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Let $X,Y$ be Borel spaces and $A\subseteq X\times Y$ be an analytic set. Let $\pi:X\times Y \to X$ denote the projection map onto $X$. Does there always exist a set $B$ such that $\pi(B) = X\setminus \pi(A)$ and such that the union $A\cup B \subseteq X\times Y$ is still analytic?

In particular, $B$ can't be an analytic set (let along Borel) since in such case $\pi(B)$ would be analytic, whereas in some cases $X\setminus \pi(A)$ is co-analytic and not Borel, hence not analytic.

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  • $\begingroup$ I have asked this question on MSE, however so far I have not received an answer or a hint determining the direction for the solution. I am not sure what is the current policy regarding posting questions on MO if they have not received enough attention on MSE. I've posted it first on MSE as I was not sure the question is difficult enough, but so far my descriptive set theory question received more answers on MO, so I decided to give it a try here as well. $\endgroup$ – Ilya Jul 17 '14 at 18:43
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    $\begingroup$ I think the question is fine for MO. It's quite interesting. $\endgroup$ – Joel David Hamkins Jul 18 '14 at 1:54
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The answer is no, not necessarily. It can happen that there is no such set $B$.

A counterexample is provided whenever $A$ is a Borel subset of the plane, while the projection $\pi(A)$ is analytic, but not co-analytic. In this case, if $A\cup B$ were analytic, it would follow that $B$ itself is analytic, since $B=(A\cup B)-A$, in light of the fact that $A$ and $B$ must be disjoint since they project to complements. But if $B$ is analytic, as you point out, then $\pi(B)$ would also be analytic, although it is the complement of $\pi(A)$, and this contradicts our assumption that $\pi(A)$ is not co-analytic.

There are numerous concrete counterexamples in the reals $\mathbb{R}$, since every analytic set there is the projection $\pi(A)$ of a Borel set in the plane $\mathbb{R}\times\mathbb{R}$. (Indeed, in the Baire space $\omega^\omega$, every analytic set is the projection of a closed subset of $\omega^\omega\times\omega^\omega$.)

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  • $\begingroup$ No, only $F_\sigma$-sets are projections of closed sets in the real plane. Logicians sometimes use $\mathbb N^{\mathbb N}$ and call it "the reals", and there it would work. $\endgroup$ – Gerald Edgar Jul 18 '14 at 2:36
  • $\begingroup$ Yes, indeed, I was using the set-theorists "reals", which really means Baire space, even though they often denote it by $\mathbb{R}$. But I'll edit to make this more clear. But there are counterexamples in the real reals, however, where you take $A$ to be merely Borel rather than closed. $\endgroup$ – Joel David Hamkins Jul 18 '14 at 2:56
  • $\begingroup$ Thanks, Joel - apparently I was missing a little but important case that if $A$ is Borel then $B$ is analytic. $\endgroup$ – Ilya Jul 18 '14 at 8:00

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