5
$\begingroup$

I don't known the correct credit for the following: every non-empty $\Pi^1_1$ set of reals contains some $X \in L_{\alpha}$ for some $X$-recursive $\alpha$. (Addison-Kondo?)

So, my question is: what is the least $\beta$, s.t. $\mathcal{P}(\omega) \cap L_\beta$ is a basis for all non-empty $\Pi^1_1$ sets?

$\endgroup$
5
$\begingroup$

I can't access the full paper at the moment, but I'm pretty sure this is exactly the question addressed in the paper "A Note on the Kondo-Addison Theorem" by D. Guaspari

$\endgroup$
  • $\begingroup$ Available online at www.jstor.org/stable/10.2307/2272898 $\endgroup$ – Trevor Wilson Jul 21 '12 at 14:15
  • $\begingroup$ Thanks a lot! The last theorem on obtainable ordinals is quite interesting. $\endgroup$ – Wei Wang Jul 22 '12 at 11:06
4
$\begingroup$

The least such ordinal is the least ordinal which cannot be a $\Delta^1_2$-well-ordering over natural numbers.

Let $$\delta^1_2=\mbox{ supremum of the }\Delta^1_2 \mbox{ wellorderings of } \omega,$$ and $$\delta=\min\{\alpha\mid L\setminus L_{\alpha}\mbox{ contains no }\Pi^1_1 \mbox{ singleton}\}.$$

We claim that $\delta=\delta^1_2$.

$\mathbf{Proof}$: If $\alpha<\delta$, then there is a $\Pi^1_1$ singleton $x \in L_{\delta}\setminus L_{\alpha}$. Since $x\in L_{\omega_1^x}$ and $\omega_1^x$ is a $\Pi^1_1(x)$-wellordering, it must be that $\alpha<\omega_1^x<\delta^1_2$. So $\delta\leq \delta^1_2$.

If $\alpha<\delta^1_2$, there is a $\Delta^1_2$ wellordering relation $R\subseteq \omega\times \omega$ of order type $\alpha$. So there are two arithmetical relations $S, T\subseteq (\omega^{\omega})^2\times \omega^2$ so that $$R(n,m)\Leftrightarrow \exists f \forall g S(f,g,n,m), \mbox{ and}$$ $$\neg R(n,m)\Leftrightarrow \exists f \forall g T(f,g,n,m).$$ Define $\Pi^1_1$ sets $$R_0=\{(h,\langle n,m\rangle)\mid h(0)=0\wedge \exists f\forall g (S(f,g,n,m)\wedge \forall n(f(n)=h(n+1)))\}$$ and $$R_1=\{(h,\langle n,m\rangle)\mid h(0)=1\wedge \exists f\forall g (T(f,g,n,m)\wedge \forall n(f(n)=h(n+1)))\}.$$ By $\Pi^1_1$-uniformization Theorem, they both can be uniformized by $\Pi^1_1$ partial functions $p_{R_0}:\omega\to \omega^{\omega}$ and $p_{R_1}:\omega\to \omega^{\omega}$. Let $p=p_{R_0} \cup p_{R_1}$. Then $p$ is a $\Pi^1_1$ total function and can viewed as a $\Pi^1_1$-singleton. Then $R$ is recursive in $p$ and so $\alpha<\omega_1^p<\delta$.

Thus $\delta^1_2=\delta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.