Consider $\omega+1$ with the interval topology, that is $U\subseteq (\omega+1)$ is open if and only if $U\subseteq\omega$ or $(\omega+1)\setminus U$ is finite.
We write $(\omega+1)^\omega$ for the collection of all functions $f:\omega\to(\omega+1)$. Let $\Box (\omega+1)^\omega$ be the set $(\omega+1)^\omega$ endowed with the box topology, where each factor $(\omega+1)$ carries the interval topology.
For each $f\in(\omega+1)^\omega$ we consider the following clopen neighborhood $U_f$ of $f$:
$U_f = \{g\in(\omega+1)^\omega:\forall n\in\omega\big((f(n) < \omega \Rightarrow g(n) = f(n)) \text{ and } (f(n) = \omega \Rightarrow g(n) \in [n,\omega])\big) \}$
Let ${\cal U} = \{U_f: f\in(\omega+1)^\omega\}$. Is there an open covering ${\cal V}$ of $(\omega+1)^\omega$ with the following properties?
- each member of ${\cal V}$ is contained in some member of ${\cal U}$;
- for $V\neq W \in {\cal V}$ we have $V\cap W = \emptyset$.
Note: This is one special case of this question, see Joel David Hamkins' comment below.
