# proof that every filter is contained in an ultrafilter

Let $Y$ be the set of all non-empty subsets of $X$ which are not contained in $\mathcal{F}$. By Zermelo’s well-orderding theorem, there exists a relation$\succ$’ which well-orders $Y$. Define $Y^{\prime}=\{0\}\cup Y$ and extend the relation ‘$\succ$’ to $Y^{\prime}$ by decreeing that $0\prec y$ for all $y\in Y$.

We shall construct a family of filters $S_{i}$ indexed by $Y^{\prime}$ using transfinite induction. First, set $S_{0}=\mathcal{F}$. Next, suppose that, for some $j\in Y$, $S_{i}$ has already been defined when $i\prec j$. Consider the set $\bigcup_{i\prec j}S_{i}$; if $A$ and $B$ are elements of this set, there must exist an $i\prec j$ such that $A\in S_{i}$ and $B\in S_{i}$; hence, $A\cap B$ cannot be empty. If, for some $i\prec j$ there exists an element $f\in S_{i}$ such that $f\cap j$ is empty, let $S_{j}$ be the filter generated by the filter subbasis $\bigcup_{i\prec j}S_{i}$. Otherwise $\{j\}\cup\bigcup_{i\prec j}S_{i}$ is a filter subbasis; let $S_{j}$ be the filter it generates.

Note that, by this definition, whenever $i\prec j$, it follows that $S_{i}\subseteq S_{j}$; in particular, for all $i\in Y^{\prime}$ we have $\mathcal{F}\subseteq S_{i}$. Let $\mathcal{U}=\bigcup_{i\prec j}S_{i}$. It is clear that $\emptyset\notin\mathcal{F}$ and that $\mathcal{F}\subseteq\mathcal{U}$.

It is easy to see that $\mathcal{U}$ is a filter. Suppose that $A\cap B\in\mathcal{U}$. Then there must exist an $i\in Y^{\prime}$ such that $A\cap B\in S_{i}$. Since $S_{i}$ is a filter, $A\in S_{i}$ and $B\in S_{i}$, hence $A\in\mathcal{U}$ and $B\in\mathcal{U}$. Conversely, if $A\in\mathcal{U}$ and $B\in\mathcal{U}$, then there exists an $i\in Y^{\prime}$ such that $A\in S_{i}$ and $B\in S_{i}$. Since $S_{i}$ is a filter, $A\cap B\in S_{i}$, hence $A\cap B\in\mathcal{U}$. By the alternative characterization of a filter, $\mathcal{U}$ is a filter.

Moreover, $\mathcal{U}$ is an ultrafilter. Suppose that $A\in\mathcal{U}$ and $B\in\mathcal{U}$ are disjoint and $A\cup B=X$. If either $A\in\mathcal{F}$ or $B\in\mathcal{F}$, then either $A\in\mathcal{U}$ or $B\in\mathcal{U}$ because $\mathcal{F}\subset\mathcal{U}$. If $A\in Y$ and $A\in S_{A}$, then $A\in\mathcal{U}$ because $S_{A}\subset\mathcal{U}$. If $A\in Y$ and $A\notin S_{A}$, there must exist $x\in\mathcal{U}$ such that $A\cap x$ is empty. Because $B$ is the complement of $A$, this means that $x\subset B$ and, hence $B\in\mathcal{U}$.

This completes the proof that $\mathcal{U}$ is an ultrafilter — we have shown that $\mathcal{U}$ meets the criteria given in the alternative characterization of ultrafilters.

Title proof that every filter is contained in an ultrafilter ProofThatEveryFilterIsContainedInAnUltrafilter 2013-03-22 14:41:44 2013-03-22 14:41:44 rspuzio (6075) rspuzio (6075) 17 rspuzio (6075) Proof msc 54A20