This question is about compact (but not necessarily effectively compact) $\Pi^0\_1$-classes in Baire space. If I am given an index for a $\Pi^0\_1$-class, and assured that it is compact, is determining whether the class is nonempty a $\Sigma^1\_1$ question of maximal difficulty?


To be more precise, let $A$ be the set of indices of compact nonempty $\Pi^0\_1$-classes, and let $B$ be the set of indices of empty $\Pi^0\_1$-classes. Is $(\Sigma^1\_1, \Pi^1\_1) \le\_1 (A, B)$?


Relatedly, does every compact nonempty $\Pi^0\_1$-class have a hyperarithmetic path? These questions are morally negations of each other, in that a yes answer for one gives a no answer for the other, and a no answer for one very likely gives a yes answer for the other.



Is checking whether a compact $\Pi^0_1$-class is nonempty $\Sigma^1_1$-hard?
I think that every compact $\Pi^0\_1$-class has a hyperarithmetic infinite path. 



> 
> **Proof**: Given a compact $\Pi^0\_1$-class $P$. So there is a recursive tree
>  $T\subseteq \omega^{<\omega}$ so that $[T]=P$. Since $P$ is compact,
>  there is a function $f$ dominating all of the members of $P$. We will
>  find a hyperaritmetic one.
> 
> 
> We use Spector-Gandy's theorem to recursively work on
>  $L\_{\omega\_1^{CK}}$.
> 
> 
> For $n=0$, there is some number $m$ and stage $\alpha<\omega\_1^{CK}$
>  so that for every $k\geq m$, $T\_k=\{k\sigma\mid k\sigma\in T\}$ is
>  well founded witnessed at stage $\alpha$. Find the first such stage
>  $\alpha\_0$ and corresponded number $m\_0$. Let $f(0)=m\_0$.
> 
> 
> Generally for any $n+1$, there is some number $m$ and stage
>  $\alpha<\omega\_1^{CK}$ so that for every $k\geq m$, $T\_{f(0)\cdots
> > f(n)k}=\{k\_0\cdots k\_n k\sigma \mid k\_0\cdots k\_n k\sigma\in T \wedge
>  \forall i\leq n (k\_i\leq f(i))\}$ is well founded witnessed at stage
>  $\alpha$. Find the first such stage $\alpha\_{n+1}$ and corresponded
>  number $m\_{n+1}$. Let $f(n+1)=m\_{n+1}$.
> 
> 
> So $f$ is a total $\Pi^1\_1$ function and so hyperarithmetic. Thus $P$ must contains a hyperarithmetic member since every member of $P$ is dominated by $f$. QED
> 
> 
> 


