Ultrafilters

# Ultrafilters

 Definition: Let $E$ be a set. An Ultrafilter is a filter that has no proper refinement.

Using the Maximal axiom it can be shown that every filter has a refinement that is an ultrafilter. We now look at some important properties of ultrafilters.

 Proposition 1: $E$ be a set. If $\mathscr{F}$ is an ultrafilter and $A \subseteq E$ then either $A \in \mathscr{F}$ or $A^c \in \mathscr{F}$.
• Proof: If $A \in \mathscr{F}$ then we are done.
• Assume that $A \not \in \mathscr{F}$. Then if $B \subseteq A$ observe that $B \not \in \mathscr{F}$ (for if $B \in \mathscr{F}$ then since $\mathscr{F}$ is a filter we would have that $A \in \mathscr{F}$). Hence:
(1)
\begin{align} \quad B \cap A^c \neq \emptyset \end{align}
• for all $B \in \mathscr{F}$. Consider the collection of sets $\{ B \cap A^c : B \in \mathscr{F}$. This collection is a filter base, for if $B_1 \cap A^c$ and $B_2 \cap A^c$ are in this set and:
(2)
\begin{align} \quad (B_1 \cap B_2) \cap A^c = (B_1 \cap A^c) \cap (B_2 \cap A^c) \end{align}
• with $B_1 \cap B_2 \in \mathscr{F}$ as $\mathscr{F}$ is a filter. The filter generated by the filter base $\{ B \cap A^c : B \in \mathscr{F} \}$ is a refinement of $\mathscr{F}$. Indeed, if $B \in \mathscr{F}$ then $B \cap A^c \subseteq B$ so that $B$ is in the filter generated by the filter base $\{ B \cap A^c : B \in \mathscr{F} \}$. Since $\mathscr{F}$ is an ultrafilter, we must have that the filter generated by the filter base $\{ B \cap A^c : B \in \mathscr{F} \}$ is $\mathscr{F}$. But a filter base for the ultrafilter $\mathscr{F}$ is $\mathscr{F}$ itself.
• Hence $A^c \in \mathscr{F}$. $\blacksquare$
 Corollary 2: Let $E$ be a set and let $\mathscr{F}$ be an ultrafilter. If $\displaystyle{\bigcup_{i=1}^{n} A_i \in \mathscr{F}}$ then $A_i \in \mathscr{F}$ for some $1 \leq i \leq n$.
• Proof: Suppose instead that $A_i \not \in \mathscr{F}$ for all $1 \leq i \leq n$. Then from the previous proposition we must have that $A_i^c \in \mathscr{F}$ for all $1 \leq i \leq n$.
• Since $\mathscr{F}$ is an ultrafilter and since $\bigcup_{i=1}^{n} A_i \in \mathscr{F}$ and since $A_1^c, A_2^c, ..., A_n^c \in \mathscr{F}$ we have that:
(3)
\begin{align} \quad \left ( \bigcup_{i=1}^{n} A_i \right ) \cap (A_1^c) \cap (A_2^c) \cap ... \cap (A_n^c) \in \mathscr{F} \end{align}
• However:
(4)
\begin{align} \quad \left ( \bigcup_{i=1}^{n} A_i \right ) \cap (A_1^c) \cap (A_2^c) \cap ... \cap (A_n^c) = \left ( \bigcup_{i=1}^{n} A_i \right ) \cap \left ( \bigcap_{i=1}^{n} A_i^c \right ) = \left ( \bigcup_{i=1}^{n} A_i \right ) \cap \left ( \bigcup_{i=1}^{n} A_i \right )^c = \emptyset \end{align}
• Therefore $\emptyset \in \mathscr{F}$ which is a contradiction by the definition of $\mathscr{F}$ being an ultrafilter. Thus we must have that $A_i \in \mathscr{F}$ for some $1 \leq i \leq n$. $\blacksquare$