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.