Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

up vote 8 down vote accepted

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.

share|improve this answer
    
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
2  
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.