1983•Transactions of the American Mathematical SocietyOpen access

Persistently finite theories with hyperarithmetic models

Terrence S. Millar

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Abstract

Nerode asked if there could be a complete decidable theory with only finitely many countable models up to isomorphism, such that not all of the countable models were decidable. Morley, Lachlan, and Peretyatkin produced examples of such theories. However, all the countable models of those theories were decidable in 0 ′ 0’ . The question then arose whether all countable models of such theories had to be, for example, arithmetic. In this paper we provide a negative answer to that question by showing that there are such examples with countable models of arbitrarily high hyperarithmetic degree. It is not difficult to show that any countable model of a hyperarithmetic theory which has only finitely many countable models must be decidable in some hyperarithmetic degree.

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Nerode asked if there could be a complete decidable theory with only finitely many countable models up to isomorphism, such that not all of the countable models were decidable. Morley, Lachlan, and Peretyatkin produced examples of such theories. However, all the countable models of those theories were decidable in 0 ′ 0’ . The question then arose whether all countable models of such theories had to be, for example, arithmetic. In this paper we provide a negative answer to that question by showing that there are such examples with countable models of arbitrarily high hyperarithmetic degree. It is not difficult to show that any countable model of a hyperarithmetic theory which has only finitely many countable models must be decidable in some hyperarithmetic degree.

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Available abstract

Nerode asked if there could be a complete decidable theory with only finitely many countable models up to isomorphism, such that not all of the countable models were decidable. Morley, Lachlan, and Peretyatkin produced examples of such theories. However, all the countable models of those theories were decidable in 0 ′ 0’ . The question then arose whether all countable models of such theories had to be, for example, arithmetic. In this paper we provide a negative answer to that question by showing that there are such examples with countable models of arbitrarily high hyperarithmetic degree. It is not difficult to show that any countable model of a hyperarithmetic theory which has only finitely many countable models must be decidable in some hyperarithmetic degree.

Key concepts: Countable set, Decidability, Mathematics, Isomorphism (crystallography), Discrete mathematics, Algorithm, Crystallography, Chemistry

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