# Is the space of countable closed covers of the Cantor set analytic?

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 $K(X)^\omega$.

Problem. Is the space $cov(X)$ analytic (equivalently, Borel in $K(X)^\omega$)?

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?

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 $K_\sigma$.

--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 $A_x=\bigcup_{n \in \omega} (K_n)_x=\bigcup_{n \in \omega} \overline{K}_n(x)$.

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.

• Thank you very much for the answer. This is exactly what I need (for the proof that the universal quasi-uniformity on an uncountable metrizable space does not have an $\omega^\omega$-base). – Taras Banakh Aug 23 '16 at 5:57
• In the first line of the answer "Polish" should be corrected to "compact Polish" or at least "$\sigma$-compact Polish", otherwise $cov(X)=\emptyset$. – Taras Banakh Aug 23 '16 at 6:01
• By the way, there is no necessity to attract the powerful Saint-Raymond Theorem: it suffices to observe that the complement of the $G_\delta$-set $B$ can be written as the countable union $\bigcup_{n\in\omega}F_n$ of compact sets and then the maps $K_n:2^\omega\to \mathcal K(2^\omega)$, $K_n:x\mapsto\{y\in 2^\omega:(x,y)\in F_n\}$, do the job. – Taras Banakh Aug 24 '16 at 17:26
• You are absolutely right, I have added it to the original answer. – vzoltan Aug 27 '16 at 13:43
• I would like to thank you in a paper for the help with the proof of non-analycity of $cov_\omega(X)$. How to write you name? At the moment it is a Mathoverflow's user vzoltan :) – Taras Banakh Aug 27 '16 at 14:31