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The Journal of Symbolic Logic Volume 00, Number 0, XXX 0000

HINDMAN’S , ULTRAFILTERS, AND REVERSE MATHEMATICS (TO APPEAR IN JSL)

JEFFRY L. HIRST APPALACHIAN STATE UNIVERSITY

Abstract. Assuming , Hindman [2] showed that the existence of certain ultralters on the power of the natural numbers is equivalent to Hindman’s Theorem. Adapting

this work to a countable setting formalized in , this article proves the equivalence of the existence of certain ultralters on countable Boolean algebras and an iterated form of Hindman’s Theorem, which is closely related to Milliken’s Theorem. A computable restriction of Hindman’s Theorem follows as a corollary.

Throughout this paper, proofs are carried out in the RCA.For a full exposition of RCA and reverse mathematics, Simpson’s book [6] is the best source. For the following discussion, the salient features of RCA are that it is a 0 subsystem of second order arithmetic with induction restricted to 1 formulas and comprehension restricted to relatively computable sets. The language of second order arithmetic is remarkably expressive, and the following concepts are easily formalized. A countable eld of sets is a countable sequence of of which is closed under intersection, , and relative complementation. We use Xc to denote {x ∈ | x/∈ X}, the of X relative to . Given a set X and an integer n ∈ , we use X n to denote the set {x n | x ∈ X ∧ x n}, the translation of X by n.Adownward translation algebra is a countable eld of sets which is closed under translation. Given a countable collection {Gi | i ∈ } of subsets of , RCA suces to prove the existence of the downward translation algebra generated by {Gi | i ∈ }, which is denoted by h{Gi | i ∈ }i and consists of all nite unions of nite intersections of translations of elements and complements of elements of {Gi | i ∈ }. RCA can also prove that h{Gi | i ∈ }i is a downward translation algebra. Note that in RCA, we encode h{Gi | i ∈ }i as a countable sequence of countable sets, and that each set in h{Gi | i ∈ }i may be repeated many times in the sequence. By allowing this repetition, we can organize the encoding sequence so that given the indices of any elements of the algebra we may deterministically compute indices for their unions, intersections, complements, and translations.

1991 Mathematics Subject Classication. 03B30, 03F35, 03D80, 05D10. Key words and phrases. Hindman’s Theorem, Milliken’s Theorem, reverse mathematics, computability.

c 0000, Association for Symbolic Logic 0022-4812/00/0000-0000/$00.00

1 2 JEFFRY L. HIRST APPALACHIAN STATE UNIVERSITY

Suppose that F = {Xi | i ∈ } is a countable eld of sets. In RCA, we dene an ultralter on F as a set U satisfying the following four properties:

If Xi = ∅ then i/∈ U. If i, j ∈ U and Xk = Xi ∩ Xj, then k ∈ U. If i ∈ U and Xj Xi, then j ∈ U. c ∈ ∈ If Xj = Xi , then i U or j U. Where no confusion will arise, we will abuse notation by writing U for {Xi | i ∈ U}, so that we may write statements using the standard notation for ultralters. For example, consider the following denition and lemma. An ultralter U is an downward translation invariant ultralter if ∀X ∈ U ∃x ∈ X (x =6 0 ∧ X x ∈ U). While the denition only requires one translating x for each set, the existence of many is shown by the following lemma.

Lemma 1. (RCA) If U is an almost downward translation invariant ultralter and X is an of U, then the set {x ∈ X | X x ∈ U} is unbounded. Proof. Suppose by way of contradiction that x is the largest number in X such that X x ∈ U. Since X x ∈ U, there is a y ∈ X x such that y =06 and (X x) y ∈ U. Since y ∈ X x, we have x + y ∈ X. Additionally, (X x)y = X (x+y), so we have x+y ∈ X, x+y>x, and X (x+y) ∈ U, contradicting our choice of x. a If X , then the notation FS(X) denotes the set of all nonrepeating sums of nonempty nite subsets of X. For example, FS({1, 2, 5})={1, 2, 3, 5, 6, 7, 8}. Using this notation, it is easy to state Hindman’s Theorem. Theorem 2 (Hindman’s Theorem). If f : → k is a mapping into the natural numbers less than k, then there is an innite set X such that f is constant on FS(X). Proof. The original non-formalized proof appears in [3]. For a proof of Hind- + a man’s theorem in the subsystem ACA , see [1]. The set X in the statement of Theorem 2 is called an innite homogeneous set for the partition. Given any set G , we refer to the statement of Theorem 2 with f(x) as the characteristic function for G as Hindman’s Theorem for G. 1. Equivalence results. We begin with the central result of the section, linking an iterated form of Hindman’s Theorem to the existence of almost down- ward translation invariant ultralters.

Theorem 3. (RCA) The following are equivalent:

(1) IHT (Iterated Hindman’s Theorem) If {Gi | i ∈ } is a collection of subsets of , then there is an increasing sequence hxiii∈N such that for every j ∈ , {xi | i>j} satises Hindman’s Theorem for Gj. (2) Every countable downward translation algebra has an almost downward translation invariant ultralter.

Proof. To prove that (2) implies (1), suppose {Gi | i ∈ } is a collection of subsets and let U be an almost downward translation invariant ultralter on h{Gi | i ∈ }i, the downward translation algebra generated by {Gi | i ∈ }. Dene a sequence of (indices for) nested sets hXiii∈N and an increasing sequence HINDMAN’S THEOREM, ULTRAFILTERS, AND REVERSE MATHEMATICS (TO APPEAR IN JSL) 3

h i c of integers xi i∈N as follows. Let X0 be whichever of G0 and G0 is in U. Let x0 = 0. Suppose that xn has been chosen and that Xn has been chosen so that ∈ c Xn U and either Xn Gn or Xn Gn. Since U is an almost downward translation invariant ultralter, by Lemma 1 we can nd a least xn+1 ∈ Xn such b that xn+1 >xn and Xn xn+1 ∈ U. Let Gn+1 denote whichever of Gn+1 and c ∩ ∩ b Gn+1 is in U, and dene Xn+1 by Xn+1 = Xn (Xn xn+1) Gn+1. h i We will show that FS( xn n>t) Xt. Let xi ,...,xik be a nite sequence of elements of hx i . Note that i > 0, so X exists, and x ∈ X . n n>t k ik 1 P ik ik 1 k Treating this as the base case in an induction, we have x ∈ X .For P m=k im ik 1 k the induction step, suppose that x ∈ X . Since i i 1, P m=j+1 im ij 1 j j+1 k x ∈ X . Since i 1, by the denition of X , X X ∩ m=j+1 im ij P j Pn ij ij 1 b k k ∩ ∈ ∈ (Xij 1 xij ) Gij ,so m=j+1 xim Xij 1 xij . That is, m=j xim Xij 1, completing theP induction step. By induction on quantier-free formulas, we have k shown that x ∈ X . Since tt, this suces to show that FS(hxnin>t) Xt. c To complete the , note that for each t, Xt Gt or Xt Gt .Thus h i h i c h i for each t, FS( xn n>t) Gt or FS( xn n>t) Gt , so the sequence xn n>0 satises the iterated version of Hindman’s Theorem presented in (1). To prove that (1) implies (2), suppose that {Gi | i ∈ } is a countable down- ward translation algebra. Since {Gi | i ∈ } is a countable collection of subsets, we may apply (1) to nd an increasing sequence of integers hxiii∈N such that h i c for all j, FS( xi i>j) Gj , where either Gj = Gj or Gj = Gj. We will show { | ∈ } that U = Gi i is an almost downward translation invariant ultralter on {Gi | i ∈ }. ∈{ | ∈ } c If X Gi i , then for some j, Gj = X or Gj = X , so either ∈ c ∈ h i 6 ∅ ∅ ∈ X U or X U. For each j, Gj FS( xi i>j) = ,so / U. Suppose ∈ ∈ X1,X2 U. Then for some k, j , X1 = Gk and X2 = Gj . Also, for ∩ ∩ c { } some m, Gm = X1 X2 or Gm =(X1 X2) . Let t = max j, k, m . then h i ∩ ∩ ∩ ∩ ∩ ∩ 6 ∅ FS( xi i>t) Gm Gk Gj = Gm X1 X2. Since Gm X1 X2 = , ∩ ∈ ∩ ∈ ∈ Gm = X1 X2. Thus, if X1,X2 U then X1 X2 U. Finally, if X U and Y is an element of the downward translation algebra satisfying X Y , { } then for some j and k, X = Gj and Y = Gk. Let t = max j, k and note that ∩ h i 6 ∅ ∈ X Gk FS( xi i>t) = ,soGk = Y and Y U.ThusU is an ultralter on the downward translation algebra. To show that U is an almost downward translation invariant ultralter we must show that for each X ∈ U we can nd an x =6 0 such that X x ∈ U. ∈ h i ∈ Suppose X U, and nd j so that X = Gj FS( xi i>j). For every y FS(hxiii>j+1), we have xj+1 + y ∈ X,soX xj+1 FS(hxiii>j+1). Because {Gi | i ∈ } is a downward translation algebra, there is a k such that Gk = { } h i ∩ 6 ∅ X xj+1.Ift = max k, j +1 , then Gk FS( xi i>t), so Gk Gk = .Thus ∈ Gk = Gk,soX xj+1 U as desired. Summarizing, we have used (1) to nd an almost downward translation invariant ultralter on a given downward translation algebra, completing the proof of the theorem. a 4 JEFFRY L. HIRST APPALACHIAN STATE UNIVERSITY

The remainder of this section extends the equivalence of the preceding theorem to a version of Milliken’s Theorem. This proof will use the functional variant of the iterated Hindman’s Theorem which appears as statement (2) in the following result.

Lemma 4. (RCA) The following are equivalent: (1) IHT (See Theorem 3 for a full statement.) (2) Given a collection of functions {fi | i ∈ }, each dened on with a bounded range, there is a sequence hxiii∈N such that for each j, fj is con- stant on FS({xi | i>j}).

Proof. To prove that (2) implies (1), just let fj be the characteristic function → for Gj. To prove the converse, if fj : k, for each i

Lemma 5. (RCA) The following are equivalent: (1) MT(n) (Milliken’s theorem for n-.) If f :[]n → k then there is an increasing sequence X = hxiii∈N such that f is constant on FSn(X). (2) Suppose S and f :[]n → k. Then thereP is an increasing sequence hAiii∈N of nite subsets of S such that for A = { Ai | i ∈ }, the function f is constant on FSn(A). Proof. The statement of (1) is a special case of (2) where S is set equal to . Consequently, we need only prove that (1) implies (2). Fix S, f, n, and k as in the statement of (2). Given any innite sequence of integers and any nite collection of moduli, we can nd a nite subsequence which sums to a value which is congruent to 0 for each of the given moduli. Con- sequently, without loss of generality, we may assume that S = hsiii∈N satises the congruence conditions i

Apply (1) to g to nd a homogenous set for g and denote it by hbiii∈N. As with S, we may assume that for i

Theorem 6. For each standard n 3, RCA can prove that the following are equivalent: (1) IHT (Iterated Hindman’s Theorem.) See Theorem 3 for a full statement. (2) MT(n) (Milliken’s Theorem for n-tuples.) See Lemma 5 for a full statement.

Proof. To prove that (2) implies (1), we assume RCA and Milliken’s Theo- rem for triples. Let {Gi | i ∈ } be a collection of subsets of as in the statement of (1). For k

Theorem 7. (RCA) Fix G N. If there is an almost downward translation invariant ultralter on the downward translation algebra hGi, then Hindman’s Theorem holds for G.

Proof. Assume RCA, suppose that G N, and let U be an almost down- ward translation invariant ultralter on hGi. Dene a sequence (of indices) of nested sets hXii and an increasing sequence of integers hxii as follows. Set X0 c to be whichever of G and G is in U, and let x0 = 0. Suppose that Xn ∈ U and xn have been chosen. Since U is an almost downward translation invari- ant ultralter, applying Lemma 1 we can nd a least xn+1 ∈ Xn such that xn+1 >xn and Xn xn+1 ∈ U. Let Xn+1 = Xn ∩ (Xn xn+1). Emulating the induction argument from the proof that (2) implies (1) in Theorem 3, show that FS(hxnin>0) X0, thereby proving that Hindman’s Theorem holds for G. a

Lemma 8. (RCA) If the downward translation algebra hGi contains no sin- gletons, then Hindman’s Theorem holds for G. Proof. Suppose hGi contains no singletons. Let U be the principal ultralter generated by 0, so U = {X ∈hGi|0 ∈ X}. Pick X ∈ U. Since X is not a singleton, there is an x =6 0 such that x ∈ X. Since x ∈ X, we have 0 ∈ X x, so X x ∈ U.ThusU is an almost downward translation invariant ultralter. By Theorem 7, Hindman’s theorem holds for G. a The preceding argument may be extended to a broader of algebras, but 0 the modications use the induction scheme on 2 formulas, which is denoted by 0 2 IND. Lemma . h i 9 (RCA + IND) If the downward translation algebra G contains nitely many singletons, then Hindman’s Theorem holds for G. Proof. Suppose that hGi contains nitely many singletons. We will show that there is a set H diering from G only on a nite initial interval, such that hHi contains no singletons. If hGi contains no singletons, let H = G. Otherwise, let {m} be the largest h i singleton in G . Let G0,...G2m1 be an enumeration of all subsets of diering from G only at or below m. Note that by applying translations and Boolean operations to G and {m}, we may construct each Gi. Thus for each i, hGiihGi and in particular, the singletons of each hGii are a subset of the singletons of hGi. Furthermore, the assertion “{j} is an element of hGii” holds c if and only if there is a nite collection of translations of Gi and Gi such that for every n, n is in the intersection of the translations if and only if n = j. This 0 0 assertion is expressible by a 2 formula. By bounded 2 comprehension (which 0 is provable in RCA from 2 induction [6]), there is a set X such that for every m+1 i<2 1 and j m,(i, j) ∈ X if and only if {j}∈Gi. If there is an i such that Gi contains no singletons, pick the rst such i and let H = Gi. Otherwise m+1 use X to dene the set Y = {mi | i<2 1}, the collection containing the maximum singleton from each Gi. Let p be the minimum of Y , and let H be the rst Gi for which p is the largest singleton. We will now show that hHi contains no singletons. If hHi contains a singleton, then it contains {p} as constructed above. Since hHi = hHci we may relabel as needed to insure p ∈ H. HINDMAN’S THEOREM, ULTRAFILTERS, AND REVERSE MATHEMATICS (TO APPEAR IN JSL) 7

Because hHi is a generated by downward translations of H and Hc, {p} is expressible as a union of intersections of translations of H and Hc. Furthermore, because {p} is a singleton, we may discard all but the rst nonempty element of the union, and write {p} as an intersection of translations of H and Hc. By way of contradiction, suppose {p} is expressed as the intersection of a collection of such translations, and H itself is not in the collection. Since p/∈ Hc, the collection must consist entirely of non-zero translations of H and Hc. Adding the value of the smallest translation to each of these sets yields a collection of elements of hHi whose intersection is a singleton which is greater than p, contradicting the fact that {p} is the largest singleton in hHi. Thus, H must be included in any intersection dening {p}, and we may write \ {p} = H ∩ (Hdk k) k∈F where F is a nite set of positive natural numbers and dk indicates whether or dk c not to take the complement of H. Note that ∩k∈F (H k) H ∪{p}. Now consider hH0i where H0 = H {p}. Suppose by way of contradiction that {p}∈hH0i. As argued above, we must be able to express {p} as the intersection of a collection of translations of H0 and H0c. Since p/∈ H0, we know H0 is not in the collection. If H0c is not in the collection, then hH0i contains a singleton larger than p. However, H0 = H ∩{p}c ∈hHi,sohH0ihHi, implying that hH0i can contain no singletons larger than p. Thus any intersection dening {p} in hH0i includes H0c, and we may write \ {p} = H0c ∩ (H0dk k) k∈W

0dk where W is a nite collection of positive integers. Note that ∩k∈W (H k) (H0c)c ∪{p} = H. Combining the preceding two centered equations, we have \ \ 0d {p} (Hdk k) ∩ (H k k) (Hc ∪{p}) ∩ H = {p}. k∈F k∈W Since H0 ∈hGi, this shows that {p} is expressible as an intersection consisting entirely of nonzero translations of H and Hc, yielding a contradiction. Thus, {p} ∈h/ H0i. Since hH0i is closed under downward translation, this shows that any singletons in hH0i must be strictly less than p. However, H0 diers from G only at or below m p,sohH0i must contain a singleton greater than or equal to p, yielding a nal contradiction and proving that hHi contains no singletons. We have shown that H diers from G only at or below m, and H has no singletons. By Lemma 8, Hindman’s Theorem holds for H. Given an innite homogeneous set Z for H, the set {n ∈ Z | n>m} is an innite homogeneous set for G. Thus Hindman’s Theorem holds for G. a Theorem . h i 10 (RCA + IND) If the downward translation algebra G doesn’t contain all the singletons, then Hindman’s Theorem holds for G. Proof. By closure under downward translation, if hGi contains a singleton {p}, it contains all singletons {n} such that n

REFERENCES

[1] Andreas R. Blass, Jeffry L. Hirst, and Stephen G. Simpson, Logical analysis of some of combinatorics and topological dynamics, Logic and combinatorics (Arcata, Calif., 1985), Contemp. Math., vol. 65, Amer. Math. Soc., Providence, RI, 1987, pp. 125–156. [2] Neil Hindman, The existence of certain ultra-lters on and a conjecture of Graham and Rothschild, Proc. Amer. Math. Soc., vol. 36 (1972), pp. 341–346. [3] , Finite sums from sequences within cells of a partition of , J. Combinatorial Theory Ser. A, vol. 17 (1974), pp. 1–11. [4] Jeffry L. Hirst, Combinatorics in subsystems of second order arithmetic, Ph.D. the- sis, The Pennsylvania State University, 1987, pp. vii+146 pages. [5] Keith R. Milliken, Ramsey’s theorem with sums or unions, J. Combinatorial Theory Ser. A, vol. 18 (1975), pp. 276–290. [6] Stephen G. Simpson, Subsystems of second order arithmetic, Perspectives in Math- ematical Logic, Springer-Verlag, Berlin, 1999.

DEPARTMENT OF MATHEMATICAL SCIENCES APPALACHIAN STATE UNIVERSITY BOONE, NC 28608 USA E-mail: [email protected] URL: www.mathsci.appstate.edu/jlh