Hello everybody!

I'm interested in $\Delta_{2}^{1}$ subsets of Polish spaces, i.e. those sets that are both $\Pi_{2}^{1}$ and $\Sigma_{2}^{1}$ in the boldface hierarchy of Polish spaces.

There is a notion of "being complicated" for a subset of a Polish space. A set $A$ is less-or-equally complicated that a set $B$ iff $A\leq B$, where the preorder $\leq$ is the so-called Wadge pre-order.

Given a set of subsets $\Gamma$, $A$ is said to be $\Gamma$-complete iff $\forall B\in \Gamma. B\leq A$.

There are "simple" examples of $\Sigma_{1}^{1}$-complete sets, for instance the set of all ill-founded trees over the natural numbers.

**QUESTION 1:** Is there a $\Delta_{2}^{1}$-complete set? If so, can you provide references?

$\Delta_{2}^{1}$ sets can be quite complicated. For example $ZFC+V=L\vdash$"there exists a non-measurable $\Delta_{2}^{1}$ set". Of course it is also consistent that every $\Delta_{2}^{1}$ set is measurable (and indeed universally-measurable.). So I guess I'm a bit in a unstable territory.

The real problem I have is that I have a collection $\mathcal{C}\subseteq\Delta_{2}^{1}$ of sets, and I would like to prove that it is consistent with ZFC, that one $C\in \mathcal{C}$ is not universally-measurable.

My "strategy" for proving this, is (given an answer to question 1) find a $C\in \mathcal{C}$ and prove that $A\leq C$. This implies that if $C$ is universally measurable, so is $A$ (and so is every $\Delta_{2}^{1}$-set). Under $ZFC+ V=L$, this implies that $C$ is not universally measurable.

However this proof-technique (assuming an answer to question 1) would work only if the $\Delta_{2}^{1}$-complete set $A$ has a reasonable description (like the one I gave for the $\Sigma_{1}^{1}$-complete set above)

**QUESTION 2:** Do you have any other proof-method in mind to get the goal i discussed?

thank you in advance,

bye

matteo