The class of recursively enumerable subsets of a recursively enumerable set

Hay, Louise

doi:10.2140/pjm.1973.46.167

Cited by 10 publications

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References 6 publications

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“…(The context of [2] was not given as a "halting problem", but as the classification of 6AB={e\ We<=B} where IFe=domain{e}; but evidently 6AB=HS, where B denotes the complement of B.) In both [2] and [7] it is shown that it is possible for HA and HA to have different Turing degrees, and in [7] the author conjectures that there exists a set A such that HA and HÄ have incomparable Turing degrees. The purpose of this paper is to prove this conjecture true; such sets occur, in fact, in all Turing degrees a satisfying one of the following conditions : (i) a is an r.e.…”

Section: Introductionmentioning

confidence: 99%

“…It is shown in that paper that the difficulty of the halting problem relativized to A is not a simple function of the "complexity" of A, in the sense that the Turing degree of HA={e\domain{e}C\Aj£0} is, in general, independent of the Turing degree of A ; the same result was obtained independently in [2] for the special case where A is r.e., nonrecursive. (The context of [2] was not given as a "halting problem", but as the classification of 6AB={e\ We<=B} where IFe=domain{e}; but evidently 6AB=HS, where B denotes the complement of B.) In both [2] and [7] it is shown that it is possible for HA and HA to have different Turing degrees, and in [7] the author conjectures that there exists a set A such that HA and HÄ have incomparable Turing degrees.…”

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The halting problem relativized to complements

Hay¹

1973

Proc. Amer. Math. Soc.

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Abstract.Let HA={e\domain{e}r\A^0}. It is shown that there exists a set A of Turing degree a such that HA is Turingincomparable to H-& whenever a is an r.e. degree with a'>0', or ag:0" or a^O' and a is r.e. in 0'. This contrasts with the fact that HA is comparable to H¿ for almost all A.1. Introduction.In [7], the "relativized halting problem" is defined to be the problem of deciding, for a fixed set A, whether domain {e} O A 5¿ 0, where {e} denotes the eth partial recursive function. It is shown in that paper that the difficulty of the halting problem relativized to A is not a simple function of the "complexity" of A, in the sense that the Turing degree of HA={e\domain{e}C\Aj£0} is, in general, independent of the Turing degree of A ; the same result was obtained independently in [2] for the special case where A is r.e., nonrecursive. (The context of [2] was not given as a "halting problem", but as the classification of 6AB={e\ We<=B} where IFe=domain{e}; but evidently 6AB=HS, where B denotes the complement of B.) In both [2] and [7] it is shown that it is possible for HA and HA to have different Turing degrees, and in [7] the author conjectures that there exists a set A such that HA and HÄ have incomparable Turing degrees. The purpose of this paper is to prove this conjecture true; such sets occur, in fact, in all Turing degrees a satisfying one of the following conditions : (i) a is an r.e. degree with a'>0';

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Section: Introductionmentioning

confidence: 99%

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confidence: 99%

The halting problem relativized to complements

Hay¹

1973

Proc. Amer. Math. Soc.

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“…THEOREM [8]. Let ~ be a nonrecursive recursively enumerable 7" -degree and let ,~, =lc: t_q,,t,(A~'~,,t c)}, It follows from these two theorems that each nonrecursive recursively enumerable ~degree ~ ~0 r is covered by an infinite number of recursively enumerable ~ -degrees, none of which is contained in ~ .…”

Section: It Is Obvious That A-~ a Qmentioning

confidence: 99%

“…Hay [8] has remarked that Jockusch has constructed an example of an effectively simple set ~ such that A ~£ 0 r, COROLLARY. Let ~ be a recursively enumerahle set such that the ~ -degree of ~ is The proposition is proved.…”

Section: Then Andmentioning

confidence: 99%

Upper semilattice of recursively enumerable 124-1124-1124-1-degrees

Omanadze¹

1984

Algebra and Logic

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“…A where H A 5 0'. This last result is a special case of HAY'S[5] analogue to SACKS' Jump Theorem [13] where weak jump ( H A ) replaces the usual jump (A'). .…”

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Degrees of Non α‐Speedable Sets

Homer¹,

Jacobs

1981

Mathematical Logic Qtrly

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Q 0. Introduction I n [ll] MARQUES showed, using a simple priority argument, that every recursively enumerable degree contains a nonspeedable recursively enumerable set. SOARE [ 181 discovered a recursion theoretic characterization for nonspeedable sets and used this to extract several properties of speedable and nonspeedable sets. I n particular, SOARE was able to provide a shorter nonpriority based proof of MARQUES' theorem.Complexity oriented properties of computations on natural numbers do not always generalize to computations on admissible ordinals. For instance, it was shown in [7]that no natural analogue to GRZEOORCZYK'S well studied complexity hierarchy exists for admissibles greater than W . Also, for many admissibles (e.g. 01 = N , ) the only generalized speedable sets are the complete ones. This last result demonstrates that properties discovered by MARQUES [ll] and SOARE [18] fail to generalize. The question of whether an analogue of MARQUES' theorem could be obtained for a-recursion theory was first proposed by JACOBS [8]. Here the notion of nonspeedable is generalized in two different ways to yield two versions of non a-speedable &-recursively enumerable set. It was observed in [8] that SOARE'S semilow characterizationfor nonspeedable sets ( A is nonspeedable iff HX = ( E I R, r\ A } 5 0') generalized only in case A was regular. Consequently, a strategy guided by SOARE'S proof of MARQUES' theorem had to be abandoned. I n this paper we follow the direction originally proposed by MARQUES. In fact, we show that for either notion of non a-speedable, there is an non a-speedable set in every @-degree.The content of the paper is a follows: I n $ 1 the basic definitions of a-complexity theory are presented. We prove in $ 2 our main theorem which is essentially another version of the splitting theorem for a-recursion theory. Several possible strengthenings of this main theorem are discussed. Following a series of lemmas, we prove in 0 3 that every a-r.e. a-degree contains a non a-speedable a-r.e. set for both notions of non a-speedable. We conclude the paper in $ 4 with several open questions. Q 1. Preliminary Notions more detailed discussion we refer the reader to [14] and [17].We present here a brief account of the basic notions of a-recursion theory. For a l )

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The class of recursively enumerable subsets of a recursively enumerable set

Cited by 10 publications

References 6 publications

The halting problem relativized to complements

The halting problem relativized to complements

Upper semilattice of recursively enumerable 124-1124-1124-1-degrees

Degrees of Non α‐Speedable Sets

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