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Jun 17, 2015 at 5:54 comment added Theodore Slaman Sorry, I didn't realize that I would run out of time on the comment. Try again: We may assume that $x$ is an element of $2^\omega$. If $x$ is $\Delta^1_2$, then the statement about $z$ that $z$ is equal to $x$ is $\Sigma^1_2(z)$: for every $n$, $n\in z$ implies $n\in x$ and $n\not\in z$ implies $n\not\in x$. Both $n\in x$ and $n\not\in x$ are $\Sigma^1_2$ statements about $n$ and $\Sigma^1_2$ statements are closed under universal number quantifiers. Then, for $\theta\in\Sigma^1_2$, $\theta(x)$ iff $\exists z(z=x \wedge \theta(z))$.
Jun 17, 2015 at 5:45 comment added Theodore Slaman Say that $x$ is defined as $\{n:\psi(n)\}=\{n:\phi(n)\}$, where $\psi$ is $\Sigma^1_2$ and $\phi$ is $\Pi^1_2$. Then,
Jun 17, 2015 at 4:16 comment added Noah Schweber I'm probably just tired, but: why is $\mathcal{O}^x_2$ $\Sigma^1_2$? Since $x$ is $\Delta^1_2$ the best I can see is $\Sigma^1_2(\Delta^1_2)=\Sigma^1_3$.
Jun 16, 2015 at 23:57 history answered Theodore Slaman CC BY-SA 3.0