For a topological space $X$ (which I'll identify with its underlying set of points), we define the Wadge preorder $Wadge(X)$: elements of the preorder are subsets of $X$, and the ordering is given by $A\le_{W(X)}B$ iff there is a continuous map $f:X\rightarrow X$ with $f^{-1}(B)=A$.

Actually, it is sometimes better to tweak the definition; if this affects the answer, feel free to use it instead, or indeed any other improvement on the naive Wadge hierarchy.

There is a natural substructure of $Wadge(X)$, namely its wellfounded part: $$WWF(X)=\{A\subseteq X: \neg\exists (B_i)_{i\in\omega}(B_0=A, B_i>_{W(X)}B_{i+1})\}.$$ (Here "$<_{W(X)}$" means "$\le_{W(X)}$ and not $\ge_{W(X)}$," as expected.)

For example, Borel determinacy implies that $WWF(\omega^\omega)$ contains the class of Borel sets, and under AD we in fact get $WWF(\omega^\omega)=\mathcal{P}(\omega^\omega)$.

Now, faced with a natural well-founded preorder, my instinct is that it should consist of the objects which can be "built up from below" by some natural procedure. But I don't see that this is the case here. So I want to ask:

Main question: Is there a way to think of $WWF(X)$ this way? Phrased a bit more abstractly, is $WWF(X)$ the least fixed point of some reasonable operator on $\mathcal{P}(\mathcal{P}(X))$, at least for "reasonable" $X$?

Even for $X=\omega^\omega$, this isn't clear to me. Indeed, it's not even clear to me that the usual "tame part" of $Wadge(\omega^\omega)$ is all of $WWF(\omega^\omega)$. So maybe the following is worth resolving on its own:

Secondary question: Is there an $A\in WWF(\omega^\omega)$ which is Wadge incomparable with (say) both the set of reals coding well-orderings and the complement of that set? (That is, the Wadge degrees $\Pi^1_1$ and $\Sigma^1_1$.)


Long comment:

It is difficult to answer "no" to the main question, given the unlimited potential interpretations of "reasonable". Still, I would be very surprised by a positive answer.

The reason is an observation by Peter Hertling that already on $\mathbb{R}$ we can build a descending chain of sets $(A_m)_{m < \omega}$ where each $A_m$ is a finite union of half-open intervals and open. Precisely, let $$A_n := [0,1) \cup (2,3) \cup \ldots \cup (2n,2n+1) \cup [2n+2,2n+3)$$

Since I believe that $\mathbb{R}$ ought to be reasonable space, this would leave us with needing a reasonable construction of sets that on $\mathbb{R}$ avoids already finite unions of (half)-open intervals, while creating at least all Borel sets on $\omega^\omega$.

Of course, the main question seems to be already very interesting on $\omega^\omega$ alone; so this example only serves to show that the scope should probably be limited a bit.

Going for Pequignot's tweaked definition would remove this obstacle, so that might be another approach..


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