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Sequences in Topological Spaces* ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Seria I: PRACE MATEMATYCZNE XI (1968) a n n at.e s s o c ie t a t is mathematicae p o l o n a e Series I: COMMENTATIONES MATHEMATICAE X I (1968) С. E . A u ll (Blacksburg, Virginia) Sequences in topological spaces* Recently there has been a renewed interest in sequences in con­ nection w ith «^-classes of Frechet p) and w ith topological spaces and their interrelations. See for instance Kisyński [1 1 ], Dudley [4] and Franklin [5] (2). We w ill be particularly interested in topological spaces (X,&~) with either of the following properties. (a) If M is a subset that contains the sequential limits of all convergent sequences, then M is closed. (b) If xeX, M <= X such that xeM', then there exists a sequences {Xnf, xn e M such that x n converges to x. I t is known that (b) (a). Topological spaces satisfying (b) and such that sequences converge to at most one point (espace 8 of Fróchet) have been studied by Frechet [6] and Urysohn [13]. Topological spaces satisfying (a) and their relations to topological spaces satisfying (b) have been studied by Hausdorff [9], Kisyński [1 1 ], Dudley [4] and Franklin [5]. The concept of a side point of a sequence plays an important role in th is paper. (A side point is an accumulation point of the set of values of a sequence such that no subsequence of the sequence converges to the point.) It w ill be proved that in topological spaces satisfying the condition that sequences converge to at most one point that each of the following conditions follow from the previous ones; (b), (a), every countably compact subset is closed, every sequentially compact subset is closed, convergent sequences lack side points. I f compact subsets are closed, the la st con­ dition is satisfied. Locally sequentially compact spaces that satisfy the condition that sequentially compact subsets are closed sa tisfy (a). Fu rth e r - * Most of the research was done at Kent State University. Most of the results were announced in the American Mathematical Society Notices (1965). P) For a discussion of basic properties of ^-classes, see Kuratowski [12]. (2) Recently at the second Prague Topology Symposium a paper on this subject was given by V. Koutnik relating to work of J. Novak. 330 С. Е. A u li more in locally sequentially compact spaces, highly divergent sequences (sequences without convergent subsequences) lack side points. Definitions of various types compactness are the same as in Kelley [10], i.e., a topological space is sequentially compact if every sequence has a convergent subsequence (the sequential lim it is in the space). A topological space is countably compact if every countable open cover has a finite subcover. In a 1\ space this is equivalent to every infinite subset having an accumulation point. A topological space (X,&~) is locally sequentially compact (countably) compact if fo r every xeX, there is a sequentially (countably) compact neighborhood of x. Side points D e f in it io n 1. A point у is a side point of a sequence [Xn\ if у is an accumulation point of the set of values of {жп} but no subsequence of {xn} converges to y. In a T x space this is equivalent to saying that the sequence {xn} is frequently in every neighborhood of у but not eventually in every neighborhood of y. D e f in it io n 2. A sequence is highly divergent if it has no convergent subsequence. Th eo r em 1. A topological space (X,ST) is sequentially compact iff it has no highly divergent sequence. AT x space is countably compact iff every highly divergent sequence has a side point. Proof. The first statement follows from Definition 2. Let (X,2T) be countably compact and let {xn} be highly divergent. { xn} takes on any value a finite number of times. Let M = [#„]. There is a point ye 31', since (X, У) is countably compact; у is then a side point of {;v On the other hand, let every highly divergent sequence have a side point. Le t 31 be an infinite set and P a countably infinite subset of M. I f among the sequences of distinct points formed from points of P there is a con­ vergent sequence, then P has an accumulation point. If all sequences are highly divergent, then 31 has an accumulation point in this case. So (X,&~) is countably compact. T h eo r em 2. In a T x space {X,2T) every countably compact subset is sequentially compact if every sequence has a subsequence without a side point. Proof. Le t 31 be countably compact. It w ill be sufficient to consider the case when M is infinite. Let {xn} be a sequence of distinct points of 31. {xn} has a subsequence {yn} without side points. Set N = U [yn] ; N must have a lim it point у e 31 and since у is not a side point of {yn}, {yn} must have a convergent subsequence. I t follows that any sequence of points of M w ill have a convergent subsequence. Sequences in topological spaces 331 T h eo rem 3. In locally sequentially compact spaces no highly divergent sequence has a side point. Proof. Le t у be a side point of a highly divergent sequence {xn}. Le t Ny be a sequentially compact neighborhood of y. There is a subsequence of {xn} in N y taking on distinct values which must have a convergent subsequence which contradicts that {xn} is highly divergent. The referee has pointed out that as a consequence of Theorems 1 and 3 locally sequentially compact, countably compact T x spaces are sequentially compact. Axioms related to sequences D e f in it io n 3. A topological space (X,&~) satisfies S 0: I f sequences converge to at most one point. I f S 0 is satisfied and every convergent (highly divergent) sequence has a subsequence without side points (3). S 2(H 2): I f S 0 is satisfied and no convergent (highly divergent) sequence has a side point. S3(S4): If every sequentially (countably) compact subset is closed. S5: If S0 is satisfied and every sequentially closed set is closed. A set is sequentially closed if it contains the sequential lim its of a ll convergent sequences. S6: If S0 is satisfied and for xeX, M a X such that xe M', there is a sequence {x^} of points of M converging to x. Z: If S0 is satisfied and for xeX, M a X such that xe M', there is a subset P с M such that [x] = P '. The next theorem relates these axioms. An ELS,- space is a topological space satisfying both Hi and A ->■ В means that a topological spaces satisfying axiom A satisfies axiom B. T h eo rem 4. (a) H 1S3 ->S4, (b) S5 Ha-> H 4, (c) S n+1-^ S W, (d) S 6 -* Z -> S4, (e) Z H x, (f) ZS5 = S6. Proof, (a) Follows from Theorem 2, (b) H 2 -> H 4 is immediate. Le t {jxn} be highly divergent. Let M = (J |]хп]. M ~ [yiyeX] is se­ quentially closed and hence closed. So S5-> H 2. (c) S2 S i -> S0 is immediate from D e finitio n 3. Le t {xn} be a convergent sequence, con­ verging to a point x. Set M — U [xn]. \xn~\ w M r**j [У‘У Ф x] is se­ quentially compact. So S3 -> S2. Since a sequentially compact subset in a T x space is countably compact S4 -> S3. S5 -> S3 since a sequentially compact set in an S„ space is sequentially closed by (a) and (b) S 5 -> S 4. S6 -> S5 is a known result, (d) S6 Z is immediate. Let M be countably compact. Le t xe M'. If Z is satisfied, there is a subset P с M, such that (3) For S0 spaces H^Sx is equivalent to the statement that every sequence has a subsequence without side points. 332 С. Е. A n il [ж] — Р'. Since М is countably compact, хе 31. So Z ->S4. (e) Le t {xn} be highly divergent and let w be a side point of {xn}. Since Z is satisfied there is a subsequence {yn} of which w is the only side point; since {yn} does not converge to w, there is a subsequence of { yn} without any side points. Hence Z H x. (f) Le t (X, ZT) be ZS5. Le t же 31'. There is P a 31 such that [ж] = P'. If P ~ [ж] is sequentially closed, by the S5 property P r'-' [#] is closed. Hence there is a sequence in P converging to ж. So S 6 is satisfied. By (c) and (d) ZS5 = S6. Clearly in Z H 2 spaces, no sequence has a side point. A countable space is S6 iff no sequence has a side point. B y Theorem 3 locally sequen­ tia lly compact topological spaces sa tisfy H 2. Th eo r em 5. Locally sequentially (countably) compact S3(Z) spaces satisfy S 5(S6). Locally countably compact H x spaces satisfy H 2. Proof. Le t 31 be sequentially closed. Le t же 31'. Le t N x be a sequen­ tia lly compact neighborhood of x. N x ^ M is sequentially compact and by the S 3 property it is closed; so хеЖх ^ M and M is closed. The f irs t statement follows.
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