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:



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> **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 $\omega\_1^{y\_0}=\omega\_1^z$?
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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 $\omega\_1^{y\_0}=\omega\_1^z$?
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A partial relativization of Gandy's basis theorem
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 $\omega\_1^{y\_0}>\omega\_1^{CK}=\omega\_1^x$.


