Difference between revisions of "Complexity and classification of countable models"

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Let $T$ be a completion of $PA$. It is not hard to see that the isomorphism problem for finitely generated models of $T$, $\cong^{fg}_T$, is Borel.
 
Let $T$ be a completion of $PA$. It is not hard to see that the isomorphism problem for finitely generated models of $T$, $\cong^{fg}_T$, is Borel.
  
Coskey and Kossak proved that $\cong^{fg}_T$, is essentially countable and  $E_0\leq_B \cong^{fg}_T$ i.e. $\cong^{fg}_T$ is not  smooth. Is $\cong^{fg}_T$ hyperfinite? In other words, is $\cong^{fg}_T$ Borel reducible to $E_0$?
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Coskey and Kossak <cite> coskeykossak2010:thecomplexity </cite> proved that $\cong^{fg}_T$, is essentially countable and  $E_0\leq_B \cong^{fg}_T$ i.e. $\cong^{fg}_T$ is not  smooth. Is $\cong^{fg}_T$ hyperfinite? In other words, is $\cong^{fg}_T$ Borel reducible to $E_0$?
  
 
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{{References}}
Reference: Coskey, Samuel, Kossak, Roman ''The complexity of classification problems for models of arithmetic'', Bull. Symbolic Logic 16 (2010), no. 3, 345–358
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Revision as of 08:48, 21 January 2013

Borel classification questions

Let $T$ be a completion of $PA$. It is not hard to see that the isomorphism problem for finitely generated models of $T$, $\cong^{fg}_T$, is Borel.

Coskey and Kossak [1] proved that $\cong^{fg}_T$, is essentially countable and $E_0\leq_B \cong^{fg}_T$ i.e. $\cong^{fg}_T$ is not smooth. Is $\cong^{fg}_T$ hyperfinite? In other words, is $\cong^{fg}_T$ Borel reducible to $E_0$?

References

  1. Samuel Coskey and Roman Kossak. The complexity of classification problems for models of arithmetic. Bull. Symbolic Logic 16(3):345--358, 2010. www   MR   bibtex
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