Let $\mathscr{F}$ be a filter on a set $I$. Prove that $\mathscr{F}$ is an ultrafilter if and only if there does not exist any filter $\mathscr{F^*}$ on $I$ such that $\mathscr{F^*} \supsetneq \mathscr{F}$.
Proof: Let $\mathscr{F}$ be an ultrafilter and suppose there exist a filter $\mathscr{F^*}$ on $I$ such that $\mathscr{F^*} \supsetneq \mathscr{F}$. Then by definition of an ultrafilter, for each $E \subset I$, either $E \in \mathscr{F}$ or $I \setminus E \in \mathscr{F}$. But that would imply that for each $E \subset I$, either $E \in \mathscr{F^*}$ or $I \setminus E \in \mathscr{F^*}$ which implies $\mathscr{F^*}$ is an ultrafilter.
I am stuck after this. I was going to conclude that an ultrafilter cannot contain an ultrafilter but I can't find anything that proves my statement. Is this case or can an Ultrafilter on a set $I$ contain another ultrafilter on the set $I$?
I am also having a really hard time proving the other direction.
Edit:
Okay I was pointed to some questions similar to this and I am still a little confused. So from what I read I should conclude that this contradicts the maximality of ultrafilters, but I am trying to prove maximality so why can I use this?
I'm also still confused how to start the other direction?
Best Answer
Suppose that $E\in\mathscr{F}^*\setminus\mathscr{F}$. Then in particular $E\notin\mathscr{F}$, so $I\setminus E\in\mathscr{F}$. And $\mathscr{F}\subseteq\mathscr{F}^*$, so $I\setminus E\in\mathscr{F}^*$. But $E\cap(I\setminus E)=\varnothing$, so $\varnothing\in\mathscr{F}^*$, which is impossible.