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.](http://en.wikipedia.org/wiki/Projective_hierarchy)


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.](http://en.wikipedia.org/wiki/Wadge_hierarchy)


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.](http://en.wikipedia.org/wiki/Universally_measurable_set)). 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



$\Delta_{2}^{1}$-hard set?
There can't be a $\Delta^1\_2$-complete set. Suppose that $B$ is any $\Delta^1\_2$ set, and let $A$ consist of the points $b$ with $f\_b(b)\notin B$, where $f\_b$ is the continuous function canonically indexed by $b$. Thus, the set $A$ is also $\Delta^1\_2$, but it cannot be that $A=f^{-1}B$ for any continuous function $f$, since any such $f$ is $f\_b$ for some $b$, and we arranged that $b\in A\iff f\_b(b)\notin B$. Thus, we've diagonalized against all the continuous functions without leaving $\Delta^1\_2$. The same argument works for any $\Delta$ class. 


