F - filter over I. Please, prove, F ultrafilter <=> when $\forall$ X, Y $\subseteq$ I, if X $\notin$ F, Y $\notin$ F, $\Rightarrow$ X $\cup$ Y $\notin$ F.
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
Hint: If $\cal F$ is an ultrafilter, pick $X,Y\notin\cal F$ and use the fact that their complements are in $\cal F$ and it is closed under intersections; in the other direction, if $\cal F$ is not an ultrafilter there is some $X$ such that $X,I\setminus X\notin\cal F$. Use these two sets to contradict the statement you have on the RHS.