I trying to study the Coding Lemma (in descriptive Set Theory) and there is a small point in the proof that I don't understand. Let me first recall the version I'm studying ( there are different version of the Moschovakis Coding lemma).


Assume **AD**. Let $\Gamma$ be a non-self-dual pointclass closed under real quantification and conjunction. Suppose that $\prec$ is a $\Gamma$ wellfounded relation on $\omega^{\omega}$. Then for any $R\subseteq dom(\prec)\times \omega^{\omega}$ such that $\forall x \in dom(\prec) \exists y R(x,y)$, there is an $A\subseteq dom(\prec)\times \omega^{\omega}$, $A\in \Gamma$, which is a choice set for $R$. That is, 


$\forall \alpha < |\prec| \exists x \in dom(<) \exists y [|x|\_{\prec}=\alpha \wedge A(x,y)]$


$\forall x, y[A(x,y) \rightarrow R(x,y)]$.


Now here's what I don't get. The proof starts as follows: Since $\Gamma$ is non-self-dual, let $U \subseteq (\omega^{\omega})^3$ be a universal set in $\Gamma$ for the $\Gamma$ subsets of $\omega^{\omega} \times \omega^{\omega}$. Let $\delta$ be the least length of $\prec$ such that the theorem fails. Then $\delta$ is a limit ordinal. 


Why is $\delta$ a limit ordinal?



Moschovakis Coding Lemma
I might be missing something, but I believe the answer is the following. First suppose $\Gamma$ isn't present. Then this just amounts to showing that, if $R\subseteq X\times Y$, $Z$ is a cofinite subset of $X$, and I have a choice set $C$ for the induced $R'\subseteq Z\times Y$ (that is, $R'=R\cap (Z\times Y))$, then I can extend that choice set to a choice set for $R$. To see that I can do this, we just use the fact that $X-Z$ is finite. Write $X-Z=\{z\_1, . . . , z\_n\}$, and let $y\_1, . . . , y\_n$ be such that $R(z\_i, y\_i)$ holds for all $i$. Then the set $C\cup\lbrace (z\_i, y\_i): 1\le i\le n\rbrace$ is a choice set for $R$.


(This is basically just showing that the Axiom of Choice holds for finite collections of sets.)


Now in the case of the theorem in question, our choice sets need to be taken from some class $\Gamma$. This means that we'll need to rely on some closure property of $\Gamma$; fortunately, we appear to have that, so (unless I'm missing something) the above proof works.


