In set theory Borel sets are important, but we don't actually care about the sets. We can about the Borel codes. Namely, the algorithm to generate a given Borel set starting with the basic open sets (intervals with rational end points, or some other canonical basis) and taking countable unions, complements, and so on.

There are many proofs and arguments that look like "A Borel set $X$ is in a certain ideal if and only if it does not contain a real in a certain generic extension". And while that does give us some useful information about ground model sets via generic extensions, it is not $X$ itself that contains that generic real, but rather the "re-computation of $X$ from its code".

But what about the set $X$? Clearly you cannot add elements to sets with forcing. When does $X$ remain Borel? Or just generally, when does $\Bbb R$ itself stay Borel in its generic extension?

Is there some relatively simple and non-trivial condition on a generic real $c$ for which ground model Borel sets remain Borel? What about when adding uncountably many reals?

Clearly, if we collapse $2^{\aleph_0}$ to $\aleph_0$, then all the sets of reals become countable and thus Borel. But I'm looking for something a bit more robust, such as a simple criterion to check against Cohen, Sacks, Random, etc.

  • $\begingroup$ Since many of the "reals of interest" can be given as "avoiding an ideal of sets of reals", phrasing an answer in terms of properties of said ideals is also an interesting (albeit partial) answer. $\endgroup$
    – Asaf Karagila
    Jun 10, 2018 at 8:19
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    $\begingroup$ This paper of Velickovic and Woodin is relevant : arxiv.org/abs/math/9501203 They show that if M is an inner model of set theory and the set of reals in M is analytic then either all reals are in M or else $\aleph_{1}^{M}$ is countable. This doesn't quite answer your question, but I think there's other stuff in there that's relevant. $\endgroup$ Jun 10, 2018 at 11:50
  • $\begingroup$ Thanks Paul, that's interesting. I remembered something like this about $L$, but a general statement is even better. I'll look at the paper later today. $\endgroup$
    – Asaf Karagila
    Jun 10, 2018 at 14:21

1 Answer 1


I think that Groszek and Slaman's result (see https://www.jstor.org/stable/421023?seq=1) gives a satisfying answer to your question.

Groszek and Slaman's result says that given any inner model $M$ of $ZFC$, $\mathbb{R}\subset M$ if and only if there is a perfect set $A\subset M$.

A immediate conclusion of the result is that for any inner model $M$ of $ZFC$ and an $M$-Borel set $A$, $A$ remains to be Borel in any extension $N$ of $M$ with $\mathbb{R}^M\neq \mathbb{R}^N$ if and only if $A$ is countable in $N$.

  • $\begingroup$ That's perfect. Thanks! I only knew this result where $M=L$. $\endgroup$
    – Asaf Karagila
    Jun 10, 2018 at 15:53
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    $\begingroup$ I think this is not quite what Groszek and Slaman prove, and, unless I'm missing something, what you have written is not true as stated. (I assume you are quoting Theorem 2.4 in their paper, but this theorem has an extra combinatorial hypothesis, and does not apply to arbitrary pairs of models.) Velickovic and Woodin prove in arxiv.org/pdf/math/9501203.pdf that there are models $M,N$ of ZFC such that $M \subseteq N$, $N$ contains reals not in $M$, and yet the reals of $M$ form an uncountable $F_\sigma$ subset of reals in $N$. (See theorems 7 and 8 in the linked paper.) $\endgroup$
    – Will Brian
    Jun 11, 2018 at 12:20
  • $\begingroup$ It depends on the definition of ``inner model". Groszek-Slaman condition is not just a pure combinatoric one but a kind of inner model condition. Essentially it says that if $M=\bigcup_{\alpha}M_{\alpha}$, i.e. $M$ is built level by level, and master codes cofinally range over $\theta$, where $\theta$ is the least ordinal so that $M\setminus M_{\theta}$ does not contain a real, then G-S conclusion holds. $\endgroup$
    – 喻 良
    Jun 11, 2018 at 13:53
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    $\begingroup$ I see -- thanks for explaining. Does this condition always hold between a ground model and one of its forcing extensions? $\endgroup$
    – Will Brian
    Jun 11, 2018 at 17:20
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    $\begingroup$ @WillBrian it does not. Forcing with Namba over a model of $\mathbb{c}= \aleph_2$ adds a countable sequence of reals which cannot be covered by countable ground-model sets. The countable covering condition in that paper is designed to exclude the example of Velickovic and Woodin (which was a refinement of the rough "forcing with Namba over the failure of CH" idea.) $\endgroup$
    – Not Mike
    Jun 11, 2018 at 23:56

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