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I am looking for any information about the following property for a compact Hausdorff space $K$: For any sequence $\left(U_{n}\right)$ of nonempty open sets (not necessarily distinct) there is a disjoint sequence $\left(V_{n}\right)$ of nonempty open sets such that $V_{n}\subset U_{n}$, for every $n$.

In particular, are there "nice" properties which imply it?

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  • $\begingroup$ Do you know of any compact Hausdorff spaces without this property? I can't think of one. $\endgroup$
    – Will Brian
    Apr 19 at 11:52
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    $\begingroup$ @WillBrian For any space with an isolated point $p$, let $U_n=\{p\}$ for all $n$. $\endgroup$
    – Wojowu
    Apr 19 at 11:54
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    $\begingroup$ @Wojowu: That's cheating. :) Can you do it without cheating? $\endgroup$
    – Will Brian
    Apr 19 at 12:04
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    $\begingroup$ @WillBrian well, depends how you define cheating ;) Off the top of my head I can't think of an example without isolated points. $\endgroup$
    – Wojowu
    Apr 19 at 12:06
  • $\begingroup$ @WillBrian The Stone space of the measure algebra for Lebesgue measure is one such example $\endgroup$
    – erz
    Apr 20 at 5:11

2 Answers 2

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Update:

Here is an exact characterization of the regular spaces having the OP's property. Recall that a $\pi$-base for a space $X$ is a collection of nonempty open subsets of $X$ such that every nonempty open subset of $X$ contains a member of the collection.

Observation: A regular space $X$ fails to have the OP's property if and only if some nonempty open $U \subseteq X$ has a countable $\pi$-base.

Proof: The "if" direction is easy — it's basically the idea from Iosif Pinelis's answer and Wojowu's comment under it. If $(U_n)$ enumerates a $\pi$-base for some nonempty open $U \subset X$, then no sequence $(V_n)$ can meet the requirements of the OP's property.

For the other direction, let us assume that no nonempty open $U \subseteq X$ has a countable $\pi$-base. Let $(U_n)$ be a sequence of nonempty open subsets of $X$. We construct the $V_n$ by recursion. First, because $U_0$ does not have a countable $\pi$-base, in particular the family $\{ U_n \cap U_0 :\, n \in \mathbb N \}$ is not a $\pi$-base for $U_0$. Thus there is some $V \subseteq U_0$ such that $U_n \not\subseteq V$ for all $n$. Using the regularity of $X$, we can shrink $V$ a little and find a set $V_0 \subseteq U_0$ such that $U_n \not\subseteq \overline V_0$ for all $n$. Next, we use the fact that $U_1 \setminus \overline V_0$ does not have a countable $\pi$-base. In particular the family $\{ U_n \cap U_1 \setminus \overline V_0 :\, n \in \mathbb N \}$ is not a $\pi$-base for $U_1 \setminus \overline V_0$. Thus there is some $V \subseteq U_1 \setminus \overline V_0$ such that $U_n \not\subseteq V$ for all $n$. Using the regularity of $X$ to shrink $V$ a little, there is some $V_1 \subseteq U_1 \setminus \overline V_0$ such that $U_n \not\subseteq \overline V_1$ for all $n$. Generally, at step $k$ we use the fact that $U_k \setminus (\overline V_0 \cup \dots \cup \overline V_{k-1})$ does not have a countable $\pi$-base. In particular the family $\{ U_n \cap U_k \setminus (\overline V_0 \cup \dots \cup \overline V_{k-1}) :\, n \in \mathbb N \}$ is not a $\pi$-base for $U_k \setminus (\overline V_0 \cup \dots \cup \overline V_{k-1})$. Thus there is some $V \subseteq U_k \setminus (\overline V_0 \cup \dots \cup \overline V_{k-1})$ such that $U_n \not\subseteq V$ for all $n$. Using the regularity of $X$ to shrink $V$ a little, there is some $V_k \subseteq U_k \setminus (\overline V_0 \cup \dots \cup \overline V_{k-1})$ such that $U_n \not\subseteq \overline V_k$ for all $n$. QED

One example of a space with the OP's property is $\mathbb N^*$, the Stone–Čech remainder of the countable discrete space $\mathbb N$. (I suppose one could argue this is not a very "nice" space, but one would be wrong. :))

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Original answer:

Building off of the nice answer of Iosif Pinelis, and Wojowu's comment on it, we have:

Observation: A space $X$ fails to have the OP's property if some nonempty open $U \subseteq X$ is second countable.

(Proof: Let $(U_n)$ enumerate a basis for $U$.)

In the other direction, let me observe that there is at least one nice(ish) properties that do imply the OP's property. But, by the observation above, any such property must contradict metrizability (some kind of "bigness" property). Recall that a space is ccc if it does not admit an uncountable family of pairwise disjoint open subsets.

Observation: A space $X$ has the OP's property if no nonempty open $U \subseteq X$ is ccc.

Proof: Assume that no nonempty open $U \subseteq X$ is ccc, and let $(U_n)$ be a sequence of nonempty open subsets of $X$. We construct the $V_n$ by recursion. First, fix an uncountable family $\mathcal U_0$ of pairwise disjoint nonempty open subsets of $U_0$. All but countably many $V \in \mathcal U_0$ have the property that $U_n \not\subseteq \overline V$ for all $n$. Let $V_0$ be some such $V \in \mathcal U_0$. Next, fix an uncountable family $\mathcal U_1$ of pairwise disjoint nonempty open subsets of $U_1 \setminus \overline V_0$. All but countably many $V \in \mathcal U_1$ have the property that $U_n \not\subseteq \overline V$ for all $n$. Let $V_1$ be some such $V \in \mathcal U_1$. Generally, at step $k$ fix an uncountable family $\mathcal U_k$ of pairwise disjoint nonempty open subsets of $U_k \setminus (\overline V_0 \cup \dots \cup \overline V_{k-1})$. All but countably many $V \in \mathcal U_k$ have the property that $U_n \not\subseteq \overline V$ for all $n$. Let $V_k$ be some such $V \in \mathcal U_k$. QED

One example of a space with this property is $\mathbb N^*$, the Stone–Čech remainder of the countable discrete space $\mathbb N$. (I suppose one could argue this is not a very "nice" space, but one would be wrong. :))

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  • $\begingroup$ Thank you, this is exactly what I was looking for! $\endgroup$
    – erz
    Apr 20 at 5:12
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Apparently, there are no such nice properties.

Even $K=[0,1]$ will not be such a compact Hausdorff space. Indeed, consider the countable double-indexed family $(U_{n,k}\colon n\in\Bbb N,k\in\{0,\dots,n-1\})$ of nonempty open subsets of $K$, where $U_{n,k}:=(\frac kn,\frac{k+1}n)$. Then any nonempty open subset $V_{1,0}$ of $U_{1,0}=(0,1)$ will contain entirely (infinitely many) other $U_{n,k}$'s. So, no nonempty subsets of those other $U_{n,k}$'s will be disjoint from $V_{1,0}$.

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    $\begingroup$ By a similar argument, no second countable (compact Hausdorff or not) space will not have this property - let $U_n$ enumerate any base of the topology. This excludes for instance all metrizable compact Hausdorffs. $\endgroup$
    – Wojowu
    Apr 19 at 12:57
  • $\begingroup$ @Wojowu : Good point! $\endgroup$ Apr 19 at 12:59

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