Let $K$ be compact Hausdorff, let $U\subset K$ be open and dense, and let $x\in K\backslash U$. Can we find a disjoint collection $\{U\_i,~ i\in I\}$ of open subsets of $U$ and a collection $\{K\_i,~ i\in I\}$ of compact sets such that $K\_i\subset U\_i$, for every $i$, and $x\in \overline{\bigcup K\_i}$?



Can a point of a compact set be approximated by a disjoint union?
Yes. This is possible. Recall that a subset of a topological space $X$ of the form $f^{-1}[\{0\}]^{c}$ for some continuous function $f:X\rightarrow \mathbb{R}$ is known as a cozero set.


Let $\mathcal{V}$ be a maximal collection of disjoint cozero subsets of $K$ subject to the condition that $\bigcup\mathcal{V}\subseteq U$. Such a set $\mathcal{V}$ is guaranteed to exist by Zorn's lemma, and $\bigcup\mathcal{V}$ is necessarily dense in $K$.


Then for each $V\in\mathcal{V}$, let $f\_{V}:K\rightarrow[0,1/3]$ be a continuous function with
$f\_{V}^{-1}[(0,1/3]]=V$. Then for each $n\geq 1$, let


1. $A\_{V,n}=f\_{V}^{-1}[[\frac{1}{3n+2},\frac{1}{3n}]]$,
2. $B\_{V,n}=f\_{V}^{-1}[[\frac{1}{3n+3},\frac{1}{3n+1}]]$,
3. $A\_{V,n}^{+}=f\_{V}^{-1}[(\frac{1}{3n+2.5},\frac{1}{3n-0.5})]$, and
4. $B\_{V,n}^{+}=f\_{V}^{-1}[(\frac{1}{3n+3.5},\frac{1}{3n+0.5})]$.


Observe that each set $A\_{V,n},B\_{V,n}$ is necessarily compact. The sets
$A\_{V,n}^{+},B\_{V,n}^{+}$ are open in $X$ with $A\_{V,n}\subseteq A\_{V,n}^{+},B\_{V,n}\subseteq B\_{V,n}^{+}$. Furthermore, the sets
$(A\_{V,n}^{+})\_{n=1}^{\infty}$ are disjoint subsets of $V$, and the sets
$(B\_{V,n}^{+})\_{n=1}^{\infty}$ are also disjoint subsets of $V$. Therefore,
$\{A\_{V,n}^{+}\mid V\in\mathcal{V},n\geq 1\}$ is a collection of disjoint open sets, and $\{B\_{V,n}^{+}\mid V\in\mathcal{V},n\geq 1\}$ is another collection of disjoint open sets. However, we have $V=\bigcup\_{n=1}^{\infty}A\_{V,n}\cup B\_{V,n}$, so
$$\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}A\_{V,n}\big)\cup\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}B\_{V,n}\big)=\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}A\_{V,n}\cup B\_{V,n}\big)=K.$$ Therefore, if $x\in X$, then
$x\in\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}A\_{V,n}\big)$ or
$x\in\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}B\_{V,n}\big).$


