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. A T 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. How let (X , be locally countably compact and satisfy Z. Since Z -^ Н -^ , by Theorem 2, (X,^~) is locally sequentially compact. Le t же 31'. There is a subset P c i such that [ж] — P'. Le t N x be sequentially compact. N x гл P is sequentially compact and infinite with ж as the sole lim it point; hence there is a sequence of distinct points of P and hence of M converging to ж. S6 is then satisfied. Let a locally countably compact space sa tisfy H x. Le t w be a side point of a highly divergent sequence {xn}. There is a countably compact neighborhood N w containing a subsequence of distinct points of {xn}, {yn}. {yn} has a subsequence of distinct points without any side points contradicting the countable compactness of Nw. So H 2 is satisfied. I t might be said that local sequential compactness has the property of improving the sequential behavior of topological spaces; local-countably compact spaces tend to behave either very poorly or reasonably well in regard to sequences. As we w ill see later the Stone-Cech compactification of the integers does not even satisfy H x.

Properties ofS 0-S 2 spaces. In a previous paper, the author (1) proved that an S 0 space is metrizable if f it has a (7-locally fin ite base and is coun­ tably paracompact. T h is la tter condition may be replaced by local countable paracompactness; see A u ll [2]. Analogous to a theorem about T 2 spaces we have the following theorem about countable filte rs. See Gaal [7, 261]. T h e o r e m 6. A topological space is S 0 iff every countable filter has at most one limit point. Sequences in topological spaces 333

Proof. B y constructing the Fróchet filte r fo r a given convergent sequence, the necessity is established. Let {Bn} be a countable base for a convergent filte r converging to a set A containing at least 2 distinct points x and y. Ev e ry N x гл N y contains B n for some n. Construct a sequence {xn} such that xn eB n. {xn} w ill converge to x and y, contrary to the S 0 property. We now turn to intersection properties of sequentially compact and countably compact subsets. T h eo rem 7. I n S 0 spaces the intersection of sequentially compact subsets is sequentially compact. In spaces the intersection of a sequentially compact and a countably compact subset is sequentially compact. Proof. Le t M = П M a where each M a is sequentially compact. Let {xn} be a sequence such that xn e M. {xn} has a convergent subsequence {yn} converging to y. By the S0 property у e Ma for each a. So у e M. Le t Ж be sequentially compact and N countably compact in an S x space. Le t {xn} be such that xne M r\ N. {xn} has a convergent subse­ quence {yn} converging to у and yeM. {yn} has a convergent subsequence without a side point. Since N is countably compact, yeN. Clearly in or S4 spaces the intersection of an arbitrary family of countably compact subsets is countably compact. The S 2 spaces include a ll T 2 spaces and there are T 2 spaces that do not satisfy any of the other axioms discussed here. T h eo rem 8. A topological space such that any compact set is closed is an S 2 space. Proof. H. Cullen [3,123] has proved that such spaces are S 0. Le t {Xn\ be a sequence convergent to x. Let Ж = U [a?n] and let у Ф x. (Ж kj [ж]) ^ [y] is compact and hence closed so {xn} has no side points. Note this same argument can be used to show that (X is S 0. It follows that T 2 spaces satisfy S 2 since in T 2 spaces compact subsets are closed (4).

Further remarks on classification.The F-spaces discussed in Gillman and Jerison [8] are a class of spaces w ith no in fin ite convergent sequence and hence sa tisfy S 8. Some like (JN do not satisfy S4. The E spaces discussed by the author [2], spaces such that every point is the intersection of countable neighborhoods is a large class of spaces including countable T 2 spaces and perfectly normal spaces which satisfy S4 but not necessarily S5 or Z. For a discussion of classes of spaces satisfying S5 and S6, see Dudley [4] and Franklin [5].

(4) For some further relations involving S„ and the condition that any compact set is closed see Wilansky [14]. 334 С. Е. A u li

I t is interesting to note that if a given topology satisfies S 0, S 4, S 2, S 3, or S 4 any fin e r topologies sa tisfy these axioms.

Examples. A series of examples are given to show the independence of the axioms. Some of the properties, particularly the compactness properties of several of the examples are well known results. The following example due to S. F ra n k lin [5], p. 110 and 1 1 3 , satisfies S5 but not S6. E xam ple 1. Le t X be the real numbers with the topology generated by the usual topology and all sets of the form [0] w V where V is a usual open neighborhood of the sequence {1/ri}. Th is example also illustrates that a space may satisfy S5 and hence have no side points fo r either convergent or highly divergent sequences and yet have a sequence w ith side points. In th is example 0 is a side point of a sequence of all the relations excluding points of the form 1 jn.

E x am ple 2. Le t (X be the uncountable product of closed u n it intervals with the usual topology. Since (X,&~) is T 2, it satisfies Sx by Theorems 4 and 8. But (X, 3~) is known to be countably compact without being sequentially compact; so by Theorem 2, {X,3~) does not satisfy H x.

E xam ple 3. Le t X be the real line and iff ~ T is countable or if T = 0. This example satisfies S4 and H 2 but not Z or S5. Proof. (X , J7") satisfies S4 since all countably compact subsets are fin ite . I t satisfies H 2 since a ll countable subsets are closed. (X,&~) does not satisfy Z since every point of the space is an accumulation point of every uncountable subset and countable subsets have no accumulation points. S 5 is not satisfied as a ll proper uncountable subsets are sequen­ tia lly closed but not closed. The next example is due to Arens. See Ke lle y [10], p. 77. E xam ple 4. Le t X be the set of all pairs of non-negative integers w ith the topology described as follow s: F o r each point (m, n) other than (0, 0) the set {(m, n)} is open. A set U is a neighborhood of (0, 0) if f for a ll except a fin ite number of integers m the set {n: (m, ri)4~U} is finite. This example satisfies Z and consequently and S4 but it does not satisfy S5 or H 2. Proof. [0, 0] is the only lim it point in (X,^~) so Z is satisfied; yet [0, 0] is not the sequential lim it of any convergent sequence so H 2 is not satisfied and by Theorem 4, S5 is not satisfied. [0, 0] is a side point of any sequence with range consisting of all points of the space.

E x am ple 5. Le t X be an uncountable discrete space. Add a new point c giving a new space X c. Let the topology °U for X c consist of all sets open in X and the complements of closed and countable sets of X. Sequences in topological spaces 335

(Esse n tia lly the Lindelóf analogy to the one point compactification). (Xc,%) satisfies Z and H 2 but not S5. Proof. {ХС,Щ satisfies Z since there is only one lim it point in the space and it satisfies H 2 since a ll countable sets are closed. I t does not sa tisfy S 5 since X is sequentially closed but not closed in X c. Example 6. The Stone-Cech compactification of j3N of the positive integers is S3 but not S4 or H1. Proof. Every sequentially compact subset is finite. See Gillman and Jerison [8], p. 208 and 215. There is a countably compact subset in fiN which is not compact and not closed so (3N is not S4. See Gillman and Jerison [8], p. 135. The author is in debt to S. Franklin for pointing out some interesting properties of the above example. The next example due to S. F ra n k lin is an example of an S 3 space w ith a compact subset that is not closed. E xample 7. Modify ftN by duplicating one point x of (3N — N and keep the same neighborhoods except both points are closed. Le t у be the new point. The complement of the positive integers and у is compact but not closed. In contrast to the above example, the next example shows a topolo­ gical space in which compact subsets are closed but not all sequentially compact subsets are closed. E xample 8. Le t Q' be the set of ordinals which are less than or equal to the f irs t uncountable ordinal Q, w ith the order topology. T h is topology is H 2, S2 and T 2 but not S3. For Q' ~ [i2] is sequentially compact without being closed. Example 9. Le t (X be (IN with a new point c added. Le t consist of the open sets of fiN and the complements of fin ite sets of @N w ith respect to X . The resulting topology satisfies S0 and H 2 but not S x. Proof. The resulting space is T 4 and a ll sequences taking on an infinite number of distinct values converge to c and to no other point of X; so (X, 2Г) is S 0 and H 2. Any side point of a sequence (highly divergent in the space ftN) is a side point of the sequence in X (sequence is con­ vergent in X). So (X,#~) does not satisfy S4. E xample 10. The free union of Example 9 with a copy of (3N is a S 0 without being H 4 or S x. E xample 11. The one point compactification of Example 4 satisfies Sx and H 2 but not S2. It is interesting to note that there are S6 spaces that are not T 2. See Fróchet [6], p. 213. 336 С. Е. A u li

References

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