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?

$\begingroup$ Does analytic uniformization means that graph is an analytical set, or that the map is analytically measurable? $\endgroup$ – Ilya Jul 18 '13 at 20:04

$\begingroup$ I think it means that the graph is an analytical set. $\endgroup$ – jonathanverner Jul 19 '13 at 6:39

2$\begingroup$ A function with analytic graph is actually a Borel function. To see this, note that $f^{1}(B)=\pi\big((X\times B)\cap\textrm{Graph} f\big)$, is analytic when $B$ is Borel. But for the same reason, $f^{1}(B^C)$ is analytic. Since $f^{1}(B)^C=f^{1}(B^C)$, this shows that the preimage of a Borel set is both analytic and coanalytic, hence Borel. $\endgroup$ – Michael Greinecker Jul 19 '13 at 7:46
There is an arithmetical set $A\subseteq 2^{<\omega}\times \omega^{<\omega}$ so that for any $x\in 2^{\omega}$, $A(x)=\{\sigma\mid \exists n(xn,\sigma)\in A\}$ 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.

$\begingroup$ I assume that the first formula should read $A\subseteq 2^{<\omega}\times\omega^{<\omega}$... Also, does the existence of $A$ follow from some theorem or should it be obvious (sorry, I am not very fluent in descriptive set theory let alone effective descriptive ST). $\endgroup$ – jonathanverner Aug 27 '13 at 15:01

$\begingroup$ You are right. I corrected the typo. The existence of $A$ follows from a well known fact that there is a $\Sigma^1_1$ set which does not contain a hyperarithmetic real. $\endgroup$ – 喻 良 Aug 27 '13 at 22:37