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

[I'll ask this in the one-dimensional setting, and (if the answer is yes) I'll leave as open what possible extensions to more general settings one might wish to discuss.]

Let $$K \subset \mathbb{R}$$ be a compact nowhere dense set. Suppose that for each $$x \in K$$ we have a sequence $$(U_n^x)_{n \in \mathbb{N}}$$ of connected neighbourhoods of $$x$$ such that the length of $$U_n^x$$ tends to $$0$$ as $$n \to \infty$$.

Does there necessarily exist a finite set $$S \subset K$$ and a list of integers $$(n_x)_{x \in S}$$ such that $$K \subset \bigcup_{x \in S} U_{n_x}^x\,$$ and for all distinct $$x,y \in S$$, $$\,U_{n_x}^x \cap U_{n_y}^y = \emptyset$$?

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

(My vague intuition on the last bit is that requiring $$K$$ to be a null set won't make a difference to the answer, as there's probably some homeomorphic or "nearly homeomorphic" transformation of $$\mathbb{R}$$ that will turn a positive-measure nowhere dense set into a null set.)

If the answer is yes (with or without the null set requirement), is there a reference for this? Even just an exercise from a textbook to prove this, or prove something that easily implies this, would suffice.

## 1 Answer

The answer is no, even if $$K$$ is countable.

For example, let $$K = \{0\} \cup \{1/k: k \in \mathbb N\}$$. Let $$U^0_n = (-1/n, 1/n)$$, and choose $$U^{1/k}_n$$ so that $$U^{1/k}_n \cap K = \{1/k\}$$. In order for $$0$$ to be covered you need $$0 \in S$$, and $$U^0_n \cap U^{1/n}_m \ne \emptyset$$ for all $$m$$ and $$n$$.