# 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 $\mathrm{\varnothing}\notin \mathcal{F}$ 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 |
---|---|

Canonical name | ProofThatEveryFilterIsContainedInAnUltrafilter |

Date of creation | 2013-03-22 14:41:44 |

Last modified on | 2013-03-22 14:41:44 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 17 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 54A20 |