# Can we inductively define Wadge-well-foundedness?

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 $$\Sigma^1_1$$.)

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.