Assume "$0^\#$ exists".

We know that $0^\#$ is a $\Pi^1_2$-singleton. That means, there is a Shoenfield tree $S$ on $\omega \times (\omega \times \omega_1)$ so that $$x = 0^\# \leftrightarrow S_x \text{ is wellfounded},$$ where $S$ has the following definiton: for some recursive $$f : (\omega\times\omega)^{<\omega} \to \text{linear orders} $$ we have

• $|s|=|t|=k \to f(s,t)$ is a linear order $k+1$ with greatest element $0$
• $s \subset s' \wedge t \subset t' \to f(s,t)$ is a suborder of $f(s',t')$.
• $(s,(t,\vec{\alpha})) \in S \leftrightarrow (\{\alpha_0, \dots,\alpha_k\};<) \simeq (k+1, f(s,t)) \wedge t(0)=0$. (The requirement $t(0)=0$ is only for convenience in the question to follow)

The question is: How wellfounded is $S_{0^\#}$? Since 0 is always the greatest element of $f(s,t)$, the rank of $\emptyset$ in $S_{0^\#}$ must $\omega_1+1$. But what is the rank of a length-1 node $(\langle 0 \rangle, \langle \alpha \rangle )$? Let $S'$ be the "stretch" of $S$ onto some $\kappa>\omega_1$. That is, $S'$ is a tree on $\omega \times (\omega \times \kappa)$ with the same definition. What is the smallest possible value of $$ \beta_{S,0^\#}=\text{rank}_{S'_{0^\#}} (\langle 0 \rangle, \langle \omega_1 \rangle )$$ for all possible choices of $S$ satisfying $p[S] = \omega^\omega \setminus \{0^\#\}$?

It is not hard to come up with a definition of $S$ so that $\beta_{S,0^\#} = (\omega_1^V)^{+L}$. It seems that $\beta_{S,0^\#} < (\omega_1^V)^{+L}$ is impossible.

The ordinal $(\omega_1^V)^{+L}$ appears to be some kind of closure ordinal, or the higher level analog of $\omega_1^{CK}$.

In general, if $x$ is a $\Pi^1_2$-singleton and there is a choice of $S$ so that $\beta_{S,x} < (\omega_1^V)^{+L}$ and $p[S] = \omega^\omega \setminus \{x\}$, must $x \in L$?

Assume "$0^\#$ exists".

We know that $0^\#$ is a $\Pi^1_2$-singleton. That means, there is a Shoenfield tree $S$ on $\omega \times (\omega \times \omega_1)$ so that $$x = 0^\# \leftrightarrow S_x \text{ is wellfounded},$$ where $S$ has the following definiton: for some recursive $$f : (\omega\times\omega)^{<\omega} \to \text{linear orders} $$ we have

• $|s|=|t|=k \to f(s,t)$ is a linear order $k+1$ with greatest element $0$
• $s \subset s' \wedge t \subset t' \to f(s,t)$ is a suborder of $f(s',t')$.
• $(s,(t,\vec{\alpha})) \in S \leftrightarrow (\{\alpha_0, \dots,\alpha_k\};<) \simeq (k+1, f(s,t)) \wedge t(0)=0$. (The requirement $t(0)=0$ is only for convenience in the question to follow)

The question is: How wellfounded is $S_{0^\#}$? Since 0 is always the greatest element of $f(s,t)$, the rank of $\emptyset$ in $S_{0^\#}$ must $\omega_1$. But what is the rank of a length-1 node $(\langle 0 \rangle, \langle \alpha \rangle )$? Let $S'$ be the "stretch" of $S$ onto some $\kappa>\omega_1$. That is, $S'$ is a tree on $\omega \times (\omega \times \kappa)$ with the same definition. What is the smallest possible value of $$ \beta_{S,0^\#}=\text{rank}_{S'_{0^\#}} (\langle 0 \rangle, \langle \omega_1 \rangle )$$ for all possible choices of $S$ satisfying $p[S] = \omega^\omega \setminus \{0^\#\}$?

It is not hard to come up with a definition of $S$ so that $\beta_{S,0^\#} = (\omega_1^V)^{+L}$. It seems that $\beta_{S,0^\#} < (\omega_1^V)^{+L}$ is impossible.

The ordinal $(\omega_1^V)^{+L}$ appears to be some kind of closure ordinal, or the higher level analog of $\omega_1^{CK}$.

In general, if $x$ is a $\Pi^1_2$-singleton and there is a choice of $S$ so that $\beta_{S,x} < (\omega_1^V)^{+L}$ and $p[S] = \omega^\omega \setminus \{x\}$, must $x \in L$?