Equivalent Definitions of Ultrafilter

Theorem
Let $$X$$ be a set and $$\mathcal{F}$$ a filter on $$X$$. The following are equivalent.


 * 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 $$A^C := X \setminus A \in \mathcal{F}$$.

Proof

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

Let $$A \subseteq X$$.

Assume that $$A \not \in \mathcal{F}$$ and $$A^C \not \in \mathcal{F}$$.

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

Let $$U \in \mathcal{F}$$. Since $$A^C \not \in \mathcal{F}$$ this implies that $$U \cap A \ne \emptyset$$.

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

Since $$A \not \in \mathcal{F}$$ 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 $$A^C \in \mathcal{F}$$.


 * Assume now that for any $$A \subseteq X$$ either $$A \in \mathcal{F}$$ or $$A^C \in \mathcal{F}$$ 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 $$\emptyset \not \in \mathcal{G}$$ this implies that $$A^C \not \in \mathcal{G}$$.

As $$\mathcal{F} \subsetneq \mathcal{G}$$, it follows that $$A^C \not \in \mathcal{F}$$.

Therefore neither $$A \in \mathcal{F}$$ nor $$A^C \in \mathcal{F}$$, a contradiction to our assumption.

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