This is a rather technical question. I cannot find my mistake in a proof of the (obviously wrong) following sentence: Every countable ordinal is $\Sigma\_2$-definable in $J\_{\omega\_1 + 1}$ by a formula with no parameter.


I first recall the definition of $J\_\alpha$, closely following the notations and ideas of "Fine Structure and Class Forcing" by Sy Friedman. Suppose that $J\_\alpha$ is defined (with $J\_0 = \emptyset$). Let $\{\phi\_e(x)\}\_{e \in J\_\alpha}$ be an enumeration of first-order formulas with one free variable and parameters in $J\_{\alpha}$. We define $J\_{\alpha+1}$ by first defining sets $X(n, e)$ for $n \in \omega$ and $e \in J\_\alpha$, where each $X(n+1, e)$ is meant to be the set of level $n+1$ and index $e$, having as elements the sets of level $n$ whose indices satisfies the formula $\Phi\_e$ in $J\_\alpha$.
$$
\begin{array}{rcl}
X(0, e)&=&e\\
X(n+1, e)&=&\{X(n, x)\ :\ J\_\alpha \vDash \Phi\_e(x)\}\\
\end{array}
$$


We then define $J\_{\alpha+1} = \bigcup \{X(n, e)\ |\ n \in \omega,\ e \in J\_{\alpha}\}$. For $\gamma$ limit we define $J\_{\gamma} = \bigcup\_{\beta < \gamma} J\_\beta$.


In the following, everytime we mention $\Sigma\_2$-definable, we mean $\Sigma\_2$-definable by a formula with no parameter. Fix some ordinal $\alpha$. Suppose some ordinal of $J\_{\alpha+1}$ is not $\Sigma\_2$-definable in $J\_{\alpha+1}$, and let $a$ be the smallest of them. Then $a$ has the following definition:


$$a \text{ is not } \Sigma\_2\text{-definable and } \forall b < a\ b \text{ is } \Sigma\_2\text{-definable}$$


Let us try to formalize this properly in $J\_{\alpha+1}$. To do so, we use what Friedman calls a universal $\Sigma\_1$ predicate, say $W(n, x, y)$. The predicate $W(n, x, y)$ is a $\Sigma\_1$ formula which is true in $J\_{\alpha+1}$ iff $\phi\_n(x, y)$ if true in $J\_{\alpha+1}$, where $\phi\_n(x, y)$ is the $n$-th $\Sigma\_1$ formula of two free variables. Friedman shows that such predicates exist at every level of the hierarchy.


We can now write "$a \text{ is not } \Sigma\_2\text{-definable}$" as a $\Delta\_3$ formula:
$$\forall n \in \omega\ \ \ \forall x\ W(n, x, a) \text{ or } \exists b \neq a\ \exists x\ \neg W(n, x, b)$$


It follows that "$a \text{ is } \Sigma\_2\text{-definable}$" is also a $\Delta\_3$ formula:
$$\exists n \in \omega\ \ \ \exists x\ \neg W(n, x, a) \text{ and } \forall b\ b \neq a \rightarrow \forall x\ W(n, x, b)$$


Suppose now that $\alpha$ is $\Sigma\_2$-definable in $J\_{\alpha+1}$. Then the following argument shows that we can lower the complexity of the above formulas to $\Sigma\_2$: Now the universal quantifications can be replaced by existential quantifications in $J\_{\alpha+1}$, given that we can use $J\_{\alpha}$. Indeed, "$\forall x \dots$" (meaning $\forall x \in J\_{\alpha+1} \dots$) becomes equivalent to:


$$\forall e \in J\_\alpha\ \forall n \in \omega\ \exists x\ x=X(n, e)\text{ and } \dots$$


It seems that both $J\_\alpha$ and $X(n,e)$ have a $\Sigma\_1$ definition (using $\alpha$ as a parameter) in $J\_{\alpha+1}$. So it follows that if $\alpha$ is $\Sigma\_2$ definable in $J\_{\alpha+1}$, then the smallest ordinal not $\Sigma\_2$ definable in $J\_{\alpha+1}$ is $\Sigma\_2$ definable in $J\_{\alpha+1}$. Then there cannot be such ordinals.


Now it seems that $\omega\_1$ is $\Sigma\_2$ definable in $J\_{\omega\_1+1}$ (which contains all the ordinals strictly smaller than $\omega\_1 + \omega$), with the following formula of $\alpha$:


$$\forall \beta\ \exists n<\omega\ \beta < \alpha + n\ \text{ and } \forall \alpha' < \alpha\ \exists \beta\ \forall n<\omega\ \beta > \alpha' + n$$


It follows that every countable ordinal is $\Sigma\_2$ definable in $J\_{\omega\_1}$, by formula with no parameter. This cannot be correct by a cardinality argument. What is wrong ?



Problem with definability in the constructible hierarchy
I may have found where is the trick hidden: I mean two different things by "being $\Sigma\_2$-definable". When I write "the smallest ordinal which is not $\Sigma\_2$-definable" I mean being definable by a formula of the form 
$\exists \forall \psi$ where $\psi$ only have bounded quantifiers. 


But when I define this smallest ordinal with a $\Sigma\_2$ definition, I define it with a formula where the bounded quantifiers are spread everywhere in the formula, including (for example) in between two existential quantifiers.


Of course in a model of sufficiently many ZF axioms, these two notions of being $\Sigma\_2$-definable are equivalent. You probably need your model to be $\Sigma\_1$ or maybe $\Sigma\_2$ admissible. But $J\_{\omega\_1+1}$ is certainly not $\Sigma\_1$-admissible...


I think this is what is wrong with this proof.


