Maharam s Problem. Omar Fouad Selim. School of Mathematics

Transcription

1 Maharam s Problem A dissertation submitted to The University of Manchester for the degree of Master of Science in the Faculty of Engineering and Physical Sciences 2008 Omar Fouad Selim School of Mathematics

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3 O. F. Selim Contents Abstract 5 Declaration 6 Copyright 7 Acknowledgements 8 1 Introduction 9 2 Maharam Submeasures and Maharam s Problem A First Formulation On Souslin s Hypothesis and von Neumann s Problem Without Loss of Generality The Bridge 54 4 Talagrand s Construction The Boolean Algebra The Submeasures ψ and (ν k ) k N The Importance of ψ Non-Triviality Exhaustivity Concluding Remarks 108 A Background Material 112 A.1 Some Set Theory A.1.1 Partial Orders A.1.2 Ordinals and Cardinals A.1.3 Filters A.2 Some topology A.2.1 Topological Spaces via Closure Operators A.2.2 Metric Spaces A.2.3 More on Closure Operators A.2.4 Topological Groups A.3 Measures on Boolean Algebras A.3.1 Boolean Algebras and Stone s Representation Theorem A.3.2 Distributive Laws

4 Contents A.3.3 Measures and Submeasures B Roberts and Farah 136 B.1 Roberts B.2 Farah References 155 Index Font size: line spacing. Margins: 40mm 15mm. Words in text: Number of maths inlines: (Counted by TeXcount version 2.0 at on 06/09/08) Last page:

5 O. F. Selim Abstract One formulation of Maharam s problem asks: Is every exhaustive submeasure uniformly exhaustive? We present a detailed account of this problem with particular emphasis on its beginnings in 1947 and also its (negative) solution which was published in the form of a counter example in

6 Declaration No portion of the work referred to in this dissertation has been submitted in support of an application for another degree or qualification of this or any other university or other institute of learning. 6

7 O. F. Selim Copyright i. Copyright in text of this dissertation rests with the author. Copies (by any process) either in full, or of extracts, may be made only in accordance with instructions given by the author. Details may be obtained from the appropriate Graduate Office. This page must form part of any such copies made. Further copies (by any process) of copies made in accordance with such instructions may not be made without the permission (in writing) of the author. ii. The ownership of any intellectual property rights which may be described in this dissertation is vested in the University of Manchester, subject to any prior agreement to the contrary, and may not be made available for use by third parties without the written permission of the University, which will prescribe the terms and conditions of any such agreement. iii. Further information on the conditions under which disclosures and exploitation may take place is available from the Head of the School of Mathematics. 7

8 Acknowledgements I would like to thank Jeff Paris for agreeing to supervise this dissertation and for all his help, comments and corrections. Many thanks to Mirna Džamonja for her enthusiasm in suggesting this topic and for kindly going out of her way to provide me with some unpublished material. I would also like to thank Mark Henrion and acknowledge his help with some difficult translations. 8

9 O. F. Selim 1 Introduction In 1947 D. Maharam provided three necessary and sufficient conditions for a complete Boolean algebra to carry a strictly positive countably additive measure. 1 The third of these conditions was a strengthening of the second and so it was natural to ask, as did Maharam, whether just the first two would alone guarantee the existence of such a measure. This became known as Maharam s problem, also known as the control measure problem, and it was only in 2006 that it was settled in the negative by M. Talagrand. For the experts it seems that the importance of this classical problem of measure theory is in the many different forms it assumes. Moreover, the anticipated negative solution of Maharam s problem has provided what Talagrand himself has described as radically new Boolean algebras. According to D. Fremlin in [12], such algebras are of particular interest in themselves and may also find a wider application in mathematical logic and in particular set theory. For the student the very multidisciplinary nature in dealing with Maharam s problem provides a very fruitful exercise in many different mathematical disciplines, for example in algebra, general topology, measure theory and set theory. The aim of this thesis is to present, in detail, Maharam s original investigations which led to this outstanding question and also to provide a full account of Talagrand s construction which eventually solved Maharam s problem. We aim to give a self contained presentation of these two pieces of work that is also user friendly to the student. On Talagrand s construction there is of course Talagarand s original publication ([33]). Fremlin has also given his account in [13]. However, perhaps understandably, neither of these references is presented in a way suitable for someone beginning in the subject; many definitions and notions are not fully explained and many of the arguments are left as obvious loosing the less experienced reader. On Maharam s work and Maharam s problem in general, the literature is much more accommodating. In particular Fremlin [12] provides a very full and crystal clear account of the myriad forms equivalent to Maharam s original question. Also a survey of a related problem, posed by John von Neumann in 1937, by Balcar and Jech ([4]) has a very good discussion on Maharam s problem and also on Maharam s original work in 1947, albeit in considerably less detail than [12]. Although [12] is very well presented and is undoubtable self contained, there is little direct reference to Maharam s original work, and though all of the material is surely present in [12], it is lost amongst the other topics presented in this giant volume. As for [4], the discussion on Maharam s original work provides only a sketch of the results and is lacking in any proper detail. 1 See appendix A for definitions. 9

10 1 Introduction We may divide this dissertation into three main sections. Sections 2 and 4 describe Maharam s and Talagrand s work, in that order, and section 3 aims to bridge the gap between Maharam s original question and the form in which it was eventually solved. We note that this bridge is given a full, thorough and very clear treatment in [12] and so in this section we excuse ourselves from providing full and complete details. 2 It is difficult to gauge what theory is to be assumed of the reader and what should form part of the main discussion. For this reason we have, as a rule of thumb, relegated all the theory that we feel is not specific to Maharam s problem to appendix A. There the reader may find all the auxiliary theory that will be needed. Since one of our main objectives is to provide a self contained presentation, appendix A is just as much a part of this thesis as the main text but never the less should (and will) be treated only as a toolbox. We have provided an index of definitions and symbols on page 157. This may be referred to if ever the reader feels lost in the absence of a definition which should, we hope, appear somewhere in this dissertation. At times it has seemed easier to justify some theory ourselves rather than search for a direct reference. Where we have unashamedly lifted from a reference we have done our best to bring this to your attention. Any unreferenced material is a result of our own efforts. 2 Note that Fremlin has made [12] available for free. 10

11 O. F. Selim 2 Maharam Submeasures and Maharam s Problem The aim of this section is to present the main ideas of Maharam s 1947 paper An Algebraic Characterization of Measure Algebras ([24]) and introduce Maharam s problem in its original context. Maharam s splits naturally into two parts. The first, which we deal with in (sub)section 2.1, is concerned with the actual characterisation given by Maharam of those Boolean algebras that carry a strictly positive countably additive measure and it is here that we will formulate and motivate Maharam s problem. The second part, dealt with in section 2.2, concerns itself with a question posed in 1937 by John von Neumann who asked whether or not every complete weakly distributive ccc Boolean algebra carries a strictly positive countably additive measure. Maharam investigated the relationship between von Neumann s question and the famous Souslin s hypothesis. The results of [24] were concerned with atomless Boolean algebras but it was remarked at the beginning of [24] that this restriction is not necessary; generalising Maharam s results to arbitrary (complete) Boolean algebras is dealt with in section A First Formulation The main objective of [24] was to find purely algebraic, necessary and sufficient conditions under which a σ-complete atomless 3 Boolean algebra carries a strictly positive countably additive measure. The strategy employed to arrive at this characterisation was to first give a topological characterisation of those Boolean algebras that carry a weaker type of measure; a strictly positive Maharam submeasure. Having done so these topological conditions are translated into algebraic or combinatorial ones and then, in turn, strengthened to give a full characterisation of those Boolean algebras that carry a strictly positive countably additive measure. Let us first give some definitions. Throughout this subsection fix a σ-complete atomless Boolean algebra B = (X, +,, 0, 1) but note that the following definitions apply also to σ-complete Boolean algebras that are not necessarily atomless. Definition ([24, pages ]) We say that a sequence (x n ) n N in B converges strongly to a point x B if and only if lim sup(x n + x) = 0. In this case we write x n s x. A strictly positive submeasure μ on B is a continuous outer measure if and 3 Maharam used the term non-atomic which could possibly be confused with not atomic, however, it is clear from the proof of Theorem 4 in [24] that we must assume that the Boolean algebra in question contains no atoms. See footnote 8. 11

12 2 Maharam Submeasures and Maharam s Problem only if for every sequence (x n ) n N in B and point x B we have x n s x μ(x n ) μ(x), where of course corresponds to convergence in (R, ). A submeasure μ on B is a Maharam submeasure if and only if lim n μ(x n ) = 0 for every decreasing sequence (x n ) n N in B with inf n x n = 0. Let us immediately state some properties of strong convergence that we shall need throughout this section ([24, page 155]). (E.1) If a sequence (x n ) n N is decreasing (increasing) then it is strongly convergent to inf n N x n (sup n N x n ). (E.2) If x n s x then any subsequence of (x n ) n N converges strongly to x. (E.3) If x n s x and y n y then x n + y n s x + y and x n y n x y. (E.4) If (x n ) n N is a disjoint sequence then x n s 0. On a σ-complete Boolean algebra the notions of a continuous outer measure and a strictly positive Maharam submeasure (strictly positive as a submeasure) are equivalent. Lemma ([24, page 156]) μ : B R is a continuous outer measure if and only if it is a strictly positive Maharam submeasure. Proof (Sketch). That every continuous outer measure is a strictly positive Maharam submeasure follows immediately by (E.1). For the converse let μ be a strictly positive Maharam submeasure and let (x n ) n N be a sequence in B such that x n s x B. Let y n = sup k n x n and z n = inf k n x n. Then sup n N z n = x = inf n N y n so that μ(y n ) μ(x) and μ(z n ) μ(x). Since z n x n y n we have μ(z n ) μ(x n ) μ(y n ). By taking the limit as n we see that μ(x n ) μ(x) as required. The following observations about Maharam submeasures will be useful. (M.1) Every strictly positive Maharam submeasure μ : B R is countably subadditive, that is, if (x n ) n N is a sequence in B then μ(sup x n ) μ(x n ). (2.1) n n N Indeed by setting u n = sup{x 1, x 2,..., x n } it is easy to see that (u n ) n N is an increasing sequence and sup n x n = sup n u n. By (E.1) u n sup n x n so that μ(sup n x n ) = lim n μ(u n ) lim n n l=1 μ(x l) = l=1 μ(x l) as required. 12

13 O. F. Selim (M.2) Every strictly positive countably additive measure is a strictly positive Maharam submeasure. (M.3) If B carries a Maharam submeasure then B is ccc ([4, page 245]). Let X B \ {0} such that for all x, y B, x y = 0 if x y. For each x X let F (x) = {y X : μ(y) μ(x) > 0}. We claim that each F (x) is finite. Indeed, say F (x) is infinite for some x X. Let {x 1, x 2,...} F (x) such that x i x j for all i, j N with i j. Let y n = sup m n x n. By (E.4) we have inf n y n = 0 and clearly y n+1 y n always. Thus μ(y n ) 0 so that for n large enough μ(x n ) μ(y n ) < μ(x), contradicting x n F (x). Thus F (x) is always finite. Since clearly x F (x) always, we have X = x X F (x), which is of course countable. With the above terminology in place we can immediately state Maharam s problem. At the very end of [24] Maharam asks...does the existence of a continuous outer measure on E [B] imply the existence of a [strictly positive countably additive] measure on E? Thus Maharam s problem asks Is (P) 1 true? where we are considering the statement (P) 1 Every σ-complete atomless Boolean algebra that carries a strictly positive Maharam submeasure carries a strictly positive countably additive measure. To motivate this question let us begin to investigate [24]. We first consider the topological properties of those algebras carrying a strictly positive Maharam submeasure. The topology under investigation will be provided by the following operator. Definition ([24, page 155]) Let C : P(X) P(X) be the operator defined by x C(Y ) Φ(x, Y ), (2.2) where Φ(x, Y ) reads There exists a sequence (x n ) n N in Y such that x n s x. 4 Also 4 We have chosen to describe the topological concepts here via closure operators since this is closer to Maharam s presentation. It is interesting to note however that in a more general setting s actually defines a -convergence. This leads to some interesting consequences. For example, as a result of this -convergence, there must exist a unique topological space whose closure coincides with the operator C and that if x n s x then x n converges to x with respect to this unique topology. For a slightly fuller account of these ideas see [12, 3A3P, page 640]. 13

14 2 Maharam Submeasures and Maharam s Problem let T C = {X \ Y : Y X and C(Y ) = Y }. Note that (X, C) is a general topological space (definition A.2.1) and (X, T C ) is a topological space proper (definition A.2.10) according to the definitions of section A.2.1. Note also that C satisfies conditions (K1), (K2) and (K3) of a closure operator (definition A.2.9). so that if C satisfied (K4) then (X, C) would be a topological space. The notion of strong convergence and that of convergence in the topology T C are related by the following crucial observations ([4, page 247]). (O.1) For any sequence (x n ) n N in X, if x n s x then x n x with respect to T C. Assume 5 for a contradiction that x n s x but that x n x. By definition there must exist a set Y T C (open) such that x Y but {n N : x n Y } is infinite. Thus there exists a subsequence (x nk ) k N in the closed set X \ Y. Now x nk s x by (E.2) so that x C(X \ Y ) = X \ Y and this is a contradiction since x Y. (O.2) If (X, C) is metrisable with metric ρ then for any sequence (x n ) n N in X, ρ(x n, x) 0 only if there exists a subsequence (x nk ) k N such that x nk s x. If x = x n infinitely often in n then we are done for we can choose the subsequence x nk = x s x. Otherwise let N be such that x n x for all n N and let y n = x N+n for each n N. Of course we still have ρ(y n, x) 0. It follows from (O.1) and lemma A.2.10 that C({y n : n N}) Cl ρ ({y n : n N}). By lemma A.2.12, Cl ρ ({y n : n N}) = {y n : n N} {x} so that if x C({y n : n N}) {y n : n N} then {y n : n N} C({y n : n N}) {y n : n N} and so {y n : n N} is a fixed point of C and so is a closed set. By lemma A.2.11, Cl ρ ({y n : n N}) = {y n : n N} x which is a contradiction. Hence x C({y n : n N}). Thus there exists a function F : N N such that y F (n) s x as n. Note that F satisfies n N m > n(f (m) > F (n)), for given any n N the set {k : F (k) F (n)} must be finite and so we may choose m = max{k : F (k) F (n)} + n + 1. Indeed if {k : F (k) F (n)} were infinite then F (k) = u N infinitely often in k and so the constant subsequence (z n = y u ) n N would converge strongly to y u x contradicting (E.2). Now let n 1 = F (1). If n k = F (j) has been chosen let n k+1 = F (l) for any l > j such that F (l) > F (j). The sequence (x nk ) k N is then a subsequence of both (y F (n) ) n N and (y n ) n N. Finally (x nk ) k N, as a subsequence of (y n ) n N, is also a subsequence 5 This proof is taken from ([12, 3A3Pa(ii), pages ]). 14

15 O. F. Selim of (x n ) n N and as a subsequence of (y F (n) ) n N we have x nk required. s x as k as It follows from these two observations that if (X, C) is metrisable with metric ρ, say, then the operators Cl ρ and C coincide. Indeed we have already seen that C(Y ) Cl ρ (Y ). Conversely if x Cl ρ (Y ), by lemma A.2.10, there exists a sequence in Y such that ρ(x n, x) 0. By (O.2) we may find a subsequence (x nk ) k N such that x nk x and so by definition x C(Y ). Definition ([24, page 159]) Y B is of type (C) if and only if 0 C(Y ). The following lemma goes some way to justifying the introduction of type (C) sets; in the correct context they are the compliments of neighbourhoods of 0. Lemma ([24]) If (X, C) is a topological space then Y X is of type (C) if and only if X \ Y is a neighbourhood of 0. Proof. Let Y X be of type (C). Since Y C(Y ) 0, 0 X \ C(Y ) X \ Y. But C is a closure operator so C(Y ) is closed (a fixed point of C) so that X \ C(Y ) is open and X \ Y is therefore a neighbourhood of 0. Now assume that Y is a neighbourhood of 0 and, for a contradiction, that X \ Y is not of type (C). By definition there must exists a sequence (x n ) n N in X \ Y such that x n s x. By (O.1) x n x with respect to the topology T C. Let Y Y be open such that 0 Y. Since x n x, Y must contain all but finitely many of the x n which is a contradiction since Y (X \Y ) =. The following theorem gives is the first step towards the intermediate topological characterisation previously mentioned of those Boolean algebras carrying a continuous outer measure. Theorem ([24, page 157]) B carries a strictly positive Maharam submeasure if and only if (X, C) is metrisable. Proof. For the only if direction assuming the existence of a strictly positive Maharam submeasure μ on B, we obtain a metric by setting ρ(x, y) = μ(x + y). Maybe the only difficulty in seeing that μ actually is a metric is the triangle inequality. However this easily follows from the fact that x, y, z B(x + y (x + z) (z + y)). With this metric in place we claim that for each Y X we have Φ(x, Y ) ρ(x, Y ) = inf{ρ(x, y) : y Y } = 0. (2.3) Indeed if Φ(x, Y ) holds then by definition there exists a sequence (x n ) n N in Y such that x n s x. Hence by (E.3) x n + x s 0 and since μ is a continuous outer measure 15

16 2 Maharam Submeasures and Maharam s Problem this implies that ρ(x n, x) = μ(x n + x) μ(0) = 0. It follows by lemma A.2.10 that x ρ(x, Y ). If ρ(x, Y ) = 0 then again by lemma A.2.10 there exists a sequence (x n ) n N in Y such that ρ(x n, x) 0. Thus we may find a subsequence (x nk ) k N such that ρ(x nk, x) < 2 k always. Write y k = x nk μ(z k ) (M.1) lim l l i=k μ(y i + x) < lim l and let z k = sup l k (y l + x). We have l i=k 2 i = lim l 2 k (2 2 k l ) 2 1 k 0 as k. (2.4) Now (z k ) k N is a decreasing sequence and so by (E.1) z k s inf k z k. Hence μ(z k ) μ(inf k z k ) and so by (2.4) and the uniqueness of limits in (R, ), μ(inf k z k ) = 0. However μ(inf k z k ) = 0 if and only if inf k z k = 0 and so in fact z k s 0. By definition lim sup k z k = 0 and since for each k N we have y k + x sup l k (y l + x) = z k we must also have lim sup k (y k + x) lim sup k z k = 0. Thus y k + x s 0. By (E.3) y k s x so that x Φ(x, Y ) as required and so proving (2.3). By (2.3) the operator C coincides with the closure operator Cl ρ. Hence T C = {A X : C(A) = A} = {A X : C ρ (A) = A} so that by lemma A.2.11 (X, C) is metrisable. For the if direction of theorem assume that (X, C) is metrisable with corresponding metric ρ so that (X, C) is a metrisable topological space. Let s show that (X, T C, +) is a topological group. Indeed, since x + x = 0 for any Boolean algebra, x = x and the map X X : x x is in fact the identity map which is clearly continuous. By theorem A.2.3 and lemma A is continuous if and only if given an arbitrary sequence (x n, y n ) n N in X X that tends to (x, y) with respect to the product topology of X X we have x n + y n x + y with respect to T C. By corollary A.2.2 (x n, y n ) (x, y) if and only if x n x and y n y with respect to T C. Assume then for a contradiction that there exists two sequences (x n ) n N and (y n ) n N, in X, such that ρ(x n, x) 0 and ρ(y n, y) 0 but ρ(x n + y n, x + y) 0. It follows that we may find an ɛ > 0 such that ρ(x n + y n, x + y) > ɛ (2.5) infinitely often in n. Since ρ(x n, x) 0 by (O.2) we may find a subsequence (x nk ) k N such that x nk s x. Since we still have ρ(y nk, y) 0 as k we may find a (sub)subsequence (y nkl ) l N such that y nkl s y as l and x nkl s x still. Hence by (E.3), x nkl + y nkl s x + y and so ρ(x nkl + y nkl, x + y) 0 by (O.1). Thus for all l large enough ρ(x nkl + y nkl, x + y) < ɛ which contradicts (2.5). The operation + must therefore be continuous and so (X, T C, +) is indeed a topological group. Since + 16

17 O. F. Selim is clearly commutative (X, T C, +) must be a metrisable Abelian topological group. By corollary A.2.5 there exists an invariant metric ˆρ defined on X. The X-valued function defined by μ(x) = sup{min{ˆρ(y, 0), 1} : y X, y x} is then the desired Maharam submeasure. 6 The key observation now is that if (X, C) is metrisable with metric ρ, say, then the following must be true. (C.1) (X, C) is a topological space. Indeed by (O.1) and (O.2) above, the operators C and Cl ρ coincide and Cl ρ is a closure operator. (C.2) (X, C) has a countable local base at 0. Since (X, C) is metrisable, (X, T C ) is a metrisable topological space, by lemma A.2.7, it must be first countable and a countable local base at 0 is guaranteed. Note that the only extra information we would obtain from (C.1) is that C satisfies (K4) of a closure operator. In fact not only are these conditions necessary for the existence of a continuous outer measure on B, they are also sufficient and it will be these two conditions that will interpret algebraically. Proposition ([24, pages ]) (X, C) is a topological space with a countable local base at 0 if and only if there exists a strictly positive Maharam submeasure on B. Proof. The if direction is taken care of by theorem 2.1.1, (C.1) and (C.2). For the only if direction assume that (X, C) is a topological space with a countable local base (U n ) n N at 0. Note that we may assume U n+1 U n (2.6) for each n N ([19, page 138]). Indeed if not then we let U n = U 1 U 2...U n. Clearly the U n satisfy (2.6) and that they also form a countable local base at 0. Let 6 That this is indeed a Maharam submeasure is not so difficult but not so straightforward. Maharam first proves a technical result asserting that: If there exists a F : B R 0 such that for all x, y B and decreasing sequences (x n ) n N in B with inf x n = 0 we have F (x) = 0 x = 0, F (x+y) F (x)+ F (y) and F (x n ) 0 then there exists a continuous outer measure on B. Given such a function one may assume that it is bounded for we can just consider the function min{f (x), 1} which also satisfies the above properties. The continuous outer measure is then given by μ(x) = sup{f (y) : y B, y x}. Maharam shows that μ is a Maharam submeasure and then induces its equivalence with a continuous outer measure. See [24, Lemma 2, page 156]. For our purposes, F (x) = ˆρ(x, 0) clearly satisfies the conditions of this result which leads to the desired submeasure. 17

18 2 Maharam Submeasures and Maharam s Problem C n = X \ U n for each n N. It is enough to show that (X, C, +) is a topological group for then, by lemma A.2.13, (X, C) would be first countable and a first countable topological group (X, C, +) is metrisable by theorem A.2.7. By theorem (X, C) would admit a continuous outer measure as required. To this end, since x x is the identity map, it is enough to show that + is continuous. Let (x, y) X X and Y be a neighbourhood of x + y. By lemma A.2.1 it is enough to show that there exists a neighbourhood L X X of (x, y) such that {u + v : (u, v) L} Y. Z = Y + x + y is a neighbourhood of 0. Let Y Y be open and such that x+y Y. Let z C(X \(Y +x+y)) so that there exists a sequence (z n ) n N in X \(Y +x+y) such that z n s z. We will show that z X \(Y +x+y). By (E.3) z n +x+y s z +x+y. Now if z n +x+y Y then z n Y + x + y which is a contradiction so z n + x + y X \ Y always. Thus z + x + y C(X \ Y ) = X \ Y, since Y is open. Since z + x + y X \ Y we must have z X \ (Y + y + x). Thus C(X \ (Y + x + y)) = X \ (Y + x + y) and so X \ (Y + x + y) is closed. Since 0 = (x + y) + (x + y) Y + x + y Y + x + y, Y + x + y is a neighbourhood of 0. U n + U n Z for some n N Otherwise U n + U n X \ Z for each n N and so we can find two sequences in X, (x n ) n N and (y n ) n N, such that x n, y n U n and x n + y n Z. Now let U be any neighbourhood of 0 and let n be such that U n U. By (2.6), we must have x k, y k U for each k n. This shows that x n 0 and y n 0 with respect to the topology T C. If {n N : x n = 0} < then let N = max{n N : x n = 0}. The sequence (x N+n ) n N still converges to 0 but does not contain 0 amongst its members. Now since x n 0, each open set of 0 contains a member of {x N+n : n N} so that 0 is an accumulation point of {x N+n : n N}. By corollary A.2.1, 0 C({x N+n : n N}). Thus there exists a function F : N N >N such that (x F (n) ) n N converges strongly to 0 as n. It is clear, since each x N+n 0, that we can choose F to satisfy, 7 n < m F (n) < F (m), so that (x F (n) ) n N is actually a subsequence of (x n ) n N. Now (y F (n) ) n N is a subsequence of (y n ) n N and since y n 0 we must also have y F (n) 0 as n. 7 One may argue as in the proof of (O.2). 18

19 O. F. Selim If {n N : y F (n) = 0} < then let N = max{n N : y F (n) 0}. The sequence y F (n)+f (N ) still converges to 0 with respect to T C as n and so as above 0 C({y F (n)+f (N ) : n N}). Thus there exists a subsubsequence (y F (nk )) k N that converges strongly to 0. Since the sequence (x F (nk )) k N also converges strongly to 0 we have x F (nk ) + y F (nk ) s 0 as k. But then x F (nk ) + y F (nk ) X \ Z always and so 0 C(X \ Z), which is a contradiction. If {n N : y F (n) 0} = then the sequence 0, 0, 0,... is a subsequence of (y F (n) ) n N and we proceed as above. If {n N : x n = 0} = then the sequence 0, 0, 0,... is a subsequence of (x n ) n N and we proceed as above. Finally let n N be such that U n +U n Y +x+y and set Y 1 = U n +x and Y 2 = U n +y. A similar argument to the one used to show that Z is a neighbourhood of 0 can be used to show that Y 1 and Y 2 must neighbourhoods of x and y, respectively. Moreover, Y 1 + Y 2 = U n + U n + x + y Y. By taking L = L 1 L 2 where the L i Y i are open and x L 1 and y L 2, we complete the proof that + is continuous. The last proposition concludes the topological characterisation of those algebras that carry a continuous outer measure. Consider the following two conditions. (I) For each sequence (x m,n ) m,n N of B such that for each m N the sequence (x m,n ) n N is decreasing with inf n x m,n = 0, there exists a function F : N N N such that inf sup i N m x m,f (i,m) = 0. (II) There exists a sequence (C n ) n N of subsets of B, each of type (C), such that if X B is countable and of type (C) then X C n for some n N. We now set about interpreting the conditions (C.1) and (C.2) into the algebraic properties (I) and (II). Let us start with (C.1) and (I). Theorem ([24, page 158]) (I) holds if and only if (X, C) is topological space. Proof. Assume (I) holds and let us show that C is a closure operator. Since C already satisfies (K1), (K2) and (K3) of a closure operator it only remains to show that that C satisfies (K4). By (K1), C(Y ) C(C(Y )) for any Y X. For the opposite inclusion let x C(C(Y )) so that there exists a sequence (x n ) n N in C(Y ) such that x n s x. Since each x n C(Y ), there exists for each n N a sequence (x n,m ) m N in Y such 19

20 2 Maharam Submeasures and Maharam s Problem that x n,m s x n as m. Now clearly the sequence y n,m = sup k m (x n,k + x n ) is decreasing in m and since x n,m s x n as m, inf m y n,m = inf m sup k m (x n,k + x n ) = lim sup m (x n,m + x n ) = 0. By (I) there exists an F : N N N such that inf i N sup n y n,f (i,n) = 0. Define a new G : N N N as follows. For each n N we let G(1, n) = F (1, n) and for each k N we set G(k + 1, n) = G(k, n) + F (k + 1, n). Note that F (k, n) G(k, n) G(k + 1, n) always so that y n,g(i,n) y n,f (i,n) for all i, n N and therefore inf i sup n y n,g(i,n) inf i sup n y n,f (i,n) = 0 so that we still have inf i sup y n,g(i,n) = 0. (2.7) n Now for any integers m n we have x n,g(n,n) + x n y n,g(n,n) y n,g(m,n), where the first inequality follows by definition and the second since G is increasing in its first variable. Thus sup n m x n,g(n,n) + x n sup n m y n,g(m,n) sup n y n,g(m,n), where this last inequality is obvious. It follows that lim sup x m,g(m,m) + x m = inf sup x n,g(n,n) + x n inf m m n m m sup n y n,g(m,n) (2.7) = 0, so that x m,g(m,m) +x m s 0. Now x m s x so that x m,g(m,m) = x m,g(m,m) +x m +x m s 0 + x = x. Since each x m,g(m,m) Y we indeed have x C(Y ) as required. Assume that (X, C) is a topological space. We want to show that (I) holds so let (x m,n ) m,n N be a decreasing sequence in B with respect to n and such that inf n x m,n = 0. For each n N set y m,n = sup i m x i,n. Clearly y m,n is decreasing with respect to n and increasing with respect to m. Note that by a repeated application of (D.5) of lemma A.3.6. for each m N we have inf n y m,n = sup i m inf n x i,n = 0 so that, since y m,n is decreasing with respect to n, we have y m,n s 0 as n. (2.8) Let (z n ) n N be a sequence in B such that z n+1 < z n and let z = inf n z n. Such a sequence exists because B is atomless. Set w n = z n + z for each n N. Clearly 0 < w m+1 < w m always and since z m s z we have w m = z m + z s z + z = 0. (2.9) Set z m,n = w m y m,n and let Z = {z m,n : m, n N}. By (2.8) and (E.3) we have z m,n = w m y m,n s w m 0 = w m as n. 20

21 O. F. Selim Thus for each m N we have w m C(Z) and so, by (2.9), 0 C(C(Z)) (K4) = C(Z). By definition there must exist a sequence (u n ) n N in Z such that u n s 0. (2.10) Let G, H : N N be defined by G(n) = m and H(n) = n if and only if u n = z m,n so that u n = z G(n),H(n). G satisfies n N m > n(g(m) > G(n)). We show that given any n N the set {k : G(k) G(n)} must be finite and so we can always choose m = max{k : G(k) G(n)} + n + 1. Indeed if for some n N, {k : G(k) G(n)} was not finite then we could find a strictly increasing sequence (n k ) k N in N such that G(n k ) = m N always. Now if for some k N, {l : H(n l ) H(n k )} were finite then we could find a subsequence (n kj ) j N such that H(n kj ) = p N so that z G(nkj ),H(n kj ) = z m,p s z m,p 0 as j contradicting (2.10). Thus {l : H(n l ) H(n k )} is always finite and by choosing, for each k N, q = max{l : H(n l ) H(n k )} + k + 1 we see that k N q > k(h(n q ) > H(n k )). Thus we can find a subsequence (n kj ) j N such that i < j H(n ki ) < H(n kj ) and so (z G(nkj ),H(n kj ) = z m,h(nkj )) j N is a subsequence of both (z m,n ) n N and (u n ) n N. As a subsequence of (z m,n ) n N we have z m,h(nkj ) s w m > 0 as j and as a subsequence of (u n ) n N we have z m,h(nkj ) s 0 j which is a contradiction. Thus {k : G(k) G(n)} is always finite. Now let G (1) = G(1). If G (k) = G(j) has been chosen let G (k + 1) = G(l) for any l > j such that G(l) > G(j). Let H (n) = H(l) where l N is such that G (n) = G(l). Now set F (i, m) = H (m) + i. We claim that inf sup i N m x m,f (i,m) = 0. (2.11) To see this first note that G (i) i always so that z G (i),h (i) y G (i),h (i) y i,h (i). Taking lim sup i across this last inequality and recalling that u n s 0 so that (z G (i),h (i)) i N as a subsequence of (u n ) also converges strongly to 0 we get that lim sup i y i,h (i) = 0 or equivalently y i,h (i) s 0. Recalling the distributivity laws of lemma A.3.5 and lemma 21

22 2 Maharam Submeasures and Maharam s Problem A.3.6we have inf k sup y m,h (m)+k = inf m k = inf k (D.1) = inf k (D.2) = sup h (D.4) = sup h = sup h ( ) 1 = sup h sup y m,h (m)+k (1 \ 0) m sup m sup m (inf k y m,h (m)+k (1 \ inf h y m,h (m)+k (sup h sup m sup y i,h (i)) i h 1 \ sup y i,h (i)) i h y m,h (m)+k 1 \ sup y i,h (i)) i h inf ((sup y m,h k (m)+k sup m<h m h inf k ((sup m<h inf k (sup m<h y m,h (m)+k 1 \ sup i h y m,h (m)+k 1 \ sup y i,h (i)) i h y m,h (m)+k) 1 \ sup y i,h (i)) i h y i,h (i)) (sup m h y m,h (m)+k 1 \ sup i h y i,h (i))) sup inf (sup y m,h h k (m)+k) ( ) 2 m<h sup h inf (sup y m,k ) ( ) 3 k m<h sup h inf k y h,k = 0. Here ( ) 1 follows since sup m h y m,h (m)+k 1 \ sup i h y i,h (i) = 0. Indeed we know that y m,h (m)+k y m,h (m) so that sup m h y m,h (m)+k sup m h y m,k from which ( ) 1 follows. ( ) 2 follows since y m,n is decreasing in n and ( ) 3 follows since y m,n is increasing in m. The final equality follows since with inf n y m,n = 0. Finally since x m,n y m,n always we have inf sup i N m x m,f (i,m) inf sup i N m Thus (2.11) is proved and so B satisfies (I). y m,f (i,m) = inf sup i N m x m,h (m)+i = 0. In the context of a topological space (X, C), the existence of a countable local base at 0 is equivalent to condition (II). Theorem ([24, page 160]) If (X, C) is a topological space then condition (II) holds if and only if there exists a countable local base at 0. Proof. Assume (X, C) is a topological space. If (II) holds then any set (countable or not) Y of type (C) must be contained in some C n. For otherwise (X \ C n ) Y for each n N so that we can find a sequence (x n ) n N such that for each n N, x n C n and x n Y. The set {x n : n N} must be of type (C) as a subset of Y and is countable but is not strictly contained in any of the C n s, a contradiction. By lemma the sets U n = X \ C n are all neighbourhoods of 0. If U us another neighbourhood of 0 then by lemma again, X \ U is of type (C). But we have just shown that X \ U C n for some n N so that U n U. This shows that (U n ) n N is a countable local base 22

23 O. F. Selim at 0. Conversely if (U n ) n N is a countable local base at 0 then (X \U n ) n N satisfy (II). Our algebraic characterisation of those Boolean algebras that carry a strictly positive Maharam submeasure now follows. Theorem ([24, page 159]) There exists a continuous outer measure on B if and only if conditions (I) and (II) hold. Proof. Assume that there exits a continuous outer measure μ on B. By theorem (T, C) must be metrisable and therefore, as we have already seen, (X, C) must be a topological space. By theorem (I) must hold. Observe that a set Y is of type (C) if and only if δ > 0 x Y μ(x) > δ. (2.12) Indeed, for the if direction if there exists such a δ but 0 C(Y ) then there would exists a sequence (x n ) n N in Y such that x n s 0. Since μ is a continuous outer measure it follows that μ(x n ) 0 so that for n large enough μ(x n ) < δ which is a contradiction. For the only if direction assume Y is of type (C) but that no such δ exists. We can then find a sequence (x n ) n N such that μ(x n ) < 2 n. Let y n = sup l n x n. As in the only if direction of (2.3) it follows that μ(y n ) 0 and z n s inf n z n. Thus μ(z n ) μ(inf n z n ) also so that inf n z n = 0 by uniqueness of limits and the fact that μ is strictly positive. But lim sup n x n = inf n z n = 0 and so x n s 0. Thus 0 C(Y ) which is a contradiction. From (2.12) the sets C n = {x B : μ(x) n 1 } are of type (C) and that any other set of type (C) must be contained in one of these C n. Hence (II) holds. Finally assume that (I) and (II) hold. By theorems and (X, C) is a topological space that has a countable local base at 0. By proposition B admits a continuous outer measure as required. We have now shown that conditions (I) and (II) are equivalent to the existence of a continuous outer measure on B. By adding a third condition, a strengthening of (II), Maharam was able to provide a characterisation for those algebras carrying a strictly positive countably additive measure. More precisely Maharam proved the following theorem. Theorem ([24, page 161]) B carries a strictly positive countably additive measure if and only if it carries a Maharam submeasure and in addition (III) The sequence (C n ) n N of (II) satisfies C n + (X \ C n+1 ) C n+1 and x y C n for every disjoint pair x, y C n+1. 23

24 2 Maharam Submeasures and Maharam s Problem We will only sketch the proof of this result (below) for as far as Maharam s problem is concerned, it is conditions (I) and (II) that we are really interested in. Indeed by theorem Maharam s problem is asking whether or not conditions (I) and (II) are sufficient to provide a strictly positive countably additive measure on B. In light of theorem this is a very natural question, for we are asking whether or not the strengthening of (II) to (III) is actually necessary. Proof of Theorem (Sketch). Assume B carries a strictly positive countably additive measure μ. Then μ is also a continuous outer measure by (M.2). By theorem 2.1.4, (I) and (II) must hold and as in the proof of theorem 2.1.4, using (2.12), we may choose the sets C n = {x B : μ(x) 2 n } to satisfy (II). From this explicit definition of the C n it is straight forward to also see that (III) holds. Assume that (I), (II) and (III) hold and let (C n ) n N be the type (C) sets satisfying (II) and (III). Note that by theorems and 2.1.4, (X, C) is a metrisable topological space. Let V 0 = {0} and set for each n N, V 1 2 n = X \ C n. For n N, k {1, 2,..., 2 n 1 1} and l {2 n, 2 n + 1,...} let V 2k = V, V 2 n n 1 k 2k+1 = V + V 2 n 2k 2 n 1, V = X. 2 n l 2 n Let F = { k 2 n : k ω} and define f : X [0, 1] by f(x) = 2 { sup{r F : x V r }, if x 0 0, if x = 0 The function f is continuous (with respect to T C ) and satisfies, for each x, y X and N N such that 2 N f(1), the following. (F.1) f(x) = 0 if and only if x = 0 (F.2) f(x + y) f(x) + f(y) (F.3) If x y = 0 and f(x) + f(y) 2 N then f(x y) = f(x) + f(y). (F.4) If x y and a [f(x), f(y)] then there exists a z [x, y] such that f(z) = a. 8 8 (F.3) and (F.4) are not so easy. To prove (F.3) we need the intermediate result: Given x 0 and ɛ > 0 there exists a y x x such that 0 < f(y x ) < ɛ. This is proved using Maharam s non-atomic property (actually Maharam quotes lemma A.3.3 with atomic instead of atomless ). But given an x 0 and choosing ɛ = f(x) > 0 we see that if y x = x then f(y x ) = ɛ which is a contradiction. Since f(y x ) > 0, y x > 0 also. Thus given any x B \ {0} we must be able find a 0 < y x < x so that we must assume that B is atomless; not atomic is not enough. See [24, 6.7, 6.8, 6.9, page 162]. 24

25 O. F. Selim Call x X irreducible if and only if for each y < x, f(y) f(x). We further have (F.5) For each x X there exists an irreducible y x such that f(x) = f(y). Let N N be such that 2 N f(1) and let Z be a collection of mutually disjoint irreducible members of X such that f(z) 2 N 1 for each z Z and assume that Z is maximal in the sense that any other set of irreducible elements with these two properties does not strictly contain Z as a subset. By (M.3) Z is countable so that we may write Z = {z n : 1 n < α} for some α ω + 1. Let Then μ(x) [0, 1]. μ(x) = 1 n<α f(x z n ) 2 n. Since f(y) [0, 1] for each y X we clearly have μ(x) 0. Also μ(x) = f(x z n) 1 n<α 2 n 1 n<α 2 n n N 2 n = 1. μ(x) = 0 if and only if x = 0. The if direction follows by (F.1). For the only if direction we note that if x 0 then there exists a z Z such that z x 0. Indeed, if f(x) 2 N 1 then by (F.5) we may find an irreducible element y x such that f(y) 2 N 1. By maximality of Z there must exist some z Z such that z y so that z x. If f(x) > 2 N 1 then by (F.4) we may find 0 y x such that f(y) = 2 N 1. By (F.5) we may find an irreducible element y y such that f(y ) = 2 N 1. By maximality of Z there must exist a z Z such that z y and so z x as required. Now given any x X \ {0}, let z Z be such that z x 0 so that we have μ(x) f(x z)/2 n > 0, as required. If x y = then μ(x y) = μ(x) μ(y). Let z Z. We have (z x) (z y) =. Also since z is irreducible, f(z x) f(z) 2 N 1 and similarly for f(z y). Hence f(z x) + f(z y) 2 N so that by (F.3) f(z (x y)) = f((z x) (x y)) = f(z x) + f(z y). Since z Z was arbitrary the result easily follows. Thus μ is a measure on B. It remains to show that μ is countably additive. By letting f n (x) = f(x z n )/2 n, c n = 2 n and applying theorem A.2.5 (with g = μ) we see that μ is itself continuous. Now let (x n ) n N be disjoint collection in X, y n = n i=1 x i and y = i=1 x n. We have y n s y so that y n y with respect to the topology T C. By continuity of μ, μ(y n ) μ(y). Thus i=1 μ(x n) = lim n n i=1 μ(x n) = lim n μ(y n ) = μ(y) = μ( i=1 x i) as required. 25

26 2 Maharam Submeasures and Maharam s Problem 2.2 On Souslin s Hypothesis and von Neumann s Problem The Scottish Book ([26]) is a collection of approximately two hundred problems compiled between the years 1935 and During this time a group of mathematicians, which included the notable Stefan Banach, 9 frequently augmented by visitors met regularly to drink and talk mathematics in the Scottish Café in the then Polish, now Ukranian, town of Lwów. The Scottish Book was kept by the café waiters and produced on request where any troublesome mathematical problems were recorded. Problem 163 of the Scottish Book, dated July 4, 1937, was entered by John von Neumann who asked whether every complete weakly distributive ccc Boolean algebra necessarily carries a strictly positive countably additive measure. 10 The notion of weak distributivity (condition (V) below) is as yet undefined but we have seen it before. Lemma ([24, page 160], [4, page 251]) Let B be a σ-complete ccc Boolean algebra. Then the conditions (I) of theorem 2.1.4, (IV) and (V) are equivalent where (IV) For every double sequence (x m,n ) m,n N such that x m,n x m,n+1 and sup n x m,n = 1 we have sup f N N inf i N x i,f(i) = 1. (V) Weak distributivity. For every double sequence (x m,n ) m,n N such that x m,n x m,n+1 we have sup inf x i,f(i) = inf f N N i N i sup x i,j. j Proof. For the if direction assume that (IV) holds and let (x m,n ) n,m such that x m,n x m,n+1 and inf n x m,n = 0. We want to find a function F : N N N such that inf i N sup m x m,f (i,m) = 0. Let y m,n = x c m,n so that the sequence (y m,n ) m,n N satisfies the assumptions of (IV). 11 Thus sup f N N inf i N y i,f(i) = 1. Let Y = {inf i N y i,f(i) : f N N }. By lemma A.3.2 we may find a countable Y Y such that sup z Y z = sup z Y z = 1. Let Y = {z 1, z 2,...} if Y is infinite or {z 1, z 2,..., z N, z N+1 = z 1, z N+2 = z 1,...} if it isn t. Let g l N N be the f N N such that z l = inf i N y i,f(i) and set F (l, m) = g l (m). Clearly sup i N inf m y m,f (i,m) = 1 and by considering compliments we have inf i N sup m x m,f (i,m) = 0 as required. The only if direction follows easily once we note the general (V) fact that if Y Y are index 9 In fact according to [7] it was Bananch s wife who introduced the journal after scribbling on tables became unacceptable! 10 A bottle of whiskey of measure > 0 is offered as a prize for solving this problem; perhaps this gives us some indication of how valuable Talagrand s construction really is! See below. 11 We need lemma for this. 26

27 O. F. Selim sets and (x n ) n Y is a subsequence of B then sup n Y x n sup n Y x n. Considering the statement (V) Every complete ccc Boolean algebra that satisfies (V) carries a strictly positive countably additive measure. von Neumann asked Is (V) true? It is true that (V) implies (P) 1. To see this we need the following lemma. Lemma ([24, page 160]) If (II) holds then B is ccc. Proof. Indeed let Y X be such that 0 Y and x y = 0 for all x, y Y with x y. Since singletons are fixed points of C the sets {x} for each x Y are all of type (C) so that if (II) holds then Y n N C n. If Y C n were infinite for some n then C n would contain a sequence of mutually disjoint elements which, by (E.4), would converge strongly to 0, a contradiction. Hence Y C n is always finite and since Y = n N Y C n, Y must be countable. Now then assuming (V), let B be a σ-complete atomless Boolean algebra carrying a strictly positive Maharam submeasure. By theorem B satisfies condition (I) and (II). By lemma B is ccc and so by lemma B is weakly distributive. Also every σ-complete ccc Boolean algebra is complete by corollary A.3.1. Thus B is a complete ccc weakly distributive Boolean algebra and so by (V) must carry a strictly positive countably additive measure. Therefore we indeed have (V) (P) 1. (2.13) By showing that (P) 1 is false Talagrand not only settled Maharam s problem but also von Neumann s. In this section we show that well before Talagrand s result a positive solution to von Neumann s problem was already effectively ruled out by, amongst others, Maharam in [24]. Definition ([24, page 164]) Let S be a set of sets. We say S is a Souslin system if and only if S and the following conditions are satisfied. (S1) x, y S(x y or y x or x y = ). (S2) If T S and x, y T (x y x y = ) then T is countable. 27

28 2 Maharam Submeasures and Maharam s Problem (S3) If T S and x, y T (x y ) then T is countable. It is easy to see that if S is a Souslin system then any subset of S must also be a Souslin system. Souslin s hypothesis reads: (S) Every Souslin system is countable. We prove the following theorem given by Maharam in [24]. 12 Theorem ([24, page 164]) If (V) then (S). It is well known that the negation of Souslin s hypothesis is relatively consistent with the usual axioms of mathematics (ZFC). 13 This means that by arguing within these usual axioms it will not be possible to prove (S). Thus it then follows that one is not able to prove (V), for a proof of (V) would, by theorem 2.2.1, produce a proof of (S). Souslin s hypothesis is more commonly known as an assertion about the uniqueness of the real line. The set theoretical form given above is stated in terms of Souslin systems so let us take a moment to tie these ideas together. According to the literature 14 the following question was originally posed in 1920 by M. Souslin: Is every ccc complete dense unbounded linearly ordered set isomorphic to the real line? 15 Note that the real line is the unique (up to isomorphism) complete separable (dense) unbounded linearly ordered set. 16 The following became known as Souslin s hypothesis. (S) 1 Every ccc complete dense unbounded linearly ordered set is isomorphic to the real line. Maharam offered her version of Souslin s hypothesis (S) in [24] but it was only a year later in [25] that she explained that this equivalence is deducible from two papers, [27] 12 Actually Maharam proved the more general result (M) (S) where (M): Every σ-complete atomless ccc Boolean algebra that satisfies condition (I) carries a non-trivial outer measure. The proof however remains the same. See footnote See for example [8]. 14 See for example [8], [3]. 15 Of course this wasn t the terminology used! According to [3, pages ] Souslin asked: Is a linearly ordered set, with no jumps and no gaps, and such that any collection of non overlapping intervals is at most countable, necessarily an (ordinary) linear continuum?. Souslin s question was considered by W. Sierpiński, in a publication eight years later, equivilant to the above form. 16 See [18, page 38]. It is easy to see that in this context separability implies ccc (just choose a member of this countable subset from each interval) and so Souslin s question was in fact a question of equivalence between the properties of separability and ccc ([3, page 183]). 28

29 O. F. Selim and [9], the first by W.E Miller and the second by D. Dushnik in collaboration with Miller. 17 Following Maharam s hint in [25] let us consider these two references by Miller and Dushnik. In [27] Miller proves the following. Miller s Theorem: There exists a linearly ordered set that is unbounded complete dense ccc and not separable (and therefore not isomorphic to the real line) if and only if there exists a partially ordered set (X, ) such that the following conditions hold. (M1) X = ℵ 1 (M2) If Y X then x, y Y (x y or y x) or x, y Y (x y x = y) Y < ℵ (M3) x, y, z X(x z and y z x y or y x). Thus Miller showed that Souslin s hypothesis (in the form (S) 1 ) is false if and only if there exists a partial order that satisfies the above three properties. Let us call for the purpose of this discussion such a partial order a Miller system. With this theorem we are already starting to see traces of Maharam s form of Souslin s hypothesis, indeed if one could show that a Miller system exists if and only if an uncountable Souslin system exists (2.14) then it follows immediately by Miller s theorem that (S) 1 (S). To this end let us first consider the following result from [9]. 19 Dushnik and Miller s Theorem: Any partially ordered set (X, ) with the property (M4) x, y, z X(z x and z y x y or y x), is isomorphic to a partial order of sets 20 (Y, ) and such that (Y, ) satisfies (S1) of a Souslin system. Note that under such an identification condition (M 2) is equivalent to conditions (S2) and (S3) of a Souslin system. For (2.14) then assume that a Miller system (X, ) exists. Define a new relation R X X by xry y x. It is not difficult to see that (X, R) remains a partial order and moreover satisfies conditions (M1), (M2) and, since (X, ) satisfies (M3), also (M4). By Dushnik and Miller s theorem we may 17 In [3, page 225] C. Alvarez says: Certainly this version of Souslin s hypothesis was obtained by her [Maharam] from Miller s paper of In [25] Maharam comments on the equivalence of her form of Souslin s hypothesis first stated a year earlier in [24]: This [(S) (S) ] follows from [3] [[27]], together with some results in [l] [[9]]. 18 That is if all elements in Y are comparable or all distinct elements in Y are in-comparable then Y is countable. 19 See Theorem 3.71, page (X, ) is then called representable according to [9]. In fact this is a straight forward result; for the collection of sets one just takes for each x X the lower segment {y X : y x}. 29