The constructible universe $L$ has some nice properties:


1. $L$ has a $\mathit{\Delta}^1\_2$-good well-ordering of $\mathbb{R}$. (Gödel, Addison)
2. For any $\mathit{\Sigma}^1\_2$ formula $\varphi(x)$ and a real $r \in \mathbb{R}\cap L, \ \varphi(r) \iff \varphi(r)^L$. (Shoenfield)
3. For any $\mathit{\Sigma}^1\_2$ set $A$, either $A\subseteq L$ or $A$ contains a perfect set. (Mansfield, Solovay)


My questions are:


* Is there, in $\mathsf{ZFC}$, an inner model that satisfies all three statements above with $\mathit{\Delta}^1\_2$ and $\mathit{\Sigma}^1\_2$ replaced by $\mathit{\Delta}^1\_3$ and $\mathit{\Sigma}^1\_3$, respectively?
* What about $\mathit{\Delta}^1\_n$ and $\mathit{\Sigma}^1\_n$ with $n > 3$?
* What if instead we work in $\mathsf{ZFC}$ + large cardinals?



Inner model with a $\mathit{\Delta}^1_3$-good well-ordering of the reals
The situation is a bit more complicated than you might hope because of the periodicity phenomena in the projective hierarchy. For odd $n$, assuming $\mathbf{\Delta}^1\_{n-1}$-determinacy, the set $Q\_n$ of reals that are $\Delta^1\_n$ in a countable ordinal is $\Pi\_n^1$ definable, so its complement is a $\Sigma\_n^1$ set that has no element in any inner model with a good $\Delta^1\_n$ wellorder. So no such model is $\Sigma\_n^1$-correct.


The set $Q\_n$ is the set of reals in $M\_{n-2}$, the minimum inner model with $n-2$ Woodin cardinals, which does have a good $\Delta^1\_n$-wellorder of the reals. The union of all countable $\Sigma^1\_n$ sets is going to be the set of all $\Delta^1\_n$ reals by the effective perfect set theorem, so $M\_{n-2}$ will satisfy the version of the perfect set theorem you want. But really this is not quite the right theorem, I think, since $Q\_n$ is $\Pi^1\_n$ not $\Sigma^1\_n$. On the other hand, there is a largest countable $\Pi^1\_n$ set (analogous to the largest countable $\Pi^1\_1$ set, the mastercodes of $L$), but this is actually strictly larger than $Q\_n$.


On the other hand, if there is no inner model with a Woodin cardinal and every set has a sharp, then the core model $K$ is $\Sigma^1\_3$-correct and has a $\Delta^1\_3$-good wellorder. Well, I'm not sure if this has been proved in such generality, the version I know assumes there is a measurable cardinal. Maybe an expert can fill this in.


For the even levels, under large cardinal assumptions, the situation should be very similar to the one for $L$, so probably exactly what you want is true. I think this cannot be done without some large cardinal assumptions, or really determinacy/mouse existence assumptions. I'm getting a little tired of writing for now, hopefully I'll update later.


Some references: Steel's "Projectively wellordered inner models," Steel's "Ordinal definability in models of determinacy," Steel's *Core Model Iterability Problem* (for the correctness of $K$), Kechris-Martin-Solovay's "Introduction to Q-theory."


