# $\Delta_{2}^{1}$-hard set?

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?

bye

matteo

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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.
Thank you Joel, great answer! I was thinking that something on these lines was the expected outcome. But now I wonder if it is possible nevertheless to have a sequence $A_{n}\in \Delta_{2}^{1}$, for $n\in \mathbb{N}$ such that for every $B\in\Delta_{2}^{1}$, $\exists n. B<A_{n}$. Let us call $\{A_{n}\}$ a $\Delta_{2}^{1}$-complete sequence. To to prove that $\mathcal{C}$ contains a non-measurable set i just have to prove that for every $n$ i can find a $C_{n}\in\mathcal{C}$ such that $A_{n}\leq C_{n}$. –  Matteo Mio Apr 22 '11 at 12:38
Since you using the boldface class, $\Delta^1_2$ will be closed under countable amalgamations of its members, so if there were $A_n$ like that, we could find a copy of their disjoint union in $\Delta^1_2$, and such a set would have to be complete under your assumptions. So there can be no such sequence. –  Joel David Hamkins Apr 23 '11 at 12:48
In general the structure of Wadge ordering for the classes closed under complement is very complicated. For example, even at the level of $\Delta^0_2$-sets the Hausdorf difference hierarchy allow increasing $\omega_1$-chains in the Wadge degrees. –  Dave Marker Jul 12 '11 at 14:27