On thin $\Sigma^1_2$ equivalence relations This question is regarding Hjorth's paper "Some applications of coarse inner model theory", J. Symbolic Logic 62 (1997), no. 2, 337–365. 
Hjorth claims that if $E$ is a thin $\Sigma^1_2$ equivalence relation, then $E$ is $\Delta^1_2(m)$ for some $\Delta^1_3$ real $m$. One of the key steps is Lemma 2.5. The problem is, I don't see why the "Claim" in the proof of Lemma 2.5 holds. In the "Claim", $M^*$ is a countable model with a Woodin cardinal without a top measure, so I don't understand the sentence 
"..., all iterates of $M^*$ with respect to the topmost measure must be wellfounded."
So I can only see that $E$ is $\Delta^1_2(x)$ for some $x$ coding $M_1|(\delta^+)^{M_1}$, where $\delta$ is the Woodin of $M_1$. Am I missing something?
 A: Lemma 2.5 of Hjorth's paper is wrong. Here is a counterexample produced from a discussion with Philipp Schlicht. 
There is a thin $\Sigma^1_2$ equivalence relation which is not $\Pi^1_2(a)$ for any $a \in \mathbb{R}\cap M_1$. 
Proof: Let $xEy$ iff there is a countable transitive model $M$ of enough fragment of ZFC containing $x,y$ such that $M$ satisfies either


*

*neither $x$ nor $y$ is a standard code for a 1-small, $\Pi^1_2$-iterable, sound premouse projecting to $\omega$, or

*$x,y$ are both standard codes for 1-small, $\Pi^1_2$-iterable, sound premice projecting to $\omega$ and these two mice have a common nondropping iterate.


The equivalence classes of $E$ look like this: If $M_1|\alpha$ projects to $\omega$, then the $E$-class of the standard code for $M_1|\alpha$ has only one element. All the other elements are in one $E$-class. $E$ is $\Sigma^1_2$. The reason why $E$ is an equivalence relation is the following: if $P,Q$ have a common nondropping iterate and $P$ is iterable, then $Q$ is iterable.
However, $E$ is not $\Pi^1_2(a)$ for any $a \in \mathbb{R} \cap M_1$. Otherwise, let $\alpha$ be the least such that $M_1|\alpha$ projects to $\omega$ and $a \in M_1|\alpha$. Let 
$$\varphi(v,a)$$
be the statement: $v$ is the standard code for a 1-small, sound premouse $P$ projecting to $\omega$, $P$ looks like the sharp of $Q$ where $Q$ is some proper initial segment of $P$ projecting to $\omega$, $a \in Q$, but for any proper initial segment $Q'$ of $Q$ that projects to $\omega$, $a \notin Q'$.
Note that 


*

*If $v$ codes $(M_1|\alpha)^\#$, i.e. $v$ codes $M_1|\beta$ where $M_1|\beta$ projects to $\omega$ and $\mathbb{R} \cap L(M_1|\alpha) = \mathbb{R} \cap M_1|\beta$, then $\varphi (v,a)$ holds.

*If $v$ codes $M_1|\beta'$ and $\beta' \neq \beta$ ($\beta$ as in 1) then $\varphi(v,a)$ doesn't hold.

*There are $v \neq v'$ such that $v E v'$ and $\varphi(v,a)$, $\varphi(v',a)$ both hold. The premice coded in $v,v'$ are not iterable.


Therefore, 
$$v \text{ codes }(M_1|\alpha)^\#$$
iff $\varphi(v,a)$ holds and there are $v' \neq v''$ such that $v' E v'' \wedge \neg(v E v') \wedge \neg(v E v'') \wedge \varphi(v',a) \wedge \varphi(v'',a)$.
This means $(M_1|\alpha)^\#$ is a $\Sigma^1_2(a)$ singleton. So $(M_1|\alpha)^\# \in L(M_1|\alpha)$. Contradiction!
