1 Introduction. Math Sci Net numbers(msc2000) Primary = 06-02; Secondary = 06A05, 28A12, 54F05 draft of

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1 Well-ordered Sets, Ulam Matrices, and Probability Measures by Harold Bennett (Texas Tech University, Lubbock, TX 79409) and David Lutzer (College of William and Mary, Williamsburg, VA 23187) Abstract In this expository paper we show how well-ordered sets, stationary sets, and Ulam matrices were used to study the domains of non-atomic probability measures. Along the way we present some classical results on CUB, Borel, and stationary subsets of ω 1. Key words and phrases: well-ordered set, Ω, ω 1, CUB, stationary sets, Ulam matrices, probability measures Math Sci Net numbers(msc2000) Primary = 06-02; Secondary = 06A05, 28A12, 54F05 draft of Introduction At the beginning of the last century, Cantor s set theory was still a relatively new idea, and many mathematicians were engaged in a long-term project to re-cast classical parts of mathematics in a set-theoretic format. Their project was largely complete by mid-century, and along the way they developed new fields of mathematics and discovered new ideas and techniques that are part of the modern mathematical world-view. The purpose of this paper is to weave together some of these ideas from sixty-plus years ago in a way that, we believe, is understandable to many undergraduates and all graduate students. We make no claim about the novelty of the results in this paper. Most of them are scattered around various undergraduate and graduate textbooks. Our goal is to unify them to tell a good mathematical story. Readers who look in any modern set theory book (say [3] or [11]) will find that the set we call Ω in this article is now denoted by ω 1 and is the first of many uncountable ordinals, and that most of our results can be proved for sets that are much larger than Ω. But for beginners, there are more than enough mysteries and surprises in Ω. It would be helpful if readers know a few basic ideas from topology. A topology on a set X is a collection U of subsets of X such that, X U, and U is closed under finite intersections and under arbitrary unions. Sets that belong to U are called U-open sets (or simply open sets) and a set if closed provided its complement is open. A point p X is a limit point of a set S X provided U S has at least two points whenever p U U and it is easy to prove that a set is closed if and only if it contains all of its limit points. If U is a topology on X and Y X, then the relative topology or subspace topology on Y is the collection U Y := {U Y : U U}. Given two topological spaces (X, U) and (Z, V), a function f : X Z is continuous if f 1 [V ] U for each V V. Finally a function f : (X, U) (Z, V) is a homeomorphism if f is 1-1 and onto, and both f and f 1 are continuous. 1

2 In our paper, the set of real numbers is denoted by R while Z and N denote the set of all integers and the set of positive integers, respectively. 2 Well-ordered sets and their subsets A well-ordered set is a nonempty set W with a linear or total order that has the additional property that any non-empty set S W has a first element. Certainly the most familiar well-ordered set is {0, 1, 2, }, the set of non-negative integers. The fact that this set is well-ordered is exactly what makes traditional induction work. There are other well-ordered sets, e.g., the set {1 1 : n N} {2 2 : n N}, ordered n n using the usual ordering of R. Many much more complicated well-ordered sets can be constructed based on the usual ordering of R, but they all share one property they are small because R has a countable dense set, so we have: Lemma 2.1 Any subset of R that is well-ordered by the usual ordering of R must be countable. Could there be larger well-ordered sets? The answer is Yes, and the quickest way to get one is to use what Paul Halmos described as transfinite trickery [8]. What Halmos meant was any one of the statements in the next theorem. Theorem 2.2 The following statements are equivalent: a) the Axiom of Choice; b) Zorn s Lemma; c) Every set can be well-ordered. For a discussion of this theorem see [3], and for an exhaustive discussion see [13]. Statements (a) and (b) of Theorem 2.2 are the most frequently used versions today; what we will need is assertion (c). Warning: when one applies assertion (c) to a given set X, the resulting well-ordering probably has no relation to any ordering that might already exist on the set X. Today transfinite trickery has become part of almost every mathematician s set-theoretic toolkit. That toolkit is described as ZFC, meaning Zermelo-Fraenkel set theory with Choice. Corollary 2.3 In ZFC there is an uncountable well-ordered set (Ω, ) with the special property that for each α Ω, the set {β Ω : β α} is countable. Proof: Start with any uncountable set, say the set R of real numbers, and use Theorem 2.2 to find a well-ordering of it. (Warning: and the usual linear ordering of R are definitely not the same.) If (R, ) has the special property mentioned in the corollary, we let Ω be the entire set R with the new ordering, and the proof is done. If (R, ) does not have the desired special property, 2

3 then the set A := {α R : α has uncountably many predecessors}, and so A must have a first element α 1. Then the set Ω := {β R : β α 1 } will be the set that we want 1. No matter what uncountable set one begins with in the proof of Corollary 2.3, members of the set Ω are traditionally called ordinals, or ordinal numbers. The set Ω in Corollary 2.3 is non-empty, so it has a first element, traditionally called 0 (which might or might not be the additive identity of R). The set Ω {0} = so it has a first element, traditionally called 1. In this fashion, we see that our uncountable set Ω begins with the set {0, 1, 2, }, which is a copy of the non-negative integers. Because Ω is uncountable while {0, 1, 2, } is countable, the nonempty set Ω {0, 1, 2, } must have a first member, traditionally called ω. This element ω is fundamentally different from the elements 1, 2, 3,, each of which has an immediate predecessor in Ω, while ω does not. Any member of Ω that does not have an immediate predecessor is called a limit ordinal. There are many limit ordinals. Start with any α Ω. The set [0, α] := {β Ω : β α} is countable, so Ω [0, α] has a first element, traditionally called α + 1. Then Ω [0, α + 1] has a first element, usually called α + 2. By recursion, we define α + n for each n ω. The set B := {[0, α + n] : n ω} is countable, being the countable union of countable sets, so the set Ω B must have a first element, usually called α + ω, and that is a limit ordinal beyond α in the ordering. We say that the set of limit ordinals in Ω is cofinal, or unbounded, in Ω, meaning that beyond any α Ω lies some limit ordinal. In (Ω, ), infima and suprema are defined exactly as in the set of real numbers (R, <) with the usual ordering. But (Ω, ) is unlike R in many ways, starting with: Lemma 2.4 Suppose C is a non-empty countable subset of Ω. Then some δ Ω has δ = sup(c). Proof: For each γ C the set [0, γ) is countable, and therefore so is D = {[0, γ) : γ C}. Consequently Ω D has a first element, and that element is sup(c). We now define a topology on Ω. We say that a subset S Ω is closed if for each countable C S we have sup(c) S and we say that a set is open if its complement is closed. It is easy to check that the collection of all open subsets of Ω contains both Ω and, is closed under finite intersections, and also under arbitrary unions, so that the collection of all open sets (as defined above) is a topology on the set Ω. The most interesting closed subsets of Ω are the uncountable closed subsets of Ω, e.g., the set of all limit ordinals in Ω. It is easy to see that a subset S Ω is uncountable if and only if S is an unbounded, or cofinal, subset of Ω, and such closed sets are most frequently called CUB (for closed, unbounded) sets, or CUBs. The key reason that CUBs are so interesting is the following: Proposition 2.5 If {C(n) : n ω} is a countable family of CUB sets, then {C(n) : n ω} is also a CUB set. 1 One does not need the full force of the Axiom of Choice to construct Ω, and one does not need to start with an uncountable set. In his undergraduate honors thesis, Jacob Hill [10] showed how to construct (Ω, ) as the family of equivalence classes of countable well-ordered subsets of the set of rational numbers, using only the axiom of countable dependent choice. 3

4 Proof: It is easy to show that {C(n) : n ω} is a closed set. We use an interlacing argument to show that it is nonempty and cofinal in Ω. Fix any α Ω. We will choose an increasing sequence α β(1) β(2) β(3) of elements of Ω where β(n) is chosen from the n th entry in the following list: C(1), C(2); C(1), C(2), C(3); C(1), C(2), C(3), C(4); C(1),, C(5); C(1),, C(6); C(1),. Then the supremum γ of the sequence β(n), which exists by Lemma 2.4, will belong to each C(k) because infinitely may terms of the sequence β(n) belong to C(k). If you write down a subset of Ω as the set of all elements of Ω with a certain specified property, chances are that you will describe a subset of Ω that contains a CUB, or whose complement contains a CUB. One of the first really amazing results about Ω is that other kinds of subsets of Ω must exist. Mary Ellen Rudin gave the following elegant proof in her article [14]. Theorem 2.6 There is a subset S Ω such that neither S nor Ω S contains a CUB set. Proof: Suppose not. Then for every subset S Ω, either S or Ω S contains a CUB set. Recall that we constructed Ω inside of the set R. For each integer n 1 there is a collection J (n) of subsets of R satisfying a) J (n) is countable; b) J (n) = R; c) each J J (n) has diameter 1 n For example, we could let J (1) = {[n, n + 1] : n Z} and J (n) be the collection of closed intervals with consecutive points in {k, k + 1, k + 2, : k Z, n N} as endpoints. n n Fix n. We claim that some J J (n) contains a CUB set. If not, then for each J J (n), Ω J contains a CUB set C J (apply our supposition that each subset of Ω, or its complement, contains a CUB set to the set Ω J). By Proposition 2.5 the set D := {C J : J J (n)} is a CUB set. Consider any δ D. Then δ Ω R and yet δ does not belong to any J J (n), even though J (n) = R. That is impossible, and therefore our claim is established. Therefore, for each integer n 1 we may choose J n J (n) such that J n contains some CUB set D n. Once again applying Proposition 2.5, we see that the set E := {D n : n 1} is a CUB set, so we can choose distinct members α, β E. But then, for each integer n we have 0 < α β diam(j n ) 1 and that is impossible. Consequently, Theorem 2.6 is proved. n The set in Theorem 2.6 has a special property: even though it does not contain any CUB set, it has a non-empty intersection with every CUB set. Any set that intersects every CUB set is called a stationary subset of Ω. The set in Theorem 2.6 has a second property: its complement also intersects every CUB set. Any set S Ω with the property that both S and Ω S intersect every CUB set is called a bistationary set. Stationary and bistationary subsets of Ω are the most important types of subsets of Ω for two main reasons. The first shows that, in some sense, stationary sets in Ω are analogous to second category sets in the usual space of real numbers: 4

5 Corollary 2.7 If S Ω is stationary and S = {A n : n 1}, then some set A n is stationary. Proof: Otherwise for each n there would be a CUB set C n with A n C n =. By Proposition 2.5, the set D := {C n : n 1} is a CUB set. But D A n = for each n so that D S =, and that is impossible because S is stationary. The most important reason to think about stationary subsets is a result known as Fodor s Lemma or the Pressing Down Lemma. It is a central tool in logic and set theory. For a proof of Lemma 2.8 see [3], [11]. Lemma 2.8 Suppose S is a stationary subset of Ω and suppose f : S Ω satisfies f(α) α for each limit ordinal α S {0}. Then f is constant on a stationary subset, i.e., there is some β Ω and some stationary set T of Ω with T S and with f(α) = β for all α T. A function of the type mentioned in Lemma 2.8 is called a pressing down function. For many applications it is enough to know that a function f : Ω Ω with f(λ) λ for each non-zero limit ordinal λ will be constant on an uncountable set, and that result is easier to prove. We will illustrate the utility of Lemma 2.8 by giving a few of its topological consequences. Corollary 2.9 Suppose S is a stationary subset of Ω and suppose that f : S Ω is a continuous 2, one-to-one function. Then the set T = {f(α) : α S} is also a stationary subset of Ω. Proof: Suppose T is not stationary. Then there is a CUB set C with T Ω C. The set Ω C, like any open subset of Ω, breaks into a union of pairwise disjoint intervals of the form (α, β) := {γ Ω : α γ β}. Because C is cofinal in Ω, each such (α, β) will be countable, showing that T is contained in a pairwise disjoint union of countable open intervals (α, β) Ω. Let V be the collection of all of those pairwise disjoint countable open intervals that cover T. Let U := {f 1 [V ] : V V}. Then V is a pairwise disjoint collection of relatively open subsets of S, and because f is one-to-one, each member of U is countable. It is easy to prove that because S is a stationary subset of Ω, then so is the set S d consisting of all limit points of S, and we have S d S U. (Warning: S d is not the same as the set of limit ordinals that happen to belong to S.) For each α S d choose the unique member U α U with α U α. Because α is a limit point of S, there is a point g(α) α such that [g(α), α] S U α. Then g : S d Ω is a pressing-down function with the stationary set S d as its domain, and yet the fact that the sets U α are countable and pairwise disjoint guarantees that the function g cannot be constant on any uncountable set, contradicting Lemma 2.8. How many different types of stationary subsets of Ω exist? That depends on what same and different mean. Recall that two topological spaces X and Y are homeomorphic provided there is a 1-1 onto function h : X Y with the property that both h and h 1 are continuous. If S Ω is a stationary set that contains a CUB set and T is a stationary set that does not, then it is an easy matter to prove that S and T cannot be homeomorphic (because S has an uncountable subspace that contains the limit of each of its sequences, while T does not). But what if neither S nor T contain a CUB? In the most extreme case, what if S T = as in Theorem 2.6? Could such sets be homeomorphic to each other? 2 Recall that to say that f is continuous means that if O is an open subset of Ω, then f 1 [O] is a relatively open subset of S, i.e., that f 1 [O] = S G for some open subset G of Ω. 5

6 Proposition 2.10 Suppose S and T are stationary subsets of Ω and that S T is stationary (e.g., in case S T = ). Then there cannot be a continuous one-to-one mapping from S into T, so that S and T cannot be homeomorphic. Proof: For contradiction, suppose there is a continuous one-to-one mapping h : S T. Break S into three subsets A := {α S : h(α) α} B := {α S : h(α) = α} C := {α S : α h(α)}. Note that B S T T so that B (S T ) =. Therefore S T S = A B C gives S T A C. Corollary 2.7 yields that either A is stationary, or else C is stationary. If the set A were stationary, we would have a violation of Lemma 2.8 because h A is a one-to-one, pressing-down function. If the set C is stationary, then so is h[c], by Proposition 2.9. But then the function h 1 restricted to the stationary set h[c] would violate the Pressing Down Lemma (Lemma 2.8). The proof of Proposition 2.10 shows that the key to the existence of a homeomorphism between stationary sets S and T is the nature of S T. For a fuller discussion, see [4]. More recent work has extended Proposition 2.10 in surprising ways. For any topological space X, the set of continuous real-valued functions on X is denoted by C(X). There are many reasonable topologies that one might use for C(X), and one of them is the topology of pointwise convergence. In this topology, basic neighborhoods of a function g C(X) are specified by a finite set F X and a positive real number ɛ and have the form N(g, F, ɛ) := {h C(X) : g(x) h(x) < ɛ for all x F }; see [5]. We indicate that the pointwise convergence topology is being used by writing C p (X). This space C p (X) is usually not metrizable, but it is a locally convex topological vector space and its properties are determined by the topological properties of X. The next result is due to R. Buzyakova in [1]: Proposition 2.11 Suppose S and T are stationary sets in Ω such that S T is stationary. Then there cannot exist any continuous, one-to-one functions from C p (T ) into C p (S). Here is one last result that follows from Lemma 2.8. For each limit ordinal λ Ω, we can choose an increasing sequence α(λ, 1) α(λ, 2) α(λ, 3) whose supremum is λ. Perhaps surprising, this cannot be done in any uniform way, as the next result shows. We leave the proof to readers who want to exercise their Pressing Down skills. Corollary 2.12 Suppose for each non-zero limit ordinal λ Ω, we have a strictly increasing sequence α(λ, n) whose supremum is λ. It is not possible that α(λ, n) α(µ, n) for all non-zero limit ordinals λ, µ with λ µ and for all n ω. 6

7 3 Ulam matrices and probability measures Recall that a probability measure on a set X consists of two things, namely, a collection A of subsets of X and a function p : A [0, 1] satisfying (i) X A with p(x) = 1 ; and (ii) the collection A is a σ-algebra, i.e., A is closed under the formation of countable unions and complements; and (iii) the function p is countably additive, i.e., p( {A n : n ω}) = Σ{p(A n ) : n ω} whenever A n : n ω is a sequence of pairwise disjoint members of A. A probability measure is non-atomic provided p({x}) = 0 for each x X. Property (iii) shows that for any non-atomic probability measure, p(c) = 0 for any countable subset C X provided {x} A for each x C. We can give an example of a non-atomic probability measure defined for all Borel subsets of Ω. Recall that Borel(Ω) is the smallest σ-algebra of subsets of Ω that contains all closed subsets of Ω. Lemma 3.1 A subset S Ω is a Borel set in Ω if and only if either S or Ω S contains a CUB set. Proof: Let B := {S Ω : either S or Ω S contains a CUB set}. Then B is a σ-algebra containing all closed sets, so that Borel(Ω) B. We can complete the proof by showing that if S Ω has the property that S contains some CUB set C, then S is a Borel set. Consider the set Ω C. This set is the union of a pairwise disjoint collection D of countable sets of the form (α, β) with α β Ω. Suppose for each D D we choose a single point p(d) D. Then the set {p(d) : D D} is the intersection of its own closure with the open set Ω C, so that the set {p(d) : D D} is a Borel set. For each D D index the set S D as S D = {p(d, n) : n ω}, possibly with repetitions. For each fixed n the set E(n) := {p(d, n) : D D} is a Borel set, and S = C {E(n) : n ω}, showing that S is a countable union of Borel sets, and hence S is a Borel set. Example 3.2 For any S Borel(Ω), define p(s) = 1 if S contains some CUB set, and define p(s) = 0 otherwise. This gives a non-atomic probability measure on Ω with Borel(Ω) as its domain. Early in the last century, mathematicians thought about the question Given a space X, how large can the domain of a non-atomic probability measure on X be? Could there be a non-atomic probability measure whose domain is the collection of all subsets of X? In 1905 Vitali [16] had shown that no translation-invariant probability measure on R or on [0, 1] R could be defined for all subsets of R (respectively, all subsets of [0, 1]), and his proof is the standard approach in most analysis textbooks today ([9], [12]). However, what about probability measures that are not translation-invariant? In [15], Ulam took a different approach to the problem, starting with the uncountable well-ordered set Ω. In that 1930 paper he introduced a combinatorial object now called 7

8 an Ulam matrix [6]. An Ulam matrix is a collection of subsets {U(n, α) : n ω, α Ω} of Ω with three special properties. To remember the properties, it helps to display the Ulam matrix in a row and column format with countably many columns, one for each n ω, and uncountably many rows, one for each α Ω.... U(0, α) U(1, α) U(n, α)... U(0, 1) U(1, 1) U(n, 1) U(0, 0) U(1, 0) U(n, 0) a) each column is pairwise disjoint, i.e., for each fixed n ω and distinct α, β Ω, U(n, α) U(n, β) = ; b) each row is pairwise disjoint, i.e., for each fixed α Ω, if n m, then U(m, α) U(n, α) = ; c) for each fixed α Ω, {U(n, α) : n ω} = {β Ω : α β}. (To simplify notation, we denote that last set by (α, ).) To define the sets U(n, α) we proceed as follows. For each γ Ω the set [0, γ) is finite or countable, so there is a one-to-one function f γ : [0, γ) [0, ω). Define U(n, α) = {γ : α γ and f γ (α) = n}. Fix n ω and suppose γ U(n, α) U(n, β). Then α, β γ and f γ (α) = n and f γ (β) = n. Because f γ is one-to-one, we know that α = β. This proves (a). Fix α Ω and suppose γ U(n, α) U(m, α). Then α γ and f γ (α) = m and f γ (α) = n. Because f γ is a well-defined function, we see that m = n. This proves (b). Finally, fix α. From the definition of U(n, α) we know that α γ for each γ U(n, α), showing that {U(n, α) : n ω} (α, ). Next, consider any γ (α, ). Then α [0, γ) so that f γ (α) [0, ω), say f γ (α) = n. But then γ U(n, α) so that we have (α ) = {U(n, α) : n ω}. This establishes (c) and we can now use our matrices to prove Ulam s theorem. Theorem 3.3 No non-atomic probability measure on Ω can be defined for every subset of Ω. Proof: Suppose there is a non-atomic probability measure defined for all subsets of Ω. For each α Ω the set [0, α] is countable so that p([0, α]) = 0. Hence p((α, )) = p(ω) p([0, α]) = 1. Because (α, ) = {U(n, α) : n ω} by property (c), property (b) gives us that 1 = p((α, )) = p( {U(n, α) : n ω}) = Σ{p(U(n, α)) : n ω} so that there must be some n(α) ω with p(u(n(α), α) > 0. Because Ω is uncountable while [0, ω) is countable, there must be an uncountable subset A Ω and a fixed k ω with n(α) = k for all α A. Hence p(u(k, α)) > 0 for all α A. 8

9 For each α A there is a positive integer j(α) with p(u(k, α)) > 1. Then there must be an j(α) uncountable B A and a fixed positive integer j with j(α) = j for each α B. Choose j + 1 members of the set B, say α(1), α(2),, α(j + 1). The sets U(k, α(i)) are pairwise disjoint by property (a) above so we must have 1 = p(ω) p( {U(k, α(i)) : 1 i j + 1}) = Σ{p(U(k, α(i)) : 1 i j + 1} > (j + 1) 1 j > 1 and that is impossible. What could Ulam s theorem have to do with probability measures on more familiar spaces such as [0, 1] or R? Recall that our set Ω was a subset of R so that Ω R. If Ω = R = [0, 1] (which is a statement called the Continuum Hypothesis) then there is a one-to-one function g from Ω onto [0, 1] and the function g can be used to transfer the sets of the Ulam matrix into [0, 1]. The reason that we need the function g to be surjective is to insure that for each fixed α, the set [0, 1] g( {U(α, n) : n ω}) is countable so that we can claim p( {g(u(α, n)) : n ω}) = 1. But could it be that Ω < [0, 1]? This was unknown in 1930 when Ulam s work was published. Today we know that Ω < [0, 1] is possible because as Gödel [7] and Cohen [2] have shown, the Continuum Hypothesis is neither provable nor disprovable using only the tools of ZF C. Therefore, the best conclusion about [0, 1] that we can get from Ulam s theorem is: Theorem 3.4 If the Continuum Hypothesis holds, then there is no non-atomic probability measure on [0, 1] that is defined for all subsets of [0, 1]. We close with three other consequences of the existence of Ulam matrices. Rudin s proof in Theorem 2.6 shows that we can get two disjoint stationary subsets of Ω. In fact, one can get many pairwise disjoint stationary subsets of Ω. Corollary 3.5 There are uncountably many pairwise disjoint stationary subsets of Ω. Proof: For each fixed α Ω we know that {U(n, α) : n ω} is the stationary set (α, ). Therefore Corollary 2.7 assures us that there is some n(α) ω such that U(n(α), α) is stationary. Because [0, ω) is countable while Ω is uncountable, there must be a k ω and an uncountable A Ω such that n(α) = k for all α A. Therefore, for α A, the sets U(k, α) are all stationary and because they all lie in column number k of the Ulam matrix, they are pairwise disjoint by property (a). From Proposition 2.10 we know that disjoint stationary subsets of Ω cannot be homeomorphic, so we have the following answer to our earlier question How many different stationary sets exist in Ω? Corollary 3.6 There is an uncountable family of stationary subsets of Ω, no two of which are homeomorphic. Recall Proposition 2.5: any countable intersection of CUB sets is a CUB set. Our final result shows how different CUB sets and stationary sets can be. 9

10 Corollary 3.7 There is a sequence S n : n ω of stationary subsets of Ω with S n+1 S n for each n and {S n : n ω} =. Proof: From Corollary 3.6 we can find an infinite sequence T 1, T 2, of pairwise disjoint stationary subsets of Ω. Let S n := {T k : n k ω}. Then each S n is stationary, and {S n : n ω} =, as required. References [1] Buzyakova, R., Injections into function spaces over ordinals, Topology and its Applications 157(2010), [2] Cohen, P., Set Theory and the Continuum Hypothesis, W.A. Benjamin, New York, [3] Devlin, K., The Joy of Sets, Springer-Verlag, New York, [4] van Douwen, E., and Lutzer, D. On the classification of stationary sets, Michigan Mathematics Journal 26(1979), [5] Engelking, R., General Topology, Heldermann Verlag, Berlin, [6] Erdos, P., Ulam, the man and the mathematician, Journal of Graph Theory 9(1985), [7] Gödel, K., Consistency proof for the generalized continuum hypothesis, Proceedings of the National Academy of Sciences 25(1939), [8] Halmos, P., Finite Dimensional Vector Spaces, Springer-Verlag, New York, [9] Halmos, P., Measure Theory, van Nostrand, New York, [10] Hill, J. The Set of Countable Ordinals: An Inquiry into its Construction, Properties and Hereditary Subcompactness, Undergraduate Honors Thesis, College of William and Mary, Williamsburg, VA, 2009 [11] Kunen, K., Set Theory: An Introduction to Independence Proofs, North Holland, Amsterdam, [12] Royden, H., Real Analysis, Macmillan, New York,1963. [13] Rubin, H., and Rubin, J., Equivalents of the Axiom of Choice, North Holland, Amsterdam, [14] Rudin, M., A subset of the countable ordinals, American Mathematical Monthly 64(1957), [15] Ulam, S., Zur Masstheorie in der allgemeinen Mengenlehre, Fundamenta Mathematicae 16(1930), [16] Vitali, G., Sul problema della misura dei gruppi di punti di una retta, Nota, Bologna,