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Nonisomorphism of lattices of recursively enumerable sets

Published online by Cambridge University Press:  12 March 2014

John Todd Hammond*
Affiliation:
Department of Mathematics, University of Chicago, Chicago, Illinois 60637, E-mail: hammond@zaphod.uchicago.edu

Extract

Let ω be the set of natural numbers, let be the lattice of recursively enumerable subsets of ω, and let A be the lattice of subsets of ω which are recursively enumerable in A. If U, Vω, put U =* V if the symmetric difference of U and V is finite.

A natural and interesting question is then to discover what the relation is between the Turing degree of A and the isomorphism class of A. The first result of this form was by Lachlan, who proved [6] that there is a set Aω such that A. He did this by finding a set Aω and a set C ϵ A such that the structure ({W ϵ AWC},∪,∩)/=* is a Boolean algebra and is not isomorphic to the structure ({W ϵ WD},∪,∩)/=* for any D ϵ . There is a nonrecursive ordinal which is recursive in the set A which he constructs, so his set A is not (see, for example, Shoenfield [11] for a definition of what it means for a set Aω to be ). Feiner then improved this result substantially by proving [1] that for any Bω, B′B, where B′ is the Turing jump of B. To do this, he showed that for each X ⊆= ω there is a Boolean algebra which is but not and then applied a theorem of Lachlan [6] (definitions of and Boolean algebras will be given in §2). Feiner's result is of particular interest for the case B = ⊘, for it shows that the set A of Lachlan can actually be chosen to be arithmetical (in fact, ⊘′), answering a question that Lachlan posed in his paper. Little else has been known.

Type
Research Article
Copyright
Copyright © Association for Symbolic Logic 1993

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References

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