Characterisation of ultrafilters: every set or its complement

Let $X$ be a set and $\mathcal{U}$ a filter on $X$ (Filter on a set). The following are equivalent:

1. $\mathcal{U}$ is an ultrafilter on $X$ (Ultrafilter);
2. for every $A \subseteq X$, either $A \in \mathcal{U}$ or $X \setminus A \in \mathcal{U}$.

Moreover, for any filter the two alternatives are exclusive: never both $A \in \mathcal{U}$ and $X \setminus A \in \mathcal{U}$. So an ultrafilter decides every subset of $X$, containing exactly one of $A$ and $X \setminus A$.

• This is the working form of the definition. Maximality is a statement about the whole poset of filters; deciding complements is a statement about $\mathcal{U}$ alone, and it is what one actually checks and uses. It is the route taken by Ultrafilters are prime: a union in $\mathcal{U}$ has a member in $\mathcal{U}$, for instance. It is not the only route: FALSE, once the ultrafilter lemma is available: every ultrafilter is principal is refuted from properness and the ultrafilter lemma directly, without this characterisation, and does not list it among its dependencies.
• The two directions run as separate threads, and the mechanical stratification interleaves them, so each step is labelled Forward or Converse. They share nothing except the filter $\mathcal{U}$ and the exclusivity of step 1.1: the forward thread builds a specific filter $\mathcal{G}$, the converse thread argues about an arbitrary filter $\mathcal{H}$ above $\mathcal{U}$.
• The filter $\mathcal{G}$ built at step 1.3 is the filter generated by $\mathcal{U} \cup \{A\}$ (A family lies in a filter exactly when it has the finite intersection property): its members are the supersets of the sets $C \cap A$ with $C \in \mathcal{U}$. Those sets are not all the intersections of finite lists drawn from $\mathcal{U} \cup \{A\}$, since a list omitting $A$ intersects to a member $C$ of $\mathcal{U}$ and not to a set of the form $C \cap A$. The two families generate the same filter all the same: finitely many members of $\mathcal{U}$ intersect to a single $C \in \mathcal{U}$, so every such finite intersection either is a $C$ or is a $C \cap A$, and in the first case it contains $C \cap A$; conversely each $C \cap A$ is itself one of the finite intersections. The upward closures therefore coincide, which is all that "generated by" asks. Step 2.1 is the check that $\mathcal{U} \cup \{A\}$ has the finite intersection property, and that check is exactly where $X \setminus A \notin \mathcal{U}$ is used.
• Exclusivity is a property of filters, not of ultrafilters, and it holds for the trivial reason that $A$ and $X \setminus A$ are disjoint. What maximality buys is only the "at least one" half. Reading the characterisation as a two-valued decision on $\mathcal{P}(X)$ makes an ultrafilter a finitely additive $\{0,1\}$-valued measure on $X$, which is the standard way it is used.
• The characterisation makes ultrafilters visibly rigid: $\mathcal{U}$ is determined by which of each complementary pair it takes, so an ultrafilter is a consistent choice on all of $\mathcal{P}(X)$ at once. This is why producing one in general is done here from a choice principle rather than by a construction (The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter).