Analytic equivalence relations satisfying hyperarithmetic-is-recursive
classification
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keywords
equivalenceanalyticclassequivalenthyperarithmetic-is-recursiveclassescomputescone
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We prove, in ZF+$\bf\Sigma^1_2$-determinacy, that for any analytic equivalence relation $E$, the following three statements are equivalent: (1) $E$ does not have perfectly many classes, (2) $E$ satisfies hyperarithmetic-is-recursive on a cone, and (3) relative to some oracle, for every equivalence class $[Y]_E$ we have that a real $X$ computes a member of the equivalence class if and only if $\om_1^X\geq\om_1^{[Y]}$. We also show that the implication from (1) to (2) is equivalent to the existence of sharps over $ZF$.
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