Yol. XXIX, i977 315

Groups cannot be Souslin ordered

By

SAB Il~E KOPPELBEI~G

A totally ordered set X is said to be Souslin ordered if it is densely ordered (for x, y ~ X s.t . x < y there is some z ~ X s.t. x < z < y) without endpoints, satisfies c. e.c. (the countable chain condition; every set of pairwise disjoint non-void open intervals of X is countable) but is not separable (has no countable subset which is dense in the order topology). I t was shown, e.g. by Jensen [3], that the existence of a Souslin continuum (a complete Souslin order) is consistent with the axioms of set theory. Note that the existence of a Souslin continuum is equivalent to the existence of a Souslin order, since the completion by cuts of a Souslin order is a Souslin continuum. Hence, in certain models of set theory there are completely and densely ordered sets satisfying c. c.e. which are not isomorphic to the real line. So it may be interesting to ask whether a Souslin order may be the underlying order of an ordered field or of some more general algebraic structure.

This is certainly impossible if the Souslin order is complete, since every completely ordered field is order isomorphic to the field of real numbers. By a theorem of Loonstra [2], even every completely ordered group is isomorphic to the additive group of real numbers. Thus it seems that the hypothesis of completeness is too strong and should be dropped. Recently, U. Felgner showed that an abelian group cannot be Souslin ordered. Here we give a simple proof obtained independently of Feigner's work that an arbitrary group cannot be Souslin ordered.

I f G is a group with multiplication., we denote by e the unit element of G and by a -1 the inverse of a. G is said to be ordered by =< ff =< is a total ordering of G s.t. for x, y, z e G, x < y implies zx < zy and xz < yz. A subset H of G is convex if, for x, y ~ H and z e G, x --< z --< y implies z e H. Our main tool is

Lemma 1. Let G be an ordered group ~atis/ying c.c.c, and H a non-trivial convex subgroup o/G. Then G/H, the set o / le / t cosets o / G w. r.t. H, is countable.

Proo f . For every a e G, the left translation mapping x e G onto a x ~ G is an order preserving permutation o f G. Since H is convex and non-trivial, H and every coset a H are convex open subsets of G. So G/H is a partition of G into disjoint non-void open subsets which must be countable by c. c. c.

316 S. KOPPELBEI~G ARCH. MATH.

Lemma 2. Let G be an ordered group satls/ylng c.c.c. Suppose that G is not densely ordered. Then G is separable.

Proo f . There are a, b e g s.t. a ~ b and there is no c e G s.t. a ~ c ~ b. Put h ---- a-1 b. Then h is the smallest positive element of G, and H, the cyclic subgroup of G generated by h, is convex. H and, by Lemma 1, G/H are countable, so G is countable, hence separable.

Lemma 3. Let G be an ordered group satis/y~ng c.c.c, and H a non-trivial convex subgroup o/G. I] H is separable, G is separable, too.

Proof . By Lemma 2, G and hence H may be assumed to be densely ordered. Suppose S is a dense countable subset of H. Let R be a set of representatives of G/H. Thus, by Lemma 1,

D---- ( r s [ r e R , s e S }

is countable. D is a dense subset of G: let a, b ~G and a < b. Choose r ~ B n a l l . I f a H ---- bH, it follows that r- la , r - lb ~ H and r - l a < r- lb. So there is some s ~ S s.t. r - l a < s < r- lb , and a < rs < b. I f a H 4= bH, a H < b H (which means that x < y for arbitrary x ~ a l l , y ~ bH) since a l l , b H are convex disjoint subsets of G. There is some s ~ S s.t. r - l a < s. So a < rs, and rs < b follows by rs ~ r H ---- a l l , b e b H .

Lemma 4. I / H is an archimedian ordered group (i. e. /or every a, b ~ H s.t. e < a, b there i8 some Tositive integer n s.t. b ~ an), H is separable.

Proo f . By a theorem of H51der [2], H is, up to an order isomorphism, a subgroup of (R, -F), the additive group of real numbers. The order topology of ]R has a count- able base, hence H has, in the topology induced by R, a countable dense subset D. D is dense in the order topology of H, too.

Theorem. Let G be an ordered group satis/ylng c.c.c. Then G is separable.

Proo f . By the preceding lemmas, we may assume that G is densely ordered and that G has no non-trivial archimedian convex subgroup. Thus every non-trivial convex subgroup of G has a non-trivial proper convex subgroup. Choose a strictly decreasing sequence of convex subgroups H of G for all ordinals ~ less than a certain ordinal 20: put Ho ---- G; if 2 is a limit ordinal and H= is non-trivial for every ~ ~ 2. put HA = ( ~ H~; if H= is non-trivial, let H=+I be a proper non-trivial convex sub-

group of H=. Let 20 be the least ordinal :r s.t. H~ is trivial ; clearly, 20 is a limit ordinal. 2o must

be countable, for ortherwise, N1 ~ 20 and, for every :r ~ N1, we may choose some positive a~ from H~\H~+I. For :r < ~ < b~l, a s < a~ < a~ -( a~ holds, so we have b;1 pairwise disjoint non-void open intervals (a~, a~) in G, contradicting c. e. c.

Vol. XXIX, 1977 Groups cannot be Souslin ordered 317

Let, for a < 20, R~ be a set of representat ives of G/H~ and pu t

D = U R~. r < ) -o

By Lemma 1, D is countable. We claim t h a t D is a dense subset of G. For, let a, b e G and a < b. So e < a- lb and there is some ~ < 2o s.t . a-lb ~ H:r Thus, aH~ < bH~. Since H~+t is a proper convex subgroup of H~, there is some h e H~ s.t. H~+I < h. Choose some d ~ ahH~+l, d < b, since d ~ ahH~+l C ahH~ = aH~ and b ~ bH~. And a < d, since H~+I < hH~+l, aH~+t < ahH~+l, a ~ aH~+l and d ~ ahH~+l.

References

[1] U. Fv.LG~,~, ])as Problem yon Souslin fiir geordnete algebraische Strukturen. To appear in: Second Conference on Set Theory and Recursion Theory, Bierutowicze 1975, LNM.

[2] L. I~vcHs, Partially ordered algebraic systems. Oxford-London-New York-Paris 1960. [3] R. B. J~s~N, The fine structure o f the constructible hierarchy. Ann. Math. Logic 4,

229--308 (1972).

Eingegangen am2.6.1976

Anschrift des Autors:

Sabine Koppelberg II. Mathematisches Institut K6nigin-Luise-Stral~e 24--26 D-1000 Berlin 33 West Germany