Basic question on complexity of a set

Question

Let $A\subseteq X\times Y$, with $X$ and $Y$ Polish spaces.

Suppose that $A=\bigcup_{c\in C}A_{c}$, where $C\subseteq X$ is a closed set and each $A_{c}$ is Borel.

Can we conclude that $A$ is analytic ?

Answer 1

Even under the sensible assumption that each $A_c$ is included in $\{c\}\times Y$ (which I presume you intended) and, moreover, that $C=X$, the conclusion is false.

For one example, take $C=X=Y=\mathbb{R}$, let $V$ be any non analytic subset of $\mathbb{R}\times\{1\}$ (for instance, a non Lebesgue measurable linear set times {1}), and finally let $A:=V\cup (\mathbb{R}\times\{0\})$. Then $A$ is not analytic, and $A_c$ actually consists of one or two points, therefore Borel.

EDIT: After second thought, there is an example where $A_c$ is always a singleton, as suggested by Andrés. If $V$ is as above and with cardinality equal to that of $\mathbb{R}$ (e.g., a Vitali set), take $A$ to be the graph of any bijection from $C=\mathbb{R}$ onto $V$. Then the non analytic $V$ is the projection onto the second coordinate of $A$; this implies that $A$ is not analytic.