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Hmmm, it seems the answer 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 $\omega_1^{y_0}>\omega_1^{CK}=\omega_1^x$.

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 $\omega_1^{y_0}>\omega_1^{CK}=\omega_1^x$.

Hmmm, it seems the answer is no.

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

To see that $A_x$ is $\Sigma^1_1(x)$, just notice that the isomorphism between $\mathscr{O}^x_n$ and $\mathscr{O}^y_m$ must be hyperarithmetic in $x\oplus y$.

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 $\omega_1^{y_0}>\omega_1^x$.

Hmmm, it seems the answer 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 $\omega_1^{y_0}>\omega_1^{CK}=\omega_1^x$.

Hmmm, it seems the answer is no.

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

To see that $A_x$ is $\Sigma^1_1(x)$, just notice that the isomorphism between $\mathscr{O}^x_n$ and $\mathscr{O}^y_m$ must be hyperarithmetic in $x\oplus y$.

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 $\omega_1^{y_0}>\omega_1^x$.

Hmmm, it seems the answer is no.

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

To see that $A_x$ is $\Sigma^1_1(x)$, just notice that the isomorphism between $\mathscr{O}^x_n$ and $\mathscr{O}^y_m$ must be hyperarithmetic in $x\oplus y$.

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 $\omega_1^{y_0}>\omega_1^x$.