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By Vitushkin A. G.

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The next lemma justifies our present interest in trees, and the terminology we employed. Lemma 3 $H iff there is no Souslin tree. Proof: Suppose T is a Souslin tree. is prolific. By the above lemmas and remark, we may assume For each ~ < ~I' T is countable, and for ~ > O, infinite. So, for each non-zero e < ml, let < that of the rationals. be a linear ordering of T Let X be the set of maximal branches of T. linear ordering on X by b < d +-+ ( ~ ) [ ( ~ x (b(e) < of order-type Define a c TI~)(x c h +-+ x e d) & d(~))], where b(e) denotes the unique element of r N b, and similarly for d(e)o It is easily seen that is a dense linear ordering without end-points.

T* = ~a<~ I T"f(a+l). Set Then T*~ Is an (~I , ~l)-tree with no cofinal branch and no uncountable antichain, such that T* satisfies (i) - (iii) in the normality definition. ). One simply picks (by induction on the levels) one limit extension of each limit branch which has an extension. The lemma is proved. Remark By examining the above proof, one can easily amend the definition of the function f so that in the resulting tree, every point has infinitely many successors on the next level. Such a tree is said to be prolific.

In parameter (i) - (vii) and the schema < rank(y). r. by case of function. p iff f is generated by -30- (viii) f(~) = p. Similarly for f rud in parameter p. r. in A iff f is generated by (i) (vii) and the schema (ix) f(~) = x. ~ A, I $ i ~ n. i Similarly for f rud in A. r. (resp. r. (resp. rud) function f: V n ÷ V such that R = {<~>If(~) + ~ }. , Xn>}}; f(~) =~ y ~ x +-+ {y} - x +~ . = x i - x i. Hence, the relation ¢ is rud, since And if f(y, ~) is rug (resp. ), then so is g(y, ~) = , since g(y, ~) = 0zey{}, of course.

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