Set of infinite well founded trees if complete coanalytic?

Question

I know that the set $WF$ of well founded trees in $\omega^\omega$ is a coanalytic complete set and I was wondering if I restrict to the set of all inifinite (as sets) trees, the set of all infinite well founded trees would also be coanalytic complete.

I've managed to prove that the set $\mathcal{T}_i$ of infinite trees in $\omega^\omega$ is a $G_\delta$ set of the set of trees (that's because its complement, the set of finite trees, is countable so a $F_\sigma$ set) so it is a polish space.

Now, as the set of well founded finite trees is equal to the set of finite trees, $\mathcal{T}_f$, we obtain that $WF_i$ cannot be Borel in $\mathcal{T}$ (the set of all trees in $\omega^\omega$), because $WF$ is not Borel in $\mathcal{T}$ and $$WF=WF_i \cup WF_f = WF_i \cup \mathcal{T}_f$$ Furthermore, if $WF_i$ is Borel in $\mathcal{T}_i$, then there exists some Borel subset $A \subset \mathcal{T}$ such that $$WF_i=A \cap \mathcal{T}_i$$ but as $\mathcal{T}_i$ is $G_\delta$ in $\mathcal{T}$ this will imply that $WF_i$ is Borel in $\mathcal{T}$ obtaining a contradiction.

Lastly, as $$\mathcal{T}_i \setminus WF_i = (\mathcal{T} \setminus WF) \cap \mathcal{T}_i$$ an $\mathcal{T} \setminus WF$ is analytic in $\mathcal{T}$ we obtain that $\mathcal{T}_i \setminus WF_i$ is analytic in $\mathcal{T}_i$, so $WF_i$ is coanalytic.

That is, if everything is correct, I know that $WF_i$ is coanalytic non-Borel in $\mathcal{T}_i$, but I don't know if it is coanalytic complete (without assuming determinancy).

Answer 1

Yes, it's still coanalytic complete.

In my opinion, the fastest way to see this is via an explicit "translation" map: given a tree $T\subseteq \omega^{<\omega}$, let $$T'=\{\sigma\in\omega^{<\omega}: (\sigma(0)=0\wedge\vert\sigma\vert<42)\vee (\sigma\in T)\}.$$ Basically, we "infinitize" $T$ without adding any new branches. The map $f:T\mapsto T'$ is continuous, preserves/reflects well-foundedness, and only outputs infinite trees, so we're done:

Given a coanalytic set $Y$ and a reduction $r$ of $Y$ to $WF$, the map $f\circ r$ is a reduction of $Y$ to $WF_i$.

(More generally it is indeed true that throwing away a countable set from a set of "high complexity" - say, worse than $F_\sigma$ - won't change the complexity of that set. In particular, if $X\subseteq\omega^\omega$ is coanalytic-complete and $C$ is countable then $X\setminus C$ is also coanalytic-complete.)