I have a Polish (completely metrizable and separable) space $X$ and a Borel subet $A \subseteq X\times X$, i.e. $A \in \boldsymbol{\mathcal{B}}(X\times X)$. Given a Borel set $C\subseteq X$ is it true that $$\{x \in X \mid A_x = C\} \in \boldsymbol{\mathcal{B}}(X)?$$ By $A_x$ I mean the section of $A$ along $x$, i.e. $A_x = \{y\in X \mid (x,y) \in A\}$
In case it is, is it also the case for projective sets?
Thanks
On
Let $X = \mathbb R$. There is a Borel set $A \in \mathcal B(\mathbb R^2)$ whose projection onto the first coordinate $$ \pi_1(A) = \{x : \exists y, (x,y) \in A\} $$ is not a Borel set. (Result due to Suslin.)
But that means the complement is also non-Borel $$ \pi_1(A)^c := \mathbb R^2 \setminus \pi_1(A) \notin \mathcal B(\mathbb R). $$ We have $$ A_x = \{y \in \mathbb R : (x,y) \in A\} $$ so \begin{align} A_x = \varnothing \quad&\Longleftrightarrow\quad\not\exists y, (x,y) \in A \\&\Longleftrightarrow\quad x \notin \pi_1(A) \\&\Longleftrightarrow\quad x \in \pi_1(A)^c \end{align} But take $C = \varnothing$. Thus $$ \{x : A_x = C\} = \pi_1(A)^c \notin \mathcal B(\mathbb R) $$
No. If $B \subseteq \mathbb{R} \times \mathbb{R}$ is a closed set, then $\{x : B_x \neq \emptyset\}$ is an analytic set and all analytic sets take such a form. Thus $\{x : B_x = \emptyset\}$ is a coanalytic set. There are coanalytic subsets of $\mathbb{R}$ which are not Borel (for instance the set of reals coding wellorderings on $\omega$.)
Given $B \subseteq \mathbb{R} \times \mathbb{R}$ and $C \subseteq \mathbb{R}$. Let $A = \{x \in \mathbb{R} : B_x = C\}$. Note that $x \in A \Leftrightarrow (\forall y)((x,y) \in B \Leftrightarrow y \in C)$. So if $B$ and $C$ belonged to some pointclass closed under existential real quantifers and complements (like the class of projective sets), then $A$ would also be in the pointclass (i.e. be projective).