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### Complementing below recursively enumerable degrees

S.Barry Cooper, Richard L. Epstein
1987 Annals of Pure and Applied Logic
., there is some degree m < a such that m U b = a and m is minimal. We prove below two theorems related to this.  ...  Epstein showed, in fact, that in the special case of 0 < a r.e. < 0', one can construct a minimal degree complement for a.  ...  Complementing below r.e. degrees 21 This completes the verification of our claim.  ...

### Degrees of Weakly Computable Reals [chapter]

Keng Meng Ng, Frank Stephan, Guohua Wu
2006 Lecture Notes in Computer Science
Then 0.χ C is left-computable but not computable. • (Soare) A real α is called strongly recursively enumerable (s.r.e. for short) if there exists a recursively enumerable set A ⊂ ω such that α = n∈A 2  ...  The proofs use the existence of noncuppable degrees and the fact that every nonzero r.e. degree has a 1-generic complement (Slaman and Steel) and a minimal complement (Seetapun and Slaman). • (Wu, 2006  ...  There is a 1-generic degree below 0 such that every nonzero degree below it contains no weakly computable reals. 2.  ...

### Page 1367 of Mathematical Reviews Vol. , Issue 90C [page]

1990 Mathematical Reviews
Lachlan, which asserts that there do not exist such degrees a, b which are recursively enumerable.)  ...  In the paper under review, a new proof that the degrees < 0’ are complemented is given.  ...

### Page 807 of Mathematical Reviews Vol. , Issue 2003B [page]

2003 Mathematical Reviews
Any c.e. set with semilow complement is automorphic to some c.e. set below any given promptly simple degree.  ...  A Turing degree is d.c.e. if it contains a set that is the difference of c.e. (computably enumerable) sets. A d.c.e. degree d is isolated by a c.e. degree a if a is the greatest c.e. degree below d.  ...

### Enumeration reducibility and partial degrees

John Case
1971 Annals of Mathematical Logic
The following theorem complements (2.7). Theorem 2.8. Every r.e. non-recursive Turing degree contains a set A ~'uch that A le -~ and A, A lie in total partial degrees. Proof.  ...  Following [ 5 ] and [6, p. 146] a formal definition is give~ below. In this paper enumeration reducibility and a ge~eralization of enumeration reducibility, arithmetical enumerability, a~e studied.  ...

### Page 5456 of Mathematical Reviews Vol. , Issue 98I [page]

1998 Mathematical Reviews
Downey using known results: Let ap and a; be computably enumerable degrees such that a, a;, and a @a;, are contiguous, ap does not bound half of a minimal pair and every computably enumerable degree below  ...  {For the entire collection see MR 98b:00020. } Peter Cholak (1-NDM; Notre Dame, IN) 98i:03057 03D30 Kumabe, Masahiro Minimal complementation below uniform upper bounds for the arithmetical degrees.  ...

### Page 5035 of Mathematical Reviews Vol. , Issue 82m [page]

1982 Mathematical Reviews
Posner on the degrees below 0 (the latest results on complementing below h high b+ © contained in his article (“Minimal degrees and high degrees”, to appear]). Barry Cooper (Leeds) Mal'cev, An.  ...  enumerable sets there is a total recursive function f(i,j) such that TC W,, RCW,=f(i,j)€ W,U W,.  ...

### On the Degrees of Index Sets

C. E. M. Yates
1966 Transactions of the American Mathematical Society
The latter approach is not only more precise but also, as we show below, provides an alternative method for solving certain problems on recursively enumerable sets and their degrees of unsolvability.  ...  The main result of the present paper is the computation, for every recursively enumerable degree a, of the degree (in fact, isomorphism-type) of the index-set corresponding to the recursively enumerable  ...  . // b is any recursively enumerable degree, then the degree of Gib) is b(3\ Also G(fc)eS3(fe) -n3(fc). Proof.  ...

### Hereditarily retraceable isols

T. G. McLaughlin
1967 Bulletin of the American Mathematical Society
Moreover, if a has recursively enumerable complement then we can satisfy the additional requirement that /3 have recursively enumerable complement.  ...  Such, then, are the objects referred to in the title of the note; the existence of a continuum of them follows from Theorem 1 below. Our terminology is, in all other respects, that of [l], .  ...  Moreover, if a has recursively enumerable complement then we can satisfy the additional requirement that /3 have recursively enumerable complement.  ...

### On the degrees of index sets

C. E. M. Yates
1966 Transactions of the American Mathematical Society
The latter approach is not only more precise but also, as we show below, provides an alternative method for solving certain problems on recursively enumerable sets and their degrees of unsolvability.  ...  The main result of the present paper is the computation, for every recursively enumerable degree a, of the degree (in fact, isomorphism-type) of the index-set corresponding to the recursively enumerable  ...  If sé is any many-one degree that contains an infinite recursively enumerable set whose complement is nonempty, then the degree of Gísé) is 0(3). Also G0OeE3-n3. Proof.  ...

### Page 4723 of Mathematical Reviews Vol. , Issue 87i [page]

1987 Mathematical Reviews
(A Turing degree a is said to be nearly recursive, or bi-immune free, if for any set A of degree < a, either A or its complement has an infinite recursively enumerable subset.)  ...  For instance, it is shown that every e-degree b which is “D2-high” has the property that every “low” e-degree a below b is part of a minimal pair of e-degrees below b.  ...

### Page 464 of Mathematical Reviews Vol. , Issue 86b [page]

1986 Mathematical Reviews
Press, Cam- bridge, 1980; MR 83i:03071], this paper presents a result with striking negative implications for the possible jumps of degrees below high degrees.  ...  Author’s summary (translated from the Russian): “We prove the existence of a recursively enumerable independent subset in a con- structive Boolean algebra, which cannot be complemented.”  ...

### Completeness, the Recursion Theorem, and Effectively Simple Sets

Donald A. Martin
1966 Proceedings of the American Mathematical Society
., of the highest recursively enumerable degree of unsolvability.  ...  A recursively enumerable set of natural numbers is called simple if its complement, though infinite, possesses no infinite recursive subset.  ...

### Completeness, the recursion theorem, and effectively simple sets

Donald A. Martin
1966 Proceedings of the American Mathematical Society
., of the highest recursively enumerable degree of unsolvability.  ...  A recursively enumerable set of natural numbers is called simple if its complement, though infinite, possesses no infinite recursive subset.  ...

### On a Problem of G. E. Sacks

A. H. Lachlan
1965 Proceedings of the American Mathematical Society
A set F is called j-recursive if both F and its complement T arej-enumerable; such a set is to be specified by indices of F and F as /-enumerable sets.  ...  Further, their /-recursiveness is uniform in / and n in the sense that given j and n we can effectively find indices of AnJ, A\J and their complements as/-enumerable sets.  ...
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