Proof of the uniform boundedness theorem for analytic subsets of WO

Question

In the paper "A reflection phenomenon in descriptive set theory" by J.P. Burgess, it is stated:

Moschovakis [11] has proved the following important Uniform Boundedness Theorem: There is a continuous function $\Gamma: Y \rightarrow Z$ such that for all $y \in Y$, if $U_y \subseteq W$, then $\Gamma(y) \in W$ and $z \prec \Gamma(y)$ for all $z \in U_y$,

where $Y=2^{\omega \times \omega}$, $Z$ is the set of codes for linear orders on $\omega$, $W=WO$ and $U \subseteq Y \times Z$ is $Y$-universal for $\Sigma^1_1(Z)$.

But the reference [11] hadn't yet appeared and I'm able to find only the boundedness theorem in [11], but not the uniform version. How can we show this theorem? Have you ever seen this uniform version somewhere?

Answer 1

Presuming that $\Sigma^1_1(Z)$ denotes the collection of all boldface-$\Sigma^1_1$ subsets of $Z$, it's basically like the Kunen-Martin theorem. I'm going to use $Y=\omega^\omega$ is Baire space instead of $2^{\omega\times\omega}$. Let $T$ be a tree (set of finite tuples closed under initial segment) on $\omega\times\omega\times\omega$ projecting to $U$ in the first two coordinates (that is, $(y,z)\in U$ iff there is $f\in{^\omega}\omega$ such that $(y,z,f)\in[T]$, where $[T]$ is the set of infinite branches through $T$).

Given $y\in{^ \omega}\omega$, let $S_y$ be the tree of attempts to build a tuple $(x,z,f)$ with $(y,z,f)\in [T]$ (so $z\in U_y$, so $z$ codes a linear order $<_z$) and $x$ is a strictly descending sequence through $<_z$, i.e. $x\in{^\omega}\omega$ and $x(n+1)<_zx(n)$ for each $n<\omega$. (The tree $S_y$ includes some finite tuple given it is compatible with these requirements so far). Construct $S_y$ in such a way that given $y\upharpoonright n$, where $n<\omega$, we can determine $S_y\upharpoonright n$, i.e. the set of tuples in $S_y$ of length $\leq n$. Then the assignment $y\mapsto S_y$ is continuous.

Note then that $U_y\subseteq W$ iff $S_y$ is wellfounded.

Let $\Gamma(y)$ be the Brouwer-Kleene order associated to $S_y$ (a linear order). This is continuously associated to $S_y$. So $y\mapsto\Gamma(y)$ is continuous. Moreover, $S_y$ is wellfounded iff $\Gamma(y)$ is wellfounded.

Now suppose $U_y\subseteq W$, so $S_y$ is wellfounded. Then as in the proof of the Kunen-Martin theorem, $z$ has rank ${\leq\Gamma(y)}$ for each $z\in U_y$.