MONOTONE UNION PROPERTIES IN TOPOLOGICAI. SPACES Thesis for‘the Degree of Ph. D; MICHIGAN STATE UNIVERSITY TINUOY‘E MICHAEL ADENIRAN ' ' l 9 5 9 ‘ ”$381. LIBRARY Michigan State University This is to eertifg that the ‘ thesis entitled Monotone Union Properties in Topological Spaces presented by Tinuoye Michael Adeniran has been accepted towards fulfillment of the requirements for Ph 0 D . degree in Mathematic S Major profess Date 8- 9—69 ABSTRACT MONOTONE UNION PROPERTIES IN TOPOLOGICAL SPACES BY Tinuoye Michael Adeniran A topological space X has the absolute monotone union property if whenever (i)A= UA (ii) Ai 3 Ai+l(topological equivalence), for each i, where {Ail is a monotone increasing sequence indexed by the positive integers, then A is necessarily topologically equivalent to X. If for each i, Ai is open, A has the open monotone union property. The thesis investigates topological spaces having some of these properties, our attention being drawn mainly to one- and two—dimensional spaces. Given a sequence {AilAi C2Ai+l} and a property P such that each Ai has the given property, we investigate whether property P is ab- solute; that is whether the monotone union .fi Ai has the given property. Finally some results are obEiined when we look at open monotone union property in invertible locally connected plane continua. MONOTONE UNION PROPERTIES IN TOPOLOGICAL SPACES BY Tinuoye Michael Adeniran A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1969 C; (0/594 4,17- 4’0 Q; Copyright by TINUOYE MICHAEL ADENIRAN 1970 Dedicated to TINUOLA OYELADUN and ROSEMARIE ACKNOWLEDGMENTS My indebtness to Professor Patrick H. Doyle goes far beyond the courtesy usually accorded advisers in this kind of work. It can be said without any exaggeration that without his singular efforts which include excellent supervision and guidance, various suggestions, continuous encouragements, a very firm stand against some opposing forces, and a great deal of understanding, this work could not have been completed in its final form. My gratitude to him will probably be made more manifest by following his various suggestions and fine examples in the future. I would also like to thank Dr. H. Davis for his help in reading the manuscript and in giving very valu- able suggestions. It is also appropriate for me to thank all those (too numerous to be listed) who have encouraged me, posi— tively or otherwise, to determinedly go through the many pitfalls that befell me in the past four years and in the preparation of this thesis. iv LIST OF FIGURES SECTION 1. BIBLIOGRAPHY INTRODUCTION TABLE OF CONTENTS ABSOLUTE MONOTONE UNION PROPERTY RELATIVE TO A CLASS C SPACES WITH HEREDITARY ABSOLUTE MONOTONE UNION PROPERTY ABSOLUTE PROPERTIES UNDER MONOTONE UNIONS ABSOLUTE PROPERTIES IN WEAK TOPOLOGY OPEN MONOTONE UNIONS AND INVERTIBLE PLANE CONTINUA Page vi 18 24 41 47 55 LIST OF FIGURES SECTION 1. BIBLIOGRAPHY INTRODUCTION TABLE OF CONTENTS ABSOLUTE MONOTONE UNION PROPERTY RELATIVE TO A CLASS C SPACES WITH HEREDITARY ABSOLUTE MONOTONE UNION PROPERTY ABSOLUTE PROPERTIES UNDER MONOTONE UNIONS ABSOLUTE PROPERTIES IN WEAK TOPOLOGY OPEN MONOTONE UNIONS AND INVERTIBLE PLANE CONTINUA Page vi 18 24 41 47 55 Figure Figure Figure Figure Figure Figure LIST OF FIGURES vi Page 10 15 34 3'7 38 SECTION 1 INTRODUCTION A topological space X is said to have the absolute monotone union property if whenever where {AiIAi C Ai+l} is a monotone increasing sequence indexed by the positive integers with each Ai topologi- III-3 A. then A is neces- 1+l’ sarily topologically equivalent to X. cally equivalent to X and if Ai Kwun [ll] shows that there are many manifolds with related property. In particular he proves that if X is a closed PL-manifold of dimension n, n # 4, and p s X, then X - p has the Open monotone union property (defined below). Without explicit definition, Brown [2] proves that the monotone union of open n-cells is an open cell. Kapoor [10] investigates the monotone union property in complexes. More extensive work has been done by Doyle [3,4] however; it is in [4] that the concept of absolute monotone union property (as defined above) is introduced and, compared with earlier works, extensively used. Our approach, in this work, is aflchi to the treat- ment of monotone union properties in the last work cited. 1 Section 2 defines A(X, C) and investigates some topological spaces that have or do not have the absolute monotone union property relative to some classes C of spaces. A characterization of the rationals is also ob- tained (Theorem 2.8). Given a topological space X, what subsets of X have the absolute monotone union property in X? Section 3 exhibits some spaces all of whose sub- sets have this property and further investigates the nature of such spaces. We next relax the definition of absolute monotone union prOperty be requiring that only the condition . T Ai‘: Ai+l needs hold and not necessarily Ai = A Then i+1° given a prOperty P, and if each Ai has property P, does the monotone union LE)Ai have property P? Some topologi— i=1 cal properties are looked at in this perspective in Sec- A} tion 4; and in Section 5 we use weak topology to get some more prOperties, specifically the separation axioms. If each Ai is open, we call the resulting property an gpen monotone union property. This property is applied in Section 6 to invertible plane continua that are locally connected. SECTION 2 ABSOLUTE MONOTONE UNION PROPERTY RELATIVE TO A CLASS C In this section we present examples of topologi— cal spaces that possess the absolute monotone union property with respect to the class C of topological spaces to which they belong. We also give examples of spaces that do not have this property. Definition 2.1: Let C be a class of topological spaces, and let X be a member of C. X is said to have the abso- lute monotone union property with respect to C, denoted A(X, C), if whenever there exists a monotone increasing sequence of copies of X:{Mi|Mi s C, Mi C Mi+l} such that Mi c Y, Y s C, Mi 2 X, then Ill-II M. l i u- 38 l (where the symbol NE) indicates a monotone union over sets indexed by the—integers). If Mi’ X and Y are in a topological space Z, X is said to have the absolute monotone union property in Z. If X isa finite space, A(X, T) holds, where T is the class of all topological spaces. 3 rd? .4 Definition 2.2: A space X has dimension 0 at a point_p if p has arbitrarily small neighbourhoods whose boundaries are empty. A nonempty space X has dimension 0, dim X =.0, if X has dimension 0 at each of its points. We say then that X is a 0—dimensional space. Lemma 2.1: A countable metric space is O-dimensional. 2592:: If Up is a neighbourhood of any point p s X of a countable space X, let 6 > 0 be a real number such that 85(p), the spherical neighbourhood about p with radius 6, is contained in UP. Let {p1,p2,...} be an enumera- tion of X and d(x,y) be the metric. Then there exists a real number 0 < 5' < 6 such that 5' # d(pi,p) for all i, and such that Sé,(p) czU. Then Bd(86,(p)), the bound- ary of Sa,(p), is empty. Since p is arbitrary, X is O-dimensional. Theorem 2.2: Let Q be the space of rationals, M the class fi— of separable metric spaces, then A(Q,M) holds. Proof: Let {M.|Mg c.MJ } be a monotone increasing se— —————- 1 1 1+1 quence of rationals, Mi C'Y. Since each Mi is countable, each Mi is O—dimensional by Lemma 2.1. Furthermore a countable union of countable spaces is countable; hence (filMi is countable and therefore is O-dimensional. i=1 Claim: Mi 2 Q for each i, and M Mi 2 Q. To prove these, i=1 we use the General Imbedding Theorem of Hurewicz and Walman [8]: "Suppose X is an arbitrary space and dim X < n < m, then X is homeomorphic to a subset of I2n+l." That is the O-dimensional Mi (and \fi)Mi) is homeomorphic i=1 to a subset of I, the closed unit interval; hence each M. (and K§)M.) can be mapped homeomorphically onto Q, the rationals. We therefore have 1' 00 Q M., Mi cMi+l and Q — irileMi l (I§)Mi is countable and each point is a limit point) 1 and the definition of A(Q,M) is satisfied. The characterization of Q is Theorem 2.8 below. Theorem 2.3: If P is the space of irrationals, and M is as in Theorem 2.2, then P does not have the absolute monotone union property relative to M. Proof: Let Q be the rationals and therefore P = E1 - Q where E1 is the real line. Let {rl,r2,..} be an enumera- tion of the rationals Q. Then Q = O ri, ri is a rational i=1 number. Since P U {ri} is an irrational space for each i, we can successively adjoin the rationals to the space P to get a monotone increasing sequence of irrational spaces: P C.P U {r1} c.P U {rl,r2} c ...... C.P U {rl,r2...rn} cg... ii For convenience, denote P U {rl,r2,...rj} = P U ((4 rj) i=1 by Pj’ each ri e Q. Then for each j, P E Yj. This follows from the fact that each Pj is O-dimensional and we can therefore use the General Imbedding Theorem quoted in 2.2. However (fijp. = P U ( G ri) = (E1 —‘Q) U Q = E1 j 1 3 i=1 But since P is the irrational space P #T E1 and therefore A(P,M) does not hold. It is shown above that if a subset A of X has the absolute monotone union property relative to a class C, X—A does not necessarily have the property. The foregoing example shows that A(X,C) is not hereditary; that is if X possesses the absolute monotone union property with re— spect to C, it does not follow that every subspace of X has that property: Theorem 2.4: Let X be the joined curve, that is x = {(o,y)1 - 1 i y i 1} U {(x, sin n/x)[ o < x i %} U a simple arc joining the points Pl(O,—l) and P2(l/2,0) (Figure 1). Since X is unique in the class of all topo— logical spaces in X (for if there were to be any other space Y homeomorphic to X, then X would be properly im— beddable in one of its subsetsb X has the absolute mono— tone union property relative to T. We now consider a Figure 1 connected subspace Y a subarc homeomorphic to [0,W). or In El, let Y1 = {-1,«=) Y2 = {-2,w) Yn = I-n,°°) Then Y. C Y. Y. I Y for each 1. But 1 1+1’ 1 0 [fi’Yi = (-w,W)T# Y0 as is required for A(Y0,T) i=1 to hold. Since the irrationals P can be embedded in the space X of Example 2.4 above, and we have proved that A(P,M) does not hold, this is another example showing that A(X,C) does not hold. Theorem 2.5: The property of having absolute monotone union property is a topological invariant; that is if A(X,C) holds for a topological space X in a class C, and if x E y, then A(Y,C) holds. Proof: Let A(X,C) hold, and let {Xilxi c C, c s C} satisfy the definition of A(X,C). In addition suppose f:X > Y is a homeomorphism of X onto Y. Then let gi:X > X. and g:X > I§)X. be the homeomorphisms 1 i=1 1 of X onto Xi and X onto H Xi respectively. Consider the i=1 following diagram: X. l 91 -1 Y f > X g Uxi i=1 The homeomorphism of f:X > Y gives the obvious homeo— morphism f—1:Y > X. And the facts that a composition of two (or more) homeomorphisms is a homeomorphism yields as the required two homeomorphisms of Y onto Xi and Y onto 00 _ E £§ixi respectively. Thus Y — Xi’ Xi c Xi+l’ and Y T: \filxi and therefore Y has the absolute i=1 monotone union property relative to C. Theorem 2.6: The Cantor set K does not have the absolute monotone union property relative to M, the class of separable metric spaces. Proof: Let Kl be the Cantor set in the closed interval [0,1]. One of the characterizations of the Cantor set is that it is a totally disconnected, compact, perfect, metric space [6]. Thus K =T Kl' Let Ki be the Cantor set in the closed interval [0,2]. By characterizing the Cantor set as the set of all points in a closed interval 3K 3K having no units in their ternary expansion, K1 2 _ I and K1 CIK2 — K2 U Kl' We inductively construct Kn so that K; is the Cantor set in the closed interval [0,n] with K E Kn’ and K CLKn = Kg U Kn-l’ The first given characterization n-l of the Cantor set implies that for each positive integer k, k ¢ m, K is topologically equivalent to .N Ki’ the non-compactness of the monotone union lfilKilimplies that K and pfilKi are not topologically the :aie, and therefore K fails—to have the desired property. In the proof of Theorem 2.2 and the ensuing dis- cussions one is tempted to ask whether some of the dis- cussed spaces, and the space Q of rationals in particular, have the absolute monotone union property with respect to any other class C of topological spaces besides M. The 10 following example of a non-metric space shows that going outside M does not yield the desired property: Let {rl,r } be an enumeration of the 2,0... rationals, Q. Construct the rational comb space R as follows: (Figure 2) 1 R = {(0,ri)} U {E,ri)ln = l,2,3,....}, ri s Q. This gives a sequence {Ql/n} of l/n — rationals converg- ing to Q0 = {(O,ri)} as limit. We next define a tOpology on R as follows: .— — ‘ _-—- —-————-_—— l (Orri) (HI 1 '3_rri) (llri) Figure 2 11 On R — {(O,ri)}, use the usual rational topology; for a neighbourhood N of (0,rj) take a neighbourhood in 0 {(0,ri)} and all but a finite number of the vertical {(l/n,ril)}'s. Claim: R, together with the topology described above, is not a metric space. For let S1 and 82 be two points on {(0,ri)}, the rationals of convergence. Since any open subset of R containing Sl meets any containing 82, we 1 and 82 respectively; thus R is not T2. However a metric space hence the conclusion that R is not metric. cannot get two disjoint open sets containing S is T 2, So let R0 be the rationals of convergence {(0,ri)}. For Rl take {(0,ri)} U {(1,ri)} = R1. For R2, let {(0,ri)} U {(1,ri)} U {(l/2,ri)} = R2. For Rn, let {(0,ri)} U U {(1/k,ri) Ik = 1,2,...,n} be Rn. This construction gives the monotone increasing sequence: C. ROCR1CR2C.... CRn .... Because each Ri is the rationals, for i = 0,1,2,..., each R. I Q, the rationals space. But Lg) = R which is not (X) 1 Ri 1=l metric. Hence the non—metric R is not topologically equivalent to the metric space Q thus completing the proof. 12 Remark: Let X be a space such that X cannot be imbedded in itself. X has the absolute monotone union property relative to the class of all topological spaces. The proof that the Cantor set does not have the property of absolute monotone union suggests that many spaces im- beddable in themselves may not have this property. Definition 2.3a: A series, (K,g), is a nonempty nonde- generate simply ordered set. Definition 2.3b: A continuous series, (K,;), is a series with the following properties: (i) If K1 and K2 are any two nonempty subsets of K such that every element of K belongs to either Kl or K2, element of K and every element of K1 precedes every 2, then there is at least one ele- ment x in K such that a. any element that precedes x belongs to K1 and b. any element following x belongs to K2 (Dedekind's Postulate) (ii) If a and b are elements of the set K and if a < b, then there is at least one element x in K such that a < x < b. (Postulate of Density) Definition 2.3: A linear continuous series, (K,<), is a continuous seriesvflfixflisatisfies the following property: The set (K,i) contains a countable subset Q in such a way that between any two elements 13 of the set K there exists an element of Q. ‘Using the elementary properties of the real line E1 and ‘the definitions 2.3a, b and 2.3, we state the following theorem: fTheorem 2.7: El, together with the usual ordering of the .Ieals, is a linear continuous series. fTheorem 2.8: (A characterization of the Rationals in El): ZLet X be a countably infinite space in El all of whose 1points are limit points. Then X is the Rationals. IProof: Since the rationals are countably infinite and (dense in El, it will suffice to show that any two count- .able dense series having neither a first nor a last eslement are ordinally similar, thus we would have charac- ‘terized the rationals. To this end, Huntington [7] has Insed the method of George Cantor to prove precisely that. In the proof that the rationals have the absolute Inonotone union property with respect to M the prOperties Insed are, in fact, those that characterize the rationals; 'this allows us to generalize Theorem 2.2 to the following. Egorollary 2.9: If M is a countably infinite space all of inhose pointsare limit points, then M has the absolute Inonotone union property relative to the class of all separable metric Spaces. Iheorem.2.10: Let X be a subset of El such that X has the absolute monotone union property in El. Then X is 14 either an open interval, or a totally disconnected set having at most a finite number of isolated points and at least a limit point. Proof: (The proof depends upon the continuum hypothesis). There are two cases. Case I. X contains an Open nonempty subset S. Then there exists an increasing sequence {UilUi CLUi+ll of prOper Open sets in El such that S 2 Ui and El 2 LfilUi. Thus 8 = E1 has the absolute monotone union i=1 property relative to El. Before considering the second case, we need the following: Lemma 2.11: If S is a totally disconnected countably in— finite subset of E1 with infinitely many isolated points, then A(S,M) does not hold. Proof: Enumerate the isolated points of S as {p1,p2,....} and call the set P. For each pi, let U2 be a finite open 1 neighbourhood of pi such that U3 (1 S = {pi} and such that 1 ‘U0 (W U0 = g for 1 # j. Next let U1 be the finite open Pi Pj 0 Pi neighbourhood U to which has been added an extra point i pil' for each 1, since Ul is an open interval containing 1 and pil’ it is clear that US. is topologically equiva- 1 lent to U1 . Pi 2 In a similar fashion, let U be the finite open i neighbourhood U: to which have been added pil and pi2 for 1 Pi 15 each 1. By using induction, define U: as the finite 1 Open neighbourhood US to which has been added the set i {p. ,p. ,...,p. l for each i, where {p. } is the rationals 1 1 1 1 10 2 k k in U . (Figure 3) Pi U0 U0 U0 pi p2 pn O O O O l- \ 1- \ l..\ 1..\ l4.\ I':X I:.\ \' I \' I \ 'l \‘ I II I \ .1 \ 1 Pl p2 . n ‘ . U1 U1 U1 p1 p2 pn Pl p11 p2 p21 pn pn 2 2 2 U U U P1 P2 pn {. . 1} /_ 11%--..){l11 1,; . ;°'"°{...} pl pl Pll k32 p2 p2 bnpn p11 2 . 1 2 Figure 3 I l - °° l 1 . Denote the p. s and the p. s in I) by P ; 1.e., 1 l1 1 i=1 Pi P = P U {pi I1 = 1,2,...} in such a way that each point 1 is a limit point of the union. Also let Then we have P =T Pl =T P2 =T ....... =T Pk =T ........ since for each j, Pj is an enumeration of isolated points 16 just as P is. Furthermore, P C P1 c:P2 c ... c Pk C .... Thus we have s = s U P =T (S-P) U P1 =T (S—P) U P2 =T . . . =T (s U P) U Pk =T... In addition, 1 k s U P C.(S-P) U P c:(s—P) U P2 c....C (S—P) U P c... thereby getting an absolute monotone union of (S-P) U P:J [fl] (S-P) U P3. j=l However, successive addition of points pik to U: for each i i yields a dense set of points in U0 . Therefore the . i monotone union Qg)(s — P) U PJ is a dense set in U U0 . '=l i=1 1 But P, and therefore S, is not dense in G U . And so T m . i=1 i S # IM)(S — P) U P], thus showing that S does not have '=l the absolute monotone union property. We now consider the second case. Case II: X is totally disconnected. A. If X is finite, we have seen that X has the absolute monotone union property with respect to any class of tOpological spaces, and therefore A(X,M); hence A(X,El). B. If X is infinite, what is the topological structure of X? Firstly, X must have a limit point for if otherwise X is topologically Z, the integers. But Z consists only of infinitely many isolated points and by 17 Lemma 2.11, Z does not have the absolute monotone union prOperty in El. Secondly, X cannot have infinitely many isolated points. For if X has infinitely many isolated points, again Lemma 2.11 shows that the desired property fails for X. SECTION 3 SPACES WITH HEREDITARY ABSOLUTE MONOTONE UNION PROPERTY Section 2 discussed some topological spaces having the absolute monotone union properties relative to a class C; also discussed were some spaces whose subsets have or do not have these properties. An example was given to show that given a space X and A(X,C), not all subsets of X necessarily have the absolute monotone union properties in that space. This section deals with those spaces all of whose subsets have the absolute monotone union proper- ties in them. To be more precise, we being with a definition. Definition 3.1: A t0pological Space X is said to have the hereditary absolute monotone union property, denoted HA(S,X), if every proper subset S of X has the absolute monotone union property in X. Remark: Henceforth in this section all spaces shall be considered infinite unless otherwise stated. Theorem 3.1: Let Z be the integers. The Z has the HA(S,Z) prOperty. l8 1 1 19 Proof: Let S be an infinite subspace of the integers. S is homeomorphic to Z; hence there exists an infinite T subset Y0 in Z such that Z - Y0 is infinite and S = Y0. Let Y1 = Y0 U {21} where Zl ¢ Yo. It is clear that Y1 is homeomorphic to S and Y0 CLYl. Inductively form the T monotone sequence {YiIYi+l'Yi = S} after Yk-l has been got by letting Y to be Y U {zk}, zk ¢ Y0. Since k k-l m T T w . . YOIJ Zi = S we have S = lmIYi, thus sat1sfy1ng the ab- i=l i=1 solute monotone union property in Z. Definition 3.2: Let X = {aili s Z+} be a set of points indexed by the positive integers. Let Ul = {a1}. U2 = {al’aZ} U = {al,a2,...an_l,an}... The set T = {Uili e Z+} along with g and X is called the tower topology on X. Theorem 3.2: Let X = {aili a 2+} with the tower topology. Then X has the HA(S,X) property. Proof: Let S be a subset of X. If S is finite, we are done. So assume that S in an infinite subset of X. We first claim that S has the same topology as X: For if S = {bl,b 2,... bi = aj for some a3. 8 X}, suppose the first element bl of S corresponds to ak in X for some k 6 2+. 20 Then Uk = {al,a2,...ak} = {al,a2,..bl} is open in X; hence Uk [)8 = {b1} is Open in S. If b2 corresponds to say aj in X, then Uj is Open in X and Uj FIS ={bl,b2} is open in S. Thus for each j a 2+, {bl,b2,...bj} is open in S; and therefore S has the same topology as X. It is now easy to see that any infinite subset S has the absolute monotone union property in X. Theorem 3.3: A space X with the discrete topology has the HA(S,X) property. Proof: Every infinite subset S of X is open and is topo- logically equivalent to any other infinite subspace of X with the same cardinality. There exists, therefore, an infinite subspace S' of X such that S - S' is infinite, card S = card (X - S') and S T: X - 8'. Let S1 = X - S'. _ ' = For some x1 8 X S , let 82 S1 U {x1}. For some x2 8 X - 5', let S = S U {x }. Assuming that S has 3 2 2 k-l been thus obtained, let Sk = Sk—l U {Xk-l} where Xk—l }. _ , . . e X S . Th1s y1elds a monotone sequence {Silsi c.Si+l T Claim: (1) Si = S, (ii) S T=lfiJSiz For each i=1 Si is open and thus [Elsi is open and infinite. There is i=1 therefore a bijection f which carries any open set in S to an open set in R 81’ And for any open set in E Si’ i=1 i=1 its inverse image is open in S by the discrete topology, f is thus a bicontinuous bijection and the topological equivalences claimed follow. 21 Theorem 3.4: The space Q of the rationals does not have the HA(S,Q) property. Proof: Take as a subset of Q the integers Z. Let {rl,r2,...} = Q' be an enumeration of the proper fractions (rationals) in Q. Form a monotone increasing sequence {Ri} as follows: Take as R the set r Z, and observe l l’ _T _ . - - that Rl — 2. Let {rl,r2}z — R2, that 15 R2 18 the set of all points of the form {rl,z,rzzlrl,r2 e Q',z e Z}. . . _T . _ S1m11arly R2 — Z and R1 CZR2. Inductively let Rk — {rl,r2,...rk}z, ri s Q' for i = l,2,...,k. Then Rk =T Z, co = = ' =T Rk CLRk+l° However, éEjRi {rl,r2,...,}Z Q Z Q, hence Z does not have the absolute monotone union property in Q since (§)Ri #T Z. i=1 Lemma 3.5: Let M be a metric space having the HA(S,M) prOperty. Then M has at most one limit point. Proof: Suppose M has at least two limit points, p1 and p2. Then there exist in M two convergent sequences {ai} and {bi} converging to p1 and p2 respectively. Let S = {ai} U pl U p2.1 S is a convergent sequence in M hav- ing pl as 1ts l1m1t po1nt. Let S1 = {ai} U bl U pl U p2. T The only limit point of S1 is pl, and S = S Let S = l' 2 {ai} U {bl,b2} U pl U p2; similarly, the only limit point . T of $2 18 p1 and as before S — 82 and 81¢: 82. After Sk-l has been obtained, let Sk = {ai} U {bl,b2,...bk_l,bk} U pl U p2. 22 S has pl as its only limit point, S T: Sk,and Sk-l C.S . k k Next consider E S.. This is the set: {a.}IJ{b.}lJp.U p2. i=1 1 1 1 1 > p1 and bi Because ai > p2, égisi has pl, p2 as two limit points whereas S has only one, namely pl. Hence sT °°S.. 761(2)ll Theorem 3.6: Let M be a metric space having the HA(S,M) property. Then M is a compact metric space with exactly one limit point, or M is a discrete space with no limit points. 2329:: If M in Lemma 3.5 has a limit point, M has to be compact. For suppose M is not compact and has a limit point. In particular let M be the union of the nonnega— tive integers and {l/n}. The limit point of M is 0 and there is no other. Let S1 = {l/n} U {O,-l}, S1 is com- pact and has 0 as its limit point. Let 82 = {l/n} U T . . {0,-l,-2}. S1 = 82, 81:: $2, and 82 1s compact w1th the limit point 0. Thus for each k, let Sk = {l/n} U T T T {OI—lIZIOOO—k}. Sl = 82 = 000 = S each Sk is compact. k’ Sk—l’ ask and Buteflthough E Si = {l/n} U {0,—l,-2,...} has the i=1 limit point 0, this last set is not compact. There does not exist, therefore, a subset S of M such that S T= S. l and S T: (EJSi' Hence M has to be compact if it has a i=1 limit and the HA(S,X) property holds. 23 If M has no limit point, then M is topologically equivalent to Z which is discrete, and we have seen that Z has the HA(S,X) property. Corollary 3.7: The Cantor set K does not have the HA(S,K) property Proof: K has more than one limit point. SECTION 4 ABSOLUTE PROPERTIES UNDER MONOTONE UNIONS Definition 4.1: A property P is absolute under monotone unions (aumu) in a class C of topological spaces if for each Y s C, X. c Y, X. c X. and for each i, each X. has 1 1 1+1 1 property P implies that (E)Xi has property P. We here i=1 investigate some topological properties that are or are not aumu with respect to a class C of topological spaces. Theorem 4.1: Connectedness is an absolute property under monotone unions in the class C of all tOpological spaces. Proof: Let Y be a topological space having a sequence . {XiIXi C:Xi+l}Xl ¢ U of subsets With the property that for each i, X5L is connected. Then 3 xi c. X hence . 1’ 1:1 there exists at least one point in common with the family {Xi} of connected subsets of Y. The union. 3 Xi of this i=1 family is therefore connected and the assertion is proved. Theorem 4.2: Arcwise (path) connectedness is an absolute property under monotone unions in the class C of all tOpological spaces. Proof: Let Y, {Xi} be as in 4.1 and Xi is arcwise (path) connected. For any two points x,y e (E)Xi, there exists i=1 24 25 some j e Z+ such that x,y s Xj. Since Xj is arcwise (path) connected, so 15 §§%Xi' Theorem 4.3:. The property of being locally connected is not aumu in C. Proof: Modify the space X in 2.4 by letting the joined sine curve be open at the point (0,1). Taking as X1 an Open interval beginning at the point (0,1), as X2 an open interval containing X as Xn an open interval containing 1 X a monotone increasing sequence {Xilxicz Xi+l} of n-l’ open intervals each beginning at the point (0,1) is thus obtained. Each Xi is locally connected. But it is easy to see that each point x in {(x,y)|x = 0, -l < y < l} CLX has a neighbourhood not containing any connected neigh- bourhood of x. Theorem 4.4: Disconnectedness is not absolute under mono— tone unions in C. Proof: If P is the set of irrationals in El, form a mono- tone increasing sequence {PiIPi C Pi+l} as in Theorem 2.3: Each Pi is disconnected but H Pi = E1 is not. i=1 Corollary 4.5: The property of being O—dimensional is not aumu in the class C of tOpological spaces. Proof: Each Pi in 4.4 is O-dimensional but E1 is not. Lemma 4.6: A O-dimensional space Y is disconnected. Proof: Let Y be O-dimensional. In particular for each 26 p s Y there exists arbitrarily small neighbourhoods of y which are both open and closed because the boundaries of such neighbourhoods are empty. Since these neighbor- hoods are neither Y nor empty, Y cannot be connected for the only subsets of a connected space both open and closed are the empty set and the space itself. Theorem 4.7: Let CO be the class of all countable metric spaces. Then disconnectedness is an absolute property under monotone unions in C0. }, a sequence of o . I C . Proof. Let Y a C0, {x1 chx1 x1+1 disconnected subsets of Y. Every subset of a countable (metric) space Y is countable, hence for each 1, X1 is countable (and therefore O-dimensional). We need to show (§)Xi is disconnected. To do this, observe that a count— i=1 able union of countable subsets is a countable subset. Therefore (Elxi is countable and is therefore O-dimensional i=1 (by Lemma 2.1). Apply Lemma 4.6 and we have proved that (fi)xi is disconnected. i=1 Corollary 4.8: The property of being countable is absolute under monotone unions in any class C. Remark: For any cardinality c greater or equal to the cardinality of the rationals, the property of having cardinality c is absolute under monotone unions in any class C, for if for each 1 card Xi = c and Xi c:Xi+ then EH ... .. 1! card 27 Definition 4.2: A set F is called an F0 — set (or an F0) if F is the union of at most countably many closed sets. A set G is called a G6 - set (or a GO) if G is the inter- section of at most countably many open sets. Theorem 4.9: The property of being F0 is aumu in any class C of topological spaces. Proof: Let Y s C such that {XiIXi c Xi+l} 15 a sequence of subsets of Y with the property that each Xi is an F0. Then for each i, x. = ('3 C. . 1 i=1 13 where each C. 1j is a closed set. Then pxi =OI lefi‘ ij] The right side is a countable union of lcountablyl many closed sets; and as such it is the union of countably many closed sets. Therefore KE)Xi is an F0 — set. Lemma 4.10: Let x be a rational point in En. Then {x} is a G6° Proof: Let v: be a spherical neighbourhood of radius 1 with center at x; let v: be a spherical neighbourhood of radius 1/2 with center at x. In general, let v: be a spherical neighbourhood of radius l/n and center at x. Then m v; is the countable intersection of open sets Vi 00 and this intersection therefore is G5. But ()V; = {x}, j=l and therefore {x} is GO' 28 Lemma 4.11: Let R = {rl,r2,...rk} be a finite set of rationals in El. Then R is a G6 - set. Proof: For each rj e R let Vij be a spherical neighbour— hood of radius 1 and center at rj. In general VEj be the th n spherical neighbourhood of rj of radius l/n and cen- ter at rj; for each rj s R. Then 6 Uvg. q=1 i=1 3 is an open set containing R, since a (finite) union of open sets is open. Let k 00 Y= U 0V:- j=l q=l 3 By Lemma 4.10, for each j = l,2,...,k, 0 V3. = {r.}. k q=l J 3 Hence Y = U r. j=l 3 G6 — sets and is therefore a G6 = R, that is Y is a finite union of — set. Lemma 4.12: The set Q of the rationals in E1 is not a G<5 — set. Proof: Suppose Q = n V1 where each V1 is open in El. i=1 . Since Q is dense, each V1 is also dense. Let Y be the family {Vi} U {E - rlr s Q} Y is a family of open sets since E — r is open for each r, Y is countable and dense in El, and E1 is a locally 29 compact space. By.a theorem of Baire [5] the intersection of any countable family of open dense sets in a locally compact space is dense. Thus by this, F) A has to be AEY dense. However 00 oh: ovimE—rlreo} ASY i=1 is empty and therefore is not the intersection of count- ably many open sets in El. So Q is not a G6 - set. Theorem 4.13: The property of being a G6 - set is not absolute under monotone union in any class C. Proof: Let C = {El}, and let Q = {rl,r2,...} be an enumera- tion of the rationals in El. Form a monotone sequence Xi as follows: Let xl {r1} X I 2 — {rl,r2} Xk = {rl,r2,...,rk} By Lemma 4.10 and 4.11 X1 is a G6 and for each j e Z+, Xj 1 15 G5. Xi czE and for each 1 Xi C Xi+l’ however @X -0 i=11 is the set of all rationals in El and it has been seen in Lemma 4.12 that Q is not a G5. 30 Definition 4.3: A space X is said to be LindelOf if every open covering of X has a countable open subcovering. Theorem 4.14: The property of being LindelOf is aumu in any class C. Proof: Let Y be an element of C, {XiIXi C'xi+l} a sequence of subsets of Y such that for each i, Xi is LindelOf. Let X = (E)Xi, and let {Uald e A} be an arbitrary open cover- i=l w i 1ng of X. Then X = ig%xi = \dan. Therefore for each d s A there exists some j e Z such that Ua (‘IXj is not empty. Claim: {Uar)Xj|j e Z+ and a e A} is a countable open sub- covering for X. For, for each a e A, Ualfi Xj C.Xj, thus getting a countable subfamily {UQFIXj} of {Ua}. Further— more X- = X U£Uaflxj}= 11 USA IIBS 1 So X is LindelOf. Definition 4.4: A space X is first countable (or, satis- fies the first axiom of countability) if X has a countable basis at each point p of X. Theorem 4.15: The property of being first countable is aumu in any class C. Proof: Let Y s C, Xi C.Y, Xi CZXi+l for all 1, and Xi 15 first countable. Let X = (EJXLand suppose p s X. Then i=1 31 for some j e Z+, p s Xj c.X. By the first countability of Xj, there exists at most a countably infinite family {Ufi(p)lk e Z+} of neighbourhoods of p having the follow- ing property: For each open G containing the point p there is some Ui(p) c.G in Xj' For each k, Ui(p) is open in Xj implies that there exists an open set Uk in X such that Uk FIXE = Ufi(p). Since p s U£(p), p 6 Uk as well. So let {Uk(p) cxlUk(p) nxj = U];(p), k 5 2+} be the countably infinite family of neighbourhoods of p in X. Thus for any G open in X and containing p, it is easy to see that there exists some Uk(p) such that Uk(p) c G. This proves that X is first countable. Definition 4.5: A space is second countable (or, satis- fies the second axiom or countability) if it has a countable basis. Theorem 4.16: Second countability is an absolute property under monotone unions in any class C of topological Spaces. Proof: Let Y s C w1th Xi c Y, Xi C’Xi+l’ and for each 1, Xi is second countable. Let X = (E)Xi, for each i, let . i=1 {V;|j e Z+} be a countable basis for Xi. We claim that the set _ w i i . . . + U — aghvj|{vj} 1s a bas1s for xi, 3 e Z 32 is a basis for X. For let U be open in X. Then for each i S 2+, U FIXi is either empty or open in Xi. If U FIXi is empty for some i, there is nothing to prove. SO as- sume that U Frxi is not empty and therefore open. By the second countability of Xi’ the open subset (in Xi) U FIXi = UV]i for some j e: Z+. Furthermore since U c. .®xi, we 3 1:1 have U=t){Ur\xi|UflXi7£9/} 1 Thus since each U FIXi is a countable Union of the Vi's, U is a countable union of V3, for some i, j 3 2+, and therefore every open set in X is the union of members of the countable family U. Therefore (E)Xi is second i=j countable. Theorem 4.17: Compactness is not an absolute property under monotone unions in the class C of all t0pological spaces. Proof: Let Y ={1/n} U {0,1,2,...} :19, n = 1,2,... Construct a monotone sequence of compact subsets {Xi} as follows: N H {l/n} U {0,1} {l/n} U {0,1,2} N II Xk = {l/n} U {0,1,2,...k} n = 1,2,3,... 33 It is obvious to see that each Xi is compact (since it is closed and bounded). Xi C xi+l7 it is equally obvious to see that 6x. = Y = {l/n} U {0,1,2,...} is not compact for it is not a bounded set. Theorem 4.18: The property of being a Tl-space is aumu in any class C of topological spaces. Proof: Let Y be a topological space with the sequence {XiIXi C X } of subsets of Y such that each Xi is T . 1+1 1 00 . . - Let X = Imlxi, and let x,y be two arbitrary dist1nct i=1 points in X. Then there exists, for some j s Z+, th: X such that x, y e Xj' Xj being Tl’ there exist open sets U', V' in Xj such that x e U', x ¢ V', y s V', y ¢ U'. Now there exist open sets U,V in X such that Unx.=U' J vflxj=V' Since x s U', x e U; similarly y e V. x s Xj and x ¢ V' imply that x t V. Similarly, y ¢ U. We thus find sets U, V open in X with x e U, y e V, x ¢ V and y ¢ U. By definition X is T1 and the theorem is proved. Theorem 4.19: Being T is not an absolute property under 2 monotone unions in some classes(3of topological spaces. Proof: Let Y = E2, I1 the open unit interval (0,1). IT Let X1 = (0 X2 = (0 X3 = (0 Xk = (0 Each X. is 1 since each (Figure 4). H II II M M II M II II II M II M h o 34 x I ) U (1 x 11) x 1 ) U (1 x 11) U (1/2 x 11) x 1 ) U (1 x 11) U (1/2 x 11) U (1/3 x 11) x 11) U (1 x 11) U....U (l/k x 11) a finite union of Open intervals in E2 and I1 is T , each X. is T . Let X = (E)X. 2 1 2 i-l 1 l/n 1/3 1/2 1 Figure 4 35 Put the following topology on X: On X — (0 x I1) use the usual topology on El. For a neighbourhood of a point x 1, take an open interval in I1 of 0 x I1 and all in 0 x I but a finite number of the (l/k x Il)'s, k = 1,2,... Using the argument similar to the construction of a non- metric space in Section 2, we conclude that for any two distinct points x, y in 0 x Il, there do not exist open sets Ux’ Uy With UX fIUy = 9. Hence X = égixi is not T2. Corollary 4.20: Any separation axiom beyond T2 is not an absolute property under monotone unions in some classes C of topological spaces. Proof: Any space satisfying any separation axiom beyond T satisfies T hence the corollary (using the defini— 2 2' tion of separation axioms in [5]). Corollary 4.21: Let {XiIXi c;Xi+l} be a sequence of spaces such that each Xi is T2 (regular, Tychonoff, normal). Then 3 X. is at least T . i=1 1 1 Proof: A simple application of Theorems 4.18, 4.19 and Corollary 4.20. Corollary 4.22: Metrizability is not an absolute property under monotone unions in the class of all topological spaces. Proof: Each Xi in theorem 4.19 is a metric space. But X is not normal and therefore not metric. 36 Another proof of Corollary 4.22 is afforded by the following: Let Q = {rl,r2,...} be an enumeration of the rationals in the plane on or above the x-axis. Let X1 = {r1} x2 = {ri'rz} - i Xk = {rl,r2,...rk} Each Xi, being finite, is metrizable. So let X = GE)Xi i=1 with the following topology: if (x,y) is a point of X and s > 0, let . b (x,y) + {(r,0)|either|r - (a + ——)| < e or /3 Ir—(a-—b)[<€} /3 be a neighbourhood of (x,y) Geometrically, such a neighbourhood with center at (x,y) can be obtained by constructing an equilateral triangle (Figure 5) with base on the x-axis and apex at (x,y). If y = 0, let the point (x,y) be the desired triangle. Then (x,y) + all rationals on the x-axis whose distances from a base vertex of the triangle are less than s is an s-neighbourhood with center at (x,y). R. H. Bing has shown [1] that although this space X has a countable basis, it is not regular and therefore not metric. 37 .al (+2'e+) (+2“) Figure 5 Definition 4.6: A space X is said to be a nontrivial product if X = Y x Z and neither X nor Y reduces to a single point. Theorem 4.23: Being a nontrivial product is not an abso— lute property under monotone unions in C. Proof: Let C = E3 . Let T1 and T2 be two solid tori as shown below (Figure 6). There exists a homeomorphism h:E3 BEER> E3 such that h:T2 > T2 is an onto homeo- morphism, and such that h is the identity exterior to some sphere [12]. Then h(Tl) = h2(T2) is a torus such that T1 c,interior h2(T2). Then M3 = Cr'EDhnuz) n=l 38 is a 3-mainfold which is a monotone union of hn(T2). But each h3(T2) is a copy of sz Sl; hence M3 is an open- 2 1 monotone union of copies of E x S [4]. Figure 6 Claim: M3 is not a nontrivial product. Suppose the contrary. Then M3 is either (i) E1 x C1 or (ii) S1 x C2 where C1’ C2 are subsets of M3. Consider the homotopy groups of M3 and of E1 x Cl and S1 x C2. For each k, Wk(m3) = 0. Thus for NI, we have (i) “1 (E1 x C1) = 0 or (ii) n1 (51 x c2) = o. For (i) nl(El x cl) = 0 we have Hl(El x C1) = nl(El) * Hl(Cl). Since nl(El) = 0, wl(Cl) has to be 0; but nl(Cl) = 0 implies C1 = E2. And therefore for (i) M3 = E1 x E2 = E3 which is a contra- diction since M3 # E3. For (ii) Trl(Sl x C2) nlSl) * nl(C2) = 0 is impossible since nl(Sl) = Z # 0. 39 Since the only two possibilities of M3 have been proved not to apply, M3 = (filhn(T2) has to be a trivial product n-l even though each hn(T2) is a nontrivial product. Theorem 4.24: Let C' = {E2}, the property of being a connected open nontrivial product is absolute under mono— tone unions in C'. Proof: Let X. C:E2, X. CLX. and for each i, X. is a ————— 1 1 1+1 1 connected open nontrivial product. So let Xi = Y x Z where neither Y nor Z reduces to single points. Then Y and Z are locally compact connected sets, and Y x Z can be imbedded in a 2—mainfold. Hence by a theorem of Jones and Young [9], Y and Z can be either an arc, a simple closed curve (and therefore homeomorphic to 81), a ray, or an open curve (homeomorphic to an open interval). But the homogeneity of Y and Z reduces these possibilities of Y and Z to either S1 or I, where I is an open interval, and therefore to El. So we have either (i) Xi = E1 x E1 or (ii) xi = E1 x I or (iii) xi = E1 x 51, xi 6E2. (i) is not possible, since Xi # E2 = E1 x El. So we con— sider (ii). If each Xi = E1 x I, then -¥lxi = E2 which is a nontrivial product. (iii). If eadh Xi = E1 x 81, then Xi is an annulus A. And the monotone union H Xi i=1 40 gives two possibilities. a) E2 l§)Xi = i=1 1 b) an annulus A Since an annulus A is a nontrivial product (A = E2 x 81) and E2 is also nontrivial, we have thus proved the theorem. SECTION 5 ABSOLUTE PROPERTIES IN WEAK TOPOLOGY It has been shown above that there are some topo- logical prOperties that are not absolute under monotone unions. In particular any separation axiom beyond T1 is not absolute under monotone unions in some classes C of spaces. Here the concept of weak topology is used in order that these separation axioms be absolute. Definition 5.1: Let X be a set, and let S = {Aald s A} be a sequence of subsets of X such that each A“ has a topology. Assume that for any a,8 e A the following two properties hold: (i) The topologies of Ad and A agree on AaJW.A B 8 (ii) Either (a) each Ad F)A. is open in A“ and in B A8 or (b) each AOL f) A B' The weak topology in X induced (or determined)by S is B is closed in Ad and in A To = {U clefor all O, U r)Aa is open in Ad} In literature X is also said to have a topology coherent withfl if X has weak topologyfTA induced by S. Denote such a topological space by (X, To) to distinguish it from other topoloqies. 41 IP~ 42 Lemma 5.1: Let (X, TA) be a topological space having the weak topology induced by S = {Xilxi c Xi+l}' Then any subset U open in Xj is open in (X, T6). 3322:: Let U c;Xj be an open subset in Xj‘ Then for any k, U F)Xk is open in Xj f)Xk by definition 5.1 (i). Not- ing that A open (closed) in Y and Y open (closed) in Z implies that A is open (closed) in Z, and using Defini- tion 5.1 (ii) (a) and (b), observe that U H Xk is open in (x, T5). Corollary 5.2: Any subset V closed in X is closed in (x, T.)- Corollary 5.3: A subset B C.X is open (closed) in X if B n Xi is open (closed)in Xi for each Xi in 3. Notation: For the remainder of this section, w shall de- note the class of tOpological spaces that are coherent with some family S of subspaces. Theorem 5.4: T2 is an absolute property under monotone unions in w. Proof: Let {XiIXi C-X } be a sequence of T —spaces and 1+1 2 such that X = IEDXi has a topology coherent with {Xi}. Let i=1 x,y be two distinct points of X. Then for some j e Z+, x,y e Xj' Since Xj is T2, there exist open neighbourhoods Ux’ Uy of x and y respectively such that UX fl Uy = U in Xj' But by Lemma 5.1 UX and Uy are open in (X, To)' |Hx 43 Definition 5.2: A space Y is a Urysohn space if for every distinct points x,_y S Y there exist open neighbour- hoods Ux' UY of x and y respectively such that U¥4q U& = U. Corollary 5.5:: The prOperty of being a Urysohn space is aumu in class w. Definition 5.3: A Hausdorff space Y is regular if each y a Y and any closed set A not containing y have dis- joint neighbourhoods. Lemma 5.6:_ Y is regular if and only if for each y a Y and closed A not containing y, there is a neighbourhood V of y such that V (IA = U. Theorem 5.7: Regularity is an absolute property under monotone unions in w. £3923; Let {XiIXi C2Xi+l} be a sequence of regular spaces and let X be its monotone union with the weak topology induced by {Xi}. Given y s X and a closed subset A C.X not contained y we want open neighbourhoods Uy of y and V containing A with UyF)V = 0. Case I: A is contained in a finite number of elements of {Xi}. Then for some k large enough, y s X and A CLX k k’ y ¢ A. By the regularity of Xk’ there exist Uy and V open in Xk, and therefore open in X, such that y s Uy, A<: V, and Uy F)V = H. Case II: A is contained in infinitely many Xi's. Again for some j e Z+, y s Xj' 44 (i) If for this j, AFIXj = U. Then Xj being T2 (for it is regular) there exist x e-Xj, x f y, and an open neighbourhood Uy of y with x U U&. Then let V = X - U&. V is an open subset of X containing A and Vnin‘IJ- (ii) If A.FIXj # U. A closed in X implies that A F)Xj is closed in Xj and therefore closedpin X (Coro- laries 5.2 and 5.3).. Note that y ¢ Atfi Xj' By thegregu_ larity of Xj then there exists an open neighbourhOdfidey of y such that U? F)(A F)Xj) = U in Xj. (Lemma 5.6) Since U§ F)X. = U', we have Uklfi A =-U. So let 3 Y V = X - U?. V is open in X, contains A and has an empty intersection with Uy' Thus X is regular. Definition 5.3: A Hausdorff space is normal if each pair of disjoint closed sets have disjoint neighbourhoods. Lemma 5.8: Y is normal if and only if for each closed subset A and an open subset U containing A there is an open subset V such that ACVCVCU. Theorem 5.9: Normality is an absolute property under monotone unions in w. Proof: Let {XiIXi C’Xi+l} be a sequence of normal spaces with X = (§)Xi having a topology coherent with {Xi}° Let i=1 C,D be two disjoint closed sets in X. 45 Case I: C,D are each contained in a finite number of elements of {Xi}. Then there exists a positive integer k large enough so that C,D c:X By the normality of X k. there exist open disjoint subsets U,V in Xk such that kl C C U, D C.V, U r)V = U. But U and V are also open in X; hence the assertion. Case II: One of the closed sets is contained in infinitely many Xi's. Without loss of generality, let D be contained in infinitelynany Xi's. Thus for some j s Z+, C c.Xj. (i) If for the j, D FIXj = U, then the normality of Xj implies that for the closed set C c;Xj and open sub- set U' of Xj containing C, there exists an open set V withC CVCVQU' in xj, by Lemma 5.8. Vopen in xj implies V open in X. So here let U = X — VI Then C c‘V, DCU, andUnv= (X-V)(')\7=U. (ii) If Xj n D # U. Since C(\ D = U, C ijj F)D = U. But D closed in X implies D lej is closed in Xj' We therefore have two disjoint closed sets C and Xj n D in X.. By the normality of Xj there exists an open set U in J Xj with C c U and Un U be a homeomorphism of Mj onto U (the exist— ence of such an hj being asserted by the fact that each M. T= U). Then th(U) C.M. c.M J J 3 be the inclusion map and consider the following diagram. T— ‘0 j+l — U. Let i.Mj ——> Mj+l 5 A I j+l hj+1 identity :4 5" (.1 I Cé————————'z CI \ map Thus it is clear that for each j, the equations (1) hj+l . i - hgl (U) C.U and (ii) h . i - h‘l (U) s U j+l j hold since it is assumed that Mj # Mj+l° So let —1 U1 = hj+l ' 1 ° hj (U). Claim:(i) Ul has the open monotone union property in M: Since U has the omu property, Mj property. But the inclusion map equally preserves this = h;l(U) has the omu property; hence i - h;l(U) c Mj+l has the said property in M. Similarly since hj+l is a homeomorphism, hj+l|[i . h;l(U)] is a homeomorphic image of i - h;l(U) and therefore has the omu property; that is h - i . h;l(U) = Ul CtU has the open monotone union j+l property. 51 Net let {M.} be the aforesaid sequence. Then for each k, U T: Mk' So let h :M ——> M be a homeomorphism k of M onto itself such that hk(Mk) is topologically equiva— lent to U. Since Mk C Mk+l’ we have U = hk(Mk) C hK(Mk+l) (proper inclusion). So for the desired U2, pick any k such that Mk T: U, select a homeomorphism hk of M onto itself with hk(Mk) = U, and let U2 = hk(Mk+l) thus obtaining U C U2 = hk (Mk+l) ' Claim: (ii) U has the omu property: For since hk is a 2 homeomorphism, we can choose thMk as an inclusion of Mk in Mk+l’ hence a homeomorphism. Thus if U has the omu property, so does hk(Mk+l)‘ Next since U T: U by the homeomorphism of the l composite function hj+l - i - hgl and U T: U2 by a similar process as described above, and since U1 and U2 have the omu property as claimed, we can repeat the process by substituting Ul for U and U2 for U respectively to obtain the desired result. Theorem 6.4: If U C M is a nondense open set having the open monotone union property in an invertible plane con- tinuum M, then U is connected. Proof: From Theorem 6.1, there exists an x in M such that M-{x} T: U. Since U has the omu property, there exists a strictly increasing sequence {Ui} of copies of U such that P. 1th»: .. 52 U = (E)U. so that we now have i=1 1 U =i'7iU. =M—{x}. From Theorem 6.3, there exist sets 01, 02 c:M such that 01 C.U C 02, 01 and 02 have the omu property in M and 0 = U = 0 By the same theorem there exists 0i such 20 that 0' C201 with the properties just mentioned above. 1 So suppose P1 is a component of U. Then there exists a copy P2 of P1 such that Ul C'Pz. But P2 C-U3 since P2 C-U T: U2 C'U3. We thus have inductively U2i_lc.P2ic;U2i+l c.......... For each 1, U2i-l U P2i is an open (for each of Uj—l and Pj is open), connected (since Uj— C Pj’ j = 2k, k 5 2+) 1 set; therefore (U U P 21—1 21) 1. =1 is open, connected and is a strictly increasing union of copies of U, thus U is connected. Definition 6.4: A point p of X is said to be a cut point of X if X—{p} = Y U Z where Y and Z are separated; other— wise p is a non—cut point of X. Corollary 6.5: Let C, U, and x be as in Theorem 6.1. Then x is not a cut point of X. 53 Proof: If x were a cut point and C— x T: U = Y U Z, where Y and Z are separated, then U is not connected thereby contradicting Theorem 6.4. Theorem 6.6: Let S1 be the 1-sphere and suppose U :81 is a connected nondense locally connected open set in 81. Then U has the omu property in 81. Proof: Let p be any point of 81. For some 3 > 0 let (p-€,p+s) be an e-neighbourhood of p and call it Vl. Let V2 = (p-2€,p+2€) and so Vl C V2. In general, let Vn = T_ . T_ 00 (p ne,p+ne). Then Vi<: Vi+l and Vi — U. Since U —iMlVi, l _ U has the desired property in S . Remark 1: The same proof goes for any open connected non-dense locally connected set U in Sn, for n > 1. Remark 2: The property of connectedness in 6.6 cannot be dispensed with since by 6.4 it has been shown that if U is not connected, U may fail to have the omu property in sl(sn) . Definition 6.5: Let X have the open monotone union property, and suppose {AiIAi C’Ai+l} is the sequence for which Ai T: X and (filAi T: X. Then for each i, A. is 1 1: called the monotone subspace of X. Theorem 6.7: Let X have the open monotone union property in an invertible plane continuum M, and let D C M be a 54 compact subset of M, D # M. Then D can be imbedded in one of the monotone subspaces of X. PPooP: Let D C.M be compact, D # M, and let h be an in- verting homeomorphism of M. D being compact, M-D is open. We can as well assume that the homeomorphism h is such that h(D) c M—D. It is also clear that h(D) # M-D. By Theorem 6.1, there exists a point x in M such that M-{x} T: X. Since h(D) C M-D C.M-{x} for such a point x, T 00 — = = C we have h(D) c; M {x} x [MIX]: where {xilxi Xi+ } 1 i=1 satisfies the definition of X having the omu property. But now h(D) is compact, hence h(D) lies in a connected subset of M-D. By the monotonicity of the sequence {Xi} we see that h(D) lies in one of the Xi's which is a monotone subspace of X; hence the theorem holds. 10. ll. 12. BIBLIOGRAPHY Bing, R. H., "A connected countable Hausdorff space," Proc. Amer. Math. Soc. 4(1953) 474. Brown, M., "The monotone union of open n-cells is an open n-cell," Proc. Amer. Math. Soc. 12(1961) 812—814. Doyle, P. H., "On monotone unions of manifolds," Proc. Amer. Math. Soc. 19(1968) 854-856. Doyle, P. 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