# Which reals are “hyperarithmetic modulo ordinals”?

The context for this question is the theory ZFC + a measurable cardinal, although answers not in this context would also be interesting to me.

In a project I'm working on, the following class of reals has emerged, and I'd like to understand it better:

Say that a real $r$ is hyperarithmetic modulo ordinals (and write $r\in\Delta_1^{1, ord}$) if $r$ can be defined in a $\Delta^1_1$ way relative to some ordinal parameters. Formally, $r$ is $\Delta_1^{1, ord}$ if there is a tuple $\overline{\alpha}$ of ordinals and a pair of formulas $\varphi(x, \overline{y}), \psi(x, \overline{y})\in\Sigma^1_1$ in only the displayed parameters such that, whenever $\overline{c}$ is a tuple of reals coding copies of $\overline{\alpha}$, we have $$r=\{n: \varphi(n, \overline{c})\}=\{n: \neg\psi(n, \overline{c})\}.$$

Now, this class of reals is much bigger than $\Delta^1_1$. For example, Kleene's $\mathcal{O}$ is $\Delta_1^{1, ord}$: $\Phi_e$ is well-founded iff it embeds into $\omega_1^{CK}$.

EDIT: as soon as I posted this, I realized that this can be pushed further: unless I'm missing something, every constructible real is $\Delta_1^{1, ord}$. So the right question now is:

Is $\Delta_1^{1, ord}=L\cap\mathbb{R}$?

Currently I suspect the answer is "yes" - note that $\Delta_1^{1, ord}$ is forcing absolute! - but I don't see how to prove it.

A more conservative question is:

Is there a $\Delta^1_3$ real which is not $\Delta^{1, ord}_1$?

In general, any information about this class (and its obvious variations - e.g. $\Sigma^{1, ord}_n$) would be valuable to me. I suspect this is all very well-known in the descriptive set-theory community, so I've added the "reference-request" tag.

• The phrase "$c_i$ is a real coding a structure isomorphic to $\alpha_i$" needs to be clarified since surely every real is arithmetical in some well ordering $c \subseteq \omega^2$. – Ashutosh Aug 29 '16 at 22:29
• @Ashutosh Note that $\varphi$ and $\psi$ have to work independently of what codes we pick, so this isn't a problem. – Noah Schweber Aug 29 '16 at 22:30
• @Ashutosh I just realized how the language in the definition is unclear - I've edited to fix this. – Noah Schweber Aug 29 '16 at 22:36
• @Ashutosh It's worth noting that even long ordinals have copies which do not code complicated reals - e.g. for each computable ordinal $\theta$ and each real $x$ not recursive in $0^\theta$, every ordinal $\alpha$ has some copy $c$ such that the $\theta$th jump of $c$ doesn't compute $x$. This was (basically) proved by Richter. – Noah Schweber Aug 29 '16 at 23:52
• @Ashutosh Suppose $r\in L\cap\mathbb{R}$. Then $r\in L_\alpha$ for some large enough $\alpha$, and moreover $r$ is the $\beta$th element in the $L$-ordering of $L_\alpha$. The parameters $\alpha,\beta$ are now enough to hyperarithmetically define $r$. – Noah Schweber Aug 30 '16 at 0:18

In fact $\Delta^{1, ord}_1 = \mathbb{L} \cap \mathbb{R}$.
For suppose $(\phi(x, \overline{y})$, $\psi(x, \overline{y})$, $\overline{\alpha})$ is as you describe, defining $r \subset \omega$. We show $r \in \mathbb{L}$. For convenience we suppose $\overline{\alpha} = \alpha$ is a single ordinal; this is no loss, by coding.
Now if $\alpha$ is countable in $\mathbb{L}$ we are done by Shoenfield. If $\alpha$ is not countable in $\mathbb{L}$ we proceed as follows.
Let $\mathbb{P}$ be the usual forcing notion for adding a surjection $f: \omega \to \alpha$. Let $G$ be $\mathbb{P}$-generic over $\mathbb{V}$; then $G$ is also $\mathbb{P}$-generic over $\mathbb{L}$ so we can consider $\mathbb{L}[G] \subset \mathbb{V}[G]$. Now, by Shoenfield, we have that in $\mathbb{V}[G]$, whenever $c$ codes $\alpha$, then $\{n: \phi(n, c)\} = r$. By Shoenfield again, this remains true in $\mathbb{L}[G]$.
Henceforth we work in $\mathbb{L}$. Since $G$ could have been chosen to contain any given $p \in \mathbb{P}$, we have shown that for each $n$, $\mathbb{P}$ forces that $\phi(n, \dot{c})$ holds iff $n \in r$ (where $\dot{c}$ is a generic code for $\alpha$). Hence $r$ is definable in $\mathbb{L}$.