For an uncountable compact metric space $X$ denote by $K(X)$ be the hyperspace of non-empty compact subsets of $X$, endowed with the Vietoris topology (which is generated by the Hausdorff metric). 


In the countable power $K(X)^\omega$ consider the subspace $$cov(X)=\{(K\_n)\_{n\in\omega}\in K(X)^\omega:\bigcup\_{n\in\omega}K\_n=X\}$$ of closed countable covers of $X$. It can be shown that $cov(X)$ is a coanalytic subspace of $K(X)^\omega$.



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> **Problem.** Is the space $cov(X)$ analytic (equivalently, Borel in $K(X)^\omega$)?
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I expect that the answer is negative and moreover, $cov(X)$ is $\Pi^1\_1$-complete. But how to prove this? Maybe it has been published somewhere? 



Is the space of countable closed covers of the Cantor set analytic?
For $X=2^\omega$ (and similarly for every uncountable $\sigma$-compact Polish space) this set is indeed $\Pi^1\_1$-complete:
Take a $G\_\delta$ set $B \subset 2^\omega \times 2^\omega$ so that $proj\_1(B)$ is $\Sigma^1\_1$-complete. Let $S=2^\omega \setminus proj\_1(B)$, then of course $S$ is $\Pi^1\_1$-complete and the set $A=2^\omega \times 2^\omega \setminus B$ is $K\_\sigma$. 




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--Added later following the suggestion of Taras Banakh--


It is not hard to show that if $K \subset 2^\omega \times 2^\omega$ the map $x \mapsto K\_x$ from $2^\omega$ to $\mathcal{K}(2^\omega)$ is Borel (where $K\_x$ denotes $\{y:(x,y) \in K\}$). Let us denote this map for a compact $K$ by $\overline{K}$. Let $A=\bigcup\_{n \in \omega} K\_n$ with $K\_n$ compact. Thus, we get Borel functions $\overline{K}\_n:2^\omega \to \mathcal{K}(2^\omega)$ for $n \in \omega$ such that for every $x \in 2^\omega$ we have $A\_x=\bigcup\_{n \in \omega} (K\_n)\_x=\bigcup\_{n \in \omega} \overline{K}\_n(x)$. 




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Alternatively, one can use a more general theorem of Saint Raymond (Theorem 35.46 in Kechris' book) to derive the same conclusion. 


Letting $K=(\overline{K}\_0,\overline{K}\_1,\dots)$ we get a Borel map $K:2^\omega \to (\mathcal{K}(2^\omega))^\omega$. Then $K(x) \in cov(2^\omega)$ if and only if $A\_x =2^\omega$ which is equivalent to $x \in S$. Thus, $S=K^{-1}(cov(2^\omega))$, so by the completeness of $S$ the set $cov(2^\omega)$ is also $\Pi^1\_1$-complete.


