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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.
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