> [!definition] > > Let $X$ be a [[Set|set]] and $\mathfrak F \subset \pow{X}$ be a [[Filter|filter]], then $\mathfrak F$ is an **ultrafilter** if it is a maximal filter with respect to inclusion. > 1. For every filter $\mathfrak F_0$ on $X$, there exists an ultrafilter $\fF$ containing $\fF_0$. > 2. If $\fF \subset \pow{X}$ is an ultrafilter, then for any such that $\bigcup_{j = 1}^n E_j \in \fF$, there exists $1 \le j \le n$ such that $E_j \in \fF$. > 3. A subbase $\fS \subset \pow{X}$ for a filter is an ultrafilter if and only if for every $E \subset X$, either $E \in X$ or $E^c \in X$. > > *Proof*. $(1)$: Let $\seqi{\fF}$ be a chain of filters containing $\fF_0$, with indices ordered by inclusion of the filters. Since for every $\seqf{i_j}$ and $E_j \subset \fF_{i_j}$, $\bigcap_{i \in I}E_j \in \fF_{\max_j i_j}$ and $\bigcap_{i \in I}E_j \ne \emptyset$, there exists a filter $\fF$ generated by $\seqi{\fF}$ that also contains $\fF_0$. By [[Zorn's Lemma]], there exists a maximal filter $\fF$ containing $\fF_0$ with respect to inclusion, which is a desired ultrafilter. > > $(2)$: Let $A, B \subset X$ such that $A \cup B \in \fF$. Suppose that $A \not\in \fF$ and let > $ > \fF_A = \bracs{E \subset X: E \cup A \in \fF} > $ > then $\emptyset \not\in \fF_A$. For any $E, F \in \fF_A$, $(E \cap F) \cup A = (E \cup A) \cap (F \cup A) \in \fF$, so $E \cap F \in \fF_A$. For any $E \in \fF_A$ and $F \supset E$, $F \cup A \supset E \cup A$, so $F \cup A \in \fF$ and $F \in \fF_A$. Thus $\fF_A$ is a filter that contains $\fF$. If $B \not\in \fF$, then $\fF_A \supsetneq \fF$, which contradicts the assumption that $\fF$ is an ultrafilter. > > $(3)$: Let $\fF \supset \fS$ be a filter, then for every $E \in \fF$, $E^c \not\in \fF$, $E^c \not\in \fS$, so $E \in \fS$ and $\fS \supset \fF$. Since this applies to every filter containing $\fS$, $\fS$ is an ultrafilter. > [!definition] > > Let $\fU$ be an ultrafilter, then $\fU$ is **principal** if it contains a minimum element.