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I am trying to solve Exercise 7.1.24 (i) of Computability Theory by Rebecca Weber. $A^{(n)}$ denotes the $n$-th Turing jump and $A^{(\omega)}=\{\langle x,n\rangle: x\in A^{(n)}\}$ the $\omega$-jump.

I am able to prove that $A^{(n)}\leq_T A^{(\omega)}$. Since $x\in A^{(n)}\Leftrightarrow \langle x,n\rangle\in A^{(\omega)}$ we get that $\chi_{A^{(n)}}(x)=\chi_{A^{(\omega)}}(x,n)$, which is computable. Hence, $A^{(n)}$ is $A^{(\omega)}$-computable, i.e. $A^{(n)}\leq_T A^{(\omega)}$.

I now want to show that $A^{(\omega)}\nleq_T A^{(n)}$. My idea was to use the Jumping Theorem which says that $A^{(n+1)}\nleq_T A^{(n)}$ but I couldn't find a way to do so.

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    $\begingroup$ You've shown that $A^{(n)}\leq_TA^{(\omega)}$ for arbitrary $n$, so you've shown that $A^{(n+1)}\leq_TA^{(\omega)}$. Now just use transitivity of $\leq_T$. $\endgroup$ – Steven Stadnicki May 4 '20 at 15:27
  • $\begingroup$ Ahh, thank you very much! $\endgroup$ – thehardyreader May 4 '20 at 15:29
  • $\begingroup$ Here, let me make that a proper answer... :P $\endgroup$ – Steven Stadnicki May 4 '20 at 15:29
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This can also be done from first principles, but the easiest way is to use what you've already shown. Since you've shown that $A^{(n)}\leq_TA^{(\omega)}$ for arbitrary $n$, you also know that $A^{(n+1)}\leq_TA^{(\omega)}$. Now, if it were the case that $A^{(\omega)}\leq_TA^{(n)}$, then by transitivity of $\leq_T$ you would have $A^{(n+1)}\leq_TA^{(n)}$, which you already know to be false.

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