Does a “flexibly fine” open-interval cover of a compact nowhere dense set admit a disjoint finite subcover?

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:

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".)

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 $$b_x lie in different intervals $$I_n$$ then there is a point of $$P$$ between them. Given $$x_1<\dots 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 $$\mathcal V\cup\mathcal V'\subset\mathcal U.$$

• 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.) – Julian Newman Mar 3 at 21:34