7
$\begingroup$

Let $\omega+1$ be endowed with the interval topology, that is $U\subseteq (\omega+1)$ is open if $U\subseteq\omega$ or $(\omega+1)\setminus U$ is finite. We call $U\subseteq (\omega+1)$ basic if either $U = \{n\}$ for some $n\in \omega$ or if $U = (\omega+1)\setminus m$ for some $m\in \omega$. (Don't confuse this with $(\omega+1)\setminus\{m\}$: I use the fact that $m=\emptyset$ or $m=\{0,\ldots, m-1\}$ for $m\in\omega\setminus\{\emptyset\}$.)

Endow $(\omega+1)^\omega$ with the box topology. We call a subset $B\subseteq (\omega+1)^\omega$ basic if $B = \prod_{n\in\omega} U_n$ with $U_n\subseteq (\omega+1)$ is basic for every $n\in \omega$.

Suppose $\mathcal{C}$ is a covering of $(\omega+1)^\omega$ by basic sets. Can $\mathcal{C}$ be refined to a covering by basic sets that are mutually disjoint?

$\endgroup$
4
  • $\begingroup$ It seems bad to me to write this like an ordinal power. But what do I know. $\endgroup$ Jan 19, 2015 at 22:08
  • $\begingroup$ Please feel free to edit the question and write it in a more understandable way $\endgroup$ Jan 20, 2015 at 6:49
  • 2
    $\begingroup$ You probably know that whether $\Box (\omega + 1)^\omega$ is paracompact (in ZFC), is a long-standing open problem. $\endgroup$ Jan 20, 2015 at 14:01
  • 1
    $\begingroup$ That's right, Ramiro, and thanks for your remark. My question should have been: is there any reason why this sharper version (ultraparacompactness as opposed to paracompactness) doesn't work? Maybe ultraparacompactness is also open... $\endgroup$ Jan 21, 2015 at 7:57

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.