# complete ultrafilter and partitions

*If $U$ is an ultrafilter ^{} on a set $S$, then*

*$U$ is $\kappa $-complete ^{} $\mathrm{\iff}$ there is no partition^{} of $S$ into $\kappa $-many pieces for which each piece ${X}_{\alpha}$ of the partition is not in $U$.*

We prove the case of $\sigma $-completeness; the case of arbitrary infinite^{} cardinality follows closely. For the $\Rightarrow $ direction, let $P$ be a partition of $S$ into $\omega $ many pieces, all of which do not belong to $U$, and write $S={\bigcup}_{n=1}^{\omega}{X}_{n}$ to illustrate this partition. Now, $\mathrm{\varnothing}={S}^{\mathrm{\complement}}={\bigcap}_{n=1}^{\omega}{X}_{n}^{\mathrm{\complement}}$. Since, by our assumption^{}, each of the ${X}_{n}$ do not belong to $U$, we have ${X}_{n}^{\mathrm{\complement}}\in U$ for each $$ as $U$ is an ultrafilter. Thus, $\left({\bigcap}_{n=1}^{\omega}{X}_{n}^{\mathrm{\complement}}\right)\in U$ by $\sigma $-completeness. This, however, means $\mathrm{\varnothing}\in U$, contradicting the definition of a filter.

Note that the converse^{} states that every partition $P$ of $S$ into $\omega $-many pieces has a (unique) piece ${X}_{1}\in U$. To prove this, let ${Y}_{n}$ be a collection^{} of $\omega $ many members of $U$ and let $Y={\bigcap}_{n=1}^{\omega}{Y}_{n}$. Now consider the partition $\{{P}_{\iota}:\iota \le \omega \}$ of $S\setminus Y$:

for each $s\in S\setminus Y$, put $s\in {P}_{\iota}$ if $\iota $ is the least index for which $s\notin {Y}_{\iota}$.

It is easy to verify that each $s\in S\setminus Y$ belongs to a *unique* ${P}_{\iota}$, the collection of ${P}_{\iota}$’s is indeed a partition of $S\setminus Y$.

Along with $Y$, $\{{P}_{\iota}:\iota \le \omega \}$ partitions $S$ into ${\mathrm{\aleph}}_{0}=\omega $ many pieces. A (unique) piece of this partition belongs in $U$: ${P}_{\iota *}\in U$ or $Y\in U$. But, ${P}_{\iota}\cap {Y}_{\iota}=\mathrm{\varnothing}\notin U$ by the definition of ${P}_{\iota}$. This excludes the possibility for the former to belong in $U$ (cf. alternative characterization of filter) and so $Y\in U$.

Thus, starting from an arbitrary collection $\{{Y}_{n}\}$ of $\omega $-many members of $U$, we have identified a partition of $S$ for which the unique piece which belongs to $U$ is $\cap {Y}_{n}$. Therefore, $U$ is $\sigma $-complete.

Title | complete ultrafilter and partitions |
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Canonical name | CompleteUltrafilterAndPartitions |

Date of creation | 2013-03-22 18:55:52 |

Last modified on | 2013-03-22 18:55:52 |

Owner | yesitis (13730) |

Last modified by | yesitis (13730) |

Numerical id | 4 |

Author | yesitis (13730) |

Entry type | Definition |

Classification | msc 03E02 |