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?



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




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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.


