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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.

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    $\begingroup$ If $X$ is a finite discrete space with $|X|=n$ and $\mathcal{U}=\{X\}$, then $\kappa=B_{n}$ where $B_{n}$ denotes the $n$-th Bell number. However, there are only $2^{n}$ subsets of $X$ and $B_{n}$ grows faster than $2^{n}$. For infinite discrete spaces $X$, there are $2^{|X|}$ many partitions of $X$ into two sets, but there are only $|X|$ many ways to define the $Z_{\alpha}$ when $\mathcal{V}_{\alpha}$ has two blocks. $\endgroup$
    – Joseph Van Name
    Commented Aug 21, 2021 at 18:29
  • $\begingroup$ On the other hand, if $X=\{0,1\}^{\omega},\mathcal{U}=\{X\}$, then you can set $\kappa=\omega$, and in this case you can always construct sets $Z_{n}$. $\endgroup$
    – Joseph Van Name
    Commented Aug 21, 2021 at 18:32
  • $\begingroup$ Thanks Joseph. In fact $\kappa>B_n$ since the open refinement need not cover $X$. I figured it'd be unlikely to hold in full generality; I'll reword to ask "when" is it possible. $\endgroup$
    – Steven Clontz
    Commented Aug 21, 2021 at 18:55
  • $\begingroup$ Also worth noting that a pairwise-disjoint open refinement for any open cover is also one for $\{X\}$, so only considering $\{X\}$ (or really just considering an enumeration of the collections of pairwise-disjoint open sets of the space) makes sense. $\endgroup$
    – Steven Clontz
    Commented Aug 21, 2021 at 20:52
  • $\begingroup$ Is $X=\bigcup\mathcal{V}_\alpha$ by definition? (I thought so, but you write $\bigcup\mathcal{V}_\alpha$ at one point, instead of just $X$.) $\endgroup$
    – Farmer S
    Commented Aug 22, 2021 at 13:27

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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 $Z_\alpha = \{x_\alpha\}$.

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