Sets that are not $\infty$-Borel I have seen a few techinques for proving that certain sets of real numbers are $\infty$-Borel (definition) but it just occurred to me that I don't know of any way to prove that a set of real numbers is not $\infty$-Borel.
I'm afraid this may turn out to be a silly question, but here goes: can non-$\infty$-Borel sets exist?  If $\mathsf{AC}$ holds then every set of reals is trivially $\infty$-Borel, and if $\mathsf{AD}$ holds then it is an open question whether every set of reals is $\infty$-Borel.  So to get a non-$\infty$-Borel set we may need to look in some weirder model.
Here we say that a set is $\infty$-Borel if it has an $\infty$-Borel code—an object that describes how the set is built up from open sets via the operations of complementation and well-ordered union.  Such a code is essentially a set of ordinals $S$.  An equivalent definition is that a set $A$ is $\infty$-Borel if there is a set of ordinals $S$, an ordinal $\alpha > \sup(S)$, and a formula $\varphi$ such that for every real $x$ we have $x \in A \iff L_\alpha[S,x] \models \varphi[S,x]$.
In the linked Wikipedia article as well as in other places, I have seen a discussion of the potential difference between the class of $\infty$-Borel sets and the class of sets generated from the open sets under the operations of complementation and well-ordered union.  The observation is that it's not clear whether, without $\mathsf{AC}$, we can choose $\infty$-Borel codes for $\infty$-Borel sets in a sequence, even if each set in the sequence has such a code.
However, I do not recall ever seeing a proof of the consistency of 


*

*$\mathsf{ZF} + {}$"the class of $\infty$-Borel sets is not closed under wellordered union," or even of 

*$\mathsf{ZF} + {}$"some set of reals is not $\infty$-Borel."
Assuming that (2) is consistent, I would like to see a proof of this.  Or if the stronger theory (1) is consistent, a proof of this would be even better.
If $\mathbb{R}$ is a union of countably many countable sets, then every set of reals is generated from the open sets under the operations of complementation and well-ordered (countable) union. In this case (1) and (2) are equivalent.  Perhaps they are both true in this case?
 A: Henle, Mathias and Woodin showed that if every set of reals is Ramsey, then forcing with $\mathcal{P}(\omega)/\mathrm{Fin}$ adds no new sets of ordinals. Any new set of reals in the extension (e.g., the ultrafilter given by the generic filter) would fail to be $\infty$-Borel in the extension. So I suppose that answers (2). I don't know about (1). 
A: This is a nice question Trevor. This is not an answer but it is a bit too long for a comment. If we assume that every set is $\infty$-Borel (say we're just assuming $AD^+$), then there is no proper ($\subsetneq \mathcal{P}(\mathbb{R}))$ pointclass $\boldsymbol\Delta$ closed under complements, arbitrary length unions and intersections. So if we can find a selfdual pointclass closed under complements, arbitrary length unions and intersections then there exists a set which is not $\infty$-Borel. 
I don't think this can follow from just $AD$ (not really sure about this though, it is still possible), because all selfdual pointclasses $\boldsymbol\Delta$ are closed under unions of length strictly less than $\delta(\boldsymbol\Gamma)$ for $\boldsymbol\Gamma$ closed under $\forall^{\mathbb{R}}$, conjunctions, disjunctions and such that $PWO(\boldsymbol\Gamma)$ holds and $\boldsymbol\Delta=\boldsymbol\Gamma \cap  \boldsymbol{\check\Gamma}$. In general, if $\boldsymbol\Gamma$ is non-selfdual, closed under $\forall^{\mathbb{R}}$ with $PWO(\boldsymbol\Gamma)$ and $\boldsymbol\Delta$ is closed under real quatification, then for $\kappa=o(\boldsymbol\Delta)$, $\boldsymbol\Delta$ is closed under union of length strictly less than $cof(\kappa)$. But the set of $o(\boldsymbol\Delta)$ for $\boldsymbol\Delta$ satisfying these assumptions is cofinal in $\Theta$, these are the places where we are at the base of a Type II or Type III projective -like hierarchy. Maybe if we reach a pointclass $\Lambda$ closed under complements and arbitrary length unions and intersections that set of such $o(\boldsymbol\Delta)$ would be bounded (pure speculation/conjecture)
As an aside there is another nice characterization of $\infty$-Borel sets: a set $X$ is $\infty$-Borel if there is a tree $T$ on $\omega \times \lambda$ such that $x\in X$ holds iff Player I wins the games $G$ where Player I and II play a branch $f\in \lambda^{\omega}$ and the playoff is Player I wins iff $(x,f)\in [T]$. The tree $T$ is the code for the set $X$.
