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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?

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    $\begingroup$ I don't know how much this will help but it seems relevant: The hyperspace of a compact Hausdorff space $X$ contains a natural homeomorphic copy of $X$ in the form of the closed set of singletons. Your set intersect this copy of $X$ is a copy of $A$. Maybe this rules out being co-analytic? $\endgroup$
    – James E Hanson
    Commented Apr 19, 2019 at 18:31
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    $\begingroup$ It is known that the hypersubspace $K(\mathbb Q\cap [0,1])\subset K([0,1])$ is $\Pi^1_1$-complete, so cannot be analytic. $\endgroup$
    – Taras Banakh
    Commented Apr 20, 2019 at 6:18
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    $\begingroup$ The topological structure of the hyperspace of compact subsets of some (nice) coanalytic sets is described in the paper link.springer.com/article/10.1007/BF02356061 of Banakh and Cauty. $\endgroup$
    – Taras Banakh
    Commented Apr 20, 2019 at 6:23
  • $\begingroup$ I think I have found it: In "The hyperspace of an analytic metrizable space is analytic", Alberto Barbati proved that if $X$ is a metrizable analytic space and $d$ a compatible metric on $X$, then the hyperspace of closed, non-empty subsets of $X$ with the Wisjman topology induced by $d$ is an analytic metrizable space. Since the Borel structure of the Wisjman topology coincides with that of the Vietoris/Hausdorff topology (it is the Effros Borel structure), this should imply that the set is analytic. I would like to clarify this last step of the argument before posting an answer. $\endgroup$
    – Iian Smythe
    Commented Apr 21, 2019 at 16:48
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    $\begingroup$ I suppose the issue is that even though the hyperspace of (relatively!) closed subspaces of $A$ is analytic, this does not coincide with the closed subspaces of $X$ which happen to be contained in $A$... $\endgroup$
    – Iian Smythe
    Commented Apr 21, 2019 at 20:56

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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.

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