alternative characterization of ultrafilter

Let $X$ be a set.A filter $ℱ$ over $X$ is an ultrafilter if and only ifit satisfies the following condition:if $A\coprod B=X$ (see disjoint union),then either $A\in ℱ$ or $B\in ℱ$.

This result can be generalized somewhat:a filter $ℱ$ over $X$ is an ultrafilter if and only ifit satisfies the following condition:if $A\cup B=X$ (see union), then either $A\in ℱ$ or $B\in ℱ$.

This theorem can be extended tothe following two propositions about finite unions:

1. A filter $ℱ$ over $X$ is an ultrafilter if and only if,whenever $A_1,\mathrm{\dots },A_n$ are subsets of $X$ such that ${\coprod }_{i=1}^n{A}_i=X$then there exists exactly one $i$ such that $A_i\in ℱ$.

2. A filter $ℱ$ over $X$ is an ultrafilter if and only if,whenever $A_1,\mathrm{\dots },A_n$ are subsets of $X$ such that ${\bigcup }_{i=1}^n{A}_i=X$then there exists an $i$ such that $A_i\in ℱ$.