A Basis Result for 2° Sets of Reals with an Application to Minimal Covers

It is shown that every 2, set of reals which contains reals of arbitrarily high Turing degree in the hyperarithmetic hierarchy contains reals of every Turing degree above the degree of Kleene's 0. As an application it is shown that every Turing degree above the Turing degree of Kleene*s 0 is a minimal cover. In this paper we consider a particular verification of the following nebulously stated and tenuously held principle: Every easily definable set of reals with enough complicated members contains members from any sufficiently large degree of complexity. Let 0 be the Turing degree of Kleene's 0 Our result is Theorem. Any 2, set of reals, with the property that every hyperarithmetic real is recursive in some member of it, contains reals of every Turing degree above 0. As the first author noticed, an application of this theorem gives a new proof of Jockusch's result that there is a cone of minimal covers and estimates the base of that cone to be 0. 1. Preliminaries. Let tu = 10, 1, 2, . . . | be the set of natural numbers and R = co the set of all functions from a> to co or (for simplicity) reals. Letters i, j, k, I, m, , . „ will denote elements of oj and a, fi, y, 8, o, r, . . . elements of R, We shall use, without explicit reference, standard facts of recursion theory, which can be found, for" example, in [5] or [7]. The basic ingredient in the proof of our main theorem is the use of the well-known determinacy of closed games in its effective form (see, for example, [4]), which we proceed to state. For a general explanation of the connection between degrees of unsolvability and determinacy of games, see Martin [3]. Let (a,, . . . , a ), where a. € co or a.e R be a trivial recursive coding of ra-tuples by reals. For A C R consider the game in which players I, II alternatively choose natural numbers a(0), /3(0), a(l), /3(l), . . . and I wins iff (a, fi)e A. If a e R is a strategy for player I, let a * fi be the Received by the editors April 3, 1974 and, in revised form, July 19, 1974 and September 6, 1974. AMS (MOS) subject classifications (1970). Primary 02F30, 04A15. Copyright © 1975, American Mathematical Society 445 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 446 L. A. HARRINGTON AND A. S. KECHRIS result of l's moves when he follows o against II playing fi. More formally (a * f3)(n) = o(f3(n)). Similarly let a ° r be the result of IPs moves when he follows t against I playing a, so that (a o r)(n) = r(a(w + l)). Then we have Fact (Determinacy of open games). Given a II, set of reals B, either (I) there is a real o recursive in IsacA that for all reals f3, (o * (3, f3) £ B. or (II) there is a hyperarithmetic real r such that for all reals a, (a, a o T) 4 B. 2. Prool ol the theorem We are now ready to prove our main result. Theorem. Any X, set of reals, with the property that every hyperarithmetic real is recursive in some member of it, contains reals of every Turing degree > U. Prool. We first prove this for II, sets of reals. Let thus A CR be a IIj set satisfying the hypotheses of the theorem. Let \e\^, where e £ co, y £ R, denote the cth function from co into to partial recursive in y and put R = \((e, f], y), fi): e£w&.y£A& (Vra)(|c|y(n) converges in exactly rj(n) steps) & (3 = (e, 7/, y) o \e\' \. Clearly B C fi is 11°, so by the fact in §1 either (I) or (U) holds. If (II) holds, let t be a hyperarithmetic real such that for all a £ R, (a, a o T) 4 B. By assumption,there is a y £ A such that r is recursive in y. So for some e £ co, \e\ = t. Let 77(77) = number of steps in which \e\' (n) converges. Let cl (c. 77, y). Then (a, a ° r) 4 B. But clearly (a, a ° r) £ B, a contradiction, so case (I) must hold. Thus there is o recursive in L such that for all (3, (o * f3, (3) £ B. Let C be recursive in f3We shall prove that a real of the same Turing degree as fi belongs to A, Since (o */3, f3) £ B, clearly o * f3 = (e, rf, y) for some e £ co, r), y £ R such that y £ A, \e\ is total and 7/(ra) = the number of steps in which \e\ (ri) converges. Also f3 = (e, 77, y) °\e\ . Thus f3 is recursive in y. But also cr is recursive in f3, thus y is recursive in /3 i.e., y and f3 have the same Turing degree. Since y e A, we are done. The following observation will now complete the proof of the theorem: For any S set of reals Ii there is a II, set of reals A such that for all Turing degrees d, dC\B40*-*d(~}A£0. To prove this, find a recursive set R such that for all a, a £ B «-> liVfikRd, i, a(k)) and let A = (<;, a, (3): Wj(f3(j) = pkR(i, j, a(k)))\. Q.E.D. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use A BASIS RESULT FOR 2° SETS OF REALS 447 Remarks. (1) The above theorem is best possible in the sense that it cannot be improved to include all 11, sets of reals—the set A = ia: iVe)i{e\a is total — \e\a/ F)\, where F is a nonempty U . set with no hyperarithmetic member, is a Il? set of reals which is closed under Turing reducibility and which contains all hyperarithmetic reals and yet which fails to contain Kleene's L. (2) The proof of the theorem can be easily modified to show that Determinacy (2°) *=> Turing Determinacy (2n + ]), where for a collection of sets of reals Y, Determinacy (D <=> VA e Y iA is determined) and Turing Determinacy (T) «=> VA £ Y iA is invariant under Turing equivalence =-» A is determined). This result has been also proved independently (and a little earlier than us) by Ramez Sami (private communication). In fact the proof above shows that if every 2 set is determined, then given a 2 + j set of reals A which is cofinal (i.e., for every a e R there is fi e A s.t. a is recursive in fi), there is a Turing degree d such that A contains reals of every degree > d. (To see this we note first that A may be assumed to be 11° since if a e A » lmA*Um, a)) where A* e U°, clearly for any degree d, A n d £ 0 <=> A IX d 4 0 • Then we define B exactly as in the proof of the theorem above and argue again that II cannot have a winning strategy in the game determined by B since B is cofinal. Then I has a winning strategy, since every 2 game is determined, so by the argument given there, A contains reals of every Turing degree above the degree of a winning strategy a tor player I.) Using this fact, Martin [2] showed that Determinacy(E )=> Turing Determinacy(A + ,) which is best possible in analysis, since he also proved that Analysis -/-> Turing Determinacy(25) (see [2]). 3. An application to minimal covers. Using 2°-determinacy, Jockusch [1] has shown that there is a cone of minimal covers. Harrington noticed that this result follows from our main theorem above (and hence follows from just open determinacy and so, in particular, is a theorem of analysis). This also allows for computing that Kleene's C can be taken as a basis of the cone. Here is Harrington's result. Theorem. Every Turing degree > 0 contains a minimal cover. Proof. By Sacks [6], for every real a, there is a minimal cover of a which is A2 in a, uniformly. Thus there is a In predicate Pia, fi) such that VaVr3(Pa, fi) —► ifi is a minimal cover of a) and Va3!/3P(a, fi). Now A = \{a, fi): Pia, /3)i is clearly a 11° set of reals with the property License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 448 L. A. HARRINGTON AND A. S. KECHRIS that every hyperarithmetic real is recursive in some member of it. It is also a collection of minimal covers By the theorem in §2 every Turing degree >U contains a member of A, thus every Turing degree > 0 is a minimal cover. Q.E.D. We conclude with an open problem raised by Jockusch [l]. Is there a hyperarithmetic real in a cone of minimal covers? In particular, is 0" in such a cone?