In full generality, my question is:
Suppose $V$ and $W$ are transitive models of $\sf ZFC$ and $V\subset W$. Let $A\in V$ be a set of reals with $V\models A$ is Borel. What can we say about the complexity of $A$ in $W$ as a subset of $\mathbb R^W$?
Here are some more specific questions:
For example, say we're living in $L$ and put $A=(2^\omega)^L$. It's well known that, in any outer model, the set of constructible reals is $\Sigma_2^1$. Is this sharp? Of course, it's consistently clopen, since we can always assume $V=L$, but I'm wondering about other models.
More generally, say we start off in a ground model $V$ and let $X=(2^\omega)^V$. Force to add a Cohen real $c$. In $V[c]$, every real is a limit point of $X$, because $X$ has all eventually constant sequences. So $X$ is dense, but definitely not closed. So this forcing killed the property of being closed. It's also not open, because any open set in $V[c]$ contains a generic real. Now, how ugly is $X$? Is it Borel? Projective?
One possible strategy would be to closely study the proof of Laver's theorem on the definability of the ground model, and see how complex the relevant formula is, but the parameter in it is way above the reals if I recall correctly.
Going in a slightly different direction, if we start with an infinite Borel set of reals $X\in V$, is there a forcing that unborelifies it? More precisely, is there a poset $\mathbb P$ with $\Vdash_\mathbb P \check{X} $ is not Borel? Are there "indestructible" Borel sets?
A reference that deals with these matters would also be appreciated. I've looked into the usual sources (Bartoszynski-Judah, Jech, Kanamori, etc) but didn't find anything.
A partial answer:
Groszek and Slaman showed that if $M\subseteq N$ are transitive models of set theory that don't have the same reals and if $\mathbb{R}^M$ countably covers $\mathbb{R}^N$, then every perfect set in $N$ has an element which is not in $M$.
Here, $\mathbb{R}^M$ countably covers $\mathbb{R}^N$ iff for every $N$-countable $H\subseteq \mathbb{R}^M$, there is some $M$-countable $G\in M$ such that $H\subseteq G$.
So after adding a Cohen real, the set of ground model reals is not Borel, since it is uncountable but doesn't contain a perfect subset by Groszek and Slaman's result.
Similarly, any uncountable Borel set in the ground model ceases to be Borel after adding a Cohen real for the same reason.