Can a point of a compact set be approximated by a disjoint union?

Question

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 $x\in \overline{\bigcup K_i}$?

Answer 1

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 $x\in\text{Cl}_{K}\big(\bigcup_{V\in\mathcal{V}}\bigcup_{n=1}^{\infty}B_{V,n}\big).$