A partial relativization of Gandy's basis theorem

Question

Gandy's basis theorem says that any nonempty $\Sigma^1_1$ set $A$ contains a real $x$ with $\omega_1^x=\omega_1^{CK}$, the least nonrecursive ordinal.

Now the following question seems quite interesting to me:

Question: Is it true that for any real $z$ and nonempty $\Sigma^1_1(z)$ set $A$ containing a real $y>_hz$, $A$ must contain a real $y_0>_h z$ so that $\omega_1^{y_0}=\omega_1^z$?

Here $\geq_h$ is the hyperarithmetic reduction. Note that to give a positive answer to the question, it is sufficient to show that every such $A$ must contain a real $y_0\geq_h z$ so that $\omega_1^{y_0}=\omega_1^z$ since the set $\{y\mid y\not\leq_h z\}$ is $\Sigma^1_1(z)$.

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The question has a negative answer (see below). Then the question can be modified to be

Question' Is it true that for any real $z$ and nonempty $\Delta^1_1(z)$ set $A$ containing a real $y>_hz$, $A$ must contain a real $y_0>_h z$ so that $\omega_1^{y_0}=\omega_1^z$?

Answer 1

Hmmm, it seems the answer to the question is no.

Fix any nonhyperarithmetic real $x$ so that the $\Sigma^1_1(x)$ set $$A_x=\{y\mid \forall n \in \mathscr{O} ( x\not\leq_T y^{(|n|)})\}$$ is not empty, where $|n|$ is the $y$-recursive ordinal coded by $n$, and $y^{|n|}$ is the $|n|$-th Turing jump relative to $y$.

Clearly $A_x$ is $\Sigma^1_1(x)$.

To see the existence of such $x$, just let $x$ be a real Turing computing all hyperarithmetic reals but $\omega_1^x=\omega_1^{CK}$ and $x\leq_T \mathscr{O}$. Then any $\Delta^1_1$-random real $y$ with $\omega_1^y>\omega_1^{CK}$ must belong to $A_x$. Then $y\in A_x$ and $y>_h x$.

However for every $y_0\geq_h x$ in $A_x$, we have that $\omega_1^{y_0}>\omega_1^{CK}=\omega_1^x$.