6
$\begingroup$

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?

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Taras Banakh Aug 23 '16 at 5:57
  • $\begingroup$ In the first line of the answer "Polish" should be corrected to "compact Polish" or at least "$\sigma$-compact Polish", otherwise $cov(X)=\emptyset$. $\endgroup$ – Taras Banakh Aug 23 '16 at 6:01
  • $\begingroup$ 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. $\endgroup$ – Taras Banakh Aug 24 '16 at 17:26
  • $\begingroup$ You are absolutely right, I have added it to the original answer. $\endgroup$ – Zoltan Vidnyanszky Aug 27 '16 at 13:43
  • $\begingroup$ 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 :) $\endgroup$ – Taras Banakh Aug 27 '16 at 14:31

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.