Suppose I am given a subset of $2^\omega\times\omega^\omega$ of some bounded Borel rank. Can I get an **analytic** uniformization of this set?



Analytic uniformization
There is an arithmetical set $A\subseteq 2^{<\omega}\times \omega^{<\omega}$ so that for any $x\in 2^{\omega}$, $A(x)=\{\sigma\mid \exists n(x|n,\sigma)\in A\}$ is an $x$-recursive tree which has an infinite path but no infinite path hyperarithmetic in $x$.


Now let $B$ be an arithmetical set so that $(x,y)\in B$ if and only if $y$ is an infinite path through $A(x)$. $B$ has the following property:


(1). For any $x$, there is some $y$ so that $(x,
y)\in B$;


(2) No analytic function uniformizing $B$. Suppose not, then there is a real $z$ and $\Sigma^1\_1(z)$-function $f$ uniformizing $B$. Let $x\geq\_h z$, then $f(x)\leq\_h x\oplus z\leq\_h x$, a contradiction.


