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

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== Borel classification questions == | == Borel classification questions == | ||

− | + | This question is about the complexity of the classification problem for finitely generated models of a given completion $T$ of $PA$. | |

− | $ | + | Let us denote the space of finitely generated models of $T$ by $X^{fg}_T$. (More precisely: if $A$ is a fixed countable set, then we can identify the set of models of $T$ whose underlying set is $A$ with a subset of $A^{A^6}$, where each model is represented by its addition and multiplication functions.) Letting $\cong^{fg}_T$ denote the isomorphism relation on $X^{fg}_T$, it is not hard to see that $\cong^{fg}_T$ is Borel, considered as a subset of $X^{fg}_T\times X^{fg}_T$. |

− | + | The complexity of $\cong^{fg}_T$ can be analyzed more closely using the concept of Borel reducibility. Here, if $E,F$ are equivalence relations on standard Borel spaces $X,Y$, then $E$ is said to be Borel reducible to $F$ iff there is a Borel function $f\colon X\to Y$ such that | |

+ | \[x\mathrel{E}x'\iff f(x)\mathrel{F}f(x')\] | ||

+ | |||

+ | It was shown in <cite> coskeykossak2010:thecomplexity </cite> that $\cong^{fg}_T$ is Borel reducible to $E_\infty$, the universal countable equivalence relation. It was furthermore shown that $E_0$, the almost equality relation on the set of infinite binary sequences, is Borel reducible to $\cong^{fg}_T$. | ||

+ | |||

+ | '''Problem''': What is the precise complexity of $\cong^{fg}_T$? There is a wide gap between $E_0$ and $E_\infty$. Is $\cong^{fg}_T$ bireducible with $E_\infty$? Is $\cong^{fg}_T$ bireducible with $E_0$? | ||

{{References}} | {{References}} |

## Revision as of 10:26, 4 February 2013

## Borel classification questions

This question is about the complexity of the classification problem for finitely generated models of a given completion $T$ of $PA$.

Let us denote the space of finitely generated models of $T$ by $X^{fg}_T$. (More precisely: if $A$ is a fixed countable set, then we can identify the set of models of $T$ whose underlying set is $A$ with a subset of $A^{A^6}$, where each model is represented by its addition and multiplication functions.) Letting $\cong^{fg}_T$ denote the isomorphism relation on $X^{fg}_T$, it is not hard to see that $\cong^{fg}_T$ is Borel, considered as a subset of $X^{fg}_T\times X^{fg}_T$.

The complexity of $\cong^{fg}_T$ can be analyzed more closely using the concept of Borel reducibility. Here, if $E,F$ are equivalence relations on standard Borel spaces $X,Y$, then $E$ is said to be Borel reducible to $F$ iff there is a Borel function $f\colon X\to Y$ such that \[x\mathrel{E}x'\iff f(x)\mathrel{F}f(x')\]

It was shown in [1] that $\cong^{fg}_T$ is Borel reducible to $E_\infty$, the universal countable equivalence relation. It was furthermore shown that $E_0$, the almost equality relation on the set of infinite binary sequences, is Borel reducible to $\cong^{fg}_T$.

**Problem**: What is the precise complexity of $\cong^{fg}_T$? There is a wide gap between $E_0$ and $E_\infty$. Is $\cong^{fg}_T$ bireducible with $E_\infty$? Is $\cong^{fg}_T$ bireducible with $E_0$?

## References

- 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