Turing incomparability in Scott sets

For every Scott set \mathcal F and every nonrecursive set X in \mathcal F, there is a Y \in \mathcal F such that X and Y are Turing incomparable.

Saved in:
Bibliographic Details
Published in:Proceedings of the American Mathematical Society 2007-11, Vol.135 (11), p.3723-3731
Main Authors: KUCERA, Antonin, SLAMAN, Theodore A
Format: Article
Language:English
Subjects:
Approximation
Arithmetic
Degree of unsolvability
Exact sciences and technology
General mathematics
General, history and biography
Logical theorems
Mathematical logic
Mathematical sets
Mathematical theorems
Mathematics
Randomness
Recursion theory
Sciences and techniques of general use
Citations: Items that cite this one
Online Access:Get full text
Tags: Add Tag
No Tags, Be the first to tag this record!
Description
Summary:For every Scott set \mathcal F and every nonrecursive set X in \mathcal F, there is a Y \in \mathcal F such that X and Y are Turing incomparable.
ISSN:0002-9939
1088-6826
DOI:10.1090/S0002-9939-07-08871-5