THE DESCRIPTIVE COMPLEXITY OF SERIES REARRANGEMENTS

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1 THE DESCRIPTIVE COMPLEXITY OF SERIES REARRANGEMENTS MICHAEL P. COHEN Abstract. We consider the descriptie complexity of some subsets of the infinite permutation group S which arise naturally from the classical series rearrangement theorems of Riemann, Ley, and Steinitz. In particular, gien some fixed conditionally conergent series of ectors in Euclidean space R d, we study the set of permutations which make the series dierge, as well as the set of permutations which make the series dierge properly. We show that both collections are Σ 0 3-complete in S, regardless of the particular choice of series.. Introduction The goal of this paper is to establish the exact descriptie complexity of some interesting subsets of the Polish group S of permutations of ω, endowed with the topology of pointwise conergence on ω, considered as a discrete set. Our methods will inole a blending of the techniques of classical real analysis and geometry, with the descriptie set theoretic notion of continuous reducibility between Polish spaces. First we recall Bernhard Riemann s celebrated rearrangement theorem of 876 [6], now a staple of eery graduate course in real analysis, which states the following remarkable fact (presented here as in [8]): gien a conditionally conergent series of real numbers a k, and two extended real numbers α, β [, ] with α β, n it is possible to find an infinite permutation π S for which lim inf a π(k) = α n n and lim sup a π(k) = β. In other words, by arying one s choice of α and β, it n is possible to rearrange the terms of a conditionally conergent infinite series so that the partial sums conerge to any particular real number, or dierge to plus or minus infinity, or een dierge properly. Almost as famous as Riemann s original theorem is the following d-dimensional analogue: Ley-Steinitz Theorem. Let k be a conditionally conergent series of ectors in R d. Then there exists an affine subspace A R d (that is, a space of the form A = + M where R d and M R d is a linear subspace) such that wheneer a A, there is π S with π(k) = a.

2 2 MICHAEL P. COHEN The statement aboe implies that the set of all possible sums of rearrangements of a conditionally conergent series of d-dimensional ectors is at least as rich as in the -dimensional case. An incomplete proof was first gien by Ley in 905 [5], and the complete proof was furnished by Steinitz in 93 [9]. Steinitz s proof, which is the one more commonly seen today, relied on a particular geometric constant for Euclidean spaces which is now commonly called the Steinitz constant. The proof is nontriial, and an excellent concise ersion of it may be found in the paper [7] by P. Rosenthal. Our proofs will also rely heaily on the existence of a Steinitz constant. The Ley-Steinitz theorem gies rise to a natural partition of S, into the set D of all permutations π for which π(k) dierges (either properly or to, where denotes the point at infinity in the one-point compactification of R d ), and the complement set S \D of permutations π for which π(k) conerges to some ector in R d. Both D and its complement are interesting nontriial sets. For instance, it is easy to obsere, as we will do briefly in Section 4, that both D and S \D are uncountable and dense in S, and also that D is a comeager set. We wish to examine these collections from the antage point of descriptie set theory, or, loosely speaking, the study of the definable subsets of Polish spaces. Definable here may refer to Borel, analytic, projectie, or any other class of wellbehaed sets, which are typically closed under continuous preimages. Of course the Borel sets may be stratified by their relatie complexity into a Borel hierarchy indexed by the countable ordinals, whose exact definition we will recall for the reader in Section 2. It is an empirical phenomenon that a great bulk of those Borel sets which present themseles in the eeryday study of mathematics will fall into the ery bottom few leels of the Borel hierarchy. Thus there has been some industry for descriptie set theorists in finding natural examples of Borel sets which are more complex than usual. For some instances of such sets, the reader may consult the well-known references [2] and Sections 23, 27, 33, and 37 of [4], or the paper [], which produces many examples in the field of ordinary differential equations. Our objectie here will be to establish the exact descriptie complexity of our set D and its complement. In classical terminology, we will show that D is G δσ but not F σδ (and hence not F σ, G δ, open, nor closed). Using the more modern notation, we proe: Theorem. Let k be any conditionally conergent series of ectors in R d, and let D S be the set of all permutations π for which D is Σ 0 3-complete. π(k) dierges. Then Of course, it follows immediately that S \D is Π 0 3-complete. Now, for π S, say that the rearrangement π(k) dierges properly if the series dierges, but does not dierge to infinity. Our methods also gie the following result:

3 Theorem 2. Let THE DESCRIPTIVE COMPLEXITY OF SERIES REARRANGEMENTS 3 k be any conditionally conergent series of ectors in R d, and let DP S be the set of all permutations π for which Then DP is Σ 0 3-complete. π(k) dierges properly. It follows that the set S \DP of series rearrangements which either conerge to a ector in R d, or which dierge to, is also a Π 0 3-complete set in S. Notice that, remarkably, none of the aboe statements depend on the nature of the particular conditionally conergent series k that we choose! Thus, we exhibit continuummany sets in S which lie no lower on the Borel hierarchy than the third leel. 2. Definitions and terminology First we recall the definition of the Borel hierarchy. Gien a Polish space X, we let Σ 0 (X) be the family of all open subsets of X, and Π 0 (X) the family of all closed subsets of X. We set 0 (X) = Σ 0 (X) Π 0 (X), so 0 (X) consists of the clopen sets in X. The rest of the family is defined recursiely as follows: Suppose for some countable ordinal β, we hae defined the classes Σ 0 α(x) and Π 0 α(x) for all α < β. Then we set Σ 0 β(x) = { n ω A n : A n Π 0 α n for some α n < β}, Π 0 β(x) = {A c : A Σ 0 β(x)}, and 0 β(x) = Σ 0 β(x) Π 0 α(x). It is well known that 0 β(x) Σ 0 β(x), Π 0 β(x) 0 β+ for each β, and that the inclusions are all proper. Let X, Y be Polish spaces and A X, B Y. If there exists a continuous function f : X Y such that f (B) = A, then we say that A is Wadge reducible or continuously reducible to B, and we write A W B. Intuitiely, we think that A is no more complex than B. Let Γ be any of the pointclasses Σ 0 β, Π 0 β, or 0 β. A standard inductie argument through the hierarchy shows that Γ is closed under continuous preimages, i.e., wheneer X and Y are Polish, A X, B Γ(Y ), and A is continuously reducible to B, then we hae A Γ(X). The aboe comment proides a useful tool for determining the complexity of a set. We say that a subset B of a Polish space Y is Γ-hard if for eery Polish space X and eery A Γ(X) we hae A W B. It follows from the aboe comments that if B is Γ-hard, then Γ is a lower bound for the descriptie complexity of B. If in addition we hae B Γ(Y ), then we say that B is Γ-complete, and we hae determined its exact complexity in the Borel hierarchy. The most common method for showing that a set B is Γ-hard is to find a set A which is already known to be Γ-complete, and proe that A W B by constructing an explicit continuous reduction. This will be the method of our proof in Section 4, and we will make use of the following subset C of the Baire space ω ω : C = {x ω ω : lim x(n) = }. n

4 4 MICHAEL P. COHEN Exercise 23.2 of [4] asks the reader to show that C is in fact Π 0 3-complete. It necessarily follows that the complement ω ω \C is Σ 0 3-complete. Our proof will continuously reduce this complement ω ω \C, simultaneously, to both D and DP, and thus establish the Σ 0 3-hardness of the latter two sets. We regard each nonnegatie integer n as a on Neumann ordinal, i.e. we think of each n as the set {0,..., n }. If a function π : n ω is injectie, then we call π a finite partial permutation. We use the notation dom(π) to refer to the map s domain n = {0,..., n } and ran(π) to refer to its range {π(0),..., π(n )}. If σ S or if σ is a finite partial permutation, then we say σ extends π if σ dom(π) = π. 3. The Bounded Walk lemma In this section we will deelop the main technical lemma on which our proof is built. An intuitie explanation for the Bounded Walk lemma is as follows: Consider a conditionally conergent series as an abstract infinite collection ( k ) k ω of ectors in R d from which we may build finite paths. Let α and β be two points in R d. Suppose we hae already walked ery close to α and we now wish to walk ery close to β. Then if all the remaining ectors to choose from are sufficiently small where ρ and are some constants to be determined later), and α and β are sufficiently far apart (say further than 3ρ), then it is possible to build a finite path which () extends the path we hae already walked; (2) uses up all except arbitrarily small remaining ectors; (3) takes us arbitrarily close to β; and (4) does not wander arbitrarily far from the straight-line path connecting α and β. In addition we may (5) use up any particular ector we wish. (Note that conditions () and (2) allow us to repeat this bounded walk process between as many points as we like, as often as we like.) (say less than ρ Now we aim to establish such a lemma. Before we do so, we first recall the following classical result as stated in [7], which is attributed to Steinitz, and which asserts the existence of a ery useful bounded rearrangement constant in Euclidean space, now referred to as the Steinitz constant: Lemma 3 (Polygonal Confinement Theorem). Let d be any integer. Then there exists a constant which satisfies the following statement: Wheneer 0,,..., m are ectors in R d which sum to 0 and satisfy i for each i m, then there is a finite permutation P S m with the property that for eery j. 0 + j P (i) i= The Polygonal Confinement Theorem is the basis for the remaining lemmas in this section. Lemma 4. Let α,,..., m be ectors in R d which sum to β R d, let ρ > 0, and let be as in the Polygonal Confinement Theorem. Further suppose we hae i ρ for each i m and β α ρ. Then there is a finite permutation P S m with the property that

5 for eery j. THE DESCRIPTIVE COMPLEXITY OF SERIES REARRANGEMENTS 5 j P (i) 2 β α i= Proof. Without loss of generality we may assume α = 0, for if not, replace α with 0 and β with β α. We may also without loss of generality take ρ =, for if not, replace i with i Cd Cd and β with β ρ ρ. In this case we hae i for each i and β. Now let s be an integer sufficiently large so that β s, and set m+ = m+2 =... = m+s = β/s. Then α,,..., m, m+,..., m+s are a collection of ectors which satisfy the hypotheses of the Polygonal Confinement Theorem, and hence there exists a permutation P S m+s for which j α + P (i) = j P (i) i= i= for eery j m+s. Let P S m be the unique permutation which arranges,..., m in the same order as P. Now let j m be arbitrary. Let j j be the least integer for which {P (),..., P (j)} {P (),..., P (j )}. Note that since P and P arrange,..., j in the same order, then for any i j, we must hae either (P ) (i) {,..., j} or (P ) (i) {m +,..., m + s}. Let I = {i j : P (i) {m +,..., m + s}}. Then we hae: as required. j j P (i) = P i= (i) i i= i I j P (j) i= + i i I + i I β s β + s β s = 2 β Lemma 5. Let σ be any finite partial permutation of ω, and let k be a series of ectors. If π S is any permutation for which exists another permutation π S for which π(k) conerges, then there

6 6 MICHAEL P. COHEN () π extends σ, and (2) π (k) = π(k). Proof. This may be accomplished by simply finding a finitely supported permutation τ S for which τ π dom(σ) = σ, and setting π = τ π. Lemma 6 (Bounded Walk Lemma). Let k be a conditionally conergent series of ectors in R d. Let ρ > 0 and ɛ > 0 be arbitrary and let be as in the Polygonal Confinement Theorem. Let n ω be arbitrary. Suppose π is a finite partial permutation with dom(π) = J + ω and such that k < ρ wheneer k / ran(π). Further suppose that α R d J satisfies α π(k) < ρ, and β R d satisfies β α 3ρ, and τ(k) = β for some τ S. Then there exists a finite partial permutation σ with dom(σ) = I + ω which satisfies the following properties: () σ extends π; (2) k < ɛ wheneer k / ran(σ); I (3) β σ(k) < ɛ; i (4) σ(k) 6 β α wheneer J + i I; and k=j+ (5) n ran(σ). Proof. By applying Lemma 5, we may assume without loss of generality that the permutation τ for which τ(k) = β is also such that τ extends π. Choose I ω to be so large that τ I (n) I, β τ(k) < min(ρ, ɛ), and that τ(k) < ɛ for all k > I. Set α = J τ(k) = J π(k) and set β = I τ(k). Now notice that we hae

7 THE DESCRIPTIVE COMPLEXITY OF SERIES REARRANGEMENTS 7 β α = β α (β β ) + (α α ) β α β β α α 3ρ ρ ρ = ρ. Moreoer the images τ(j + ),..., τ(i) do not lie in the range of π, since τ is a bijection extending π and dom(π) = {0,..., J}. Hence τ(j+),..., τ(i) all hae length less than ρ by our hypothesis, and therefore we may apply Lemma 4 to find a bijection P : {τ(j + ),..., τ(i)} {τ(j + ),..., τ(i)} which satisfies i k=j+ P (τ(k) 2 β α wheneer J + i I. If we define σ : I + ω by σ(k) = τ(k) for k J and σ(k) = P (τ(k)) for J < k I, then σ is a finite partial permutation with domain I + which clearly satisfies () aboe. Note that if k / ran(σ), then k / {σ(0),..., σ(j), σ(j+),..., σ(i)} = {τ(0),..., τ(j), P (τ(j+ )),..., P (τ(i)} = {τ(0),..., τ(i)}. So τ (k) > I, and hence by our choice of I, we hae k = τ(τ (k)) < ɛ. Thus (2) is also satisfied. Since P is a bijection, we hae ( I J β I σ(k) = β τ(k) + I = β τ(k) < ɛ, k=j+ P (τ(k) ) so (3) is satisfied. Lastly note that β α β β + β α + α α < ρ+ β α +ρ 3 β α by our hypothesis. Hence, we hae i σ(k) = k=j+ i P (τ(k)) k=j+ 2 β α 6 β α wheneer J + i I. So (4) is also satisfied. (5) holds simply because τ (n) I = dom(τ), and hence n ran(τ) = ran(σ). So the lemma is proed.

8 8 MICHAEL P. COHEN Remark. For simplicity, from now on when we apply the Bounded Walk lemma, we will say that we use it to walk from α to β, where α and β are as in the statement of the theorem. 4. Proof of Theorems and 2 First let us make a few simple obserations. We will describe our sets of interest using logical notation, with the assumption in place that all quantified ariables range oer ω. First notice that by Cauchy s criterion for conergence, the following equialence holds: [ ] j π D m n i j i, j n π(k) m. Since the latter predicate is an open condition in S, a count of quantifiers erifies that D indeed lies in Σ 0 3(S ). Now fix any m, and consider the set D m defined by the following rule: [ ] j π D m n i j i, j n π(k) m. Then D m is a nonempty Π 0 2 (G δ )-subset of D which is inariant under multiplication by finitely supported permutations, and hence dense in S. This shows that D is a comeager set. The complement S \D is also nonempty and inariant under finitely supported permutations, and hence dense as well. In fact, if we let T S be the set of all permutations whose only action is to transpose (perhaps infinitely many) consecutie integers, then it is easy to see that T is uncountable, and that both D and S\D are inariant under multiplication by elements of T. Thus both sets are uncountable dense, i.e. in some sense they are large nontriial sets in S, as promised in the introduction. The reader may consult [3] for similar obserations about some other sets in S which are closely related to our D and DP. Next define a set I S by the rule [ ] i π I m n i i n π(k) m. Then I is a Σ 0 3 set, which consists exactly of those permutations whose corresponding series rearrangements π(k) do not dierge to infinity. Since DP = D I, so too we hae DP Σ 0 3(S ). Proof of Theorems and 2. In light of our comments aboe, it suffices to show that D and DP are Σ 0 3-hard. Recall that the set C = {x ω ω : lim x(n) = } n is known to be Σ 0 3-complete. We will build a function f : ω ω S that will be a continuous reduction from ω ω \C to both D and DP simultaneously. That is, both of the following will hold: k=i k=i x ω ω \C f(x) D,

9 THE DESCRIPTIVE COMPLEXITY OF SERIES REARRANGEMENTS 9 x ω ω \C f(x) DP. Fix an arbitrary x ω ω. Let = k. We will recursiely construct a sequence of integers (J n ) n ω and a sequence of finite partial permutations (π n ) n ω, each with domain {0,..., J n }, which satisfy the following seen conditions for each n 0: (I) π n extends π n ; (II) n {π n (0),..., π n (J n )}; (III) the definitions of π n (J n + ),..., π n (J n ) depend only on the alues of x(n) and x(n + ); Jn (IV) πn(k) < x(n + ) + ; j (V) πn(k) k=j n + 36 x(n) + for eery j {J n +,..., J n }; j (VI) there exist i, j {J n,..., J n } for which πn(k) > x(n)+ ; and (VII) k < x(n+)+ for all k / ran(π n ). After this construction is finished, we will let π be the unique permutation which extends all the π n s, and set f(x) = π. Conditions (I) and (II) will guarantee that π is indeed a permutation, while (III) will guarantee that the map f is continuous. Conditions (IV) and (V) will ensure that if x C, then π(k) will conerge to k=i, while condition (VI) will guarantee that if x / C, then π(k) will dierge properly. (Condition (VII) is just a technical requirement to facilitate our recursie definition.) We will now proceed with our construction. Here for the sake of conenience J our base case will be n =. Let J 0 be so large that k < x(0)+, and that k < x(0)+ for all k > J. Let π : J 0 + J 0 + be the identity permutation. Note that π and J triially satisfy (IV) and (VII) aboe; this will be enough to facilitate our induction. Now we assume that J i and π i are defined for all i < n, and satisfy at least (IV) and (VII), and we proceed with the inductie step of defining J n and π n. As we go we will erify that J n and π n in fact really do satisfy all of conditions (I)-(VII). By the Ley-Steinitz theorem, the set of all points β R n for which some rearrangement of k conerges to β is an affine subspace of R d ; in particular,

10 0 MICHAEL P. COHEN there is at least a line of possible points to which the rearrangements conerge. So we may choose some β R d for which β = 3 x(n)+, and a permutation τ S for which τ(k) conerges to β. Now we will define π n and J n by applying the Bounded Walk lemma twice: first, we will use the lemma to walk out to a point near β, and then we will use the lemma to walk back in to a point near. To walk out : apply the Bounded Walk lemma to walk from to β, extending. Thus we obtain an index I > J n and a finite partial permutation σ : I + I + which satisfies properties () (5) of the lemma. In particular, condition (3) ensures that we hae the finite partial permutation π n and with ρ = ɛ = x(n)+ I k=j n + σ(k) J = I β β + n σ(k) + I β β σ(k) > 3 x(n) + x(n) + x(n) + = x(n) + σ(k) J n σ(k) i while condition (4) guarantees that σ(k) k=j n + 6 β = 8 x(n)+ wheneer J + i I. Next we walk back in. Apply the Bounded Walk lemma to walk from β to, extending the finite partial permutation σ, with ρ = x(n)+ and ɛ = x(n+)+. Then we obtain an index J n > I and a finite partial permutation π n : J n + J n + which again satisfies properties () (5). By the preious inequality, and applying condition (4) for π n, we see that i k=j n + πn(k) I k=j n + I k=j n + πn(k) σ(k) + i k=i+ + 6 β 8 x(n) x(n) + = 36 x(n) + πn(k)

11 THE DESCRIPTIVE COMPLEXITY OF SERIES REARRANGEMENTS wheneer I + i J n. Thus we hae shown that (V) holds for π n. (I), (III), and (IV) obiously hold from our definition of π n, and (II) holds if we utilize condition (5) in either of our two applications of the Bounded Walk lemma to ensure that n ran(π n ). We hae shown that (VI) holds if we take i = J n and j = I, and (VII) follows from condition (2) in our second application of the Bounded Walk lemma. So our construction is complete and we may let π S be the unique permutation which extends all of the π n s. Define the map f : ω ω S by f(x) = π, where π is as we hae constructed aboe. The function f, as a map between the Polish space ω ω and its Polish subspace S, is continuous by condition (III). We claim that f is in fact the continuous reduction we desire. To see this, suppose x C, so lim x(n) = and hence lim n n x(n)+ = 0. For any i ω, let n i be the greatest integer for which J ni < i J ni. Then by (IV) and (V) we hae i π(k) J n πn (k) + x(n i ) x(n i ) + = 37 x(n i ) +. i k=j n + Now taking the limit as i (and as n i ) we see that to. Hence f(x) S \D and f(x) S \DP. πn(k) π(k) conerges On the other hand, suppose x ω ω \C. Then the sequence (x(n)) is cofinally bounded, i.e. there is an M < such that x(n) M infinitely often. Hence x(n)+ > M+ infinitely often. It follows from (VI) that there are infinitely many j blocks i,..., j of integers for which π(k) > x(n)+ > M+, and hence π(k) k=i dierges by the Cauchy criterion. In addition, we hae already demonstrated that for any n i depending on i as aboe, we hae i π(k) 37 x(n 37. i)+ This implies that all partial sums of the rearranged series are bounded, and so the series must in fact dierge properly. Thus in this case we hae f(x) D and f(x) DP. So f is the reduction we seek, and D and DP are Σ 0 3-complete. References [] A. Andretta and A. Marcone, Ordinary differential equations and descriptie set theory: uniqueness and globality of solutions of Cauchy problems in one dimension, Fund. Math. 53 (997),

12 2 MICHAEL P. COHEN [2] H. Becker, Descriptie set theoretic phenomena in analysis and topology, in: Set Theory of the Continuum, H. Judah, W. Just, and H. Woodin (eds.), Math. Sci. Res. Inst. Publ. 26, Springer (992), 25. [3] R. G. Bilyeu, R. R. Kallman, P. W. Lewis, Rearrangements and category, Pacific J. Math. 2() (986), [4] A. S. Kechris, Classical Descriptie Set Theory, Springer, 995. [5] P. Ley, Sur les séries semi-conergentes, Nou. Ann. d. Math. 64 (905), [6] B. Riemann, Uber die Darstellbarkeit einer Function durch eine trigonometrische Reihe, Gesammelte Mathematische Werke (Leipzig 876): [7] P. Rosenthal, The remarkable theorem of Ley and Steinitz, Amer. Math. Monthly 94(4) (987), [8] W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, Inc., 976. [9] E. Steinitz, Bedingt Konergente Reihe und Konexe Systeme, J. Reine Angew. Math. 43 (93), Department of Mathematics, Uniersity of North Texas, Denton, TX 76203, U.S.A address: michaelcohen@my.unt.edu

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