n odd: $\bf\Delta^1_n$ wadge degrees are $< \bf\delta^1_{n+1}$ My adviser is out of town and there is a comment in the Van Wesep paper "wadge degrees and descriptive set theory" that I can't figure out.
Work in ZF+AD throughout.
As stated in the title, the comment says that for n odd, the (order type of the) $\bf\Delta^1_n$ wadge degrees are less that $\bf\delta^1_{n+1}$ and Van Wesep includes the comment "(Prewellorder the codes of of $\bf\Delta^1_n$ sets as preimages of initial segments of a $\bf\Pi^1_n$-prewellordered complete $\bf\Pi^1_n$ set)". 
This is easy enough, let U be some $\bf\Pi^1_n$ complete set with corresponding well-order $\phi$. For x and y codes of $\bf\Delta^1_n$ sets, let $x\leq y$ iff $\exists w \in U_x$ such that (x,w) is $\leq_\phi$ greater than all (y,z) such that $z\in U_y$. This will be defined for all such x,y because of the boundedness principle.
Now, in order to finish the proof, one must show that for $\bf\Delta^1_n$ codes x,y,  $U_y <_w U_x \rightarrow y < x$. While this seems intuitively true (if $y<x$ we can compute $U_y$ from $U_x$ using $<_\phi$), I cannot prove this fact.
I believe this is the correct approach. One alternative way of defining $\leq$ would be $y\leq x$ iff $(<_\phi |U_y) \leq$ (as an order type) $ (<_\phi |U_x)$ or something along those lines. But, this seems less hopeful.
Any help would be greatly appreciated. I apologize if this question is too basic for this website.
,
Cody
 A: Cody, in the 4th paragraph, when you consider a universal $\bf \Pi^1_n$ universal set $U$ (or $\bf \Pi^1_n$-complete like you wrote), for $n$ odd, I think you meant to say that $\phi$ is a norm corresponding to a $\it prewellorder$ on that universal set(since for $n$ odd, by the first periodicity theorem $\bf \Pi^1_n$ has the prewellordering property). 
Concerning the question (maybe I'm missing the point here), it seems this is basically the "covering lemma". Any $\bf \Delta^1_n$ set is going to be bounded in the norm on the $\bf \Pi^1_n$ universal set given by the prewellordering property of $\bf \Pi^1_n$. The proof is exactly the same as the proof that $\bf \Delta^1_1$ sets are bounded in any norm given on a $\bf \Pi^1_1$ set (or say in the norm on the $\bf \Pi^1_1$ complete set $\bf WO$. Actually in our case here $\bf \Sigma^1_n$ sets would also be bounded in the norm. Using an appropriate norm (i.e sending a real to a corresponding Wadge degree) gives the result mentioned by Van Wesep. (One might need to also show that the norm on the $\bf \Pi^1_n$ set is onto $\bf \delta^1_n$, See theorem 4.A.4 in Moschovakis for a full proof of this fact for $\bf \Pi^1_1$, it generalizes nicely). I can tell you more about it at UNT if you want (My answer might not be the most complete one).
