x { ( )} = K x { }, which r  r  implies that v e E.  g  r  v  I t remains to prove that E i s open. But  since g i s a quotient map, i t s u f f i c e s to show that g ^(E) i s open. If a e Y and K x {a} d f ( G ) then f ( K x { a } ) d G , which implies - 1  that K x g(a) CZ G and hence that g(a) E E. g ( E ) = {y e Y : K x { y } < C f ( G ) } . - 1  _1  open. {y^  Thus  Suppose that g ( E ) i s not - 1  Then, since Y i s sequential, there i s a sequence  : n e to} i n Y-g ^"(E) converging to some y E g "*"(E). S O  K x {y^} CZf"f "'"(G) f o r each n E to.  Hence there exists a sequence  {x : n e to} i n K with each (x , y ) £ f ^(G). Because K i s n .n n countably compact, { X : n E to} has a c l u s t e r point x E K. r  the sequence { ( * , y ) : n e to} has a c l u s t e r point  Then  (x, y) e f "'"(G).  n  Since f "'"(G) i s open, {(x » Y ) : n E to} i s frequently i n f "*"(G) n  n  contradicting (x » y^) £ f ^"(G) f o r a l l n e to. n  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 d i s c r e t e , D i s l o c a l l y compact Hausdorff and i t follows m m from (1.25) that D '* [0» 1] i s sequential. map with sequential domain  Thus g  m  i s a quotient  x [ 0 , 1] and so (2) implies (3).  _ 41 (3)  > (1)  Suppose that X i s not l o c a l l y countably Let {U^ : a e D^} be a neighbourhood  compact at-some point X . q  basis at x . o  For each a e D , c l U i s not countably compact m a  a and thus has a countable family {F  : n e W} of d i s t i n c t non-empty  n  ,,xlos.ed,,suhse.ts...sa,tiafying^the..finite...intersection...property .whose Si  cl  i n t e r s e c t i o n i s empty. Let E = D {F^ : n >_ k e W} f o r each " a a a n e W. I t 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 n  a family {E^ : n e W} of d i s t i n c t  exists a countable well-ordered non-empty closed subsets of c l U  s a t i s f y i n g the f i n i t e i n t e r -  a  '  section property whose i n t e r s e c t i o n i s empty. To e s t a b l i s h that h i s not a quotient map, a e D  m  define S a C  and define S C X  X x (D  m  x [0, 1]) by S  &  = U  for each  {E* x {( , ^)} : n e W}, a  x y  b y S = U {h(S ) : a e D }. I t s u f f i c e s to m . a m show that h ( S ) i s closed i n X x (D x [0, 1]) and that S i s not m - 1  closed i n X x y . for any x e X.  Note that ( x , y ) t S since (x, (a, 0)) I o  If U x V i s a neighbourhood of ( x  then, since X i s regular, c l  (c, i ) e g^(y)n <<c, the point (c,  Q  W CZ Q  : n E"W},  o >  y ) i n X x y^  U f o r some c e D^.  Thus choosing  0 * h(E^ x { ( c , | ) } ) C (U * V ) n S;  e x i s t s because V i s a neighbourhood of y  Q  and  - 42 . v ^ r : n e co} converges to 0 i n [0, 1].  I t follows that  (x , y ) e cl„ „ (S) and hence that S i s not-closed i n X x Y . o' o X*Y m m N  J  Since D  m  h~ (S)n 1  i s d i s c r e t e , i t remains to prove that (X * ({a} x [0, 1])) i s closed i n X x ({a} x [0, 1]) f o r  each a e D .  Assume that 'the point (x, (a, a)') i s contained  = S • a  with respect to X x ({a} x [0, 1 ] ) .  fl  - 1  (X x ({a} x [0,1]))  3.  HI  S  » 0, h ( S ) n  But since (X x {(a, 0 ) } ) D S  a i s clear that x e E-. 1  Then, since {E  Since each  a  i n the closure of i s closed, i t  a : n e W} i s well-ordered n , -  a a {E : n e W} = 0, there i s a smallest set E^ containing x.  f)  n  follows that a j> ^- because otherwise G x ({a} x (a - e, a + where  0 < e <  and It  e)n[°,  1]),  ?~ . >. and G i s a neighbourhood of x d i s j o i n t from  a Ep ^, i s a neighbourhood of (x,  (a, a)) d i s j o i n t from S .  +  g  since [0, 1] i s Hausdorff, a E (x,(a, a)) £ E  a  ot  x {(a, a)} C S  a  : n <_ p}. .  Thus h  -1  Moreover,  Therefore  (S) i s closed i n  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 i s l o c a l l y 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  c a r d i n a l such that each point of X .has a neighbourhood basis of c a r d i n a l i t y £ m* Proof  The proof that (1) implies (2) i s given i n (1.25), and  • i t i s ^obvious -that *(2-) implies (3) .«since Y^ ,is •sequential »and every sequential space i s a k-space.  By v i r t u e of (1.28), to e s t a b l i s h  that (3) implies (1) i t s u f f i c e s to prove that h = l  xg  v  A quotient map.  is a  m  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 l e t P be the f i r s t m projection map of X x Y  m  onto X.  I f (X x {y^}) r\ K = 0, h ( K ) = K _ 1  *ahd-"'K*-'±s-^Iso^-xbmpaet^ih*X'x-(. ^x«['0,-"'"x  '{y^)nYc' f''^ !  D  for any a e D  ffi  coordinate  }  the set E = (K-(X x {y })) U (P(K) x { ( , 0)}) i s a Q  a  compact subset of X x (D x [0, 1]) because any open neighbourhood of (a, 0) i n x [0, 1] i s contained i n some open neighbourhood of •y i n Y . In addition, h(E) = K. I t follows that every compact o m subset of X x y i s the image under h of a compact subset of m X x (D x [0, 1 ] ) . m  Suppose now that B i s a subset of X x Y with  m  h ( B ) closed i n X x (D x [0, 1 ] ) . _1  m  Since X x y  i s a Hausdorff  m  k-space, to prove that B i s closed i t i s s u f f i c i e n t to show that B A K i s compact i n K for every compact subspace K of X x y  - 44 •But K •== h(C) f o r some compact subset C of X x (D x [0, 1 ] ) . m E = h ^"(B) n 'C i s a compact, subset of C. in B A I .  I f .b e B n K  such that h(a) = b.  Thus  C l e a r l y h(E) i s contained  then b e K and so there exists some a e C  Furthermore, b e B implies that a e h "''(B) and  hence that a e h ( B ) n C = E. - 1  consequently h(E) = B o K.  Therefore b = h(a) e h(E) and  Then, since E i s compact i n C and h i s  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 d i s j o i n t  t o p o l o g i c a l sum of convergent  Proof sequence  sequences.  Let X be a sequential space. {X  r  For each x e X and f o r each  : n e to} i n X converging to x, l e t SCx^, x) = {x  be a Hausdorff topological space i n which each x sequence {x^ : n e to} converges to x.  n  n  : n e to} U {x}  i s isolated and the  Although the elements of  S ( x , x) need not be d i s t i n c t i n X, they are taken to be d i s t i n c t i n n  S ( x , x) . n  Thus (  order topology.  s  x n  >  x) i s homeomorphic to to + 1 provided with the  Clearly S ( x  n>  x) i s a convergent sequence i n S ( x  Let W be the d i s j o i n t topological sum of a l l possible S(x , x ) .  n >  x).  Since for ,.each x e X the convergent sequence {x  : x^ = x, n e co} U {x}  i s a .summand >of W, the natural function f : W  > X defined by  f(x) = x i s a s u r j e c t i o n . 'is'continuous  In addition, f i s continuous because i t  on each summand.  To complete the'proof,  e s t a b l i s h that f i s a quotient map. f  "\(11)  opjen i n W.  I f {.y  y e U, y e f ^ ( U ) n ^^ » n  {y  n  Let U be a subset of X with  : n e co} i s a sequence i n X converging to Y^ which i s open i n S ( y  : n e co} as a subset of S ( y  hence • {y  i t -remains to  n >  n >  y).  Then  y) i s eventually i n f ^"(U), and  : n e co} as a subset of X i s eventually i n U.  Consequently  U i s sequentially open and therefore open i n X.  1.31  Corollary  A Hausdorff space i s sequential i f and only i f i t  i s a quotient of the d i s j o i n t t o p o l o g i c a l sum of i t s convergent sequences.  Proof  The necessity of the condition i s c l e a r from (1.30).  It i s  only necessary to remark that i f X i s Hausdorff then W i s p r e c i s e l y the d i s j o i n t t o p o l o g i c a l sum of a l l the convergent sequences i n X. Conversely, (1.11) implies that each summand of W i s a metric space and hence a sequential space. and  (1.13.2).  Then X i s sequential by (1.13.5)  - 46 1.32  Corollary  Every sequential space i s the quotient of a zero-  dimensional, l o c a l l y compact, complete metric space. Proof  It s u f f i c e s to show that W i s a zero-dimensional,  Suppose that U i s a neighbourhood  compact, complete metric space. basis at y e W.  Each D e 0 i s both open and closed i n W because  U r\ S(x , x) i s bo.th open and .closed i n S(x , x ) . n n  Hence W i s  r  zero-dimensional. metric space.  locally  According to (1.11), each S(x , x) i s a compact n  Obviously W i s l o c a l l y compact.  If d . x  v is a  x  S v _ » /  n  metric on S(x , x ) , then n  f  dS(x ,x) ' ( u  V )  l f  U  n  1 i s a metric on W.  >  V  £  S ( x  n»  X )  otherwise  L a s t l y , W i s complete by v i r t u e of ([7], C o r o l l a r y  14.2.4).  1.33  Proof  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 f i r s t - c o u n t a b l e space.  In view of the preceding c o r o l l a r y , (1) implies (2).  (2) implies (3) because metric spaces are f i r s t - c o u n t a b l e .  Clearly Since  f i r s t - c o u n t a b l e spaces are sequential, (3) implies (1) by (1.13.2).  - 47 1.34  Example  There i s a separable sequential space which i s not  the quotient of a separable metric space.  Proof  Let H be the r e a l numbers provided with the half-open  i n t e r v a l topology.  Sets of the form {x : a <^ x < b} = [a, b) are  a basis for t h i s topology.  Since for each x E H the c o l l e c t i o n  {[x, ^j_) 5 n e co} i s a countable neighbourhood basis at x, H i s n  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 i s the quotient of  a separable metric space X, and l e t P : X map.  > H x H be the quotient  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 a  n  : n E co} of H x  H.  Consider the subspace K = {(x, -x) : x i r r a t i o n a l } 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, f o r 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 i s a closed subspace of the Lindelof space  H x H.  Since K i s d i s c r e t e and uncountable, ,K i s not L i n d e l o f .  However, i f {V - : a e A} i s an open covering of K, Si  {V  : a e A} U {(H x H)-K}  a  since H*x {V  a  n  i s an open covering of H x H. Then,  H i s Lindelof, there i s a countable subcovering  : n e co} U {(H x H)-K}  covered by {V  a  of H x H.  It follows that K i s  : n e co} and hence that K i s Lindelof.  n  The  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 map  Proposition  For X sequential, l e t 4^ denote the  quotient  of X* onto X, where X* i s the d i s j o i n t t o p o l o g i c a l sum  of  convergent sequences i n X as derived i n (1.30). (1) i f and only i f  A subspace 4  v  i , - l ,  v  v  Y of a sequential space X i s sequential i s a quotient  map.  'X (2)  The product of two  sequential i f and only i f <j> .x 4 Y  sequential spaces X and Y i s i s a quotient  map.  - 49 ( 1 ) Let Y  Proof  = c C ^ Y ) and cf> = <j> | ' . X • v  X  Let g : Y*  >Y  1 ^  A  be the function defined by g(x) = x f o r each x e Y*. a quotient map i f and only i f Y i s sequential.  Then g i s  I t s u f f i c e s 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.  I f 4>~ (U) i s open i n Y 1  i s open i n 'S f o r each summand S of Y^.  then <j) (U)/^\ S -1  1  From the d e f i n i t i o n of the  r e l a t i v e topology on Y i t i s clear that each convergent sequence i n Y i s a convergent sequence i n X.  Therefore each summand of Y*  i s a summand of Y^, and consequently g ^(U) = < f > ^(U)/n Y* i s open i n Y*.  Conversely, suppose that g ^(U) i s open i n Y*.  Then  4 "*"(U)/*"} S = g ^(U)n S i s open i n S f o r each summand S of Y^ which i s also a summand of Y*. then  = S ( x , x) D Y i s a n  Let S ( > ) be a summand of X*; x  summand of Y^.  i s either f i n i t e or i n f i n i t e . i s i s o l a t e d and thus  n  C\ < J > ^(U) must be open i n S^. Assume I f x t U then  : n e w} n <{> ^(U) which i s c e r t a i n l y open i n S^.  If x e U, each sequence i n subsequence {x^ \  converging to x e  : k e to} of {x  : n E to}. But since  n  k  D <f> ''"(U).  O <i> "^(U) i s a  n  * ^(U) H S(x , x) i s open i n S ( x , x ) , {x °k k in  The topological space  In the f i r s t case, each point of  now that the second case occurs. S ^ n I "*"(U) = {x  x  n  : k e to} i s eventually  n  I t follows that S f) <j> "*"(U) i s a sequentially open 1  subset of the f i r s t - c o u n t a b l e space S^. i n Y^,  Therefore  <j>  (U) i s open  and the proof of (1) i s complete. (2)  Since X* and Y* are f i r s t - c o u n t a b l e , the t o p o l o g i c a l  product space X* x Y* i s f i r s t - c o u n t a b l e and hence sequential. i f h '=  -.<(>.  X  x  A  Y  i s a quotient map,  X x y i s sequential.  .Then  To e s t a b l i s h  the converse, assume that .X x y i s sequential and l e t G be a subset of X x y with h~ (G) open i n X* x y*. 1  Suppose ( (x »' y ) : n e co}  i s a sequence i n X x Y converging to (x, y) e G.  Then (x, y) e h  ^"(G)  and there e x i s t s a basic open subset U x V of X* x y* such that (x, y) e U x V C  h (G).  Accordingly, U n S ( x  - 1  S(x , x) and V f l S(y , y) i s open i n S(y , y ) . n n n the sequences {x^  : n e co} and  {y  large.  Therefore  x) i s open i n This implies that  : n e co} are eventually i n U and  n  V r e s p e c t i v e l y , and hence that ( x  n>  y ) e h .(G) for a l l n s u f f i c i e n t l y 1  n >  (x , y ) = (<j> (x ), ^ ( y ^ ) ) = h(x n  xi  A H  x n  , y ) e G for n i l  a l l n s u f f i c i e n t l y 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 f i r s t - c o u n t a b l e space with unique sequential l i m i t s i s Hausdorff since otherwise i t i s possible to f i n d a sequence converging to two d i s t i n c t points.  That i s , i f x and y are d i s t i n c t  points  which cannot be separated by d i s j o i n t open sets and { U : n e co} and {V : n E co} are countable neighbourhood bases of x and y n  - 51 respectively, then the sequence {x converges to both x and y.  : n e to} s a t i s f y i n g  x  n  e U^HV  R  The succeeding examples show that t h i s  r e s u l t cannot be generalized to sequential spaces. The construction used i n the f i r s t example i s based on Sorgenfrey's [29] well-known r e s u l t 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  l i m i t s which i s not Hausdorff. Proof  Let H be the r e a l numbers provided with the half-open  *inter-vai ^ in  sequential. H x H, l e t A  I f A = {(x, y) : x + y = 1} i s the antidiagonal of  q  and A. be those points of A with r a t i o n a l and l  i r r a t i o n a l coordinates r e s p e c t i v e l y .  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^; s i m i l a r l y , ([x, x+e) x [y, y+e))D i f and only i f (x, y) e A ^ x + y < 1, ([x, x+6) x [ Therefore A  y >  A^ ^ 0  And f o r each (x, y) e H x H such that l - x - 5 ) ) O A = 0 whenever  0 < 6 < l - x - y .  and A. are d i s j o i n t closed subsets of' H x H.  To prove that  and A^ cannot be separated by d i s j o i n t  open s e t s , 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 z e r o .  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. — n  In the r e a l l i n e R, the t  i r r a t i o n a l s are of the second category ( [ 7 ] , pp. 249-251) and consequently there e x 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  there e x i s t s p e R such that P < ~  V of the point ( r , 1 - r ) ,  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 - , 1-s + ^ ) ) j 0, s  and hence U H V ^ 0 for every neighbourhood V of ( r , 1 - r ) . fore ( r , 1-r)  e c l U , and so A ^ / T c l U  There-  0 for every neighbourhood  U of A . ±  The set E = ( A x A ) U ( A x A ) {J { ( x , x) : x e (H x H) q  ±  ±  ( A ^ U A^)} i s an equivalence r e l a t i o n i n the f i r s t - c o u n t a b l e H x H.  space  The quotient space X = (H x H ) / „ i s s e q u e n t i a l and T, but hi  not Hausdorff.  Let 4 : H x H  I  > X be the quotient map, and l e t  q = * (A ) and i = <|>CA.). Then q and i are the only pair of d i s t i n c t points of X which cannot be separated by open sets, and consequently i f some sequence i n X converges to two d i s t i n c t points, ..they must be q and I . converges to q.  Suppose that S = "{x  : n e w}  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  from q, there must be some q^ e A^ and a subsequence S converging to q^ i n H x H. A^ d i s j o i n t  different  q  of S  But then there i s a neighbourhood of  from S , and thus S cannot converge to i . q  Hence X  h a s ^unique,..sequential ..limits...  1.37 Example  There i s a countable, compact, sequential space  with unique sequential l i m i t s which i s not Hausdorff. Proof  Let M be the Hausdorff sequential space of (1.17). = M 0 {p} with M open  Let p be some point not i n M and l e t 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 i s Hausdorff and a  convergent sequence i n M cannot also converge to p, unique sequential l i m i t s .  However, M  n  has  i s not Hausdorff because  - 54 0 and p have no d i s j o i n t neighbourhoods. i s countable and compact.  It i s also c l e a r that  To v e r i f y that  i s compact, l e t  U and V be any open neighbourhoods of 0 and p r e s p e c t i v e l y . W-(U  U V) i s ^finite .and M^-(U  U V) i s the union of the ranges of  a f i n i t e number of convergent sequences i n W x hi. "sequence i n *M^-(U XI V)  s a t i s f y i n g n e U , M -(U n 1 1  To see that  Assume now  each point i n V-{p}  U V V [ (J{U  : n e W-(U  n  U V)} of open sets  :n e W-(U U V)}]) i s f i n i t e ,  i s sequential, l e t V be a sequentially  I f p ft V, V i s sequentially open i n the  sequential space M. i n M^.  But each  converges "to some "meniber Of "W-"(U"U V) .  Hence for any f i n i t e c o l l e c t i o n {U^  open subset of M^.  Then  Then, since M i s open i n M^, that p £ V. because V-{p}  Since any subset {(m,  V i s also open  C l e a r l y , V i s a neighbourhood of i s sequentially open i n M.  n) : m e A C W, n e B C W} contains a sequence  converging to p whenever A i s 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 f i r s t - c o o r d i n a t e s . Thus V contains a basic neighbourhood of p and so p E i n t V. i s an open subset of M  1  and consequently M  1  Therefore  i s sequential.  V  1.38 Proposition  5 5  -  (1) A sequential space with unique sequential  l i m i t s i s T^. (2)  I f X i s a topological space with unique sequential  l i m i t s and X x X i s 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} i s sequentially closed and hence closed i n Y. (2)  Every sequence i n A i s of the form t ( > x  n  Since X has unique sequential l i m i t s , { ( > x  n  x n  )  x n  )  :n e -  : 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 i n X x X. It follows frqmjpart and  (2) that the product spaces X x x  x M^, where X = (H x H ) ^ and £  are the non-Hausdorff  sequential spaces of (1.36) and (1.37) respectively,are not sequential.  I f these products were sequential, X and  would be  Hausdorff. A f t e r a preliminary r e s u l t  i t w i l l be shown that a  sequential space with unique sequential l i m i t s i n which each point has a neighbourhood basis consisting of countably compact sets i s Hausdorff.  The t o p o l o g i c a l spaces X = (H x H ) ^ and  s a t i s f y t h i s compactness condition.  £  do not  For any neighbourhood U of q  - 56 i n X, let x  n  {x^  {x  n  : n e to} be a sequence i n U such t h a t  E ( n , n+e ) x (1-n, 1-n+e ) CZ U where 0 < E n n n  < -i-.  2  The  : n E 10} has no c l u s t e r p o i n t i n the neighbourhood U .  sequence  ^  There-  f o r e U i s not c o u n t a b l y compact, and c o n s e q u e n t l y X i s not even l o c a l l y c o u n t a b l y compact. hood of p i n M ^ .  Suppose now t h a t V i s any n e i g h b o u r -  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 t o 0, V-(W U {0}) i s a l s o a neighbourhood of p .  It  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} 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 .  {(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}). c o u n t a b l y compact, and  is But  So V-(WU{0}) i s not  does n o t have a c o u n t a b l y compact 1  neighbourhood b a s i s at 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 u n i q u e 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 s u b s e t i s c l o s e d (and hence sequential). Proof  L e t X be a s e q u e n t i a l s p a c e .  Suppose X has u n i q u e 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 s u b s e t of X . be a sequence i n K c o n v e r g i n g t o x . limits,  Let S = {x  n  : n e to}  S i n c e X has u n i q u e s e q u e n t i a l  {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 o r f r e q u e n t l y x  = y.  - 57 Again since X has unique sequential l i m i t s , y = x.  Thus x i s the  only c l u s t e r point of S, and consequently x e K.. Therefore  K is a  .sequentially closed subset of the sequential space X. Assume now  that S i s a sequence i n X converging to two  d i s t i n c t 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.  ,1.40 if  Corollary  A sequential space has unique sequential  limits  and only i f each sequentially compact subset i s closed.  Proof the  But then {y} i s a non-closed compact subset of X.  Since every sequentially compact set i s countably compact, necessity of the condition follows from (1.39).  .suppose that S = {x^  Conversely,  : n E .to} i s a sequence i n a sequential space  X converging to points x and y.  Then {x} U range S i s sequentially  compact because each sequence i n the set has a subsequence which i s either a subsequence of S or eventually equal to x .m .e-.io.  m  f o r some  Therefore...{x} U range .S_is .closed.,. and_consequently either  y = x or frequently x^ = y.  The l a t t e r case cannot occur since  otherwise {y} i s a non-closed sequentially compact subset of X.  1.41  Proposition  limits.  Let X be a sequential space with unique sequential  If each point has a neighbourhood basis c o n s i s t i n g of  countably compact sets, then X i s Hausdorff.  - 58 Proof  Each countably compact subset of X i s closed  sequential  (by 1.39),  (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 l i m i t s 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 f i r s t - c o u n t a b l e spaces-. The study of Frechet spaces i s c l o s e l y related to that of both f i r s t - c o u n t a b l e 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 l i m i t s 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]) i n v e s t i g a t i o n of Frechet spaces.  There are, however, several  r e s u l t s concerned with A r h a n g e l * s k i i s [2] study of pseudo-open 1  maps and a r e s u l t due to Harley [12] connected with the product of Frechet spaces.  2.1 D e f i n i t i o n  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 l i m i t s of sequences i n 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 i n 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 definition 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 i n W x W converges to 0.  2.3  Theorem (2)  (1)  Every subspace of a Frechet space i s Frechet.  The disjoint topological sum of any family of  Frechet spaces i s Frechet. (3)  Every l o c a l l y 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 i n A i d e n t i f i e d , 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. sequences i n A.  Then c l (A) i s the set of l i m i t s i n X of Hence c l (A) = c l ( A ) n v  Y  Y i s the set of l i m i t s  X  i n Y of sequences i n A. {X  a  (2) Let.X be the d i s j o i n t topological sum of the family : a e ,A} of Frechet ..spaces. Let B be a subset of X and l e t B'  be the set of l i m i t s of sequences i n B. i s the set of l i m i t s i n X c Frechet, B' n X proved.  c  (3)  of sequences i n B(1 X ; because X i s c c  i s closed i n X . c /  For each c e A , B ' n  Then B' i s closed and (2) i s  '  Let B' be the set of l i m i t points of sequences In  a subset B «of-a l o c a l l y -Frechet'<space-X. -'For-•each'-'x e X-B' ;  i s a neighbourhood G of x which i s Frechet. i s Frechet.  Let V = ( i n t G ) O  (X-B').  there  By part (1), i n t G  The i n t e r s e c t i o n ( i n t G ) O  i s the set of l i m i t s of sequences i n ( i n t G) H B.  B'  Then ( i n t G ) n B'  i s closed i n the subspace i n t G, and consequently V i s an open subset of i n t G.  I t follows that V Is open i n X.  exists a c o l l e c t i o n {G^ : x e X-B } 1  that each G  x  i s open i n X. and so B  1  of Frechet subspaces of X such  i s a neighbourhood of x. Therefore X-B' = U {V  i s a closed subset of X.  By hypothesis, there  Each \r = ( i n t G > H (X-B') x  x  : x e X-B*}  i s open i n X  Hence X i s Frechet.  (4) Let  If g : X  > Y i s the quotient map, l e t i = g(A).  B' be the set of l i m i t points of sequences i n a subset B of Y.  clyCB) = 1 ( ) = ' « C  If i I c l ( B ) then, since X i s Frechet,  B  B  X  Y  If i e c l (B), either k e c l (B) f o r some k e A or no such k e x i s t s . Y A In the second case, there i s a c o l l e c t i o n {U : x e A} of open x subsets of X s a t i s f y i n g x e U U {U  x  x  and U^H B = .0.  But then  : x e A} i s a neighbourhood of i d i s j o i n t from B.  contradiction shows that the f i r s t case must occur.  The  Thus k e c l ( B ) v  and consequently there exists a sequence i n 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 (by  (1)  The r e a l l i n e with the integers i d e n t i f i e d i s Frechet  2.3.4) but does not s a t i s f y the f i r s t axiom of c o u n t a b i l i t y  (by 1.14). (2)  I t follows from  (2.3.4) that the t o p o l o g i c a l space  £)', the r a t i o n a l s with the integers i d e n t i f i e d , i s Frechet. Q x  Then  Q.' i s the product of two Frechet spaces which, by (1.19), i s  not sequential and hence not Frechet. i s not a Frechet  space.  S i m i l a r l y , the square  x Q_'  ',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  first-  countable space i s Frechet, the proof of (1) i s the same as that of (1.16.2). (2)  Let X be the t o p o l o g i c a l space of (1.15) and l e t : n e <JJ} U  A = X-({  {0}).  n { X j , : i e w} i n A converging to  For each n e w  n +  1 ^-  l i m i t s ([16], Theorem 2.4), 0 e c l A.  there i s a sequence  By the theorem on i t e r a t e d However, every sequence i n  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 i s a quotient of  the f i r s t - c o u n t a b l e space Y.  2.6 D e f i n i t o n  A s u r j e c t i v e function f : X  > Y of the t o p o l o g i c a l  space X onto the t o p o l o g i c a l 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 i n t f ( U ) .  2.7 D e f i n i t i o n  A map  f : X  > Y s a t i s f y i n g some property C i s  said to be h e r e d i t a r i l y C i f and only i f for each subspace Y^ of f(X) and X  = f" ^) 1  1  property C.  the induced map  f|  x  :  > Y  1  satisfies  - 64 2.8 Proposition  (1)  -  Every open or closed surjection i s pseudo-  open .  Proof  (2)  Every continuous pseudo-open map i s a quotient  (3)  Every pseudo-open map i s h e r e d i t a r i l y pseudo-open.  Let f : X  map.  > Y be a s u r j e c t i v e mapping of the t o p o l o g i c a l  space X .onto .the top.oiogic.al space "Y. (1)  Let y e Y and l e t U be an open neighbourhood of  I f f i s an open map, y e f ( f ( y ) ) C f(U) = i n t f ( U ) .  f (y). _ 1  _ 1  Thus open surjections are pseudo-open. closed map.  Suppose now that f i s a  Then, since X-U i s closed and f ^ ( y ) c ^ X-U, y i s not  contained i n the closed set f(X-U).  Therefore Y-f(X-U) i s 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 i n t 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 i n t f ( f "'"(V)) = i n t V and hence V i s open. (3)  The function f induces a map h : X^  i s a subspace of Y, X^ = f "''(Y^), and h = f |^ . pseudo-open. h ( y ) i n X^. _ 1  f"" ^) = h 1  y e (int  y  - 1  Let y e Y^ and l e t = U PI X  Then  (y)C ^ C U .  f(U))n Y  1  Suppose that f i s  be an open neighbourhood of  where U i s open i n X and  Accordingly, y e i n t f ( U ) .  = (int  Hence h Is pseudo-open.  1  y  y  > Y^ where Y^  f(U))n  Y  1  = int^(f(U)n  But . = i n t hCU^).  ,2.9 Example (2)  (1) There i s a pseudo-open map which i s not open. There i s a quotient map which i s not h e r e d i t a r i l y  quotient. Proof  (1) The quotient map from the r e a l l i n e onto the r e a l  l i n e with the integers i d e n t i f i e d i s a pseudo-open map which i s .not open. (2)  Let X and Y be as i n (1.15), and l e t P be the  quotient map of Y onto X. Y  ±  If X^ = X-{^£j- : n e to}, then  = P ~ ( X ) = {(0, 1)} U {(x, 0) : 0 t x e R M ^ j i p 0) : n e to} 1  1  and P induces the map P^ = P|  Y  • But P^ i s not a quotient map  sines P^({0}) = {(0, 1)} i s open i n Y^ and yet {0} i s not open in The next r e s u l t 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 h e r e d i t a r i l y quotient. Proof f. = f L  Let f : X  > Y be a continuous surjection, and l e t  where Y. i s a subspace of Y and X  i s open i n Y^,  = f^CY-).  I f U.  = U O Y^ f o r some open subset U of Y.  Then,  since f i s continuous, f ^ C O ^ ) " f ^ C U O Y ) = f 1  f "^(U)n X- i s open i n X-.  - 1  (Uf l Y ^=  Thus continuous maps are h e r e d i t a r i l y  - 66 continuous.  It follows from (2.8) that i f f i s continuous  pseudo-open then f ^ i s a continuous pseudo-open map and therefore a quotient map.  Conversely, assume that f i s hereditarily quotient.  Suppose y e Y and V i s an open neighbourhood of f "*"(y). Let Y  = (Y-f(V))U {y>, X = f ( Y ) •= [X-f~ (f(V))].U f ( y ) , and _ 1  2  2  1  _ 1  2  "f "=;f-| . Then f~*(y) '=-f ~*(y)•;*«• V'n;X , • which :±s open- in X » 2  x  2  Since f i s a quotient map, {y} i s open in Y 2  2  2  and thus {y} = G H Y  2  for some open subset G of Y. This implies that G i s contained i n f(V) and hence that y  2.11 Proposition  Proof  e int f(V).  Every continuous pseudo-open image of a Frechet  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 ( y ) n c l f ( B ) =0, _ 1  _1  U = X-cl f ^(B) i s an open neighbourhood of f "'"(y).  Then, since f  is pseudo-open, y E int f(U)ci f(U) = f(X-cl f ' ^ ) ) C f (X-f (B)) d Y-B -1  1  contradicting y e c l B. Hence there i s some x E f ^(y)O and,  c i  f ^(B)  since X i s Frechet, there exists a sequence {x^ : n E OJ} i n  f ^"(B) converging to x. Thus {t(* ) : n E to} i s contained i n B and, n  since f i s continuous, {f(x ) : n E w} converges to f(x) = y. Theren  fore Y i s a Frechet space.  -672.12  Cl)  Corollary  Frechet  The continuous open or closed image of a  space i s Frechet. C2)  coordinate  If a product space i s Frechet, so i s each of i t s  spaces.  CD follows from C2.8.1) and  Proof  C2.H), and  Cl)  implies  C2).  The following i s a s l i g h t generalization of the necessity condition of Franklin's Proposition 2.3  C[8]).  The  Hausdorff hypothesis i s replaced by "unique sequential l i m i t s " .  2.13  Proposition  If f : X  space X onto a Frechet  > Y i s a quotient map  of a t o p o l o g i c a l  space Y having unique sequential  limits,  then f i s pseudo-open. Let y e Y and  Proof f Cy). - 1  suppose that U i s an open neighbourhood of  Assume that y I i n t fCU).  Then y e Y-int fCU)  = c l CY-fCU)),  and consequently there exists a sequence S i n Y-fCU) converging to y. Because Y has unique sequential l i m i t s , c l Crange S) = {y} U range S. If F = f C S ) then, since f i s continuous, c l F = c l C f C S ) ) C f ~ C c l S) = -1  f'Hs  - 1  U (y>) = f~*(S) U f ( y ) = F U  U n F = 0. F i s closed.  _ 1  f (y). _ 1  1  But f C y ) C U and - 1  This implies that f C y ) n c l F = 0 and therefore that - 1  Hence f CY-S) = f C Y ) - f C S ) = X-F i s open. -1  since f i s a quotient map,  - 1  Y-S  _ 1  Then,  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 d i s j o i n t topological of i t s convergent Proof  sum  sequences.  Each Frechet Hausdorff space i s sequential Hausdorff and  hence, .by (1.31), a quotient of the d i s j o i n t topological sum of i t s convergent ..sequences. must be pseudo-open.  Then, .by ,(2.13), the quotient map  Conversely, for any Hausdorff space X each  convergent sequence i n 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 C o r o l l a r y  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  (3)  X i s the continuous pseudo-open image of a f i r s t -  space.  countable space. Proof  By v i r t u e of (1.32) and (2.14), X i s the continuous pseudo-  open image of a zero-dimensional, l o c a l l y compact, complete metric space.  Since metric spaces are f i r s t - c o u n t a b l e and f i r s t - c o u n t a b l e  spaces are Frechet, (2) implies (3) and (3) implies (1) by (2.11).  -  69  -  As previously stated, f i r s t - c o u n t a b l e T^-spaces are p r e c i s e l y 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 f i r s t - c o u n t a b l e (Hausdorff) space i s the continuous open image of a d i s j o i n t 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 i n a Hausdorff space i s metrizable.  However, many spaces are f i r s t - c o u n t a b l e  Hausdorff but not Baire spaces.  The r a t i o n a l s Q_ i s an example of  such a t o p o l o g i c a l space. An unanswered question of Alexandroff asks whether or «>not there "is -a >«*fir-st-countable'ecompact- -Hausdorff - space vwith c a r d i n a l i t y > c.  The corresponding question f o r 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 d i s c r e t e  space i s a Frechet space.  Proof  Let X* = X U {<*>} be the one point compactif i c a t i o n of the  d i s c r e t e space X.  For any subset A of X*,  i f A contains i n f i n i t e l y many points. {x^ : n E io} i n A s a t i s f y i n g to . 00  = x  m  00  e ( c l A)-A i f and only  Moreover, any sequence i f and only i f n = m converges  Therefore, i f °° e e l A, A contains a sequence converging to . 00  - 70 If x E ( c l A)n  X then, since X i s d i s c r e t e , x e A.  Hence X*  i s Frechet The topological spaces X = (H  x H  )/  a n E  d M^., of (1.36)  and (1.37) respectively, are sequential spaces with unique sequential l i m i t s which are not Hausdorff.  Although  i s not  Frechet, the next r e s u l t shows that X i s .  2.17 Example  There are Frechet spaces with unique sequential  l i m i t s which are not Hausdorff. Proof  Let H and X = (H x H ) ^ be as i n (1.36).  <f> : H x H — >  R e c a l l that  X i s a quotient map of the f i r s t - c o u n t a b l e space  ^x^H^onto^t'he^non-HauBdor-f'f^sequent'ial^'sp'ace"X-'which *has unique sequential l i m i t s .  To e s t a b l i s h that X i s Frechet, i t s u f f i c e s  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 f o r any neighbourhood V of x such that V fl A = 0, x e U H V = *(U n V) C <J)(U). hood of A^ d i s j o i n t from A ^  If x = q and G i s an open neighbourq e < J > (U H G) = i n t  S i m i l a r l y , i f x = i then i E <f>(U). open and X i s Frechet.  <j>(un G) C  i n t <J>(U) .  Thus <j> i s continuous pseudo-  - 71 Let Y be a Frechet  space with, unique sequential l i m i t s .  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 l i m i t s 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)  Let each ( i , j ) e W x W be an isolated point.  j 0.  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 t o p o l o g i c a l  space Y i s compact because i f U and V are open neighbourhoods of p and q r e s p e c t i v e l y , then Y-(U U V) i s f i n i t e .  It i s also c l e a r that  Y i s not Hausdorff since p and q cannot be separated open sets.  by d i s j o i n t  Then, since Y-{p, q} i s d i s c r e t e , i f some sequence converges  to two d i s t i n c t points, they must be p and q.  However,a sequence  { ( i ^ , j ) : n E co} i n W x W can converge to p only when { i ^ : n e w} i s unbounded, and to q only when { i ^ : n e co} i s bounded. Y has unique sequential l i m i t s . Frechet.  Let A be a subset of Y.  Therefore  I t remains to prove that Y i s 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 i n 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 i n f i n i t e , t h e r e i s a sequence i n 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. even sequential.  In fact, these products are not  If these product spaces were sequential then,  by (1.38.2), X and Y would be Hausdorff. Q x  Similarly, the spaces  and £' x OJ of (2.4.2) are products of Frechet spaces  which are not sequential. not always be the case.  Example 2.19 shows that this need This example also demonstrates that the  term "sequential" cannot be replaced by "Frechet" i n (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 r e a l l i n e with the integers i d e n t i f i e d , and  l e t I = [0, 1] be the closed unit i n t e r v a l . Hausdorff.  Both X and I are  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. by A = J  To see t h a t X x I i s n o t F r e c h e t , d e f i n e A C X x I  {A : n e W} where A = {(n n n  U  k  : k E W}.  -) n  Then  (0, 0) E c l A s i n c e (0, 0) e c l {fr-1, -) : n E W} and n {(n-1,  : n e W} CL A.  because no sequence c o n t a i n e d i n A converges f o r any  2.20  i n R x I t o (k,  0)  The p r o d u c t o f two c o n t i n u o u s pseudo-open maps  may be a q u o t i e n t map w i t h o u t b e i n g c o n t i n u o u s  By  0)  k e Z.  Example  pseudo-open.  L e t X and I be the H a u s d o r f f F r e c h e t spaces o f  Proof  t o (0,  But no sequence i n A converges  ( 2 . 1 3 ) , t h e q u o t i e n t maps <|> : X* x  a r e c o n t i n u o u s pseudo-open. i m p l i e s t h a t tb  x <j)  X  : I*  Since X x I i s s e q u e n t i a l ,  • X* x I *  > X  > I (1.35.2)  x I i s a q u o t i e n t map.  i  X However, <f>  > X and cf>  (2.19).  x A  X.  cannot be c o n t i n u o u s pseudo-open because X* x I *  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 f u r t h e r i l l u s t r a t e t h e n o n - p r o d u c t i v e n a t u r e o f F r e c h e t spaces, g e n e r a l i z e H a r l e y ' s Theorem  2.21  ([12]).  Lemma  The author p r o v i d e s t h e p r o o f .  The p r o d u c t o f two F r e c h e t spaces, one o f which i s  discrete, 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 d i s c r e t e ,  U x {v} i s a neighbourhood of (u, v) f o r 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 i s a sequence '{u^ : n e t o } i n {x : *(x,v) e *B} converging to u.  The sequence { ( u  n>  v) : n e t o } i s contained i n B and  converges to (u, v ) .  2.22  Theorem  Let X be a Frechet space.  Let A be a subset of : n e to}  X s a t i s f y i n g the property : there i s a sequence {U^ ''"o'f open subsets of X such that"""(T)""U^C."* in U = U  {U  n  : n e t o } , and  (3) A U U  n  u n+1  >' (2) A i s •contained  i s not open.  Let  r  X /A  /A  denote the quotient space obtained from X by i d e n t i f y i n g the points i n A.  Then i f X i s ^  and Y i s Hausdorff, X^  Frechet i f and .only i f Y i s d i s c r e t e . hypothesis Proof  may  A  x y is  If A i s closed, the T^  be replaced by r e g u l a r i t y .  If .Y i s d i s c r e t e then, since  the t o p o l o g i c a l product space X ^  i s Frechet (by 2.3.4),  x Y i s Frechet (by 2.21).  To e s t a b l i s h the converse, l e t i = g(A) where g : X i s the quotient map,  >  and suppose that y i s not an i s o l a t e d point  - 75 of Y.  Then y e cl(Y-{y}) and consequently, since Y i s Frechet  (by 2.3.2), there i s a sequence {y^ : n E co} i n Y-{y} to y.  converging  Since Y i s Hausdorff, i t can be assumed without loss of are d i s t i n c t .  : n E co},  Let W = -U {U-(U U A) x {y }  generality that the y  n  Then ( i , y) E c l W.  Indeed, t h i s Is so because A U  implies that ( i , y^)  e c l (U-(U U A) x {y }).  n  n  n  being not open  Since X^  n  A  x Y is  Frechet, there exists a sequence {(r^, s^) : k E to} i n W converging to ( i , y) .  The Hausdorff hypothesis on Y implies that {s^ : k E to} : n E co}.  i s a subsequence of {y  n  n  J  Let y n,  k  = s, f o r each k E co. k  Thus {(r, , y ) : k E to} i s a sequence i n W converging to ( i , y) k n k  with each r, e U-(U U k k  A).  n  Since X i s T, (or X i s regular and A ±  i s closed), f o r each k E to there i s an open neighbourhood such that r, t G, . k k  Let U  k = U n n  D  of A  ( H {G. : j < k}) for each n, j —  k e co; and l e t m^ be the largest member of to s a t i s f y i n g r ^ i U^ ^_ -j« m  It follows that  i  : k E to} i s an open neighbourhood of A  U (U  ™k d i s j o i n t from the sequence {r^ : k E to}.  But then {r^ : k E to} ) : k E to}  cannot converge to i which contradicts {(r, , y K.  to ( i , y ) .  II.  converging  k 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.  A f t e r another such example, a  characterization of those sequential spaces which are also Frechet follows.  The non-Frechet sequential space M of (1.17) i s used  to give a c h a r a c t e r i z a t i o n i n the Hausdorff  2.33  Example  case.  There i s 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 d i s j o i n t  family of i n f i n i t e subsets of the natural numbers hi (Two sets U and V are almost d i s j o i n t i f and only i f U H  V is finite).  To  e s t a b l i s h the existence of F, l e t G be the c o l l e c t i o n of a l l i n f i n i t e pairwise almost d i s j o i n t families of i n f i n i t e of W. G ^ 0.  The set G i s p a r t i a l l y ordered by set i n c l u s i o n .  Note that  Indeed, for each r e a l number r there i s a sequence  {x^ : n E co} of r a t i o n a l numbers converging to r . f : Q  subsets  >  Then i f  W i s a b i j e c t i o n between the r a t i o n a l s and the natural  numbers, {{f(x ) : n e co} : r e R} e G.  Now,  l e t {E  : a £ A}  3.  Vi  be a chain i n G and l e t E =  U {E  : a E A}.  For any pair E,  cl  F £ E there i s some c e A such that E, F e E  which implies that c  EOF  is finite.  Therefore E E G  bound of the chain {E  : a E A}.  and consequently  E i s an upper  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  i s o l a t e d 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.  C l e a r l y ty i s Hausdorff.  Furthermore, ty i s l o c a l l y  compact because F U {F} i s a compact neighbourhood of F i n ...follows-.from ..(.[,16.].,. JTheorem 5.21) -that  •= 4-..U .{•«*>•}., ...t-he-one -point  compactification of ty, i s a compact Hausdorff space. of the topology on ty* i t i s clear that °° e c l W. t  {x  n  It  By d e f i n i t i o n  However, i f  : n e to} i s a sequence of d i s t i n c t points i n W then, since F  i s maximal,  {x : n E to} f l F i s i n f i n i t e f o r some F e F and hence n  {x^ : n E to} converges to F. ,„to  Therefore no sequence i n W converges  .^and—so .^*.,.is.„not,.Er.echet. l  i  It remains to show that ty* i s a sequential space.  Suppose  that V i s a sequentially open subset ofty*,and l e t x E V. If x e W, x E i n t V because {x} i s open.  If x e F then, since any  sequence of d i s t i n c t points i n x converges to x, x-V i s f i n i t e and hence {x} U {n : n e x O V} i s a neighbourhood of x contained i n V. Assume now that x = . 00  Since F U {F} i s a neighbourhood of F E F  and ty*-\] i s compact f o r each open neighbourhood U of °°, any sequence of d i s t i n c t points of F converges to °°. Accordingly, V contains a l l but  f i n i t e l y many members of F.  m E W.  Let F-V = { F  i  : i <_ m} where  I f {U^ : i _< m} i s any f i n i t e c o l l e c t i o n of open sets  F . d U. then, since F i s maximal, U-[V.U ( (J {U  satisfying  : i £m})] i s finite  - 78 and consequently IJJ*-V i s compact.  Hence TJJ* being Hausdorff  implies that ifj*-V i s a closed compact subset of V i s open whenever » e V. of  ifi.  Therefore  Thus every sequentially open subset  i s a neighbourhood of each.of i t s points, and so  is  sequential. . Since no sequence i n hi converges to •», the singleton {<»} i s sequentially open but not open i n the subspace i p * - F . Hence ^*-F  i s a non-sequential  subspace of ty*. The next r e s u l t  proves that such a subspace must always exist i n sequential spaces which are not  2.24  Frechet.  Proposition  A sequential space i s Frechet i f and only i f  i t is' h e r e d i t a r i l y sequential.  Proof  I f a sequential space i s Frechet, every subspace i s Frechet  and hence sequential.  Conversely, i f a t o p o l o g i c a l space X i s  h e r e d i t a r i l y sequential, (1.35.1) implies that 6^ quotient map  with Frechet domain.  Then -$ A  and therefore X i s Frechet.  i s a hereditarily  i s continuous pseudo-open  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 h :M  Let Y be a subspace of a sequential space X, and l e t > Y  g  be a homeomorphism of M onto the subspace Y provided  with the sequential closure topology.  I f 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 i s 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  e x i s t s a subset B of X such that c l B ^ B' where B' i s the set of l i m i t s of sequences i n B.  Since X i s sequential, there i s a  sequence S = {x_^ : i e co} i n B' converging to some point x e ( c l BVB'. The sequence S i s not frequently i n B because otherwise i t has a subsequence i n B converging to x t B'.  Consequently,  assumed that the x. are d i s t i n c t and S C B'-B.  i t can be  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 d i s t i n c t .  This i s  so because X i s Hausdorff and S converges to x; that i s , there e x i s t s a family {U. : i e to} of pairwise d i s j o i n t open sets s a t i s f y i n g  - 80 e U\.  to x.  Since x.-^.B', no sequence consisting of the x^  Thus the topological space Y  converges  where the subspace s  Y = {x} U (x. : i e W} U {x. . : i , j e hi} i s homeomorphic to M. 1 xj It i s easy to v e r i f y that the non-Frechet Hausdorff sequential space ty* of (2.23) s a t i s f i e s (2.25).  If {E^ : i c co}  i s any c o l l e c t i o n of pairwise d i s j o i n t subsets of hi then, since "F i s maximal, for each i e co there e x i s t s F. e F such that F.H ' x x is infinite.  Thus {F^H  i n F converging to °°. such that the i j (j n  E  w  E. : i e w} i s a pairwise d i s j o i n t  Let in±^  3 E co} be any sequence i n  ) are d i s t i n c t .  C l e a r l y , each i^^j  E. x sequence E^  : j e co}  converges to F^ D E^ and the n ^ ( i , j e co) are a l l d i s t i n c t .  Then,  since no sequence i n hi converges to «>,• the subspace {<»} U {F^O  E_^ : i e hi} U {n^j : i , j  closure topology i s homeomorphic to M.  e hi} provided with the sequential  Chapter 3 Generalized Sequential Space Methods  When a topology i s s p e c i f i e d 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 i n the study of topological spaces whose topologies are determined by t h e i r convergence classes.  For example, one can study  convergence subbases and convergence bases consisting of convergent sequences i n l i e u of studying sequential respectively.  spaces and Fre'chet spaces  The notion of convergence subbasis i s also u s e f u l i n  the i n v e s t i g a t i o n of generalized  sequential  spaces.  In t h i s chapter,  --•the'"topological "''spates ^ sets are examined; these spaces are c a l l e d m-sequential (m-Frechet) spaces.  I t w i l l be shown that any topological space can, f o r  s u f f i c i e n t l y large m, be so described.  3.1 D e f i n i t i o n  Let X be a set and l e t C be a class of p a i r s (S, x)  where S i s a net and x i s a point i n X.  The class C i s a  p-convergence class on X i f and only i f i t s a t i s f i e s : and E i s a c o f i n a l subset of D, then ({x  If ( {  : n e E}, x) e C.  x  : n  n  e  D},x)eC  A  n p-convergence class i n which a l l of the nets are sequences i s c a l l e d a sequential p-convergence c l a s s .  - 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 t o p o l o g i c a l l y by means of the largest topology on X i n which the C-nets ( i . e . , a l l of the nets i n 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 l i m i t s of C-nets.  Then t - c l i s a  closure operator and hence defines a topology T(t) for X ( [ 1 6 ] , Theorem 1.8).  This i s the largest topology on X i n which the C-nets  converge.  Proof  It i s 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 d e f i n i t i o n , t - c l <> f i s empty. each subset A of X. of  By d e f i n i t i o n of t - c l , A C  Then t - c l A C  t - c l ( t - c l A).  t - c l A for  Again by d e f i n i t i o n  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 l i m i t s of C-nets.  But t - c l A i s  closed with respect to the formation of l i m i t s 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).  prove that t - c l ( A U B) = ( t - c l A) (J ( t - c l B). A U B C ( t - c l A) U ( t - c l B) C t - c l (A U B) . i n c l u s i o n , l e t S = {x  It remains to  Clearly  To e s t a b l i s h the opposite  : n e D} be a net i n ( 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-cl  - 83 Since D.U A  D_ = D, either D. or D„ i s c o f i n a l i n D. B A B  assumed without loss of generality that  It can be  i s c o f i n a l i n D.  Then  S.. = {x : n e D.} i s a subnet of S i n 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 l i m i t s of C-nets. Let (S, x) e C.  If S does not converge to x i n (X, T ( t ) ) ,  there i s an open neighbourhood U of x such that S i s not eventually i n U.  Then S i s frequently i n X-U and there i s a subnet S^ i n X-U  with (S^, x) e C. x e X-U.  But since U i s open, X-U = t - c l (X-U) and hence  The contradiction shows that S must converge to x with  respect to T ( t ) , and hence that each C-net  converges i n (X, T ( t ) ) .  Suppose now that T^ i s a topology on X i n which the C-nets converge. I f V e T^ then f o r each net p a i r (S, x) e C such that S C x e X-V.  X-V,  Thus X-V = t-cl(X-V) which implies that V e T ( t ) .  /  3.3 C o r o l l a r y  Let X be the topological space provided with the  topology T ( t ) derived from a p-convergence C. only i f C s a t i s f i e s : If S = {x  Then X i s T^ i f and  : n e D} i s a net i n X such that n  x  = x f o r 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 l i m i t s , X i s T^.  Proof  If X i s  and y ^ x, S cannot converge to y and so (3.2)  implies that ( S , y) £C.  Conversely, i f S  then, since ( S , Z) £ C, z £ t-cl{x}. Q  q  i s a net i n {x} and z ^ x  Hence {x} i s closed and there-  fore X i s 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 i n 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  n  n  = x  for each n e to, then ( S , x) e C. (2)  I f S i s 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) Proof  I f ( S , x) e C and ( S , y) e C then x = y.  I t i s clear that C < l C ( T ( t ) ) .  suppose that ( S , x) £ C.  To prove the opposite i n c l u s i o n ,  By (2), there i s a subsequence S = {y^ : n e to} q  of S, no subsequence of which together with x i s a member of C. I t can be assumed without loss of generality that y  4 x f o r each n e to.  - 85 Either there exists ( S ^ , S  q  z)  -  e C such that  or no such sequence pair e x i s t s .  in S^U  In the f i r s t case, any C-net  {z} i s either a subsequence of  some point i n S^.  Then conditions  i s a subsequence of  (2) and  i s a closed subset of X d i s j o i n t from x.  or frequently equal to (3) imply that  U  {z}  In the second case, S o  i s a closed subset of X d i s j o i n t from x.  In both cases, S has a  Hence ( S , x) £ C ( T ( t ) ) .  subsequence not converging to x. Another closure operator  i s defined i n the following  proposition.  3.5 Proposition  Let C be a p-convergence class on a set X,  and  for each subset A of X l e t c - c l A be the union of A and the l i m i t s of those C-nets contained  i n A.  c - c l i s a closure operator, and  Then i f C i s a convergence c l a s s , ( 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 i s given i n ([16], Theorem 2.9).  In the presence of a  convergence c l a s s , c - c l i s p r e c i s e l y the closure operator  described  i n that theorem.  3.6  Corollary  and T  a  If C^ and C^ are convergence classes on a set X  and T„ are the associated topologies, then C C 3 a  only i f T ^> a  T .  C. i f and  3  - 86 Proof  If (S, x) e C^, S i s eventually i n each neighbourhood of x  i n (X. T ). a  Thus T.CT T implies that S i s eventually i n each 0 a  neighbourhood of x i n (X, T.) and hence that (S, x) e C . p  suppose that C C then (S, x) E C  c  C„ and l e t U e T„. and U e T  Q  Conversely,  p  If (S, x) e C  implies that x e X-U.  and S CI X-U, Therefore X-U  P  ""p  i s closed i n (X, T ) and so U e T . a a  3.7 Proposition  Let C be a p-convergence class on a set X, and  l e t 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 i n which the C-nets converge.  Then, i f T i s any topology on X whose convergence  c l a s s C(T) contains C, T O T ( t ) .  The preceding r e s u l t 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 - c l whenever c - c l ( c - c l A) = c - c l A for each subset A of X.  C l e a r l y , c - c l A i s a subset of t - c l A, and  i t can be a proper subset.  Observe that a t o p o l o g i c a l space i s  Frechet i f and only i f f o r each subset A of X, c l A = c - c l A with  respect to the convergent sequences i n X; s i m i l a r l y , 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. consider the t o p o l o g i c a l space M of (1.17). no sequence i n hi x W converges to 0.  In p a r t i c u l a r ,  As previously observed,  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 i t e r a t i o n of c - c l . Define A° = A and for each  ct successor o r d i n a l a l e t A 8.  (3+1  6 = c-cl A  denotes the o r d i n a l successor of 3).  o r d i n a l define A  a  = U {A^  : 3 < a}.  for each o r d i n a l a, and i f A 3 _> a.  where a = B + 1 for some o r d i n a l  a  If a i s a l i m i t  It i s clear that A C a  t-cl A  = t - c l A then A^ = t - c l A whenever  For any subset A of X the c a r d i n a l i t y of the number of  i t e r a t i o n s of c - c l to obtain t - c l A i s <_ 2 of the set X.  m  where m i s the c a r d i n a l i t y  Then, since the ordinals are well-ordered, for each  x e t - c l A there Is a smallest o r d i n a l n such that x belongs to the n-th i t e r a t e of c - c l on A; that i s , x e A  whenever a _> n and x i A  whenever a < n. 3.8 D e f i n i t i o n  Let C be a p-convergence c l a s s on a set X and l e t  A be a subset of X. (1)  A point x e t - c l A i s 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 o r d i n a l such that x i s a member of the n-th i t e r a t e of c - c l 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 - c l A f o r each subset A of X, which occurs i f and only i f every subset of X has Baire order <_ 1.  3.10 D e f i n i t i o n space  Let C be a p-convergence class on a t o p o l o g i c a l  X. Cl)  C i s a convergence subbasis f o r X, or f o r the  topology on X, i f and only i f the topology on X I s 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 t o p o l o g i c a l  space X. Cl)  C i s a convergence subbasis for X i f and only i f  X has topology TCt). (2) topology TCc).  C i s a convergence basis f o r X i f and only i f X has  - 89 Proof  By v i r t u e of (3.7) and (3.10), (1) i s c l e a r .  Then (1)  together with (3.9) implies (2).  3.12 D e f i n i t i o n  Let m be a n . i n f i n i t e cardinal number.  i s a net whose directed  3.13 D e f i n i t i o n  An m-net  set i s of c a r d i n a l i t y <_ m.  Let X be a topological space, and l e t 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 i n 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. - i t - ^ a s ^ a ^ 1  3.14 Proposition (2)  (1) Every m-Frechet space i s m-sequential.  I f a topological  space i s m-sequential then i t  is m^-sequential whenever m^>_ m. is m^-Frechet i f  Proof  S i m i l a r l y , an m-Frechet space  > m.  The proof of (1) i s obvious because, by d e f i n i t i o n , every  •convergence basis i s a convergence is an m.-net f o r m  n  subbasis.  >_ m, (2) i s also  clear.  Since every m-net  - 90 The next two r e s u l t s give several equivalent formulations of the d e f i n i t i o n s of m-sequential space and m-Frechet space respectively.  3.15  Proposition  The following statements about an a r b i t r a r y  - t o p o l o g i c a l space X-are-equivalent. (1)  X i s m-sequential.  (2)  X has topology T(t) with respect to some p-convergence  class consisting of m-nets i n X. (3)  A subset F of X i s closed i f and only i f no m-net  i n F converges to a point not i n F. (4)  A subset U of X i s open i f and only i f each m-net  "in X converging to a point lri""U i s eventually i n U. (5)  The class C of a l l p a i r s (S, x) where S i s an m-net  in X converging to the point x i s a convergence  Proof  subbasis f o r X.  I f X i s m-sequential then, by d e f i n i t i o n , X has a convergence  subbasis  i n which a l l of the nets are m-nets. According to  X has topology T(t) with respect to  and therefore (1)  (3.11),  implies  (2).  I f F i s a subset of X with no m-net i n F converging to a point not i n F, no C^-net i n F converges to a point not i n 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 i n U i s eventually i n U.  Let S be an m-net i n X-U converging to a point x .  Then x e X-U  - 91 Thus (3) implies that  because otherwise S i s eventually i n U. i s closed and hence that U i s open.  To e s t a b l i s h that (4)  X-U  implies  ( 5 ) , l e t T be any topology for X i n which the C-nets converge. Then i f V e T, every m-net i n X converging to a point i n V i s eventually i n V, and so V i s open by ( 4 ) .  3.16  that (4) implies (.5).  Proposition  following are  the  .C-nets-converge. .It .follows ..from  largest topology i n -which-the (3.6)  Accordingly X has  By d e f i n i t i o n , (5) obviously implies  For any a r b i t r a r y topological space X,  (1).  the  equivalent.  (1)  X i s m-Frechet.  (2)  The c l a s s C of a l l p a i r s (S, x) where S i s an m-net  "in X converging to x i s a cohvergerice"'basis ¥ or X. J  (3)  The closure of any subset A of X i s the set of l i m i t s  of m-nets i n A. (4)  X has topology T(c) with respect to some p-convergence  c l a s s consisting of m-nets i n X. If X i s m-Frechet, X i s m-sequential and hence the c l a s s C  Proof  i s a convergence subbasis for X. Baire order <^ 1. (2).  Therefore  It follows from (2)  with C.  Moreover, every subset of X has  C i s a convergence basis and  (1)  implies  that X has the topology T(c) associated  Consequently (2) implies ( 4 ) .  In addition, (2) i s equivalent  to (3) because f o r each subset A of X, x e c l A i f and only i f there e x i s t s an m-net i n A converging to x. i s c l e a r form (3.11).  The proof that (4) implies  (1)  The following c o r o l l a r y together with (3.15) and  (3.16)  shows: (1)  A t o p o l o g i c a l space i s sequential i f and only i f  i t has a convergence subbasis i n which a l l of the nets are sequences. (2)  A t o p o l o g i c a l 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, t h i s r e s u l t implies thkt.'eyiery Frechet  space i s  m-Frechet and that every sequential space i s m-sequential.  3.17  Corollary  only i f i t i s  U  A t o p o l o g i c a l space i s Frechet i f and only i f i t  -Frechet.  Proof  Since every sequence i s an  and the Frechet respectively. •non-trivial an  A t o p o l o g i c a l space i s sequential i f and  /^-sequential.  (2) is  (1)  h{o~net, the sequential spaces  spaces obviously s a t i s f y (3.15.3) and  (3.16.3)  To prove the converses, i t i s f i r s t shown that every /^-rnet has.a c o f i n a l sequence.  }-j -net and l e t g :. to  '> D be a b i s e c t i o n .  o  directed set, f o r each for every i < k . -  k  E  Then {x  Let {x  n  : n e D}  Because D i s a  to there e x i s t s n^ E to such that ,  g(n ) k  N  be  : k E to} i s a subnet of {x  gC  n  1 1  ^)  ^_  gCO  : n E D}.  Suppose that F i s a sequentially closed subset of a t o p o l o g i c a l space  X.  I f S i s an  hj-net  i n F converging to some point x, x e F since  otherwise S has a c o f i n a l sequence which i s eventually i n X-F.  Then  if X is  ^ - s e q u e n t i a l , F i s closed by (3.15).  proved. of an  Thus (1) i s  To complete the proof of (2), suppose that A i s a subset |y -Frechet space X and l e t x e c l A.  there i s an trivial  /^-net i n A converging to x.  By v i r t u e of (3.16), Then, since every non-  • rj -net has a c o f i n a l sequence, there exists a sequence i n  ,A...converging,,to x.. In view of (3.15) and (3.16), i t i s 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 i n a t o p o l o g i c a l space X.  According to (3.15), C i s a convergence subbasis for X i f  and only i f X i s m-sequential.  However, i t i s possible to have a  convergence subbasis which i s a proper subset of C. a smaller convergence subbasis may  Although using  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 o r d i n a l of c a r d i n a l i t y m .  (m denotes the cardinal successor of m)  - 94 Proof  Let to denote the least o r d i n a l of c a r d i n a l i t y m~*\  Then  a i s of the form 8 + 1 f o r some o r d i n a l 3, where to i s the l e a s t  p o r d i n a l of c a r d i n a l i t y 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 c a r d i n a l i t y of B i s _< m.  Assume that A i s a subset of X and l e t x e c l A.  Since X i s m-sequential, there exists an m-net {x that x  n  e c l A and ord x  n  < ord x.  n  : n E D}  such  Then ord x = sup {ord x : n e D} ^ 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 f o r a topology on Y which i s larger than the r e l a t i v e topology.  This induced topology coincides with  the r e l a t i v e topology on Y i f Y i s closed or open i n X.  The  two  topologies coincide f o r 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 c l e a r l y a p-convergence c l a s s on Y.  For each subset A of  Y define u - c l A to be the smallest set containing A and closed with respect to the formation of l i m i t s ' o f P-nets.  By (3.2), u - c l i s a  closure operator on Y and hence defines a topology T(u) f o r 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 r e l a t i v e topology on Y  because the P-nets converge i n the r e l a t i v e topology and T(u) i s the largest topology on Y i n which the P-nets converge. To e s t a b l i s h that these topologies on Y coincide when Y i s closed or open, or when C i s a convergence basis, i t s u f f i c e s to show that the two topologies have the same closed sets. F be a subset of Y.  Let  By d e f i n i t i o n of the u-closure operator,  u - c l F CZ ( t - c l F) p| Y.  To prove.the opposite i n c l u s i o n , assume  that x e ( t - c l F) H Y with ord x = X with respect to C and F, and proceed by t r a n s f i n i t e induction on X• ( t - c l F) H Y = F° H Y m F fVY = F C u - c l F.  If X = 0, If X = 1 then  "(t-cT F) Pi Y = ~ r n ' Y = ( c - c T T ) r T T C ti-cl F. 1  Thus "the proof i s  complete for the case i n which C i s a convergence basis.  For X: > 1  consider separately the cases Y i s closed and Y i s open.  First,  suppose that Y i s closed.  Then t - c l F C. t - c l Y = Y.  induction hypothesis, there exists a net p a i r ( { with x each x  n  n  E ( t - c l F) H Y and ord X e u - c l F and ({x  x e u - c l F.  r  J  n  n  < X for each n E D.  e  ^W,  e C  Consequently  : n e D}, x) e V, which implies that  n  Assume now that Y i s open.  there i s a net pair ({y n  By the induction hypothesis,  : n e D}, x) e C with y E t - c l F and 'n  ord y < X for each n E D. •'n {y  r  :  x  By the  Since Y i s open and x e Y, the net  : n E D) i s eventually i n Y.  Thus E = {n £ D : x  e Y} i s a  - 96 c o f i n a l subset of D and so ( { y '• n e E}, x) e V. n  Therefore  y^ e u - c l F f o r each n e E implies that x e u - c l F. It remains to show that i f the two topologies are the same then C i s a convergence b a s i s .  I f C i s not a convergence  basis f o r X, there i s at least one Baire order 2 s i t u a t i o n . That i s , there e x i s t s a subset ord x = 2.  Let Y = B U {x}.  B of X with x e t - c l B and Then x e ( t - c l B) H Y but, since  ord x = 2, ( c - c l B) n Y = B and no net i n B converges to x e Y. Thus x £ u - c l B and the topologies are d i f f e r e n t .  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 h e r e d i t a r i l y m-Frechet i f and only i f i t i s h e r e d i t a r i l y 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 for the r e l a t i v e topology on Y.  subbasis  Hence Y i s m-sequential.  It i s obvious that every h e r e d i t a r i l y 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), f o r every subspace Y of X the trace of C on Y i s a convergence topology on Y.  subbasis f o r the r e l a t i v e  Thus Y i s m-sequential and consequently every  m-Frechet space i s h e r e d i t a r i l y m-sequential.  In addition, any  subspace of an m-Frechet space i s h e r e d i t a r i l y m-sequential. I t remains to show that every h e r e d i t a r i l y m-sequential space i s m-Frechet.  Assume that X i s a h e r e d i t a r i l y m-sequential space  with a convergence subbasis C i n 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. on Y surely coincides with V.  The trace of C  The preceding theorem implies that  C i s a convergence basis for X and hence that X i s m-Frechet.  3.21 Proposition family {X  a  If X i s the d i s j o i n t topological sum of any  : a E A} of topological spaces where each X  convergence subbasis C , then C =U{C Si  subbasis for X.  If each C  SL  a  has a  : a e A} i s a convergence  i s a convergence basis, so i s C.  Si  Let (S, x) E C and suppose that U i s an open neighbourhood  Proof  of x i n X.  Then (S, x) e C  for some a E A and therefore, since  cL  Ufl X  a  i s open i n X , x a  E U H X for a l l n s u f f i c i e n t l y large, n a .  Thus the convergence c l a s s on X contains C.  Now l e t T denote the  usual topology on X and l e t T^ be any topology for X whose convergence c l a s s CCT^) in X  f o r some  c E A.  contains C.  I f V £ T, V D X  £  i s not open  From t h i s , i t follows that there e x i s t s a  - 98 C-net  i n X -V converging to a point y e V Pi X £  (S , y) e C and C C C ( I ), V £ T . X ex ct n  c >  Then, since  Consequently, T i s the largest  topology on X i n which the C-nets converge, or equivalently by ( 3 . 6 ) , T i s the topology with the smallest convergence containing C.  Hence C i s a convergence  class  subbasis for X.  - •Assume -now -that - each -C^.is- a -convergence-.basis.. complete  the proof, i t s u f f i c e s  to show that every subset of X  has Baire order <^ 1 with respect to the convergence Let F be a subset of X.  Then ( c - c l F) f) X a  ( c l F) D X . a  To  Therefore ( c - c l F) fi X  a  subbasis C.  = c - c l (F) = c l (F) = a a A  A  i s closed i n X f o r each a  a e A, and hence c - c l F i s closed i n X.  3.22 Corollary  The d i s j o i n t topological sum of any family of  m-sequential spaces i s m-sequential.  The d i s j o i n t t o p o l o g i c a l  sum of any family of m-Frechet spaces i s m-Frechet.  3.23 D e f i n i t i o n  Let C be a p-convergence class on a set X. For  any function with domain X l e t fC denote the set of a l l net p a i r s ( { f ( x ) : n e D}, f ( x ) ) f o r ( { X : n e D>, X) e C. n  r  - 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 i n the convergence class of Y. (2)  If f i s s u r j e c t i v e , fC i s a convergence subbasis  for Y i f and only i f Y i s a quotient space.  Proof  (1)  Let ({f(x ) : n e D}, f ( x ) ) e f C. n  Since C i s a  convergence subbasis for X, each C-net belongs to the convergence c l a s s of X.  Accordingly, {x^ : n e D} converges to x i n X.  Then,  since f i s continuous, ( f ( x ) : n e D} converges to f(x) i n Y. n  Conversely, "let A be a closed subset of Y and suppose that ({x^ : n e D}, x) e C with x^ e f ''"(A) f o r every n e D. each f (  x n  ) £ A.  Clearly  Because A i s closed and by hypothesis { f ( x ) : n e D} n  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) i s closed. (2)  By d e f i n i t i o n , fC i s a convergence subbasis f o r Y  i f and only i f the topology on Y i s 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. subbasis C  The space X has a convergence  i n which a l l of the nets are m-nets.  that each net "pair 'in -f C -is 'an-m-net-'pair.  I t i s obvious  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 t o p o l o g i c a l 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 i s continuous pseudo-open.  3.26 Theorem  Let f : X  > Y be a s u r j e c t i o n of the t o p o l o g i c a l  space X onto the topological space Y, and l e t C be a convergence basis f o r X.  Then f i s 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 l e t V be the trace of C on  f "^(Y^).  By v i r t u e of (3.19), V i s a convergence subbasis f o r the  r e l a t i v e topology on f ^"(Y^).  I f f i s continuous pseudo-open, then  f i s h e r e d i t a r i l y quotient and so fV i s a convergence subbasis f o r the r e l a t i v e 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 l e t U be an open neighbour-  If f i s not pseudo-open, y E c l ( Y - f (U)) and  hood of f "*"(y) i n X.  hence there i s a net pair ({f(x ) : n e D}, f(x)) E f C such that n  f(x) = y and each f ( ) e Y-f(U). x  n  n E D.  Consequently x  n  £ U for every  Then, since x E U, the net { X : n E D) does not converge r  to x i n 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 i n f i n i t e 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 cardinality <_ m. (Note that concepts.)  Ho -first-countable and first-countable are equivalent  - 102 3.29 Proposition  If X i s a topological space and m i s an i n f i n i t e  cardinal, then each of the following implies the next. (1)  X i s m-first-countable.  (2)  X i s m-Frechet.  (3)  X i s m-sequential.  (4)  For any subset A of X, each point i n c l A i s i n  c l B for some subset B of A with the cardinality of B <^ m. (5) Proof  (1)  x e c l F.  X i s 2 -Frechet. m  > (2)  Let F be a subset of X and suppose that  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 <. C  Because x e c l F, there  3.  exists x e U n F for each a e A. Then {x : a e A} i s an m-net i n a a a F converging to x, and hence X i s m-Frechet. (2)  > (3)  This i s clear since every convergence basis  i s a convergence subbasis. (3)  > (4)  If X i s m-sequential, X has a convergence  subbasis C i n 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 transf i n i t e 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}.  I f 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 i n 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 with x^ E c l A and ord x^ < X f o r each n E D. x  n  ({x^ : n E D}, X ) E C  Consequently  each  E c l B f o r some subset B of A with the c a r d i n a l i t y of B < m. n n .n — . J  But then x E c l ( U {B : n e D}), L/{B : n E D} C A, and the n n 2 c a r d i n a l i t y of . N {.B : n E J)} i s < .m = m. n (4) x E c l A.  > (5)  Let A be a subset of X and suppose that  By hypothesis, x E c l B for some subset B of A with the  c a r d i n a l i t y of B <^ m. x, U^H B  —  I f {U\ : i e 1} i s a neighbourhood basis at  0 for every i e I.  are at most 2  m  Since B has c a r d i n a l i t y <_ m, there  d i s t i n c t sets U^H B. Define an equivalence r e l a t i o n  • • -"on '-I "bytLd~eiit±fying a  . - Let D'b e-the  ;  index set I under t h i s equivalence r e l a t i o n .  The c a r d i n a l i t y of D  is £ 2 .  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 f l B. Then {x : n E D} i s a 2 -net m  m  n  n  n  i n B converging to x. For the case m = M  e  , there are examples which show that  a l l of the conditions i n the foregoing proposition are d i s t i n c t .  As  previously observed, the r e a l l i n e with the integers i d e n t i f i e d i s a Frechet space which i s not f i r s t - c o u n t a b l e and the space M of (1.17) i s a sequential space which i s not F r e s h e t . 0, x  The countable  space  of (1.19) c l e a r l y s a t i s f i e s (4) but i t i s not sequential.  - 104 F i n a l l y , the o r d i n a l space Q + 1 with the order topology i s a  H 2 °  -Frechet space which does not s a t i s f y ( 4 ) . The next r e s u l t i s a characterization of m-sequential  spaces which generalizes the characterization of sequential spaces. An i n t e r e s t i n g c o r o l l a r y to t h i s 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 i n (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 v i r t u e of (3.29) and  (3.25), a quotient of an m - f i r s t -  countable space i s m-sequential. space with a convergence For each ({x  n  Conversely, l e t X be an m-sequential  subbasis i n which a l l of the nets are m-nets.  : n E D}, x) e C, l e t S(x , x) = {x : n e D} U {x} be n n  a t o p o l o g i c a l space i n which the x^ are taken to be d i s t i n c t and x  •/ x f o r every n e D, and which has the convergence basis generated  by the C-net p a i r ( { * a neighbourhood i s <_ m.  '•  n  n  e  D}, x ) .  Each x  n  i s i s o l a t e d , and x has  basis indexed by the directed set D whose c a r d i n a l i t y  Thus each S ( > x) i s m-first-countable. x  n  The d i s j o i n t  t o p o l o g i c a l sum W of a l l such S(x , x) i s therefore m-first-countable n  and has, by (3.21), a convergence basis E formed by taking the union  - 105 of the convergence bases for the ( x > x) . s  n  from (3.24). quotient map  3.31  The surjection f : W  The theorem how  follows  > X defined by f(x) = x i s a  since fE = C.  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 f E = 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 i s Frechet.  (2)  X i s a continuous pseudo-open image of a f i r s t -  countable space. (3)  X i s a continuous pseudo-open image of a metric  space. Proof  According to (3.31) and  C l e a r l y (3) implies (1).  (3.17.2),  (1) i s equivalent to (2).  To e s t a b l i s h the opposite implication,  l e t X be a Frechet space with a convergence basis C i n which a l l of the nets are sequences.  Then W has a convergence basis E c o n s i s t i n g  - 106 of sequences, and f E = C implies that f i s a continuous pseudoopen map  of W onto X.  Each summand . S ,(x , x) of W i s a convergent  sequence i n the Hausdorff space S(x , x ) . n  Hence, by (1.32), W  i s metrizable. In general, the product of m-sequential or m-Frechet •spaces -need not -he -m-sequential. m = H  a  have already been given.  -SeveraT<examples"f or -the-case However, the product of two  m-sequential spaces, one of which i s 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.) t h i s r e s u l t i s analogous to that of (1.24).  The proof of  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 n o n - t r i v i a l t o p o l o g i c a l spaces (each space has at least one nonempty proper open s e t ) . not m-sequential.  If the c a r d i n a l i t y of A i s > m, then X i s  In p a r t i c u l a r , no uncountable product of non-  t r i v i a l spaces i s 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.  e be the function i n X whose a-th value i s 1 for each a e A, and  Let  - 107 l e t E be the subset of X consisting of a l l c h a r a c t e r i s t i c functions of f i n i t e subsets of A.  Clearly e e c l E.  It s u f f i c e s to prove  that no i t e r a t i o n of m-nets i n 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 i n 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 c a r d i n a l i t y of each coz f < m. n —  Since coz f CZ  U{coz f  : n e D}, n  J  2 the c a r d i n a l i t y of coz f i s < m  = m.  Thus by forming i t e r a t e d l i m i t s  of m-nets i n E i t i s only possible to obtain functions whose cozero sets have c a r d i n a l i t y j< m.  Consequently, i f the c a r d i n a l i t y of A  i s > m, then the c a r d i n a l i t y of coz e i s > m and hence no m-net i n ( c l E)-{e} converges to e. 3.34 Proposition  If X i s the product of any family {X  &  : a e A}  of n o n - t r i v i a l topological spaces, then each point of X has a neighbourhood basis of c a r d i n a l i t y l e s s than or equal to the maximum r of the c a r d i n a l i t y p of A and q = sup {X(X ) : a e A}.  (Write  cl  X(Y) = m i f and only i f Y i s m-first-countable.) Proof  Let x e X and assume that {U. : i e I } i s a neighbourhood i a  basis f o r the a-th coordinate of x with the c a r d i n a l i t y of I <_X(X ). a a For each a e A l e t 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 B =  ^{{P  of elements i n  - l a (U ) : i e I } : a e A} i s a neighbourhood basis n  2 for x.  Since the c a r d i n a l i t y of B i s £ pq <_ r  c a r d i n a l i t y <_  • r = r.  (See 1.18.)  = r , B has  Chapter 4 Generalized Sequential Spaces and their Properties i n Ordered Topological Spaces  The properties of convergence subbasis and convergence bases are applied, i n 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 i n 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 l e t m be an  i n f i n 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 i n which a l l of the nets  - 110 are well-ordered (2)  -  m-nets.  X i s 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 - n e t s , these 0  generalized sequential spaces and generalized Frechet  spaces  c l e a r l y contain the sequential and Frechet spaces r e s p e c t i v e l y . In p a r t i c u l a r , a topological space i s (sequential, Frechet) i f and only i f i t i s ( ^ -sequential,  H  0  -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.  S i m i l a r l y , a t o p o l o g i c a l 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 t h i s chapter i s to characterize the generalized sequential spaces of (4.2) and characterizations lead to new and  (4.3).  Their  characterizations of the Frechet  the sequential spaces i n terms of orderable spaces.  spaces  To avoid  tedious r e p e t i t i o n , the elementary properties of these generalized .sequential .spaces w i l l . n o t be formally stated..  .The. preceding  chapter's survey of convergence subbases greatly f a c i l i t a t e s study.  their  It i s easy to see that the i n v e s t i g a t i o n of these spaces i s  analogous to that of the m-sequential spaces.  - Ill 4.4 D e f i n i t i o n  (1)  A t o p o l o g i c a l space i s weakly f i r s t - c o u n t a b l e ,  or a weakly f i r s t - c o u n t a b l e space, i f and only i f each of i t s points has a well-ordered neighbourhood b a s i s .  (A c o l l e c t i o n { F  of sets i s c a l l e d well-ordered whenever A i s well-ordered  : a e A}  &  and  F C~ F i f and only i f a > c i n 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 c a r d i n a l i t y <_ m. first-countable,  H  0  (Note that  \\0 -iirst-countable  - f i r s t - c o u n t a b l e , and  are  equivalent concepts.)  4.5 Proposition  (1)  Every weakly f i r s t - c o u n t a b l e 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 f i r s t - c o u n t a b l e space X and  suppose that x e c l F. hood basis {U  : a e A}.  By d e f i n i t i o n , x has a well-ordered neighbourThen, since x e c l F, there e x i s t s a w e l l -  a ordered net {x  a  converges to x. < m.  : a e A} which s a t i s f i e s x  a  e U f) F and therefore a  If X i s m-first-countable, the c a r d i n a l i t y of A i s  - 112 4.6  Example  For any uncountable c a r d i n a l m, there i s an m - f i r s t -  countable space which i s 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  c a r d i n a l i t y i s m, and of c a r d i n a l i t y m. i n which the x  n  order D by Z2> .  Let S = {x  Then D i s a directed set  : n e D} XJ {x} be a topological space  are d i s t i n c t and x  £ x f o r every n e D, and which '  n  J  has a convergence basis generated by the net pair ({x^ : n e D}, x ) . Each x^ i s i s o l a t e d and x has a neighbourhood Consequently S i s m-first-countable.  basis indexed by D.  However, S i s not weakly  sequential because x E cl(S-{x}) and yet no well-ordered net i n S-{x}  converges to x.  To v e r i f y that t h i s i s so, suppose that  {x„,„ : k e K} i s a well-ordered subnet of {x : n e D}. N(k) n v  Choose  a countable c o l l e c t i o n {n. : i E to} of d i s t i n c t elements i n D. i  For each i E to there exists k N(k)25n^.  e K such that, i f k j> k^ then  From the description of D, i t i s obvious that there i s  no supremum of {k^ : i e to} i n K.  But then {k^ : i E to} i s a c o f i n a l  subset of K and hence {N(k_^) : i E to} i s a c o f i n a l subset of D. i s impossible since  This  U {N(k_^) : i E to} i s only a countable subset  of the given set of c a r d i n a l i t y m.  - 113 4.7 D e f i n i t i o n  An ordered topological space i s a space which,  has the order topology a r i s i n g from a t o t a l order on the set. A t o p o l o g i c a l space i s orderable i f and only i f some t o t a l order can be imposed on the set r e l a t i v e to which the given topology coincides with the order topology.  4.8 Proposition space X.  Let A be a subset of an ordered topological  If an m-net i n 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 i n A converging to  x.  Proof  It i s f i r s t shown that every t o t a l l y ordered set has a  c o f i n a l well-ordered subset. of  Let F = {F^ : i e 1} be the family  a l l well-ordered subsets of a t o t a l l y ordered set Y.  Partially  order F by defining F. < F, whenever F. = F. or F. i s an i j i j l  initial  segment of F.. J  Let C  &  Note that F. < F. implies that F . C J i  F,, . j  1  be a chain i n F and suppose that B i s a subset of  UC.  There  e x i s t s 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 t o t a l order < on C implies that  b i s the least element of B and hence that  U C e F.  lemma, F has a maximal well-ordered element Y^.  By  Zorn's  Then Y^ i s also  a c o f i n a l subset of Y because otherwise there exists y e Y~Y^ with no element of Y  greater than y; from t h i s , Y  1  \J {y} e F  - 114 contradicting the maximality Let A  q  be the i n t e r s e c t i o n of A with the range of the  m-net i n the hypothesis. Let A . = {y e A  1  o  of Y^.  The c a r d i n a l i t y of A ^ i s c l e a r l y <^m.  : y. < x} and A „ = {y E A : y > x}. ^ 2 ^ o  Since X has  the order topology, x e c l A^, f o r i = 1 or i = 2; assume the former.  The set A ^ i s directed by the t o t a l order inherited from  X, and thus the i d e n t i t y map on A ^ i s a s t r i c t l y monotone m-net converging to x. Moreover, A ^ has a c o f i n a l well-ordered  subset  A ^ and the i d e n t i t y map on A ^ i s the desired net.  4.9 Theorem , ^.^^^ topological space X are equivalent. (1)  X i s weakly sequential.  (2)  X i s the quotient of a weakly f i r s t - c o u n t a b l e  orderable space. •- (3) X i s the quotient of a weakly f i r s t - c o u n t a b l e space. (4)  Proof  X i s the quotient of an orderable space.  C l e a r l y (2) implies both (3) and ( 4 ) .  implies (1) by (4.5-1) and ( 3 . 2 4 ) .  (1)  > ( ) and (4) 2  > (1).  In a d d i t i o n , (3)  I t remains to show that  - 115  -  To e s t a b l i s h the l a t t e r implication, l e t f : y be the quotient map space X.  > X  of an ordered space Y onto a t o p o l o g i c a l  Suppose U i s a subset of X such that any well-ordered  net converging to a point i n U i s eventually i n U. to prove that f "^(U) i s open i n Y.  It s u f f i c e s  Let y e f "^(U).  I f y £ i n t f "*"(U),  there i s a net {y : n e D} which i s d i s j o i n t from f "''(U) and .n converges  to y.  well-ordered.  According to (4.8), i t can be assumed that D i s But then {f(y ) : n e D} i s a well-ordered net i n n  X-U converging to f ( y ) e U. Assume now  Hence (4)  implies (1)  by contradiction.  that X i s a weakly sequential space with a  convergence subbasis C i n which a l l of the nets are well-ordered nets.  For each ({x n  : n e D}, x) e C, l e t S(x , x) = {x : n e D'} U n n  be a topological space i n which the x x  n  n  are taken to be d i s t i n c t  and  ^ x f o r every n e D', and which has the order topology a r i s i n g  from the t o t a l order defined as follows.  Let a  be the least element  o of D and l e t D' = (w x { a » ( J ( Z * (D-{a })) with w and Z ordered i n o  the usual way.  Q  T o t a l l y order D' by specifying ( i , n) < ( j ,  whenever n < m or i < j and n = m.  m)  Each element of D' has an  immediate successor, and each element other than (0, a ) has an Q  immediate predecessor. Now  l e t X,. = x f o r each ( i , n) e D', and t o t a l l y order S(x , x) (i,n) n ' n' N  by defining x n J  (Such an order i s c a l l e d a d i s c r e t e order.)  b  J  < x m  i f n < m i n D' and x n  < x f o r each n e D'.  {x}  - 116 In the ordered space S(x , x ) , each x i s i s o l a t e d and x has a . n n well-ordered neighbourhood basis indexed by a set order isomorphic to D.  Then SCx^, x) i s weakly f i r s t - c o u n t a b l e and has a convergence  basis generated by the net pair ({x^ : n e D'}, x ) . The d i s j o i n t ...topological,,sum W .of a l l such S.(x , x) i s also weakly f i r s t - c o u n t a b l e The natural mapping of W onto X defined by x map because the net pairs (i^  n  '•  n  > x i s a quotient  e D'}, x) form a convergence  subbasis f o r X. To demonstrate that W i s orderable, l e t {S :• a e A} denote a the set of a l l S(x , x) and define a d i s c r e t e order on Z x A i n the n "same'-'way ••*as*D. -"'(In^th-^ t  1  -order with least element a .) Using the existence of a one-to-one correspondence between A and Z x A, this d i s c r e t e order can be Q  imposed.on A.  Let W be t o t a l l y ordered by specifying x < y whenever  x < y i n S where x, y e S , or a < b i n A where x e S and y E S, . a a a b J  Because of the d i s c r e t e orderings and the fact that each  J  has a  greatest element and a least element, the order topology on W c o i n cides with i t s usual d i s j o i n t t o p o l o g i c a l sum topology. orderable, and the theorem i s proved.  Thus W i s  - 117 4.10  Corollary  For any t o p o l o g i c a l space X and any i n f i n i t e  cardinal m, the f i r s t three statements are equivalent. they are also equivalent to  If m = H  0  (4).  (1)  X i s m-sequential.  (2)  X i s the quotient of an orderable m-first-countable  (3)  X i s the quotient of an m-first-countable space.  (4)  X i s the quotient of an orderable metric space.  space.  Proof  This i s analogous to  (4.9).  I t i s only necessary  to remark  that each S(x , x) i s m-first-countable and hence so i s W. n —  4.11 ,P,roppsit ion  The ..fallowing are .equivalent.  (1)  X i s weakly Frechet.  (2)  X i s the continuous  pseudo-open image of an orderable  weakly f i r s t - c o u n t a b l e space. (3)  X i s the continuous  pseudo-open image of a weakly  f i r s t - c o u n t a b l e space. (4)  X i s the continuous  pseudo-open image of an orderable  space.  Proof by  C l e a r l y (2)  implies both (3)  (4.5.1) and (3.26).  i n the same way that  The fact that  and (4),  (1)  and (3)  implies  (1)  > (2) follows from (4.9)  (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 continuous pseudo-open function o f an ordered space Yonto a topological 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 aw e l l - o r d e r e d n e t  in A converging to f ( y ) = x . 4.12 Corollary  For any topological space Xand any infinite  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 they are also equivalent to (4). (1) X i s m-Frechet. (2) X i s the continuous pseudo-open image o f an orderable m-first-countable space. (3) X i s the continuous pseudo-open image of an m-first-countable space. (4) X i s the continuous pseudo-open image of an orderable metric space. The final results are concerned with the sequential properties of ordered topological spaces and the relation between the notions of first-countable space, Frechet space, and sequential space i n products of these spaces.  I t i s now known that Fre'chet  - 119 spaces and sequential spaces are successive proper generalizations of f i r s t - c o u n t a b l e spaces, and.that the product of two Frechet spaces need not be sequential.  For t o p o l o g i c a l spaces which are  products of ordered spaces, the s i t u a t i o n i s quite d i f f e r e n t .  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^X i& m-3S.e.queatial. M  Proof —>  a  From (3.29), (1) (3) and  (4)  > (2)  > (2).  It remains to show that (2)  > (3).  Furthermore,  (2)  > (1) and (3)  Obviously (4)  > (5)  > (4) by (4.8). > (2).  Assume f i r s t  that X i s an ordered topological space. (2)  > (1).  x has a neighbourhood  Let x e X.  If x i s an i s o l a t e d point then  basis consisting of the singleton {x}.  Suppose  that x i s not i s o l a t e d 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 converging to x.  Let A  n  : n E D} i n {y : y > x}  ={y:x_<y<x}.  The c o l l e c t i o n  - 120 {A^ : n e D} i s a neighbourhood D i s _< m.  basis at x and the c a r d i n a l i t y of  S i m i l a r l y , x has a neighbourhood  <_ m i n the l a t t e r case.  basis of c a r d i n a l i t y  Suppose now that x i s not i s o l a t e d 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 f o r each n. n to x with each y  There i s also an m-net {y : n e E} converging n °  > x.  e  The open sets {z : x  (n, m) e D x E form a neighbourhood of D x E i s <  < z < y } where  basis f o r x and the c a r d i n a l i t y  = i. > (2). Suppose that the nets S = {x" : i E E } n i n  (3)  r r  converge to x  n  and S = {x  n  : n e D} converges to x, with the  c a r d i n a l i t i e s of E and D < m. n —  I t s u f f i c e s to construct an m-net  i n 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 o r d i n a l numbers. or decreasing, and the nets S  n  Either.S i s increasing  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). the nets S  There are four cases to consider.  F i r s t , assume that  are increasing and the net S i s decreasing.  Since x  > x  - 121 for each n, there i s i ( n ) e E with x > x V / > x. ' n n i(n) — N  Then {x7, : n e i(n) N  i s the desired m-net; i t converges to x because i t i s 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. of a l l i s o l a t e d ordinals i n D.  Let D' denote the set  C l e a r l y D' i s a c o f i n a l subset of D;  for each n e D the successor ordinal n + 1 i s i s o l a t e d .  Then  x  , < x f o r each n e D', and there exists i ( n ) e E s a t i s f y i n g n ' n °  x  , < x , i(n)  n-1  n  n-1  to x.  I t follows that { x V / . : n e D'} i s an m-net converging i(n) " e  The remaining two. cases are similar to the f i r s t and second  cases.  This completes the proof f o r the case i n which X consists of  one ordered space. •  -: >a -e-'-A<} a.  of ordered t o p o l o g i c a l spaces.  It i s only necessary to prove that  (3) implies (1). By v i r t u e of (3.33), the c a r d i n a l i t y 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  I f X i s the product of any family {X  : a e A} of  Si  n o n - t r i v i a l 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 c a r d i n a l i t y of A i s < i .  - 122 Proof  If X i s m-sequential then, by (3.33), the c a r d i n a l i t y of  A i s / m and hence <_m. each X  a  To e s t a b l i s h the converse, observe that  i s m-first-countable by the preceding theorem.  Then i f the  c a r d i n a l i t y of A i s < m, i n view of (3.34), X i s m-first-countable.  "4.15 "Corollary  "An ordered t o p o l o g i c a l space i s weakly Frechet  i f and only i f i t i s weakly sequential.  Proof  This i s the same as (3) <  the c a r d i n a l i t i e s of D and E  4.16  Example  n  > (2) of (4.13).  are not  In t h i s case,  important.  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  c a r d i n a l i t y m and t o * has the reverse order to that of t o . d e f i n i t i o n , X has the order : x < y i f  a)  x, y  or  b)  x, y e  or  c)  x e  a  E  a  +  +  to*  1  1  and x < y i n  cc +  and x < y i n  to*,  and y e  to*.  1,  By  - 123  -  L e t X have t h e o r d e r t o p o l o g y a r i s i n g from t h i s t o t a l o r d e r . i s c l e a r that X i s m - f i r s t - c o u n t a b l e .  It  To e s t a b l i s h t h a t X i s n o t  w e a k l y 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 t h e 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 n e i g h b o u r h o o d b a s i s a  at 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 p r e d e c e s s o r  nor an immediate s u c c e s s o r .  F o r each a e A , l e t x  element of (a + 1 ) D U , and l e t y to* n U . a {U  a  a  H  least  a be t h e g r e a t e s t element o f  : 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 E A} i s w e l l - o r d e r e d , t h e r e a r e l e s s t h a n m elements i n t h e  range o f { x {y  cl  The range of {y  be t h e  &  : a £ A}.  3-  and so {x  : a E A} a s s o c i a t e d w i t h each element i n t h e r a n g e o f The supremum o f {x  5 a E A} i s t h e r e f o r e l e s s t h a n a , cl  : a E A} cannot converge t o a .  >  -  124  -  Bibliography  A r e n s , R. "Note on Convergence i n T o p o l o g y " , Magazine,  23  (1950),  Mathematics  229-234.  A r h a n g e l ' s k i i , A. "Some Types o f F a c t o r Mappings, and t h e R e l a t i o n s between C l a s s e s S o v i e t Math. D o k l . , Boehme, T.K.  4  .of . 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