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A splitting theorem for simple Π11 Sets

  • James C. Owings (a1)

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As was first mentioned in [3, §5], if A is any set, A is the union of two disjoint sets B(0), B(1). In metarecursion theory this is proven as follows. Let ƒ be a one-to-one metarecursive function whose range is A, let R be an unbounded metarecursive set whose complement is also unbounded, and set B(0) = f(R), B(1) = f(). The corresponding fact of ordinary recursion theory, namely that any r.e. but not recursive set can be split into two other such sets, was proved by Friedberg [2, Theorem 1], using a clever priority argument. Sacks [7, Corollary 2] then showed that any r.e. but not recursive set is the union of two disjoint r.e. sets neither of which was recursive in the other, a much stronger result. In this paper we attempt to prove the analogous result for sets A, but succeed only in the case A is simple; i.e., the complement of A contains no infinite subset. As a corollary we show the metadegrees are dense, a fact already announced by Sacks [8, Corollary 1], but only proven by him for nonzero metadegrees.

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[1]Driscoll, G. C. Jr., Metarecursively enumerable sets and their metadegrees, this Journal, vol. 33 (1968), pp. 389411.
[2]Friedberg, R. M., Three theorems on recursive enumeration, this Journal, vol. 23 (1958), pp. 309316.
[3]Kreisel, G. and Sacks, G. E., Metarecursive sets, this Journal, vol. 30 (1965), pp. 318338.
[4]Owings, J. C. Jr., Topics in metacursion theory, Ph.D. thesis, Cornell University, 1966.
[5]Owings, J. G. Jr., Π11 sets, ω-sets, and metacompleteness, this Journal, vol. 34 (1969), pp. 194204.
[6]Owings, J. C. Jr., The metarecursively enumerable sets but not the Π11 sets, can be enumerated without duplication, this Journal, vol. 35 (1970), pp. 223229.
[7]Sacks, G. E., On the degrees less than O′, Annals of Mathematics, vol. 77 (1963), pp. 211231.
[8]Sacks, G. E., Metarecursion theory, Sets, models, and recursion theory, Leicester Symposium Volume, North-Holland, Amsterdam, 1965, pp. 243263.

A splitting theorem for simple Π11 Sets

  • James C. Owings (a1)

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