How can one demonstrate there is no sequence $X\_i$ of sets such that $X\_{i+1}' = X\_i$ (this is really equality as sets though Turing equivalence would be interesting too).


I know it fails if I relax equality to simply $X\_{i+1}' <\_T X\_i$ as in [this question](https://mathoverflow.net/questions/173685/infinite-decreasing-sequence-by-the-turing-jump) but I'd think there has to be some kind of simple fixed point style argument in this case. Maybe I'm just sleep deprived but I'm blanking on how to show it and even trying the Martin determinacy stuff didn't immediately give me an answer.


Specifically, the reason I care is I need to show that for any $X$ there is a maximum value of $k \in \omega$ such that $(\exists Y)(X = Y^k)$.



Infinite descending chain of Turing jumps with equality
Let $\Theta(X, n)$ be a uniform process such that $\Theta(Z^{(n)}, n) = Z$, i.e. $\Theta$ undoes jumps. Using the recursion theorem, define a functional $\Phi\_e$ such that $\Phi\_e^X(e)\downarrow$ iff $\exists n\, \Theta(X, n)(e) = 0$.


Now suppose such a sequence $(X\_i)\_{i \in \omega}$ existed. Consider the sequence $X\_0(e), X\_1(e), X\_2(e),\dots$. So $X\_i(e) = 1$ iff there is a $j \ge i$ with $X\_j(e) = 0$. If this sequence contains infinitely many 0s, then it is is entirely 1s, a contradiction. If this sequence contains only finitely many 0s, then it contains only finitely many 1s, again a contradiction.


