Equivalence of definitions of ultrafilter I'm proving that given a nonempty set $I$, and given a filter $F$, there exists an ultrafilter $D$ on $I$ such that $F \subseteq D$. I used Zorn's lemma to prove that for a given filter $F$, there exists a maximal filter $D'$, where $F \subseteq D'$. I need to prove that this maximal filter $D'$ is a ultrafilter, defined as a filter $B$  that satisfies the following condition: $\forall A \subseteq I , A \in B \lor (I -A) \in B$. I tried to use the proof by contradiction, but failed. How do I prove it?
 A: Proof by contradiction is indeed a valid method here.
You probably want to use the so-called "finite meet property" : If $C \subseteq P(I)$, we say it has the finite meet property when all finite intersections of elements of $C$ are non-empty. The important fact is : $C$ has the finite meet property if and only if it is included in a filter. For the direct implication, you can use the filter $\lbrace X \in P(I) \, | X $ contains a finite intersection of elements of C $\rbrace$.
So here, if you assume neither $A$ nor $I \setminus A$ belongs to $B$, you can check that $B \cup \lbrace A \rbrace$ has the finite meet property. It is thus included in a filter, contradicting maximality.
A: Let $\mathcal{F}$ be a filter on $I$ and take $A\subseteq I$ such that $A\notin\mathcal{F}$ and $B=I\setminus A\notin\mathcal{F}$.
Choose $C\in\mathcal{F}$. Without loss of generality, we can assume $C\cap A\ne\emptyset$ (otherwise, exchange $A$ and $B$).
We want to prove that $X\cap A\ne\emptyset$, for every $X\in\mathcal{F}$. We have
$$
X\cap C=(X\cap A\cap C)\cup(X\cap B\cap C)
$$
If $X\cap A=\emptyset$, then $X\cap B\cap C\in\mathcal{F}$, so $B\in\mathcal{F}$, contrary to the assumption.
Then $\mathcal{F}\cup\{A\}$ is a filter base, so $\mathcal{F}$ is not a maximal filter.
A: I like FiMePr's answer, but here is an alternative route which avoids invoking the finite meet property.
Either $A\in D'$ or $A\notin D'$. If $A\in D'$ then we are done so suppose $A\notin D'$. Let $$B=\{X\subseteq I\mid \exists Y\in D',\ A\cap Y\subseteq X\}$$ and show that $B$ is a filter which properly contains $D'$. By the maximality of $D'$, this implies $B$ contains all subsets of $I$. So there exist $Y\in D'$ such that $A\cap Y\subseteq\emptyset$ which implies $Y\subseteq I\setminus A$ and therefore $I\setminus A\in D'$. Thus, either $A\in D'$ or $I\setminus A\in D'$ so that $D'$ is an ultrafilter. (I ended up opting for the law of excluded middle in place of proof by contradiction)
