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Let $K \subset \mathbb{R}$ be a compact nowhere dense set. Suppose we have $K$-indexed families $(U_x)_{x \in K}$ and $(V_x)_{x \in K}$ of open sets $U_x,V_x \subset \mathbb{R}\,$ with the property that for every $x \in K$, $\,\sup(U_x)=\inf(V_x)=x$.

Does there necessarily exist a finite set $S \subset K$ and $(a_x)_{x \in S},(b_x)_{x \in S}$ with $a_x \in U_x$ and $b_x \in V_x$ for each $x \in S$, such that the collection of open intervals $\{(a_x,b_x):x \in S\}$ is mutually disjoint and covers $K$?

If not, what about if we add the assumption that $K$ is a Lebesgue-null set?

(I want to emphasise that $U_x$ and $V_x$ may have infinitely many connected components, and hence in particular, may not contain an interval having $x$ as a boundary point.)


Intuition:

In Does a "fine" open-interval cover of a compact nowhere dense set admit a disjoint finite subcover?, I asked:

Given a compact nowhere dense set $K \subset \mathbb{R}$ and a cover of $K$ by open intervals, if this cover includes an arbitrarily small neighbourhood of every point in $K$, does it necessarily admit a disjoint finite subcover?

(In the title, I referred to the cover as "fine" because it includes an arbitrarily small neighbourhood of every point in $K$.)

In reply, I was given the following beautifully simple counter-example: Take $K=\{\frac{1}{n}\}_{n \geq 1} \cup \{0\}$, cover $0$ by open intervals with supremum precisely at $\frac{1}{n}$, and take all the other intervals in the cover to intersect $K$ only at a single point.

This counter-example seems to rely on "infinitely precise fine-tuning" of the upper endpoints of the intervals about $0$. So now I am modifying my question so as to "allow some continuous leeway" in the endpoints of the intervals in the cover. (Therefore, in the title I now refer to the cover as "flexibly fine".)

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2
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No. I will show:

  1. If $K$ is an uncountable compact set, there is a flexibly-fine open interval cover of $K$ with no disjoint subcover.
  2. If $K$ is a countable compact set, then every flexibly-fine open interval cover of $K$ has a disjoint subcover.

For 1, $K$ contains a non-empty perfect subset $P.$ For a specific counterexample, take $K$ to be the Cantor set and $K=P.$ The complement of $P$ is a countable disjoint union of open intervals $I_n$ with endpoints in $P.$ I claim we can color these intervals red and green such that:

  • $P$ is the boundary of the red set,
  • $P$ is the boundary of the green set,
  • $(-\infty,\inf P)$ is red, and
  • $(\sup P,\infty)$ is green.

Just proceed in stages, starting by coloring $(-\infty,\inf P)$ red and $(\sup P,\infty)$ green. Suppose we have colored a finite number of intervals such that, from lowest to highest, the colored intervals alternate between red and green. Pick a largest uncolored interval and color it red. There are then two intervals $I,I',$ with $\sup I\leq\inf I',$ both colored red and with no green interval between them. $P$ is perfect so $\sup I\neq\inf I',$ and $P$ is nowhere dense so there is an open interval in $[\sup I,\inf I']\setminus P.$ Pick any such interval and color it green. Repeating this process for $\omega$ steps ensures that every interval gets colored.

Define $U_x$ and $V_x$ as follows. If $x\in K$ is in the closure of a red interval, take $U_x$ to be the set of points less than $x$ in red intervals, and take $V_x$ to be the set of points greater than $x$ in red intervals. Otherwise, take $U_x$ to be the set of points less than $x$ in green intervals, and take $V_x$ to be the set of points greater than $x$ in green intervals. I claim this gives a flexibly-fine cover. When $x$ is the right endpoint of a red interval, then $x$ is a limit point of $P$ so $x$ has red intervals arbitrarily close on the right-hand side. Similarly for left endpoints, and for green intervals. Points of $P$ not in the closure of an open interval in $\mathbb R\setminus P$ have green (and red) intervals arbitrarily close on both sides, and points of $K\setminus P$ lie entirely inside a colored interval.

This construction ensures that any $(a_x,b_x)$ must be monochromatic - $a_x$ and $b_x$ lie in intervals of the same color. And if $b_x<a_y$ lie in different intervals $I_n$ then there is a point of $P$ between them. Given $x_1<\dots<x_k$ in $K,$ and disjoint $(a_{x_i},b_{x_i})\in U_{x_i}\times V_{x_i},$ if $a_{x_1}<\inf P$ then $a_{x_1}$ lies in a red interval, and if $b_{x_k}>\sup P$ then $b_{x_k}$ lies in a green interval, so there must be some point of $P$ not covered by $\bigcup (a_{x_i},b_{x_i}).$


For 2, we can use induction on Cantor-Bendixon rank. Assume that for all ordinals $\alpha<\beta,$ for all countable compact $K$ of rank $\alpha$ and all flexibly-fine covers of $K$ by open intervals, there is a disjoint subcover of $K.$ Now let $K$ have Cantor-Bendixson rank $\beta>0$ and let $\mathcal U=\{(a_x,b_x\mid x\in K, a_x\in U_x, b_x\in V_x\}$ be a flexibly-fine cover. Shrinking each $U_x$ and $V_x$ if necessary we can assume that each $U_x$ and $V_x$ is a subset of $\mathbb R\setminus K.$ Since $K$ is countable and compact, $\beta$ is a successor ordinal $\beta'+1$ and $K^{\beta'}$ is a discrete set. So $K^{\beta'}$ has a disjoint cover by some $\mathcal V\subset\mathcal U.$. The set $K\setminus \bigcup\mathcal V$ has strictly smaller Cantor-Bendixson rank. So it has its own disjoint cover by $\mathcal V'\subset\mathcal U',$ where $\mathcal U'$ is $\mathcal U$ restricted to $x\in K\setminus \bigcup\mathcal V$ and restricted to intervals that do not intersect $\bigcup\mathcal V$ - this can be done by shrinking $U_x$ and $V_x.$ This gives a disjoint cover of $K$ by $\mathcal V\cup\mathcal V'\subset\mathcal U.$

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  • $\begingroup$ Wow, beautiful. And thank you for going to greater effort than requested, in working out what the necessary and sufficient condition is for $K$ to have the property that every flexibly fine open-interval cover admits a disjoint finite subcover. I will award the bounty when Math.SE lets me do so. (I didn't realise there's a waiting time before the bounty can be awarded.) $\endgroup$ – Julian Newman Mar 3 at 21:34

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