TWO TO ONE IMAGES AND PFA

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1 TWO TO ONE IMAGES AND PFA ALAN DOW Astract. We prove that all maps on N that are exactly two to one are trivial if PFA is assumed. 1. Introduction A map f : X K is precisely two to one if for each k K, there are exactly two points of X that map to k. For the remainder of the paper we are assuming that f is a precisely two to one mapping from N onto some (compact) space K. The question of whether there are non-trivial two to one maps on N is motivated y the papers of van Douwen [vd93] and R. Levy [Lev04]. In particular, Levy asks if every two to one image of N is homeomorphic to N. In fact there are several questions in [Lev04] that are consistently answered y the results in this paper. The ehavior of two to one maps on N when CH is assumed is investigated in [DT04]. It is well known that van Douwen has shown in [vd93] that there is a compact separale space which is a 2 to one image of N and this pathology motivates the current study. R. Levy showed that if f is precisely two to one on N then K will have weight equal to c and countale discrete susets of K will have closure homeomorphic to βn. In the two to one mapping context, it is natural to say that a mapping g from X to K is trivial if there are disjoint clopen susets A, B of X such that g[a] = g[b] = K. Proposition 1. If f is locally one to one (every point has a neighorhood on which f is one to one), then N can e partitioned into a such that f[a ] = f[ ] = K. Since f is two to one, f is then a homeomorphism on each of a and. Proof. If each point of N has a neighorhood on which f is one to one, then there is a finite cover y such neighorhoods. Let A e a finite partition of N such that f is one to one on each a A. Enumerate A = {a i : i n}. We will use induction on n. Consider the compact set B 0 = f 1 (f[a 0]) \ a 0 and note that f[b 0 ] = f[a 0]. Since f is two to one, f[b 0 a 1] is disjoint from f[ 1<j a j ]. Therefore there is a c 1 a 1 such that B 0 a 1 c 1 and f[c 1] also disjoint from f[ 1<j a j ]. Since f is precisely two to one, and is one to one on c 1, it follows that f[c 1] f[a 0]. That is, we have shown that B 0 a 1 = c 1. The same argument applies for each i > 0 replacing 1, hence B 0 is equal to 0 for some infinite 0 N \ a 0. It follows that the restriction of f to the union of {(a 1 \ 0 ), (a 2 \ 0 ),..., (a n \ 0 ) } is precisely two to one and is one to one on each piece Mathematics Suject Classification. Primary 54A25. Key words and phrases. two to one maps, Stone-Cech, PFA. This article was consideraly improved y the careful referee s reports. Supported y NSF grant DMS

2 2 ALAN DOW The statements of PFA, OCA, MA and MA(ω 1 ) can e found in [Tod89] and some familiarity will e assumed. Basic information aout N can e found in [Wal74]. Of course it is well known (see [Vel93]) that OCA and MA implies that the mapping, f 1 f from a to in the aove proposition will actually e a trivial mapping. For a function f, we will use f( ) when the function is applied to a memer of its domain and f[ ] when we applying to a set of elements from the domain. Definition 2. Let J e the collection of those sets a [ω] ω such that, on a, f is precisely two to one and locally one to one. Let J denote the ideal generated y J. Proposition 3. If a 0, a 1 are disjoint infinite susets of N and f is one to one on each of a 0 and a 1, then f[a 0] f[a 1] is clopen in K and is equal to f[c ] for some c a 0 a 1 in J. Proof. Set A = N \(a 0 a 1 ) and note that f[a ] is disjoint from f[a 0] f[a 1] y the two to one property of f. Thefore K \ f[a ] is open and is easily seen to e equal to the closed set f[a 0] f[a 1]. Therefore there is an c N such that c = f 1 (U) where U is the clopen set f[a 0] f[a 1]. It is now routine to verify that c is as required since f[c ] = f[(a 0 c) ] = f[(a 1 c) ] and f is two to one. Proposition 4 (OCA + MA). For each a J, there will e a permutation h a on some cofinite suset of a such that h a is never the identity, h 2 a is the identity, and for each a, f[ ] = f[h a [] ] and is clopen. Proof. By Proposition 1, a can e partitioned as a 0 a 1 such that f[a 0] = f[a 1]. It follows easily that f 1 f restricts to a homeomorphism from a 0 to a 1. By OCA and MA, this is a trivial homeomorphism and h a is the witness together with its inverse (with possily finitely many elements of a removed). For each a, each of f[( a 0 ) ] = f[(h( a 0 ) ) ] and f[( a 1 ) ] = f[(h( a 1 ) ) ] are clopen y Proposition 3. Naturally the task is to prove that J does not generate a proper ideal. The first step is to prove that J is not empty. We will proceed y first showing that if f is not locally one to one, then there is a point x such that for every countale family {A n : n ω} x, there is an A J such that A is almost contained in each A n. There are two main results. The first is Lemma 17 which is critical to estalishing that such an x exists. The second, Theorem 25, is to show that this leads to a contradiction. We will certainly need the following results from [Far00] Definition 5. [Far00, 3.3.2] An ideal I (N) is ccc over fin, if there is no uncountale family of almost disjoint susets of N such that none are in I. The following are consequences of OCA and MA (and therefore of PFA). Proposition 6 (OCA + MA). [Far00, 3.8.2] If Φ : (N)/fin (N)/fin is a homomorphism, then there is an A N and an h : A N, such that {a N : Φ(a) = h 1 (a)} is ccc over fin. It will e useful to state the topological dual (a similar ut slightly weaker formulation was given in [Far00, 3.5.5]). The kernel of Φ will form an ideal of susets of N and so the closures of the complements will intersect to a closed set

3 TWO TO ONE IMAGES AND PFA 3 K N. Therefore Φ 1 will induce an isomorphism from a sualgera of (N)/fin to the clopen susets of K. A closed set F N will e said to e ccc over fin if there is no uncountale family of pairwise disjoint clopen susets of N each meeting F. A closed set which is ccc over fin will e nowhere dense in N. Proposition 7 (OCA + MA). If H is a continuous mapping from N onto a suset K of N, then there is an A N and a function h : A N so that H[A ] is clopen, H A = βh A, and H[(N \ A) ] is ccc over fin. Corollary 8 (OCA + MA). If a nowhere dense set T of N is homeomorphic to N, then there does not exist an uncountale family of pairwise disjoint clopen susets of N each of which meets T. 2. Basic properties of f and K We will have to show that K is nowhere ccc (a space is said to e nowhere ccc if no non-empty open suset is ccc). Fix any closed suset Z of N such that f restricted to Z is irreducile (meaning no proper closed suset of Z will map onto). Lemma 9 (MA). The set f[n \ Z] is dense in K. Proof. Let W e a non-empty open suset of K and assume, for a contradiction that W f[n \ Z] is empty. Therefore, f 1 (W ) is contained in Z. Let U e a non-empty clopen suset of Z such that U f 1 (W ). Since f Z is irreducile, J a = a f 1 (f[z a]) is nowhere dense in Z for each a clopen in Z. Also, K \ f[z \ U] is a non-empty open suset of K, hence there is a 0 [N] ω such that f[ 0] K \ f[z \ U]. Since we are assuming f 1 (f[u]) Z, we have that f 1 (f[ 0]) U. Let { α : α c} enumerate [ 0 ] ω. For each α c, J α = J α is a nowhere dense suset of 0. By Martin s Axiom, it is routine to inductively choose a sequence {d α : α < c} [ 0 ] ω, descending mod finite, so that d α J α = for each α < c. Therefore there is a point x 0 such that x / {J α : α c}. Since f is precisely two to one, there is a point x x such that f(x ) = f(x). Since x x, there is an α c such that x α and x / α. Since x / J α, it follows that x / Z \ α contradicting that f 1 (f[ 0]) Z. Lemma 10 (MA). For each a [ω] ω, there is a [a] ω such that f is one to one. Proof. Since f is two to one, it will suffice to find a [a] ω so that K = f[(n \) ]. If a is not contained in Z, let a e such that Z =. Since (N \ ) Z and f[z] = K, it follows that K = f[(n \ ) ]. Otherwise we have that a Z. In this case, (N \ a) N \ Z. By Lemma 9, f[(n \ a) ] will contain a dense suset of K, and, eing compact, will contain K. Lemma 11 (MA). If f is one to one, [N] ω, then the interior of f[ ] is equal to the union of clopen sets of the form f[c ] for c J and c f 1 (f[ ]). Proof. Let e as in the hypothesis of the Lemma and let y e in the interior of f[ ]. Since f is one to one on, there is an x (N \ ) such that f(x) = y. By continuity, there is an a 1, disjoint from such that f[a 1] is contained in f[ ]. If we let a 0 =, then the Lemma now follows y Proposition 3. For N, let f + denote the mapping f and f = f (N \ ).

4 4 ALAN DOW Lemma 12. If f is one to one on, then g = f 1 f + is an emedding of into (N \ ). If g [ ] has some interior, say c, then c contains a memer of J. Proof. It clearly follows from the two to one assumption on f, that g is an emedding. If c N \ is such that c is contained in g [ ], then f is one to one on each of c and. Proposition 3 implies that c contains some memer of J (in fact c will e in J ). Proposition 13 (OCA + MA). If f is one to one on, then can e partitioned into two, 0 and 1, such that f[ 0] is clopen and f[ 1] is nowhere dense. Proof. Again let g denote the emedding of into (N \ ) given y f 1 f +. By Proposition 7, there is c N \ such that c g [ ] and g [ ] \ c is nowhere dense. There are 0 such that g [ 0] = c and 1 = \ 0 will satisfy g [ 1] is nowhere dense. It follows that f[ 0] = f[c ] and f[ 0] = K \ f[(n \ ( 0 c) ] is clopen. In addition, f[ 1] is nowhere dense in K ecause g [ 1] is equal to f 1 (f[ 1]) and is nowhere dense in (N \ ). Although we will not need this result until the next section, this still seems the most natural place to present it. Proposition 14. If a J then there is a c [a] ω such that c Z and f[c ] is disjoint from f[z \ c ]. Proof. By Definition 2 and Proposition 1, a can e partitioned as a 0 a 1 with f[a 0] = f[a 1]. Since f maps Z irreducily onto K, f[a ] is covered y f[a 0 Z] and f[a 1 Z], so assume a 0 Z is not empty. Let W = K \ f[z \ a 0] f[a 0 Z] and recall that W is non-empty open since f Z is irreducile. Choose any infinite c a 0 such that f[c ] W. Since f[a 1] contains f[c ] and f[a 1 Z] is disjoint from f[c ], it follows that f[c ] f[a 1 \ Z] f[a 0 Z]. Since f is one to one on a 0, we have that f[c ] is disjoint from f[(a 0 \ c) ] and so c a 0 Z. Finally, since f[z \ c ] f[z \ a 0] f[(a 0 \ c) ], we also have that f[c ] is disjoint from f[z \ c ]. Lemma 15 (MA). The space K is nowhere ccc. Proof. Oserve that if U, W are disjoint non-empty open susets of Z, then K \ f[z \ U] and K \ f[z \ W ] are disjoint non-empty open susets of K. Therefore, it suffices to show that Z is nowhere ccc. Let A N and A Z and assume that A Z is ccc. Since f[z \ A ] and f[a Z] meet in a nowhere dense suset of K, there is a clopen suset 0 of A Z such that f[ 0 ] f[z \ A ] is empty. Since Z \ A is compact, there is a clopen set B of N such that Z \ A B and f[ 0 ] f[b ] is empty. Let B = { α : α c} enumerate the collection of clopen susets of 0. We may view 0 as the Stone space of the Boolean algera { α : α c}. Fix an unounded set C c so that for each λ C, { α : α λ} is a sualgera of B. Further let {A α : α c} enumerate all the infinite susets of N with the property that their closures are disjoint from Z. For each clopen suset a of 0, let J a = a f 1 [f[z \ a]]. Since f is irreducile on Z, each J a is nowhere dense in Z. Also let Y α = Z A f 1 [f[a α]] for each α c. Since f is one to one on A α, each of f[a α] and Z f 1 (f[a α]) are nowhere

5 TWO TO ONE IMAGES AND PFA 5 ccc. Since we are assuming that Z A is ccc, it follows that Y α is also nowhere dense in Z. We inductively define a family, {a β : β < c} of clopen susets of A Z. Let a 0 = 0. Our inductive hypotheses are that for each α < c, (1) {a β : β < α} has the finite intersection property; (2) a β is contained in one of { β, 0 \ β }; (3) for each β + 1 < α, a β+1 is contained in a β \ (J aβ Y β ). Suppose we have chosen the family {a β : β < α}. Let Z α 0 denote the closed set {a β : β < α}. For each integer n, the selection of a n is trivial, and since we are assuming that a 0 is ccc, we can assume that α ω and that Z α is nowhere dense in Z. If α is a limit, then simply let a α equal α if α meets Z α, otherwise set a α = 0 \ α. Otherwise, α is a successor and there is a β such that α = β + 1. We must avoid J a β Y β. It suffices to show that Z α is not contained in J a β Y β ecause then we can select a α a β to meet Z α, miss J a β Y β and to e contained in one of { α, 0 \ α }. For each γ < α, J aγ Y γ Z α is a nowhere dense suset of A Z. Since 0 A Z is ccc, we may fix a countale family U γ of clopen susets of 0 so that Uγ is a dense open suset of 0 \ (J aγ Y γ Z α ). Since MA implies that c is a regular cardinal, there is a λ α C larger than α so that {a γ } U γ { ζ : ζ < λ α } for each γ < α. Let B α e the Boolean sualgera { ζ : ζ < λ α }. It follows that there is a map g α from 0 onto S(B α ), the Stone space of B α, such that gα 1 ( ζ ) = ζ for each ζ < λ α. Let F α = g α [Z α ]. It follows that F α is also equal to the intersection of the family {a γ : γ < α}, ut in the different Stone space of course. Our assumptions have guaranteed that F α is nowhere dense in S(B α ). Since B is ccc, so is B α. In addition, B α is of cardinality less than c, thus it follows from MA that B α is σ-centered and S(B α ) is separale. Recall that α = β + 1 and let U α denote the dense open suset of S(B α ) which is generated y the family U β. By construction, gα 1 (U α ) is disjoint from Z α. Let D e a countale dense suset of U α. The neighorhood filter of F α traces a filter on D which has a filter ase of cardinality less than c. Since we are assuming Martin s Axiom, there is a countale set {x n : n ω} D which is mod finite contained in every memer of that filter ase. In other words, the sequence {x n : n ω} converges to F α. For each n, let z n Z 0 e chosen so that g α (z n ) = x n. Note that since x n U α, we have ensured that z n / J a β Y β Z α and all ut finitely many are in a β. Therefore all the limit points of {z n : n ω} are in Z α. By passing to a susequence, we can assume that f(z n ) f(z m ) for n < m. For each n, let z n N e distinct from z n such that f(z n) = f(z n ). Let T = f[z α ] which is a nowhere dense suset of K. By construction, the image of {z n : n ω} is contained in {f(z n ) : n ω} T. Since K has no isolated points, and {f(z n ) : n ω} is a relatively closed suset of Z \ T, it is discrete and nowhere dense. Since f maps {z n : n ω} {z n : n ω} onto the discrete set {f(z n ) : n ω} y a two to one map, {z n : n ω} {z n : n ω} is a discrete suset of N. It follows that {z n : n ω} and {z n : n ω} have disjoint closures and f[{z n : n ω}] = f[{z n : n ω}]. Let x Z e a limit point of {z n : n ω}, hence x Z α. There is a limit point x of {z n : n ω} such that f(x) = f(x ). Clearly x a β and we claim that x / J a β Y β. To show this it is sufficient (and necessary) to show that x is not in

6 6 ALAN DOW (Z \ a β) A β. For each n, z n / Y β, hence z n / A β. Since A β is clopen and x is a limit of {z n : n ω} it follows that x / A β. The collection {z n : n ω} Z has all ut finitely of its elements contained in a β, which is clopen in Z, hence none of the limit points are in Z \ a β. We will e finished if we show that the closure of {z n : n I} = {z n : n ω} \ Z is disjoint from Z. Since {z n : n ω} is a nowhere dense suset of 0 and f[ 0 ] is disjoint from f[b ], f[{z n : n ω}] is a nowhere dense suset of K \ f[b ]. It follows now that Z {z n : n ω} is a nowhere dense suset of Z \ B. Since Z \ B is ccc, there is a collection {c n : n ω} of clopen susets of N such that each is disjoint from {z n : n ω} and the union contains a dense suset of Z \ B. For each n I, let d n e a clopen suset of N \ B such that z n d n and d n (Z {c k : k n}) is empty. For each n ω, shrink c n y removing {d k : k < n}; note that this does not change c n Z. Then we have that n I d n and B n c n are disjoint, and as is well-known, they have disjoint closures in N. Since the latter closure contains Z we have finished the proof. 3. tree-like families An emedding of N into N is said to e trivial, if the emedding lifts to an emedding of βn into βn. It is an open prolem to determine if there can e a non-trivial emedding of N into N under OCA and MA (see [HvM90, Prolem 219] and [Far00, Question ]). If there are none, then it is easy to show that that the set 1 in Proposition 13 would e empty y using Levy s proof from [Lev04, 2.4] which shows that the preimages of closures of countale discrete sets are again closures of countale discrete sets. The main result of this section is used as an alternative approach. The set of finite sequences {0, 1} <ω has a standard tree ordering y set inclusion. A family A of susets of N is said to e tree-like if there is an emedding T of N into {0, 1} <ω such that for each A A, T [A] is contained in a single ranch of {0, 1} <ω, and distinct memers of A are sent to distinct ranches (see [Far00, ]). Proposition 16 (MA(ω 1 )). [Vel93, 2.3] Let A e an uncountale almost disjoint family of infinite susets of N. Then there is an uncountale B A and for each a B a partitition a = a 0 a 1 such that the family B i = {a i : a B} is tree-like for each i {0, 1}. Lemma 17 (OCA + MA). Suppose that {a α : α ω 1 } is a tree-like family of susets of N with the property for all α there is a α [a α ] ω such that f( α) is disjoint from f[(n \ a α ) ]. Then there is an α such that, with = α, g [ ] has interior. Proof. We may assume that f is one-to-one on α y Lemma 10. For each c a α define F (c) α as follows. Since f[(a α \ α ) ] contains f( α) and f is precisely two to one, there will e a suset F (c) of α such that f[f (c) ] = f( α) f[(c \ α ) ]. The definition of F on (a α ) can also e expressed as F (c) is the clopen suset of = α which is equal to g 1 [c g ( )]. It is easily seen that F is a homomorphism from (a α )/fin onto ( α )/fin. By Corollary 8, if we find some α and some uncountale family of pairwise disjoint clopen sets each of which meets g α [ α], then this copy of N will not e nowhere dense. Equivalently, y Farah [Far00] (Proposition 6), if for some α the kernel of

7 TWO TO ONE IMAGES AND PFA 7 F (a α ) is not ccc over fin, then there is a c a α \ α such that F is a trivial isomorphism from (c) to ( α ). We proceed as in [Vel93]. Let X denote the set of all pairs c, d such that for some α, d c a α \ α, and each of F (d) and F (c \ d) are not 0. We define a set K 0 [X ] 2 according to { c, d, c, d } K 0 providing (1) c a α and c aᾱ implies α ᾱ (2) c F ( c) and c F (c) are empty; (3) c d = c d; (4) F (c) F ( d) is not equal to F ( c) F (d). The appropriate separale metric topology on X (given y considering it as emedded in (ω) 4 y the mapping sending c, d to c, d, F (c), F (d) ) will result in K 0 eing an open suset of [X ] 2 (see [Vel93]). Assume that Y is an uncountale suset of X and that [Y] 2 K 0. Let I ω 1 e the set of α such that there is c, d Y such that c a α. Also let c α, d α Y e chosen for each α I so that c α a α. Since [Y] 2 K 0 and Y is uncountale, it follows that I is uncountale. Let C = {c α : α I} and D = {d α : α I}. Let c, d and c, d e an aritrary distinct pair from Y. Note that c (F (c) F ( c)) is empty. It follows that C is disjoint from {F (c) : ( d) c, d Y}. Also, D c will equal d for each c, d Y. Hence (C \ D) c = c \ d for each c, d Y. Now consider the two families {F (d α ) : α I} and {F (c α \d α ) : α I}. Assume that E ω and that E F (c α ) = F (d α ) for each α I. Let n ω and I [I] ω1 such that (E F (c α )) F (d α ) is contained in n for all α I. Let α β oth e in I. We may assume that F (c α ) n = F (c β ) n, F (d α ) n = F (d β ) n. Also, we may assume that F (d α ) \ F (c α ) is contained in n for all α I. Since {(c α, d α ), (c β, d β )} K 0, there is some j F (c α ) F (d β ) such that j / F (c β ) F (d α ). Clearly j must e larger than n. Since j F (d β ), it follows that j F (c β ). Therefore j is in E. On the other hand, since j is in E F (c α ), it must follow that j F (d α ), contradicting that j / F (c β ) F (d α ). It follows then that there is no such E. This means that {F (d α ) : α I} and {F (cα \ d α ) : α I} do not have disjoint closures in N. Fix any x N which is in each of the closures. Notice that x / C since C is disjoint from F (c α ) for all α I. Since f[d α] is equal to f[f (d α ) ] and f[(c α \ d α ) ] = f[(f (c α \ d α )) ], it follows that f(x) is in the image of the closure of α I d α and of α I (c α \d α ). Therefore f(x) is in the image of D and of (C \ D). However this contradicts that f(x) only has two points mapping to it as we have found points in D, (C \ D), and (N \ C). Therefore y OCA, X can e expressed as a countale union n Y n such that [Y n ] 2 K 0 is empty for each n. For each n, there is a countale Y n Y n such that for each integer m and each c, d Y n, there is some c, d Y n such that c m = c m, d m = d m, F (c) m = F ( c) m, and F (d) m = F ( d) m. Fix any α ω 1 such that c a α is finite for each c, d n Y n. Construct an increasing sequence {k n : n ω} of integers so that for each n and each i n and each sequence c, d, a, of susets of k n, if there is a c, d Y i such that c k n = c, d k n = d, F (c) k n = a, and F (d) k n =, then there is a pair c, d Y i that also has this property, and in addition, a α aᾱ k n+1 where c aᾱ.

8 8 ALAN DOW Define E i to e {a α [k 3n+i, k 3n+i+1 ) : n ω} for i 3. There is an i 3 such that F (E i ) is not finite. There is a j 3 such that E j F (E i ) is not finite. Fix any c E i such that F (c) is infinite and is contained in E j modulo finite. Let d 0, d 1, d 2 e a partition of c so that F (d 0 ), F (d 1 ), and F (d 2 ) are each infinite. For simplicity we will assume that {i, j} = {1, 2}. Note that for all x d 1, the pair c, d 0 x is a memer of X (since c \ (d 0 x) contains d 2 ) and we may assume that (c F (c)) [k 3m, k 3m+1 ) is empty for all n. It can now easily e shown that F (d 1 ) is σ-borel (see [Far00, p103]) which, in the case that F is an isomorphism, was shown to imply F (d 1 ) has a Borel representation ([Vel93, 2.2]) and would complete the proof y Proposition 7. However it is pointed out in [Far00, p103] that this is not sufficient in the case that F is only a homomorphism as it is here. The proof of [Far00, ] certainly handles a very similar situation ut is not directly applicale to ours. Therefore to finish the proof we will directly produce an uncountale family of clopen susets of N each of which will meet g α [ α]. We construct an increasing sequence {m i : i ω} {k 3l : l ω} together with susets t i d 1 [m i, m i+1 ) and possily infinite sets {J i : i ω} y induction. We can let m 0 = 0 and J 1 =. For each J N, let D(J) = d 0 (d 1 {[k 3j, k 3j+3 ) : j J}). Given that m i and J i 1 have een chosen we will construct the set t i also y a finite induction and will then choose m i+1. Fix an enumeration {(e l, n l, n l ) : l < L} of (m i ) i i. We will construct {t(i, l) : l < L} such that t(i+1, l) max(t(i, l))+ 1 = t(i, l) and will let t i = {t(i, l) : l < L}. Our inductive hypotheses on J i are that F (D(N \ J i )) \ F (d 0 ) is infinite and that if j J i \ J i 1, then k 3j > m i. As we define t(i, l), we will also define J(i, l) and will set J i = {J(i, l) : l < L}. For convenience let t(i, 1) = and J(i, 1) = J i 1. Suppose we have chosen t(i, l 1) and J(i, l 1) such that F (D(N \ J(i, l 1))) \ F (d 0 ) is infinite. Let n = n l and n = n l. First we simply try to get into Y n. That is, choose, if possile, J (i, l) J(i, l 1) such that J (i, l) max(t(i, l 1)) J(i, l 1), c, D(J (i, l)) Y n, and F (D(N \ J (i, l))) \ F (d 0 ) is infinite. If there is no such J (i, l), then let J (i, l) = J(i, l 1). Next we try to get into Y n with infinite growth. That is, choose, if possile, an x d 1 \ max(t(i, l 1)) such that F (e l t(i, l 1) D(J (i, l)) x) \ F (D(J (i, l 1))) is infinite and c, d 0 e l t(i, l 1) D(J(i, l 1)) x Y n. If such an x exists, call this case one, and choose some large enough j x such that there is a k 3l < j such that l / J (i, l) and F (e l t(i, l 1) D(J (i, l)) x) \ F (D(J (i, l))) contains some element of F (c) [m i, k 3l ). We define J(i, l) = J (i, l), and t(i, l) = t(i, l 1) (x j + 1). Note that l will not e in J i. On the other hand, if no such an x exists, then choose J(l, i) J (l, i) such that each of and F (e l t(i, l 1) D(J(i, l))) \ F (D(J (i, l))) F (D(N \ J(i, l)))

9 TWO TO ONE IMAGES AND PFA 9 are infinite. Again choose an l / J (i, l) so that k 3l > max(t(i, l 1)), and ensure that J(i, l) l + 1 equals J (i, l) l. Also ensure that d 1 [k 3l, k 3l +3) is not empty and let t(i, l) = t(i, l 1) (d 1 [k 3l, k 3l +3)). Let m i+1 = k 3n e chosen so that 3n / J i = {J(i, l) : l < L} and t i = {t(i, l) : l < L} [mi, m i+1 ). Let J ω = i J i and note that y construction D(J ω ) m i+1 = D(J i ) m i+1. For each infinite I ω, set e I = {t i : i I} d 1. We have found our uncountale family since we will show that each clopen set e I meets g α [ α]. To show this, it suffices to show that F (e I ) is infinite. Since F (e I ) contains F (D(J ω ) e I )\F (D(J ω )) mod finite, it suffices to show this latter set is infinite. Since c, D(J ω ) e I X, there is an n such that c, D(J ω ) e I Y n. There is also an n such that c, D(J ω ) Y n. Let i I e any integer greater than oth n and n. We will show that F (D(J ω ) e I ) \ F (D(J ω )) is not contained in i. Set d = D(J ω ) e I and fix l such that at stage i in the construction of the m i s, e l = d m i and n l = n, and n l = n. We consider the properties of t(i, l). Let Y = D(J (i, l)). Since c, D(J ω ) is in Y n, it follows that we were ale to ensure that c, D(J (i, l)) = c, Y is in Y n. Recall that there was an l such that m i < k 3l < m i+1 and that the maximum of t(i, l) was greater than k 3l. If J(i, l) \ J (i, l) was infinite, then we know F (e l t(i, l 1) D(J(i, l))) \ F (Y ) is infinite. Therefore F (d) \ F (Y ) is infinite ecause d e l t(i, l 1) D(J(i, l)). If we set x = d \ max(t(i, l 1)), then x would e a witness to the fact that we should have een in case one when defining t(i, l). Therefore, in fact, J(i, l) does equal J (i, l) and at stage l we were ale to find some x as in case one. In addition, t(i, l) was defined as t(i, l 1) (x j + 1) for some j x \ k 3l. Since e l = d m i and i I, we have that d k 3l = (e l t(i, l 1) D(J (i, l)) x) k 3l. That is, if we let d x = e l t(i, l 1) D(J (i, l)) x, then c, d x Y n and d x k 3l = d k 3l. Now also c, d Y n, so there is a pair c, d Y n such that c k 3l = c k 3l, d k 3l = d k 3l, F ( c) k 3l = F (c) k 3l, and F ( d) k 3l = F (d) k 3l. In addition, if we let ᾱ ω 1 such that c aᾱ, we have that aᾱ a α k 3l +1. Further, recall that (c F (c)) [k 3l, k 3l +1) is empty. Oserve also that Y k 3l is equal to D(J ω ) k 3l. Each of the following are straightforward consequences: (1) c a α and c aᾱ and α ᾱ (2) c F ( c) and c F (c) are empty: ecause c F ( c) c (F ( c) a α ) c (F ( c) k 3l +1) c F (c) and similarly, c F (c) c F (c) k 3l c F (c) (3) c d = c d: ecause each of c d and c d are equal to d k 3l. Therefore, the only reason that { c, d, c, d } is not in K 0 is that F (c) F ( d) must e equal to F ( c) F (d). The same is true with d x in place of d, hence F (c) F ( d) must e equal to F ( c) F (d x ). By the choice of x, there is some integer m F (c) [m i, k 3l ) F (d x ) \ F (Y ), and therefore m F ( c) F ( d). This then means that m F (d). By the same reasoning, there is a pair c, d Y n such that c k 3l = c k 3l, d k 3l = D(J ω ) k 3l, F (c ) k 3l = F (c) k 3l, and F (d ) k 3l = F (D(J ω )) k 3l. In addition, if we let α ω 1 such that c a α, we have that a α a α k 3l +1.

10 10 ALAN DOW Further, recall that (c F (c)) [k 3l, k 3l +1) is empty. Repeating the argument aove with c, D(J ω ) and c, d in place of c, d and c, d respectively, we have that F (D(J ω )) k 3l is equal to F (d ) k 3l. Furthermore, F (Y ) k 3l will also equal F (d ) k 3l ecause Y k 3l = D(J ω ) k 3l. Since m F (c) F (c ) \ F (Y ), it follows that m / F (D(J ω )). This shows that m F (d) \ F (D(J ω )) and completes the proof. Corollary 18 (OCA + MA). For each A N such that A Z, there is a c J such that c A and c Z. Proof. Let A N and assume that A Z. Since f is irreducile on Z, f[a Z] has interior in K. Let W e an non-empty open suset of f[a Z]. Assume we find a set N such that f[ ] W and f 1 (f[ ]) contains a for some a J. By Proposition 14, there is c [a] ω such that c Z and f[c ] f[z \ c ] is empty. Since f[c ] f[a Z], it follows that c A, or y removing finitely many integers, c A as required. The remainder of the proof is to show there is such a set. Assume there is some N such that Z is empty, and f[ ] W has interior. Since Z is empty and f[z] = K, it follows that f is one to one. By Lemma 11, this set has the property that f 1 (f[ ]) contains a for some a J. Now, y Lemma 15, we may fix an uncountale family {U α : α ω 1 } of pairwise disjoint open susets of W. By Lemma 9, we may choose, for each α, some infinite c α such that c α Z is empty and f[c α] is a suset of U α. By the previous paragraph, we may assume that f[c α] is nowhere dense in K for each α ω 1. For each α, fix an a α N \ c α such that f[a α] is a suset of U α and meets f[c α]. By Proposition 16 and with re-indexing, we may assume that {a α : α ω 1 } is tree-like. For each α let w α f[a α] f[c α]. Since f is two to one, w α / f[(n \(a α c α )) ]. Let W α U α e an open set neighorhood of w α which is disjoint from f[(n \ (a α c α )) ]. Since f[c α] is nowhere dense, W α \ f[c α] is non-empty so there is an infinite α a α c α such that f[ α] W α \ f[c α]. Clearly α is also almost disjoint from c α, so we may choose it to e contained in a α. By Lemma 10, ensure that f α is one to one. We have that for each α, f[ α] f[(n \ a α ) ] is empty. Now apply Lemma 17 and let = α e chosen so that g [ ] has interior. By Lemma 12, g [ ] contains a for some a J. Since f[ ] = f[g [ ]], it follows that a f 1 (f[ ]) as required. Lemma 19 (OCA + MA). If f is not locally one to one, then there is a point x such that for every countale family {A n : n ω} x, there is an a J such that a is almost contained in each A n. Proof. Since f is not locally one to one, there is a pair {x, x } in N such that f(x) = f(x ) and neither is in a for any a J. Since f[z] = K, we may assume that x Z. This is our choice for the point x. Assume that {A n : n ω} x and, y possily shrinking, we may assume that A 0 / x and A n+1 A n for each n. Let B e any memer of x which is contained in A 0. We check that f[b ] \ f[(n \ A 0 ) ] is non-empty. If f[b ] was contained in f[(n \ A 0 ) ], then f B would e one to one. By Proposition 13, we may assume that either f[b ] is clopen or is nowhere dense. By Lemma 11, f[b ] cannot e clopen, since f(x) is not in f[a ] for any a J. However, since x Z, f[b Z] has interior and so cannot e nowhere dense.

11 TWO TO ONE IMAGES AND PFA 11 Next we show that f[z B ] \ f[(n \ A 0 ) ] is also non-empty. We have that f(x) / f[(a 0 \ B) ] since x B and x / A 0. Fix any B 1 x with B 1 B such that f[b 1] f[(a 0 \ B) ] is empty. By the previous paragraph, there is a y B 1 such that f(y) / f[(n \ A 0 ) ], and there is a z Z such that f(z) = f(y) and which is not in f[(n \ A 0 ) ] f[(a 0 \ B) ]. We check that z B Z. Clearly z / (N \ A 0 ) (A 0 \ B), hence z must e in B. We recursively construct a sequence of infinite sets B n x and n B n so that (1) B n+1 A n+1 B n \ n, (2) f[b n+1] (f[ n] f[(a 0 \ B n ) ]) is empty, (3) n J, (4) and f[ n] f[(n \ A 0 ) ] is empty. Let B 0 A 0 e any memer of x. Since f[z B 0]\f[(N\A 0 ) ] is not empty, there is some C 0 [B 0 ] ω such that C 0 (Z B 0) is not empty and f[c 0 ] f[(n \ A 0 ) ] is empty. By Corollary 18, there is an infinite 0 C 0 such that 0 J. Clearly 0 / x, and since x / 0, f(x) is not in f[ 0]. We can choose B 1 x so that B 1 B 0 A 1 \ 0 and so that f[b 1] is disjoint from each of f[ 0] and f[(a 0 \ B 0 ) ]. The induction proceeds for each n in the same way. By possily shrinking each n we can assume that f n is one to one, and we can choose c n so that n c n J. Since f[ n] f[ m] is empty for all n < m, and f[ n] = f[c n] for all n, we can assume that ( n c n ) is disjoint from ( m c m ) for n < m. Recall that, y Lemma 4, f[ ] is clopen for any which is contained in n for any n. For each n, let {(n, α) : α ω 1 } e an almost disjoint family of infinite susets of n. Inductively define an almost disjoint family {a α : ω α < ω 1 } of susets of n n so that a α ( n c n ) is almost equal (n, α) for each n. Apply Proposition 16, to find a partition a α as a 0 α a 1 α so that there is an uncountale I ω 1 \ω such that each of the families {a 0 α : α I} and {a 1 α : α I} are tree-like. For each α I, at least one of a 0 α or a 1 α will meet infinitely many of the n s. Therefore, y re-indexing, we can assume we have an uncountale tree-like family {a α : ω α < ω 1 } such that each a α meets infinitely many of the n s in an infinite set. Claim 1. there is some n n such that f is one to one, f[ ] f[(n \A 0 ) ] is empty, n is finite for each n, and f[ ] has interior. To prove the Claim, assume first there is some α, ω α < ω 1, such that f[(n \ a α ) ] contains f[a α]. Therefore f a α is one to one. By Lemma 13, there is a partition d 0 d 1 of a α so that f[d 1] is nowhere dense and f[d 0] is clopen. Again note that d 0 is contained in some memer J y Lemma 11. By the assumption on the family { n : n ω}, we must have that d 1 n is finite for all n. Let y e any point in n (d 0 n ) \ n (d 0 n ). Notice that f[(d 0 n ) ] f[c n] for each n, hence there is a point y n c n \ n c n such that f(y ) = f(y). Since n n ( n n) and n c n ( n c n), y y. Also, f[ n] = f[c n] is disjoint from f[(n \ A 0 ) ], hence each of n n and n c n are contained in A 0. We have shown then that f(y) is not in f[(n \ A 0 ) ] since f is two to one. Let Y y e chosen so that f[y ] f[(n \ A 0 ) ] is empty. Since Y (d 0 n ) is infinite for infinitely many n, there is an infinite Y d 0 such that n is finite for all n. Since d 0 and d 0 J, we have that f[ ] is clopen y Lemma 4. This proves the Claim in this case.

12 12 ALAN DOW Now we assume that, for each α ω, there is some infinite α a α such that f[ α] is disjoint from f[(n \ a α ) ]. Clearly f[ α] f[(n \ A 0 ) ] is therefore empty. It also follows that α is almost disjoint from each n since c n N \ a α. By Lemma 17, there is some α ω such that f[ α] has interior. This proves the claim. We can now complete the proof of the Lemma. Let n n e as in the Claim. By Lemma 11, there is some a J such that a f 1 (f[ ]). Clearly f[a ] f[ ], hence f[a ] f[(n \ A 0 ) ] is empty. It follows that a A 0. Let n > 0, and notice that f[a ] f[ ] f[ m>n m] f[bn+1]. Since f[bn+1 f[(a 0 \B n ) ] is empty, we have that f[a ] is disjoint from f[(n \ A n ) ]. Therefore it follows that a A n. This completes the proof of the Lemma. 4. locally one to one In this section we prove Lemma 23 in a (slightly) more general setting than we have developed in the paper and prove the main theorem, Theorem 25, as a simple consequence. The approach is almost a routine application of OCA except it is made more complicated ecause we must first add a Cohen real. Definition 20. A family Y is σ-cofinal in an ultrafilter x if for each countale family {A n : n ω} x, there is an a Y such that a A n for all n. Definition 21. A family {(c α, d α ) : α ω 1 } is a Hausdorff-Luzin family of pairs if for each α < β < ω 1, c α d α is empty and (c α d β ) (c β d α ) is not empty. Proposition 22. Suppose that {(c α, d α ) : α ω 1 } is a Hausdorff-Luzin family of pairs of susets of N, then {c α : α ω 1 } and {d α : α ω 1 } do not have disjoint closures in N. Proof. Assume, for a contradiction, that the two sets do have disjoint closures. It follows then, that there is a Y N such that c α Y and Y d α = for all α. Furthermore, there is an integer m and an uncountale set I ω 1 such that c α \ m Y and Y d α m for all α I. Fix any α < β, oth in I, so that c α m = c β m and d α m = d β m. Since c α d α is empty, it follows that (c α m) (d β m) is empty. Also, (c α \ m) Y and d β (N \ Y ), hence c α d β is empty. By the same reasoning it follows that c β d α is empty, contradicting that the family of pairs was assumed to e Hausdorff-Luzin. Lemma 23. Assume that I is a family of infinite susets of N and that for each a I there is a one to one function h a from a cofinite suset of a onto some cofinite suset of a such that h a (n) n for all n a. If I is σ-cofinal in x for some x N, then there is a proper poset P such that if G is P -generic, then in V [G] there is a partition {C 0, C 1, C 2 } of N and an uncountale family {a α : α ω 1 } I such that {(c α, d α ) : α ω 1 } forms a Hausdorff-Luzin family of pairs, where, for each α ω 1, c α = C 2 h aα (a α C 0 ) and d α = C 2 h aα (a α C 1 ) We defer the proof of Lemma 23 until after Theorem 25. The following corollary is a routine application of PFA proven y selecting the appropriate family of ω 1 many P -names and dense sets for a P -filter to meet. We use the convention that ˇv is the P -name for the ground model set v. Corollary 24 (PFA). Let I e a family of infinite susets of N and for each a I let h a e given which is a one to one function from a cofinite suset of a onto a

13 TWO TO ONE IMAGES AND PFA 13 cofinite suset of a such that h a (n) n for all n a. If there is an x N such that I is σ-cofinal in x, then there is a partition {C 0, C 1, C 2 } of N and an uncountale family {a α : α ω 1 } I such that {(c α, d α ) : α ω 1 } forms a Hausdorff-Luzin family of pairs, where, for each α ω 1, c α = C 2 h aα (a α C 0 ) and d α = C 2 h aα (a α C 1 ) Proof. Let P e the proper poset given y Lemma 23. Let Ċ0, Ċ1, Ċ2, and {ȧ α : α ω 1 } e the family of P -names so that for some p 0 P, p 0 forces {C 0, C 1, C 2 } is a partition of N, {a α : α ω 1 } is contained in Ǐ and {(c α, d α ) : α ω 1 } forms a Hausdorff-Luzin family of pairs, where, for each α ω 1, c α = C 2 h aα (a α C 0 ) and d α = C 2 h aα (a α C 1 ). By replacing P with the poset of all p P such that p p 0, we may assume that p 0 is the largest element of P. For each α, let ċ α and d α also denote the P -names for c α and d α respectively. For each α ω 1, let D α = {p P : ( a α I) p ȧ α = ǎ α } For each n N, let E n = {p P : ( i 3)(p ň Ċi)}. For each α < β < ω 1, let D α,β e {p P : ( k N, i 2)(p k Ċ2 ȧ α a β, hȧα (k) Ċi, and hȧβ (k) Ċ1 i)} Note that for each p D α,β, there is a k such that p ǩ (ċ α d β ) ( d α ċ β ). It is routine to check that all of the aove sets are dense in P, and that if G is a P -filter which meets each of them, then C 0, C 1, C 2 and {a α : α ω 1 } is our desired family where for each i 3, C i = {n : ( p G E n ) p ň Ċi} and, for each α ω 1, a α is the unique element of I such that there is a p G D α with p ǎ α = ȧ α. Theorem 25 (PFA). The function f is locally one to one. Proof. By Lemmas and 4 and 19, the hypotheses of Corollary 24 are satisfied y letting I e J from Definition 2 and the family of functions from Lemma 4. Let C 0, C 1, C 2 and {a α : α ω 1 }, {(c α, d α ) : α ω 1 } e as in Corollary 24. By Proposition 22, there is a point w N such that w c α d α C2. α ω 1 α ω 1 Let α ω 1, and oserve that y Proposition 4, f[(a α C 0 ) ] = f[(h aα (a α C 0 )) ] and f[(a α C 1 ) ] = f[(h aα (a α C 1 )) ]. Therefore f[c0 ] f[ α ω 1 c α] and f[c1 ] f[ α ω 1 d α]. It follows that f(w) has a preimage in each of C2, C0 and C1. This contradicts that f is two to one. Proof of Lemma 23. The remainder of the section is a proof of Lemma 23. Assume that I is a family of infinite susets of N and that for each a I there is a one to one function h a from a cofinite suset of a onto a cofinite suset of a such that h a (n) n for all n a. Also assume that I is σ-cofinal in x for some x N. By choosing one representative in each equivalence class in I mod finite, we may assume that if a, a are distinct memers of I, then a and a are not equal mod finite. Lemma 26. If I is covered y a countale family {Z n : n ω}, then Z n is σ-cofinal in x for some n.

14 14 ALAN DOW Proof. For each n, let A n e a countale suset of x such that if Z n is not σ-cofinal in x, then Z n has no memer which is contained mod finite in each memer of A n. Then A = n ω A n is a countale suset of x and there is some a I which is contained mod finite in each memer of A. Since there is some n such that a Z n, it follows that Z n is σ-cofinal in x. Proposition 27. If H is a function from N to N and H(n) n for all n, then there is a set X x such that H[X] X is empty. Proof. The three set lemma [Wal74, 6.25], implies there is a partition X 0, X 1, X 2 of N such that H[X i ] X i is empty for each i. Lemma 28. If Y I is σ-cofinal in x and m ω, then there are a, Y and k > m such that h 1 a (k) h 1 (k) and oth are greater than m. Proof. Let X e any memer of x such that N \ X is infinite. Let Y denote the set of integers k X so that k is in the domain and range of h ak for some a k Y. Since Y is σ-cofinal in x, Y meets every memer of x in an infinite set. Since x is an ultrafilter, Y x. For each k Y, define H(k) = h 1 a k (k). Note that H(k) k for all k Y. Extend H to N\Y aritrarily to a permutation on N so that H(n) n for all n. By Proposition 27, there is an X 1 x so that H[X 1 ] X 1 is empty. Let a Y such that a Y X 1 and choose k Y X 1 a\(h a [m] h 1 a [m] H[m] H 1 [m]). Note that since k Y X 1, h 1 a (k) H(k) = h 1 a k (k). It follows that if we let = a k then we have our desired pair a, Y. The standard countale poset for adding a Cohen real has many forcing equivalent forms. The form most useful for us is the set, <ω 3, of all functions from some integer into 3 = {0, 1, 2}. This poset is ordered y extension. Corollary 29. If G <ω 3 is a generic filter and if I is covered, in V [G], y a countale family {Y n : n ω}, then there are p G, n ω, a, Y n and k a such that p(k) = 2, p(h 1 a (k)) = 0, and p(h 1 (k)) = 1. Proof. Let {Ẏn : n ω} e <ω 3-names and assume that some p is any memer of G which forces that I is covered y n ω Ẏn. Let p < p e aritrary. For each q < p and integer n, let Y q,n e the set of a I such that q ǎ Ẏn. Since p < p, we have that q<p n ω Y q,n covers I. By Lemma 26, there is a q < p and an n such that Y q,n is σ-cofinal in x. Let m e large enough so that the domain of q is contained in m. By Lemma 28, there is a pair a, Y q,n and k a \ m such that h 1 a (k) h 1 (k) and oth are greater than m. We can extend q to a condition q <ω 3 so that q (k) = 2, q (h 1 a (k)) = 1 and q (h 1 (k)) = 2. Since p was an aritary element elow p, the set of conditions with this property of q is dense elow p, hence there will e such a q G. For the remainder of the proof we work in the extension V [G] for a generic filter G <ω 3. We have our desired partition {C 0, C 1, C 2 } of N given y g = G. That is, g is a function from N onto 3, and we let C i = g 1 (i) for i 3. We will now show that our uncountale family {a α : α < ω 1 } can e found in a susequent proper forcing extension. For each a I, let c a = C 2 h a (a C 0 ) and d a = C 2 h a (a C 1 ). Since h a is one to one and C 0 C 1 is empty, we have that c a d a is empty for all a I. We leave it as an exercise to verify that, since I V and distinct memers of I

15 TWO TO ONE IMAGES AND PFA 15 have infinite symmetric difference, c a and c a also have infinite symmetric difference for a a I. We define a separale metric topology on the set I y declaring [a; m] = { I : c a m = c m and d a m = d m} to e open for each a I and m ω. Let K 0 I e the set of pairs (a, ) I such that (c a d ) (d a c ) is not empty. To see that K 0 is an open suset of I 2 assume that (a, ā) is K 0. Let k e any integer in (c a dā) (d a cā) and let m > k. Let (, ) e an element of [a; m] [ā; m]. Clearly k (c d ) (d c ) which shows that [a; m] [ā; m] K 0. Our desired Hausdorff-Luzin family of pairs is {(c a, d a ) : a Y } for any uncountale suset Y I such that Y 2 \ Y K 0, where Y denotes the diagonal in Y 2. We are nearly ready for a standard application of OCA, however we must recall that we are no longer in a model in which OCA holds since we have added a Cohen real. Instead, we use the following result which comes from the well-known proof that PFA implies OCA. Proposition 30. [Do91, 6.2] If X is a separale metric space and K 0 is an open suset of X 2, then either X is covered y a countale collection {X n : n ω} such that Xn 2 \ X is disjoint from K 0 for all n, or there is a proper poset P which introduces an uncountale set Y X such that Y 2 \ X K 0. By Proposition 30, we will finish the proof of Lemma 23, if we show that I cannot e covered y a countale union, n ω Y n, of sets such that Yn 2 \ I is disjoint from K 0 for each n. By Corollary 29, there is an n, p G, and a, Y n and k a so that p(k) = 2, p(h 1 a (k)) = 0, and p(h 1 (k)) = 1. Let i = h 1 a (k) and j = h 1 (k). It follows that i a C 0 and j C 1. Therefore k c a d. Since this means that (a, ) K 0 Yn 2 \ I this finishes the proof. References [vd93] Eric K. van Douwen, Applications of maximal topologies, Topology Appl. 51 (1993), no. 2, [Do91] Alan Dow, Set theory in topology, Recent progress in general topology (Prague, 1991), North-Holland, Amsterdam, 1992, pp [DT04] A. Dow and G. Techanie, Two to one images of N under CH, Fund. Math. to appear. [Far00] I. Farah. Analytic quotients: theory of liftings for quotients over analytic ideals on the integers. Numer 702 in Memoirs of the American Mathematical Society, vol pp. [HvM90] Klaas Pieter Hart and Jan van Mill, Open prolems on βω, Open prolems in topology, North-Holland, Amsterdam, 1990, pp [Lev04] R. Levy. The weight of certain images of ω Topology and its Applications, to appear. [Tod89] S. Todorcevic, Partition prolems in topology, Contemporary Mathematics, [Vel93] Boan Velickovic, OCA and automorphisms of P(ω)/fin, Topology and its Applications 49 (1993), [Wal74] Russell C. Walker, The Stone-Čech compactification, Springer-Verlag, New York, 1974, Ergenisse der Mathematik und ihrer Grenzgeiete, Band 83. Department of Mathematics, UNC-Charlotte, 9201 University City Blvd., Charlotte, NC address: adow@uncc.edu URL: adow