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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?

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  • $\begingroup$ It seems bad to me to write this like an ordinal power. But what do I know. $\endgroup$
    – Gerald Edgar
    Commented Jan 19, 2015 at 22:08
  • $\begingroup$ Please feel free to edit the question and write it in a more understandable way $\endgroup$
    – Dominic van der Zypen
    Commented Jan 20, 2015 at 6:49
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    $\begingroup$ You probably know that whether $\Box (\omega + 1)^\omega$ is paracompact (in ZFC), is a long-standing open problem. $\endgroup$
    – Ramiro de la Vega
    Commented Jan 20, 2015 at 14:01
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    $\begingroup$ 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... $\endgroup$
    – Dominic van der Zypen
    Commented Jan 21, 2015 at 7:57

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