I am finding on the internet that the following is an impossible problem, yet, it is one of my homework problems.
For some fixed $y \in \mathbb{R}$. Show that if $A$ is a Borel subset of $\mathbb{R}^2$ then $A(y) = \{x \in \mathbb{R} : (x,y)\in A\}$ is Borel.
As I said, the problem is that I keep reading online that this is actually not true. Am I missing something?
This is not a projection but a "section" of $A$ and these are Borel:
The map $f_y:x \to (x,y)$ is a homeomorphism between $\Bbb R$ and $\Bbb R \times \{y\}$ and $A(y)= (f_y)^{-1}[A \cap (\Bbb R \times \{y\})]$ is thus Borel.
The projection of $A$ is $\{x \in \Bbb R: \exists y \in \Bbb R: (x,y) \in A\}$, which is in general an analytic set, and need not be Borel. The point is that we fix $y$ here, so we don't get "uncountable quantification".