Definition:Ultrafilter on Set

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Definition

Definition 1

Let $S$ be a set.

Let $\mathcal F \subseteq \powerset S$ be a filter on $S$.


Then $\mathcal F$ is an ultrafilter (on $S$) if and only if:

there is no filter on $S$ which is strictly finer than $\mathcal F$

or equivalently, if and only if:

whenever $\mathcal G$ is a filter on $S$ and $\mathcal F \subseteq \mathcal G$ holds, then $\mathcal F = \mathcal G$.


Definition 2

Let $S$ be a set.

Let $\mathcal F \subseteq \mathcal P \left({S}\right)$ be a filter on $S$.


Then $\mathcal F$ is an ultrafilter (on $S$) if and only if:

for every $A \subseteq S$ and $B \subseteq S$ such that $A \cap B = \varnothing$ and $A \cup B \in \mathcal F$, either $A \in \mathcal F$ or $B \in \mathcal F$.


Definition 3

Let $S$ be a set.

Let $\mathcal F \subseteq \mathcal P \left({S}\right)$ be a filter on $S$.


Then $\mathcal F$ is an ultrafilter (on $S$) if and only if:

for every $A \subseteq S$, either $A \in \mathcal F$ or $\complement_S \left({A}\right) \in \mathcal F$

where $\complement_S \left({A}\right)$ is the relative complement of $A$ in $S$, that is, $S \setminus A$.


Definition 4

Let $S$ be a non-empty set.

Let $\mathcal F$ be a non-empty set of subsets of $S$.


Then $\mathcal F$ is an ultrafilter on $S$ if and only if both of the following hold:

$\mathcal F$ has the finite intersection property
For all $U \subseteq S$, either $U \in \mathcal F$ or $U^c \in \mathcal F$

where $U^c$ is the complement of $U$ in $S$.


Also see