TITLE: Sets that are not $\infty$-Borel
QUESTION [12 upvotes]: 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?

REPLY [6 votes]: 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).