Embedding Trees in the Rationals

Proceedings of the National Academy of Sciences Vol. 67, No. 4, pp. 1748-1753, December 1970 Embedding Trees in the Rationals J. Baumgartner,* J. Malitz, and W. Reinhardt DARTMOUTH COLLEGE,* HANOVER, NEW HAMPSHIRE; AND DEPARTMENT OF MATHEMATICS, UNIVERSITY OF COLORADO, BOULDER, COLO. 80302 Communicated by S. Ulam, September 4, 1970 Abstract. An example is presented of a simple algebraic statement whose truth cannot be decided within the framework of ordinary mathematics, i.e., the statement is independent of the usual axiomatizations of set theory. The statement asserts that every tree-like ordering of power equal to or less than the first uncountable cardinal can be embedded-homomorphically into the rationals. In this paper, we show (see also refs. 1-3) that the following assertion A, which implies Souslin's hypothesis, is independent of the axioms of set theory (more precisely, both A and its negation are relatively consistent with ZFC, the Zermelo-Fraenkel axioms plus the axiom of choice). A. If T is a tree-like partial order in which every chain is countable and if T has cardinality N1, then there is an embeddingf: T-V. Q (Q the rational numbers) such that x < y impliesf(x) < f(y). The following fragment of A already implies Souslin's hypothesis. B. Every Aronszajn tree can be embedded in the rationals. Questions about embeddings of trees into linear orders have been discussed before (see Kurepa4 and Johnston5). Definition 1: Let P be a partial order. (a) P is tree-like iff for each q E P, P, = { x E P: x < q} is linearly ordered. P is a tree iff each Pq is well-ordered. (b) x,y E P are compatible if there is r C P with xy < r, otherwise x,y are incompatible. (c) C C P is an antichain iff for all x,y C C, if x $ y then xy are incompatible. (d) C C P is a chain iff C is linearly ordered. (e) If T is a tree, the order l(t) of t C T is the order type of { s E T: s < t}. (f) An Aronszajn tree is a tree with no uncountable chains in which each level {t: I(t) = a} (a C w') is countable, such that above each point t there are points s of every countable order > 1(t). (g) A Souslin tree is an Aronszajn tree which contains no uncountable antichains. (h) P is embeddable in the rationale iff there is a function f: P Q preserving strict order, i.e., such that x < y impliesf(x) <f(y). Note thatf is not required to be one-one. We will write card x for the cardinality of x. 1748

VOL. 67, 1970 TREE-EMBEDDING IN RATIONALS 1749 Notice that if P is tree-like then C C P is an antichain iff for all xy C C, neither x < y nor y < x. Souslin's hypothesis (SH) asserts that there are no Souslin trees. It is easy to see that if T is an Aronszajn tree then card T = Rl. Hence A implies B. Next we assert that B implies Souslin's hypothesis. Suppose T is a Souslin tree and f: T -- Q is anl embedding. Then T = U { Tr: r E Q}, where Tr = { p E T: f(p) = r}, so some Tr is uncountable. But it is easy to see that each Tr is an antichain, contradicting the fact that T is Souslin. Since the negation of Souslin's hypothesis is known to be consistent (see Jech6 and Tennenbaum7), it follows that the negation of B (and hence the negation of A) is consistent. In Theorem 4 below, we show that A follows from Martin's Axiom together with K, < 2Po. Since Martin's Axiom is known to be consistent with N, < 2No (see Solovay and Tennenbaum8), A is consistent, and hence A is independent of the axioms of set theory. Baumgartner first showed the consistency of A for trees; Malitz and Reinhardt independently of Baumgartner (but not of each other) showed this and the consistency of A. Later Baumgartner observed that A follows from the special case for trees. Several other results concerning embeddings and decompositions of trees have been obtained by Baumgartner and will appear in another paper. (Some of these results are announced in ref. 2.) Laver has shown (in ZFC) that there is a tree of power 2Ho embeddable in the reals but not in the rationals and such that each uncountable subset contains an uncountable antichain. (Subsequently Galvin found an example of such a tree which cannot be embedded in the reals. Proofs of these results may be found in ref. 1.) Thus the negation of A This, together with Jensen's recent result (un- is provable in ZFC + 2No = K. published) that SH is consistent with 21o = K, shows that A does not follow from SH. The question whether B follows from SH is open. The Laver example also shows that RI cannot be replaced by 2PO in A. The following partial order (due to Sierpifiski) shows that the condition in A that T be tree-like cannot be weakened to T an arbitrary partial order. Let P = { re,: a C coi} be an uncountable set of distinct reals. Put ra < r, iff ra < r, in the usual ordering of the reals, and also a < B3. It is easy to see that every set of pairwise incomparable elements of P is countable and hence, as in the proof that B implies SH, P cannot be embedded in the rationals. Definition 2: (a) Let P be a partial order. A set D is called P-dense if D C P and for all p P there is q > p such that q E D. (b) A subset G of P is called P-generic over M iff the following three conditions hold. (i) if p,q E G then there is r E G with p,q < r (ii) if q & G and p < q thenp E G (iii) if D C M is P-dense, then G n D $ 0. Martin's Axiom. If P is a partial order in which every antichain is countable, and if M is any set of cardinality <2NO, then there is a set G C P such that G is P- generic over M. For a general discussion of Martin's Axiom and its consequences, see Martin and Solovay9.

1750 MATHEMATICS: BAUMGARTNER ET AL. PRoc. N. A. S. If T is a partial order and p C T X Q is a finite function, then we say that p is a partial embedding if x,y C dom p and x < y implies p(x) < p(y). With any partial order T we associate the set P of partial embeddings, ordered by inclusion. If T is the order mentioned in A, then it is the partial order P to which we apply Martin's Axiom. The only difficult point is to see that the antichain condition in Martin's Axiom is satisfied. This is taken care of by Theorem 3. We first give the proof that Theorem 4 follows from Theorem 3. THEOREM 3. Let T be a tree-like partial order which contains no uncountable chains. Let P be the partial order of partial embeddings of T described above. Then every antichain contained in P is countable. THEOREM 4. Assume Martin's Axiom. Let T be any tree-like partial order in which every chain is countable, and card T < 20o. Then there is a function f: T-+Q such that x < y impliesf(x) < f(y). Proof of 4: Let P be the set of partial embeddings of T. For each x & T let D= {p C P: x & dom p}. We claim each D1 is P-dense. Suppose p & P andx Xdomp. LetX = lp(y):yedompandy<x} andlety =p(y):y E dom p and x < y}. Since p P it follows that if r1 C X and r2 C Y then r1 < r2. Since Q is densely ordered there exists r C Q such that for all ri C X, r1 < r and for all r2 C Y, r < r2. Then it is easy to see that p U { (xr)} Dz, and p U (x,r) } extends p. Hence D. is P-dense. Let M = { D.: x C T}. Clearly card M = card T < 2Po. By Theorem 3, P has no uncountable antichains, so we may apply Martin's Axiom to obtain a set G which is P-generic over M. We assert that U G is the desired embedding. It is easy to see that U G is a function since G satisfies condition (i) of the definition of generic. For each x C T we have G nf D5$ 0, so x C dom U G and hence dom U G = T. If x,y C T and x < y then there are pq G such that x C dom p and y C dom q. But then U G(x) = p(x) < q(y) = U G(y) since p and q must be compatible. This completes the proof of Theorem 4. Proof of 3: We shall prove by induction on n that * If B C P is an antichain and card p = n for all p C B, then B is countable. We assume now that * is true for m < n. We proceed by contradiction. Let B C P be an antichain with card p = n for all p C B, and suppose B is uncountable. We split the induction step of the proof into a number of steps; throughout these steps, B and n are fixed as above. Some notation will be useful. If card p = m, then p = { Po, *.., Pm-l}, where each pi is an ordered pair which we shall write as pi = (f(i), p'(i)). Thus P enumerates the domain of p, p' the range of p, and p(p(i)) = p'(i). So that P will be uniquely determined by p, we linearly order T (in any way) and require that pj enumerate dom p in increasing order. For any D C P we introduce the notation D' = { P(i): p C D}. If p C P, card p = n and i < n, we write pi = p- P = (p(j), p'(j)):j < n and j $ 4; we havep C Pandcard n-1. Notice that if x E T then { y E T: y < x} is a chain since T is tree-like, and hence must be countable. This observation will be used frequently in the rest of the proof.

VOL. 67, 1970 TREE-EMBEDDING IN RATIONALS 1751 Step 1: Let C C P be uncountable. Then there is some D C C, D uncountable, such that (a) for all p,q E D, p' = q' (b) if Ci is countable, then for all p,q E D, N(i) = q(i). Proof: {p': p E C} is countable, and to say that Ci is countable is to say that {p(i): p C C} is countable. Step 2: Suppose that C C B is uncountable. Then for each i < n, Ci is uncountable. Proof: Suppose Ci is countable, and i < n. Then step 1 allows us to choose D C C, D uncountable, and z C T such that for all p,q E D, p' = q' and p(i) = z. Evidently D is still an antichain. Let E = {pi: p C D}. Now we claim that if p,q C D then pi is incompatible with qi. From the claim it follows that the function p - pi is one-one, so card E = card D and E is an uncountable antichain with card pi = n - 1 for all pi E E. This contradicts the induction hypothesis. To prove the claim, first observe that if p,q are incompatible, then there are x E dom p, y C dom q so that either (i) x = y and p(x) $ q(y), (ii) x < y and p(x) > q(y) or (iii) y < x and p(x) < q(y). Now suppose that p,q C D. In case (i), if x = y $ z, thenx E dompi, y E dom q' and p2,q2 are clearly incompatible. If x = y = z, then case (i) does not arise because p'(i) = q'(i). In case (ii), if x,y $ z it is again clear that pi,qi are incompatible. If x = z, then we have (a) z < y and p(z) 2 q(y). But since p' = q' and z = p(i) = q(i), p(z) = p'(i) = q'(i) q(z), and we have (b) q(z) > q(y). Now (a) and (b) contradict q E P, so x = z cannot arise in case (ii) either. Similarly, y = z cannot arise. Case (iii) is similar to case (ii), so the claim is proved. Let U(C) denote the following condition on a subset C of B: U(C): For every i < n and every x E C2 there is a unique p C C such that x = Pf(i). Step 3: Suppose that C C B is uncountable. Then there is an uncountable D C C such that U(D) holds. Proof: By induction on i, suppose that E C C is uncountable and for all j < i, and all x E EJ, there is a unique p E E with x = p(j). By step 2, Ei is uncountable. For each x E E' we choose exactly one p CE E so that Px(i) = x, and we set D = I px: x E E'i. Then D is uncountable and satisfies the uniqueness condition for j < i. Step 4: Suppose that C C B is uncountable. Then there is an uncountable D C C such that for each i < n either Di is an uncountable antichain or D' contains no uncountable antichain. Proof: We construct a sequence C0 D C1 D.. D Ce of uncountable subsets of C as follows: Let C0 = C. Suppose we have constructed C1. If S C Cii is an uncountable antichain, then for each x C S choose P EC Ci so that pf(i) = x, and let Ci+i = Spa: x E S}. If Ci2 contains no uncountable antichains, let

1752 MATHEMATICS: BAUMGARTNER ET AL. PROC. N. A. S. C+, = Ca. Then C+1 is uncountable, and either (Ci+1) is an uncountable antichain or (Ci+1)1 contains no uncountable antichains. Let D = C,. It is clear that D satisfies the assertion in step 4. Step 5: Suppose R C T and R = Ro U R1 U... U Rm-, where each RJ is uncountable. Suppose further that R contains no uncountable antichains. Then there exist distinct xo, xi,..., xm-i C R such that { x>: j < m} is an antichain and for each j < m, { y C Rj: xj < y} is uncountable. Proof. We proceed by induction on m. Suppose m = 1. Let S C R be a maximal antichain. Then S is countable and every member of R is comparable with some member of S. Moreover, for each x E S the set { y C R: y < x4 is countable. Therefore, for some xo C S, { y C R: xo < y} is uncountable. Suppose m = 2. Let S' = { x E Ro: Iy E Ro: x < y} is uncountable}. Then S' is uncountable, for if not then Ro- S' is uncountable and by the case just proved for m = 1 there would exist y C Ro- S' such that { z C Ro- S': y < z4 is uncountable, and then y E S', a contradiction. Since S' is uncountable, it cannot be a chain. Hence we may choose incomparable yo,y EC S'. If { z E RIi:y < z4 is uncountable, then, letting xo = yo and xi = yi, we are done. If not, then R1' = { z e R,: z is incomparable with yi} is uncountable, and by step 5 for m = 1 there is y e R,' so that {z e R1': y < z} is uncountable. But then y and Yi are incomparable, so we let xo = Yi and xi = y. Suppose m > 2. Let R' = Ro U R1 U... U Rm-2. By inductive hypothesis there exist mutually incomparable xo', xi',..., Xm-2' e R' so that Iy e Rj: x,' < y } is uncountable for each j < m - 1. Suppose that for some j < m - 1 {y e Rm-i: x' < y } is uncountable. Then, applying step 5 for m = 2 to { y e R: X1 < yj}u{ e Rm-: x1' < y } we obtain incomparable yo,yi 2 x1' so that {y erj: yo < } y and fy e Rm-i: Yi < y } are both uncountable. Let xi = xl' for all i < m - 1, i 5 j; let xj = yo and xmi = Yi- Since T is tree-like Ix,: j < m } is an antichain. The second condition is clear. If the case above does not occur, then for all j < 1 y E Rm-: Xj' < y } is countable. Then R' = Y E Rmi-: y is incomparable with each xj' } is uncountable, and by step 5 for m = 1 there is xm-i e R' so that {y E R': xmi- < y} is uncountable. Let xj= Xj' for allj < m-1. This completes the proof. Step 6: If C C B is uncountable, then for every natural number m there is a set D C C such that card D > m and for each i < n DI is an antichain. Proof: By steps 3 and 4 we may assume that U(C) holds and that for each i < n either Ci is an uncountable antichain or else Ci contains no uncountable antichains. (By step 2, of course, each Ci is uncountable.) Let H = {i < n: CI contains no uncountable antichains }. If H is empty then it is clear that D = C works. Suppose card H = k > 0. Let ho,..., hk-i enumerate H and let m be fixed. It is now a simple matter to use step 5 repeatedly to construct simultaneously a sequence (xip: i < k, j < m) of elements of T and a sequence (Cij: i < k, j < m) of subsets of e such that the following conditions are satisfied: (1) CO1= C for each j < m (2) For each i < k, xio,...., xm-i are incomparable members of T such that for each j < m, { y C (Cij)ht: Xij < yj is uncountable, and hence {p C Cij: xj < P(hi) I is uncountable

VOL. 67, 1970 TREE-EMBEDDING IN RATIONALS 1753 (3) For each i with 0 < i < k, Coj = {p E Ci-i,j: xi-ltj <P(hK-i). Finally, for each j < m, choose pj E Cp. We claim D = {pj: j < m} works. Suppose ji < j2 < m and i < n. We must show that pf1(i) and pf,(i) are incomparable. If i X H then this follows from the fact that U(C) holds and CI is an antichain. If i E H then i = hi, for some io < k. But then pj#(i) 2 xtiil, Pj2#) > xioj, and xij and xioj2 are incomparable, so since T is tree-like ph2(i) and Pi2(i) are incomparable. Step 7: Suppose that C C B is uncountable and U(C) holds. Then there is no set D C C such that card D > n2 and DI is an antichain for each i < n. Proof: Suppose otherwise. Since U(C) holds (and since { y: y < x} is countable for all x E T) the set E = {p CE C: P(i) < q(j) for some ij < n and q E D} is countable. Letp C C - E and let q E D. Since q must be incompatible with p, it follows that there is a pair (ij) such that q(i) <P(j). But card D > n2, so by the familiar pigeonhole principle there exist ql,q2 C D and a pair (ij) with q, # q2 and ql(i),2(i) < PU). But then q1(i) and q2(i) are comparable since T is tree-like, and this is a contradiction. Steps 3, 6, and 7 are contradictory, and therefore * holds for n. This completes the induction and proves Theorem 4. The following problems are open: 1. Is it consistent with ZFC to assume that for any partial order P, if card P = R, and every uncountable subset of P contains an uncountable antichain, then P can be embedded in the rationals? 2. (Galvin, unpublished) Is it consistent with ZFC to assume that for any partial order P, if card P = K and if every P' C P with card P' C K can be embedded in the rationals, then P can be embedded in the rationals? 3. (Baumgartner) Is it consistent with ZFC + A = Rl to assume that every Aronszajn tree is embeddable in the rationals? This work was supported in part by NSF grants 2551-59, 2551-61, 2551-67. 1Baumgartner, J., Results and Independence Proofs in Combinatorial Set Theory, Ph.D. thesis, University of California, Berkeley, 1970. 2 Baumgartner, J., "Decompositions and Embeddings of Trees," Notices Amer. Math. Soc., 17, 967 (1970). 3 Baumgartner, J., J. Malitz, and W. Reinhardt, "Embedding Trees in the Rationals," Notices Amer. Math. Soc., 17, 966 (1970). 4Kurepa, G., "Ensembles Lin6aires et une Classe de Tableaux Ramifies," Publ. Math. Univ. Belgrade, 6-7, 129-160 (1937-38). 5 Johnston, John B., "Universal Infinite Partially Ordered Sets," Proc. Amer. Math. Soc., 7, 507-514 (1956). 6 Jech, T., "Non-provability of Souslin's Hypothesis," Comment. Math. Univ. Carolinae, 8, 291-305 (1967). 7Tennenbaum, S., "Souslin's Problem," Proc. Nat. Acad. Sci., USA, 59, 60-63 (1968). Solovay, R. M., and S. Tennenbaum, "Iterated Cohen Extensions and Souslin's Problem," in press. 9 Martin, D. A., and R. M. Solovay, "Internal Forcing Extensions," Ann. Math. Logic, in press. 10 Cohen, P. J., Set Theory and the Continuum Hypothesis (New York, Benjamin, 1966).