Uniquely selecting points from open pairwise disjoint refinements of an open cover Let $\mathcal U$ be an open cover of some space $X$.
Let $\{\mathcal V_\alpha:\alpha<\kappa\}$ enumerate all of its pairwise-disjoint
open refinements.
When is it possible to define sets $Z_\alpha$ such that
$Z_\alpha\subseteq\bigcup\mathcal V_\alpha$,
$|V\cap Z_\alpha|\leq 1$ for all $V\in\mathcal V_\alpha$, and
$Z_\alpha=Z_\beta$ only if $\alpha=\beta$? As noted in a comment below, this fails for $\mathcal U=\{X\}$ for finite discrete $X$, but this would hold for any finite ordinal $\gamma$ with the topology $\gamma+1$, since any pairwise disjoint collection of open sets would be a singleton $\mathcal V_\alpha=\{\alpha+1\}$ and $Z_\alpha=\{\alpha\}$ would work.
 A: As it was pointed out, the pairwise disjoint open families need not cover the space. With that in mind, I propose a partial answer (since I don't yet have enough rep to comment).
If $X$ has at least two isolated points, this cannot be done. Let $x$ and $y$ be distinct isolated points and consider the collections $\mathscr V_1 = \{\{x\}\}$, $\mathscr V_2 = \{\{y\}\}$, $\mathscr V_3 = \{\{x,y\}\}$, and $\mathscr V_4 = \{\{x\},\{y\}\}$. Avoiding trivialities, we can include $\mathscr V_0 = \{\}$ and let $Z_0 = \emptyset$. Now, $Z_1$ and $Z_2$ must be $\{x\}$ and $\{y\}$, respectively. But now $\mathscr V_3$ and $\mathscr V_4$ don't have unique $Z_j$.
If $\kappa$ is regular, every open subset of $X$ has cardinality $\kappa$, and the cardinality of the collection of pairwise disjoint open families is $\kappa$, then we can pick singletons as our representatives (this is the case, for example, when $X = 2^\omega$). Let $\{\mathscr V_\alpha : \alpha < \kappa \}$ be all collections of pairwise disjoint open families, ignoring the empty family. Initialize some $x_0 \in \bigcup \mathscr V_0$. At $\alpha < \kappa$, $\{x_\beta : \beta < \alpha\}$ has cardinality less than $\kappa$ so we can pick $x_\alpha \in \bigcup \mathscr V_\alpha \setminus \{ x_\beta : \beta < \alpha \}$. Then $Z_\alpha = \{x_\alpha\}$.
