It is well known that every $\mathbf{\Pi}^1_1$-set is a union of $\aleph_1$-many Borel sets. I wonder whether it can be improved under certain reasonable set theory axioms assumption.

For example, assuming $ZFC+CH$, then it is trivially true that every set is a union of $\aleph_1$-many closed sets. But this seems heavily depends on $CH$ since if $ZFC+\neg CH+MA$, then there is a lightface $\Pi^0_2$-set which cannot be a union of $\aleph_1$-many closed sets.

So my question is: is it consistent with $ZFC+\neg CH$ that every $\mathbf{\Pi}^1_1$-set is a union of $\aleph_1$-many closed sets?

  • 1
    $\begingroup$ - Can you please give a reference for the above fact about $\Pi^0_2$ sets? - If every lightface $\Pi^0_2$ is in boldface $\Pi^0_2$ (which is $\Pi^1_1$ then isn't this a counter-example to your question? $\endgroup$
    – Eran
    May 11, 2012 at 19:28
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    $\begingroup$ Just consider a $\Pi^0_2$, null, and comeager set, which cannot be a union of $\aleph_1$-many closed sets under the assumption $ZFC+\neg CH+MA$. Here $MA$ really matters. $\endgroup$
    – 喻 良
    May 12, 2012 at 0:47
  • $\begingroup$ One approach to the question would be to consider what various cardinal invariants of the continuum must be like in order for your $\Pi^0_2$ set not to be a counterexample. Then if these are known to be consistent you can analyze the specific models in which they are proved to hold. $\endgroup$ Jul 17, 2012 at 19:57
  • $\begingroup$ If it's true for any uncountable Polish space then it's true for Cantor space (that's probably why you say in the comments below that the question is about Cantor space.) If it's true for Cantor space (even just for $\Pi^0_2$ sets) then the Baire space $\omega^\omega$ is the union of $\aleph_1$-many Meager sets, and is also the union of $\aleph_1$-many compact sets. So you might want to look at iterated forcing extensions with these properties where CH fails. $\endgroup$ Jul 26, 2012 at 0:01
  • $\begingroup$ Howard Becker and Randall Dougherty has a nice answer to the question: sciencedirect.com/science/article/pii/… $\endgroup$
    – 喻 良
    May 24, 2020 at 1:48

2 Answers 2


Every ${\bf\Pi}_1^1$ set is the union of $\aleph_1$ Borel sets, so we only have to make sure that every Borel set is the union of $\aleph_1$ closed sets. But every Borel set is analytic and thus a continuous image of the Baire space $\omega^\omega$.
Now the Baire space is the union of $\mathfrak d$ (the dominating number) compact sets. Continuous images of compact sets are again compact and hence every analytic set is the union of $\mathfrak d$ compact (and hence closed) sets.

It is well known that $\mathfrak d=\aleph_1$ is consistent with the failure of CH. This happens for instance in the so called Sacks model.

Edit: I forgot to mention that it was Paul Larson who turned my attention to this question. Paul conjectured (or should I say, was convinced) that in the model obtained by forcing with a large countable support product of Sacks forcing over a model of CH every ${\bf\Pi}_1^1$ set is the union of $\aleph_1$ closed sets, which is true, since $\mathfrak d$ is $\aleph_1$ in that model.

  • $\begingroup$ Your answer is much better, however. $\endgroup$ Jul 8, 2014 at 21:36
  • $\begingroup$ Dilip had a similar idea to Paul's. I think the direct method should work. A further question is whether every Borel set can be decomposed into $\aleph_1$ many disjoint closed sets? $\endgroup$
    – 喻 良
    Jul 8, 2014 at 22:21

There is a theorem of my teacher Steve Jackson which says that assuming $ZFC + AD^{L(\mathbb{R})}$ every projective set is $\aleph_{\omega}$-Borel. So in particular this holds for $\Pi^1_1$ sets. The proof uses the theory of descriptions and every other technical tool from descriptive set theory (homogeneous trees, scales,...). Also, with respect to $MA$ and $CH$, $AD$ can't decide them, so maybe that result might be what you're looking for, I'm not sure. You can find the result in this survey of Jackson "A survey of Determinacy" somewhere in the end of the paper.

  • $\begingroup$ Actually $AD$ decides $CH$ (true), namely $M\models AD$ then $M\models CH$ (in the sense that every uncountable set of reals has size continuum). I believe that you wanted to say that $ZFC+AD^{L(\mathbb R)}$ cannot decide $CH$ in $V$. $\endgroup$
    – Asaf Karagila
    May 12, 2012 at 15:55
  • $\begingroup$ Yes indeed, it's true. $\endgroup$ May 12, 2012 at 21:20
  • $\begingroup$ It is a $ZFC$-theorem that every $\mathbf{\Pi}^1_1$-set is a union of $\aleph_1$-many Borel sets. The reason is (may be from a set theorist point of view) that it can be represented by an $\aleph_1$-Suslin set. Moreover it seems no way to improve the result by replacing Borel with closed via modifying the proof (the proof is by a so-called bar-induction, which use complement, union, and intersection transfinitely many times). Steve's result essentially says every projective set can be represented by an $\aleph_{\omega}$-suslin set. $\endgroup$
    – 喻 良
    May 13, 2012 at 1:53
  • $\begingroup$ Nice observation Yu. Concerning you original question, since you want $\Pi^1_1$ to be an $\aleph_1$ union of closed sets, is it possible to first start with your $\Pi^1_1$ set as a $\aleph_1$ union of Borel sets, then use that theorem which changes the topology to get a finer topology in which all these Borel sets are clopen and then say that the original $\Pi^1_1$ set is now an $\aleph_1$ union of clopen sets in the new topology? For this to happen, the space where all this takes place must be Polish. $\endgroup$ May 13, 2012 at 17:28
  • $\begingroup$ It might work. So you want to solve the question by changing the topology? My question is only for Cantor space. $\endgroup$
    – 喻 良
    May 14, 2012 at 0:38

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