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Borel partitions of infinite sequences of reals

arXiv:math/9301209v1 [math.LO] 15 Jan 1993Borel partitions of infinite sequences of realsA. Louveau, S. Shelah, and B. VelickovicIntroductionThe starting point of our work is a Ramsey-type theorem of Galvin (unpub-lished) which asserts that if the unordered pairs of reals are partitioned intofinitely many Borel classes (or even classes which have the property of Baire)then there is a perfect set P such that all pairs from P lie in the same class.The obvious generalization to n-tuples for n ≥3 is false.

For example, lookat the coloring of triples where a triple {x, y, z} with x < y < z is coloredred provided that y −x < z −y and blue otherwise. Then any perfect setwill contain triples of both colors.

Galvin conjectured that this is the onlybad thing that can happen. It will be simpler to state this if we identifythe reals with 2ω ordered by the lexicographical ordering and define for dis-tinct x, y ∈2ω ∆(x, y) to be the least n such that x(n) ̸= y(n).

Let thetype of an increasing n-tuple of reals {x0, . .

. xn−1}< be the ordering ≺on{0, .

. .

, n −2} defined by i ≺j iff∆(xi, xi+1) < ∆(xj, xj+1). Galvin provedthat for any Borel coloring of triples of reals there is a perfect set P suchthat the color of any triple from P depends only on its type and conjecturedthat an analogous result is true for any n. This conjecture has been provedby Blass ([Bl]).

As a corollary it follows that if the unordered n-tuples ofreals are colored into finitely many Borel classes there is a perfect set P suchthat the n-tuples from P meet at most (n −1)! classes.

The key ingredientin the proof is the well-known Halpern-La¨uchli theorem ([HL]) on partitionsof products of finitely many tree. In this paper we consider extensions of thisresult to partitions of infinite increasing sequences of reals.

Define a typeof an increasing sequence of reals as before and say that such a sequence{xn : n < ω} is strongly increasing if its type is the standard ordering on ω,i.e. if ∆(xn, xn+1) < ∆(xm, xm+1) whenever n < m. We show, for example,that for any Borel or even analytic partition of all increasing sequences of1

reals there is a perfect set P such that all strongly increasing sequences fromP lie in the same class. In fact, for any finite set C of types there is a perfectset P such that for any type in C all increasing sequence from P of that typehave the same color.

It should be pointed out that the same statement isfalse if C is an infinite set of types.Our result stands in the same relation to Blass’ theorem as the Galvin-Prikry theorem ([GP]) to the ordinary Ramsey’s theorem and the proof againrelies heavily on the Halpern-La¨uchi theorem. There are known several exten-sions of the Halpern-La¨uchli theorem that are relevant to this work.

Milliken([Mi]) considered partitions of nicely embedded infinite subtrees of a perfecttree and obtained a partition result in the spirit of Galvin-Prikry howeverin a different direction from ours, and Laver ([La]) proved a version of thistheorem for products of infinitely many perfect trees.The paper is organizes as follows. In §1 we introduce some notation andpresent some results on perfect trees which we will need later.In §2 wereduce the main theorem to two lemmas which are then proved in §§3 and 4.We shall present our result using the terminology of forcing.

If P is a forcingnotion we let, as usual, RO(P) denote the regular open algebra of P, i.e. acomplete Boolean algebra in which P is densely embedded.

If b is a Booleanvalue in RO(P) and p ∈P we shall say that p decides b if either p ≤b orp ≤1 −b. For all undefined terminology of forcing see, for example, [Ku].1Basic properties of perfect treesPerfect trees Let 2<ω denote the set of all finite {0, 1}-sequences orderedby extension.

T ⊆2<ω is called a perfect tree if it is an initial segment of 2<ωand every element of T has two incomparable extensions in T. Let P denotethe poset of all perfect trees partially ordered by inclusion. Thus P is thewell-known Sacks forcing ([Sa]).

For a subset C of T let TC be the set of allnodes in T which are comparable to an element of C. If {s} is a singletonwe shall simply write Ts instead of T{s}. For a tree T let T(n) denote then-th level of T, i.e.

the set of all s ∈T which have exactly n predecessors.We say that a node s in T is splitting if it has two immediate extensions.Given integers m ≤k let us say that a set D is (m, k)-dense in T provided Dis contained in T(k) and every node in T(m) has an extension in D. Given2

trees T0, . .

. Td−1 and a subset A of ω let⊗Ai

We are now ready to state a version of theHalpern-La¨uchli theorem ([HL]).Theorem 1 ([HL]) For every integer d < ω given perfect trees Ti, for i < d,and a partition⊗i

. , xd−1) ∈⊗i

Elements of A(P) are pairs (T, n), where T ∈P and n ∈ω. Saythat (T, n) ≤(S, m) iffT ≤S, n ≥m, and T ↾(m + 1) = S ↾(m + 1).

If inaddition n = m we shall say that (T, n) is a pure extension of (S, m). If Gan A(P)-generic filter over a model of set theory letT(G) =[{T ↾(n + 1) : (T, n) ∈G}.Then, by genericity, T(G) is a perfect tree and is called the A(P)-generictree derived from G.Combs An n-comb C is a tree such that there is some strongly increasingsequence of reals {xi : i < n} and some m > ∆(xn−2, xn−1) such that C isthe set of all initial segments of length < m of members of this sequence.

Aninfinite comb is a tree such that there is some strongly increasing sequence{xn : n < ω} such that C is the set of all finite initial segments of members ofthis sequence. Clearly there is a 1-1 correspondence between infinite combsand strongly increasing sequences and we shall in fact state our theorem interms of infinite combs.

For a tree T if n < ω is such that T ↾(n + 1) isa comb let Cω(T, n) denote the set of all infinite combs contained in T andextending T ↾(n + 1). Let Cω(T) = Cω(T, 0).

Note that Cω(T) has a naturaltopology as a subspace of P(T) with the Tychonofftopology. Thus we canspeak about Borel, analytic, etc.

subsets of Cω(T).3

The comb forcing C Let C be the subposet of A(P) consisting of all pairs(T, n) such that T ↾(n + 1) is a comb, with the induced ordering. Let ussay that (T, n) has width d if T ↾(n + 1) is a d-comb.

The notion of pureextension is defined as in the case of A(P). If (R, m) ≤(T, n) and if these twoconditions have the same width then we say that (R, m) is a width preservingextension of (T, n).

Note that in this case (R, n) is a pure extension of (T, n)which is equivalent in terms of forcing with (R, m). Clearly, if G is a C-genericfilter over some model of set theory the setC(G) =[{T ↾(n + 1) : (T, n) ∈G}is a infinite comb, we call it the generic comb derived from G.2The main theoremThe main result of this paper is the following partition theorem.Theorem 2 For every partitionCω(2<ω) = K0 ∪K1where K0 is analytic and K1 co-analytic there is a perfect tree T and i ∈{0, 1}such that Cω(T) ⊆Ki.The proof of the theorem will consist of two lemmas which combined yieldthe desired result.Lemma 1 Let b be a Boolean value in RO(C) and let (S, n) ∈C.

Then thereis a pure extension (T, n) of (S, n) which decides b.Lemma 2 Let T be an A(P)-generic tree over a model of set theory M.Then every infinite comb contained in T is C-generic over M.Given these two lemmas it is quite easy to prove the theorem. Take acountable transitive model M of ZFC−containing the codes of K0 and K1.Consider forcing with C as defined in M. Note that if C is a generic comb thestatement whether C belongs to K0 is absolute between M[C] and V .

Let4

b be the Boolean value that this statement is true in M[C]. Then it followsfrom Lemma 1 that there is a pure extension (S, 0) of the maximal conditionwhich decides b, let us say, for concreteness, that it forces b.

Now considerforcing over M with A(P) and take a generic filter G over M which contains(S, 0). Let T be the generic tree derived from G. Then by Lemma 2 everyinfinite comb contained in T is C-generic over M and, since it is containedin S as well, it follows that it is in K0.

Thus T is the homogeneous tree weseek. In the next two sections we prove Lemmas 1 and 2 and thus completethe proof.3Proof of Lemma 1Unless otherwise stated in this section we work with the forcing notion Cintroduced in §1.

Given a Boolean value b in the completion algebra RO(C)let us say that a condition (T, n) accepts b if (T, n) ≤b and that it rejects bif (T, n) ≤1 −b. We shall need the following auxiliary lemma.Lemma 3 Let (S, n) be a condition in C of width d and let b ∈RO(C) bea Boolean value.

Then there is a pure extension (T, n) of (S, n) such thateither (T, n) accepts b or no extension of (T, n) of width d + 1 accepts b.PROOF: Let {t0, . .

. , td−1}< be the increasing enumeration of S(n) in thelexicographical ordering.

We first find an infinite set A and a perfect subtreeS∗of S such that for any m ∈A and z0, . .

. , zd ∈S∗(m) such that zi ≥ti fori < d and zd ≥td−1, letting Z = {zi : i ≤d}, if there is a pure extension of(S∗Z, m) deciding b then already (S∗Z, m) decides b.

This can be done by astandard fusion argument. Moreover we can arrange that between any twoconsecutive levels in A there is at most one splitting node.

We now define acoloring:⊗i

. .

, xd−1) ∈⊗i

Let Z = {zi : i ≤d} andput (x0, . .

. , xd−1) in K0 if (S∗Z, m) accepts b, in K1 if it rejects b, and inK2 otherwise.

By the Halpern-La¨uchli theorem we can find (x0, . .

. , xd−1) ∈⊗i

for i < d, such that Di is (m, k)-dense in S∗xi and ⊗i

. .

, xd−1) ∈Kǫ, as well.We now build an increasing sequence (bk)k<ω of elements of A and aperfect subtree T of S∗which will have one splitting node on levels betweenbk and bk+1. To begin let b0 be the level of the xi and let T(b0) = {xi : i < d}.This uniquely determines T ↾(b0 + 1) as the set of all initial segments ofelements of T(b0).

Suppose now we have defined bk and T ↾(bk + 1). Wechoose one node y in T(bk) and we will arrange so that the only splittingnode of T on levels between bk and bk+1 is above y.

Let m be the least levelwhich is in A and such that y has two extensions, say y′ and y′′ in S∗(m).Now find some b ∈A and sets Di, for i < d such that Di is (m, b)-densein S∗xi and such that ⊗i

This uniquely definesT ↾(bk+1 + 1). During our construction we arrange the choice of the pointsy in such a way that the final tree T is perfect.

Let B = {bk : k < ω}. Itfollows that ⊗Bi

First note that if(R, l) is any extension of (T, n) then there there is m ∈A such that R hasno splitting nodes on levels between l and m and hence (R, l) and (R, m) areequivalent condition. Suppose now that some condition of width d + 1 below(T, n) accepts b and let (R, m) be such a condition with m minimal suchthat m ∈A.

Let Z = R(m) = {z0, . .

. , zd}< be the increasing enumerationin the lexicographical order and let k be the largest such that bk < m. Thensince on levels between bk and bk+1 there is at most one splitting node itfollows that R(bk) has size d. Let R(bk) = {y0, .

. .

, yd−1}< be the increasingenumeration. By the construction of T it follows that yd−1 was the pointchosen at stage k, that zd−1 and zd are the only extensions of yd−1 in T onlevel m, and that zi is the lexicographically least extension of yi in S∗(m)for i < d −1.

Thus (y0, . .

. , yd−1) is colored according to whether (S∗Z, m)accepts b, rejects b, or cannot decide.

Since (R, m) is a pure extension of(S∗Z, m) which accepts b and m ∈A by the property of S∗it follows that(S∗Z, m) also accepts b and thus (y0, . .

. , yd−1) ∈K0.

Hence we must haveǫ = 0.Now since then ⊗Bi

PROOF OF LEMMA 1: Let (S, n) be a condition in C and let b be a Booleanvalue. Assume that there is no pure extension of (S, n) which accepts b. Wefind a pure extension (T, n) of (S, n) which rejects b.

We shall build thetree T by a fusion argument. Along the way we shall construct a decreasingsequence (T (0), a0) ≥(T (1), a1) ≥.

. .

of conditions in A(P).To begin let (T (0), a0) = (S, n). Suppose now (T (k), ak), has been defined.Let {Zi : i < l} be an enumeration of all subsets Z of T (k)(ak) which generatea comb extending S ↾(n+1).

The inductive assumption is that for each suchZ the condition (T (k)Z , ak) does not have a pure extension accepting b. Toavoid excessive notation let R be a variable denoting a perfect subtree ofT (k). We initially set R to be equal T (k) and then trim it down in l stepsas follows.At step i consider Zi.Since (RZi, ak) is a pure extension of(T (k)Zi , ak) from the inductive assumption it follows that it does not have apure extension accepting b.

If the size of Zi is di then by Lemma 3 there is apure extension (Q, ak) of (RZi, ak) such that no extension of (Q, ak) of widthdi +1 accepts b. We now shrink R as follows.

For every s ∈Zi replace Rs byQs and for s ∈T (k)(ak)\Zi keep Rs the same. After all the l steps have beencompleted pick a node y in T (k)(ak).

Let ak+1 be the least a such that y hastwo extensions in R(a). Keep those two extensions of y and for every othernode in T (k)(ak) pick exactly one extension on level ak+1.

Let then T (k+1) bethe set of all nodes of R comparable to one of these nodes. If now Z is anysubset of T (k+1)(ak+1) which generates a comb extending S ↾(n+1) we claimthat there is no pure extension of (T (k+1), ak+1) accepting b.

Notice that theset of all predecessors of members of Z on level ak is listed as one of the Zi.Since between levels ak and ak+1 there is at most one splitting of T (k+1) itfollows that card(Z) ≤di + 1. If the size of Z is di then every pure extensionof (T (k+1), ak+1) is equivalent to a pure extension of (T (k), ak), but by theinductive hypothesis such a condition cannot accept b.

On the other hand ifthe size of Z is di +1 at stage i of the construction of T (k+1) we have ensuredthat no such condition accepts b. This shows that the inductive hypothesisis preserved.Finally let T = T T (k).

Throughout the construction we make the choiceof the points y above which we keep a splitting node carefully to ensure thatthe final tree T is perfect.It follows that no condition (R, m) extending(T, n) accepts b and hence (T, n) rejects b, as desired.✷7

4Proof of Lemma 2In the proof of Lemma 2 we need the following lemma whose proof is almostidentical to the proof of Lemma 3 and is thus omitted.Lemma 4 Let (S, n) ∈C be a condition of width d and let U be a set ofinfinite combs. Then there is pure extension (T, n) of (S, n) such that eitherCω(T, n) is contained in U or there is no extension (R, m) of (T, n) of widthd + 1 such that Cω(R, m) is contained in U.Now note that to complete the proof of Lemma 2 and Theorem 2 it sufficesto prove the following.Lemma 5 Let (S, n) be a condition in A(P) and let D be a dense opensubset of C. Then there is a pure extension (T, n) of (S, n) such that forevery infinite comb C in Cω(T) there is m such that (TC(m), m) ∈D.PROOF: We first show that if (S, n) ∈C there is a pure extension (T, n) of(S, n) such that for every C ∈Cω(T, n) there is m ≥n such that (TC(m), m) ∈D.

To begin find an infinite subset A of ω and a pure extension (S∗, n) of(S, n) such that for every m ∈A and every subset Z of S∗(m) which generatesa comb extending S ↾(n + 1) if there is a pure extension of (S∗Z, m) which isin D then already (S∗Z, m) is in D. Let thenU = {C ∈Cω(S∗, n) : there is m such that (S∗C(m), m) ∈D}Assume now towards contradiction that there is no pure extension (T, n) of(S∗, n) such that Cω(T, n) is contained in U. As in the proof of Lemma 1we build a decreasing sequence (T (0), a0) ≥(T (1), a1) ≥.

. .

of conditions inA(P). To begin set (T (0), a0) = (S∗, n).

Suppose now (T (k), ak) has beendefined. Our inductive assumption is that for any subset Z of T (k)(ak) whichgenerates a comb extending S ↾(n + 1) there is no pure extension (Q, ak)of (T (k)Z , ak) such that Cω(Q, ak) is contained in U.

Let {Zi : i < l} be anenumeration of all such Z. To avoid excessive notation let, as before, R bea variable denoting a perfect subtree of T (k).

To begin set R equal to T (k).We then successively trim down R in l steps as follows. Suppose that stepi has been completed.

Since (RZi, ak) is a pure extension of (T (k)Zi , ak), bythe inductive hypothesis it has no pure extension (Q, ak) such that Cω(Q, ak)8

is contained in U.Let the size Zi be di.Then by Lemma 4 there is apure extension (Q, ak) of (RZi, ak) in C such that if (Q∗, m) is an extension(Q, ak) in C of width di + 1 then Cω(Q∗, m) is not contained in U. Now trimdown R as follows.

For nodes s in Zi replace Rs by Qs and for nodes s inT (k)(ak)\Zi keep Rs the same. Finally when all the stages are completed andwe have taken care of all the Zi we choose a node y in T (k)(ak) and let ak+1be the least member of A above ak such that y has two successors in R(ak+1).Then T (k+1) is obtained from R by keeping those two successors of y and bykeeping for every other node in T (k)(ak) one successors and throwing awaythe remaining ones.

Then T (k+1) is set to be the set of all nodes of the finalR comparable to one of the chosen points. Note that in this way we arrangethat for every subset Z of T (k+1)(ak+1) which generates a comb extendingS ↾(n + 1) the set of all predecessors of members of Z on level ak is listedas one of the Zi and since between ak and ak+1 there is at most one splittingnode it follows that card(Z) ≤di + 1.

Thus it follows that if (Q, ak+1) is apure extension of (T (k+1)Z, ak+1) then Cω(Q, ak+1) \ U ̸= ∅.In then end we let T =T Tk. We make the choice of the nodes y above wechoose a splitting at each stage judiciously so that the final tree T is perfect.It follows that if (R, m) is any extension of (T, n) in C then Cω(R, m)\U ̸= ∅.Now since D is dense open we can find k and a condition (R, ak) ∈Dextending (T, n).

Let Z = R(ak). By the property of S∗it follows that(S∗Z, ak) is also in D. But then Cω(S∗Z, ak) ⊆U, a contradiction.Now to deal with the general case assume that only (S, n) ∈A(P).

Wethen proceed as in the successor stage of the previous case. We enumerateall subset Z of S(n) which generate a comb as {Zi : i < l}.

Let, as before, Rbe a variable denoting a perfect subtree of S. To begin set R to be equal toS. We then trim down R successively in l stages.

At stage i look at Zi andapply the special case of the lemma to find a pure extension (Q, n) of (RZi, n)such that for every infinite comb C extending Q ↾(n+1) there is m ≥n suchthat (QC(m), m) ∈D. Trim down R by replacing Rs by Qs for every nodes ∈Zi and keeping Rs the same for evert s ∈S(n) \ Zi.

We let T be equalto R after all the stages have been completed. It follows that (T, n) ≤(S, n)and for every comb C ∈Cω(T) there is m such that (TC(m), m) ∈D.

Thisfinishes the proof of Lemma 5 and Theorem 2.✷9

RemarksIn this paper we have only considered partitions of strongly increasing se-quences of reals and have shown that every such partition into an analyticand a co-analytic piece has a perfect homogeneous set. A similar result canbe obtained for any other type ≺of increasing sequences.

All we have to dois modify the forcing notion C so that the generic sequence produced has type≺. Consequently, if I is a finite set of types of infinite increasing sequencesof reals for every analytic partitions of infinite increasing sequences of realswe can find a perfect set P such that for every ≺in I all the sequences fromP which have type ≺have the same color.

On the other hand it is easy to seethat if I is any infinite set of types there is a partitions such that no perfectset is homogeneous for all types in I simultaneously. Namely, choose for eachs ∈2<ω a type ≺s in I such that the function which maps s to ≺s is 1-1.Now given a sequence {xn : n < ω} of type ≺s color it red if ∆(x0, x1) = sand blue otherwise.

Let now P be a perfect set and let T be the tree of allfinite initial segments of elements of P. Then for any s which is a splittingnode of T there are sequences from P of type ≺s which are colored by eithercolor.References[Bl]A. Blass, A partition theorem for perfect sets, Proc. Amer.

Math.Soc., vol.82 (1981), pp.271-77[GP] F. Galvin and K. Prikry, Borel sets and Ramsey theorem, Journal ofSymb. logic, vol.38, (1973), pp.193-98[HL] J.D.

Halpern and H. La¨uchli, A partition theorem, Transactions ofthe Amer. Math.

Soc., vol.124 (1966), pp.360-67[Ku] K. Kunen, Set theory, Studies in Logic, Philosophy, and the Method-ology of Science, vol.102, North-Holland, Amsterdam, 1982.[La]R. Laver, Products of infinitely many perfect trees, Journal of theLondon Mathematical Society, vol.

29 (2), 1984, pp.385-9610

[Mi] K. Milliken, A partition theorem for the infinite subtrees of a tree,Trans. Am.

Math. Soc., vol.

263 (1981), pp.137-48[Sa]G. Sacks, Forcing with perfect closed sets, in Axiomatic Set Theory,Proceedings of the Symp. Pure Math., vol.13(II), (D. Scott, ed.

), pp.331-355, American Math. Soc., Providence, Rhode Island 1971Universit´e Paris VI,Hebrew University, Jerusalem, andYork University, Toronto11

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