Stone-Čech compactification using ultrafilters Let $X = \omega \cup \{ x \}$ ne the Stone-Čech compactification of $\omega$. (I am viewing $X$ as a subspace of the set of ultrafilters over $\omega$). Let, $\mathcal A$, $\mathcal B$ be two disjoint countable subfamilies of $X$. And suppose that $x \in \overline {\bigcup \mathcal A}$ and $x \in \overline {\bigcup \mathcal B}$. Then, why is this a contradiction to the fact that $\mathcal A$ and $\mathcal B$ are disjoint?
My question is coming from the last line in the proof to example 1.6 in  This article.
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Thank you!
 A: The parts below are verbatim quotes from Engelking's General topology.

Two subsets $A$ and $B$ of a topological space $X$ are called completely separated if there exists a continuous function $f\colon X\to I$ such that $f(A)=0$ and $f(B)=1$. We say that $f$ separates sets $A$ and $B$.
Corollary 3.6.2. Every pair of completely separated subsets of a Tychonoff space $X$ has
disjoint closures in $\beta X$.

(This corollary also says that this property characterizes Stone-Čech compactification among all compactifications of $X$. But you only need this first part.)

Proof. Let $A$, $B$ be a pair of completely separated subsets of a Tychonoff space $X$ and let $f\colon X\to I$ be a continuous function that separates $A$ and $B$. By the last theorem, $f$ is extendable to a continuous function $F\colon\beta X\to I$, so that we have $\overline A\cap\overline B=\emptyset$, because $\overline A\subset F^{-1}(0)$ and $\overline B\subset F^{-1}(1)$. $\hspace{1cm}\square$

In your case you have subsets of $\omega$ with discrete topology. Clearly, any two subsets of a discrete space are completely separated. (I hope I understood the question correctly - I did not check the paper you linked, but from the context you provided here it seems that $\bigcup \mathcal A$ and $\bigcup\mathcal B$ are subsets of $\omega$.)
A: I'll approach this from the ultrafilter construction of $\beta \omega$, in contrast to Martin Sleziak's answer.
Note that the basic open sets are of the form $$[A] = \{ x \in \beta \omega : A \in u \}$$ for all $A \subseteq \omega$ (while identifying $\omega$ with the collection of principal ultrafilters).  Then the family $\{ [A] : A \in x \}$ is a neighbourhood base for $x \in \beta \omega$.
It follows from this that if $x \in \beta \omega$ and $B \subseteq \omega$, then $x \in \overline{A}$ iff $A \cap B \neq \varnothing$ for all $A \in x$, which occurs iff $B \in x$ (since $x$ is an ultrafilter either $B$ or $\omega \setminus B \in x$, and in the latter case it is impossible that $B \cap A \neq \varnothing$ for all $A \in x$).
Therefore, if $B , C$ are disjoint subsets of $\omega$, it is impossible that $x \in \overline{B}$ and $x \in \overline{C}$ both hold (since otherwise $x$ would contain two disjoint sets, contradicting that it is a filter, let alone an ultrafilter).
From the question, if $\mathcal{F}$ is a (countable) family of (finite) pairwise disjoint subsets of $\omega$, then letting $\mathcal{A} , \mathcal{B}$ be two disjoint subfamilies of $\mathcal{F}$, then $\bigcup \mathcal{A}$ and $\bigcup \mathcal{B}$ are disjoint subsets of $\omega$, so by the above at least one of $x \in \overline{ \bigcup \mathcal{A} }$ or $x \in \overline{ \bigcup \mathcal{B} }$ must fail.
