# complete ultrafilter and partitions

If $U$ is an ultrafilter on a set $S$, then

$U$ is $\kappa$-complete $\Leftrightarrow$ 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, $\varnothing=S^{\complement}=\bigcap_{n=1}^{\omega}X_{n}^{\complement}$. Since, by our assumption, each of the $X_{n}$ do not belong to $U$, we have $X_{n}^{\complement}\in U$ for each $n<\omega$ as $U$ is an ultrafilter. Thus, $\left(\bigcap_{n=1}^{\omega}X_{n}^{\complement}\right)\in U$ by $\sigma$-completeness. This, however, means $\varnothing\in U$, 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 $\{P_{\iota}:\iota\leq\omega\}$ of $S\setminus Y$:

for each $s\in S\setminus Y$, put $s\in P_{\iota}$ if $\iota$ is the least index for which $s\not\in Y_{\iota}$.

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 $Y$, $\{P_{\iota}:\iota\leq\omega\}$ partitions $S$ into $\aleph_{0}=\omega$ many pieces. A (unique) piece of this partition belongs in $U$: $P_{\iota*}\in U$ or $Y\in U$. But, $P_{\iota}\cap Y_{\iota}=\varnothing\not\in U$ 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 CompleteUltrafilterAndPartitions 2013-03-22 18:55:52 2013-03-22 18:55:52 yesitis (13730) yesitis (13730) 4 yesitis (13730) Definition msc 03E02