The following is a proof regarding principal ultrafilters from pg.2 of ULTRAFILTERS AND HOW TO USE THEM by Burak Kaya (https://users.metu.edu.tr/burakk/lecturenotes/village2019lecturenotes.pdf).
Proposition 2. Any ultrafilter on a non-empty finite set is principal.
Proof. Let $X$ be a non-empty finite set and $\mathscr{U}$ be an ultrafilter.
Consider the set $A = \cap_{U \in \mathscr{U}} U$. Since $X$ is finite, so is the ultrafilter
$\mathscr{U}$. It then follows from the properties of a filter that $A \in \mathscr{U}$ and
hence $A \neq \emptyset$. (Up to now, we have only used that $\mathscr{U}$ is a filter and
indeed shown that any filter on a non-empty set is principal whose
generating set is the intersection of all sets in the filter.) Let $a \in A$. Being an ultrafilter, we know that either $\{a\} \in \mathscr{U}$ or $X − \{a\} \in \mathscr{U}$.
If it were the case that $X − \{a\} \in \mathscr{U}$, then we would have *$a \in A \subseteq X −
\{a\}$* which is a contradiction. Thus, $\{a\} ∈ \mathscr{U}$. We claim that $\mathscr{U}$ is generated by $a$. If it were not, then there would be $B \in \mathscr{U}$ such that
$\{a\} \not\subseteq B$ in which case $\{a\} \cap B = \emptyset ∈ \mathscr{U}$ which is a contradiction. Therefore, $\mathscr{U}$ is generated by a and is principal.
However, I don't understand the part that is marked with *. Why would the $\subseteq$ have to hold? I tried looking into the definition of filters again, but still can't figure it out.
Best Answer
$A$ is a subset of all members of $\mathcal{U}$, hence also of $X\setminus\{a\}$.