TITLE: Complexity of the set of closed subsets of an analytic set
QUESTION [8 upvotes]: Let $X$ be a compact Polish space and $K(X)$ the hyperspace of closed subspaces of $X$ with the Vietoris/Hausdorff metric topology.
Question: If $A$ is an analytic subset of $X$, what is the complexity of the set $\{C\in K(X): C\subseteq A\}$?
This set is clearly $\mathbf{\Pi}^1_2$, but can this be reduced to $\mathbf{\Delta}^1_2$ or even to (co-)analytic?

REPLY [3 votes]: As far as I see, the set can be $\mathbf{\Pi}^1_2$-complete. Take a $\mathbf{\Pi}^1_1$ set $C \subset 2^\omega \times 2^\omega$, such that $\pi_1(C)$ is a $\mathbf{\Sigma}^1_2$-complete set. Then consider the mapping $x \mapsto \{x\} \times 2^\omega$. This is a continuous mapping from $2^\omega$ to $\mathcal{K}(2^\omega \times 2^\omega)$ and a reduction of $2^\omega \setminus \pi_1(C)$ (which is a $\mathbf{\Pi}^1_2$-complete  set) to $\{K \in \mathcal{K}(2^\omega \times 2^\omega): K \subset 2^\omega \times 2^\omega \setminus C\}$, that is, the collection of the compact subsets of $C$'s complement.