# Decomposing $\mathbf{\Pi}^1_1$ sets into closed sets

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?

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- 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? –  Eran May 11 '12 at 19:28
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. –  Liang Yu May 12 '12 at 0:47
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. –  Trevor Wilson Jul 17 '12 at 19:57
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. –  Trevor Wilson Jul 26 '12 at 0:01
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.
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 $V$. –  Asaf Karagila May 12 '12 at 15:55
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. –  Liang Yu May 13 '12 at 1:53
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. –  Carlo Von Schnitzel May 13 '12 at 17:28