I'm sure i have read that the following (or something that implies this) is true



> 
> Let $X$ be a $\Pi\_1^0$ class with top Medvedev degree. Then for every
> $x\in X$, there is $y\in X$ with $y<\_T x$.
> 
> 
> 


But i don't remember where. If this is true, do you know where can I find this or a similar statement?



Turing degrees inside the $\Pi_1^0$ class with top Medvedev degree
We can modify the proof of the Kreisel/Shoenfield basis theorem (see Theorem 3.7 in [Diamondstone/Dzhafarov/Soare](https://projecteuclid.org/journals/notre-dame-journal-of-formal-logic/volume-51/issue-1/%CE%A0-1-0-Classes-Peano-Arithmetic-Randomness-and-Computable-Domination/10.1215/00294527-2010-009.full)):


Let $T\subseteq 2^{<\omega}$ be an infinite binary tree all of whose paths are of PA-degree (this is equivalent to $[T]$, the set of paths through $T$, having top Medvedev degree amongst $\Pi^0\_1$ classes). Fix a path $f\in [T]$; we want to find a path $g\in [T]$ with $g<\_Tf$.


Consider the following computable $4$-ary tree $S$: a node on $S$ of length $n$ consists of a pair $(\alpha,\beta)$ where $\alpha$ is a node on $T$ of length $n$ and $\beta$ is a binary string of length $n$ such that for each $e<i$ it is **not** the case that $\Phi\_e^\alpha(e)[n]\downarrow=\beta(n)$. Clearly $S$ is infinite, so $S$ has a path $h=(h\_0,h\_1)\le\_Tf$. But $h\_0$ is a path through $T$ such that $h\_0\not\ge\_Th\_1$, and since $(h\_0,h\_1)\le\_Tf$ we have that $h\_0\not\ge\_Tf$.


