Equivalent Definitions of Ultrafilter

Theorem
Let $X$ be a set and $\mathcal F$ a filter on $X$.

The following are equivalent.

where $\complement_X \left({A}\right)$ is the relative complement of $A$ in $X$, i.e. $X \setminus A$.
 * 1) $\mathcal F$ is an ultrafilter, i.e. for any filter $\mathcal G$ on $X$ satisfying $\mathcal F \subseteq \mathcal G$ it holds that $\mathcal F = \mathcal G$.
 * 2) For any set $A \subseteq X$ either $A \in \mathcal F$ or $\complement_X \left({A}\right) \in \mathcal F$.

Proof

 * Assume first that $\mathcal F$ is an ultrafilter.

Let $A \subseteq X$.

Assume that $A \notin \mathcal F$ and $\complement_X \left({A}\right) \notin \mathcal F$.

Then $\mathcal B := \left\{{A \cap V: V \in \mathcal F}\right\}$ is a basis of a filter $\mathcal G$ on $X$, for which $\mathcal F \subseteq \mathcal G$ holds.

Let $U \in \mathcal F$. Since $\complement_X \left({A}\right) \notin \mathcal F$ this implies that $U \cap A \ne \varnothing$.

We know that $A \cap U \subseteq A$, thus $A \in \mathcal G$ by construction.

Since $A \notin \mathcal F$ this implies $\mathcal F \subsetneq \mathcal G$.

Thus $\mathcal F$ is not an ultrafilter, a contradiction to our assumption.

Hence either $A \in \mathcal F$ or $\complement_X \left({A}\right) \in \mathcal F$.


 * Assume now that for any $A \subseteq X$ either $A \in \mathcal F$ or $\complement_X \left({A}\right) \in \mathcal F$ holds.

Let $\mathcal G$ be a filter on $X$ such that $\mathcal F \subseteq \mathcal G$.

Assume that $\mathcal F \subsetneq \mathcal G$.

Then there exists $A \in \mathcal G \setminus \mathcal F$.

Since $\varnothing \notin \mathcal G$ this implies that $\complement_X \left({A}\right) \notin \mathcal G$.

As $\mathcal F \subsetneq \mathcal G$, it follows that $\complement_X \left({A}\right) \notin \mathcal F$.

Therefore neither $A \in \mathcal F$ nor $\complement_X \left({A}\right) \in \mathcal F$, a contradiction to our assumption.

Thus $\mathcal F = \mathcal G$, which implies that $\mathcal F$ is an ultrafilter.