Let SB denote the set of closed subspaces of $C(2^\mathbb{N})$ equipped with the Effros Borel structure, and $A\subset$ SB be analytic.
I am reading a proof that says $A_\sim = \{Z\in $SB $ \mid $ there exists $Y\in A$ such that $Y\sim Z\}$ is analytic since the isomorphism relation $\sim$ is analytic. I can't see how this follows. Could anyone please explain this to me?
The projection of $\{(X,Y)\in$ SB$^2 \mid X\sim Y\}$ onto the second coordinate is analytic. This projection is $\{Y\in$ SB$ \mid $ there exists $X\in$ SB such that $X\sim Y\}$. Does is follow on from this?
Your set $A_\sim$ is the projection onto the second coordinate of the set $$ \{(X,Y)\in \mathrm{SB}^2 \mid (X,Y)\in A\times \mathrm{SB} \text{ and } X\sim Y\}, $$ and $A\times \mathrm{SB}$ is the preimage of $A$ by the first projection, so both sets are analytic.