extension to an ultrafilter Let X be a topological space. How to show a filter T on X can be extended to a unique ultrafilter on X iff T is an ultrafilter.
All I know is that every filter can be extended to an ultrafilter. I think this can be useful here. But I cannot find the link.
Any help would be aporeciated. Thanks in advance.
The key idea is this: 


  If $\mathcal{F}$ is a filter on $X$, and $A \subset X$ is such that both $A \notin \mathcal{F}$ and $X\setminus A \notin \mathcal{F}$, then $\mathcal{F} \cup \{A\}$ has the finite intersection property.


The proof is simple: suppose not. Then there are finitely many sets of $\mathcal{F} \cup \{A\}$ that have empty intersection. As $\mathcal{F}$ is closed under finite intersections, this can only happen if there is some $F_1 \in \mathcal{F}$ such that $F_1 \cap A = \emptyset$. But this is equivalent to $F_1 \subseteq X\setminus A$, which implies (by the enlargement axiom for filters) that $X\setminus A \in \mathcal{F}$, quod non. This contradiction shows that $\mathcal{F} \cup \{A\}$ has the fip, so can be extended to a filter $\mathcal{F}_A$ which can be explicitly defined as $$\mathcal{F}_A := \{B \subseteq X: \exists F \in \mathcal{F}: F \cap A \subseteq B\}$$

One easily shows this is a filter on $X$ that contains $A$ and all of $\mathcal{F}$.

So if for some filter $\mathcal{F}$ we have that there is some set $A$ such that neither $A$ nor $A^c = X\setminus A$ are not in $\mathcal{F}$, we can also apply the above idea (by symmetry) to get a filter $\mathcal{F}_{A^c}$ that properly extends $\mathcal{F}$ and then $\mathcal{F}_{A^c}$ and $\mathcal{F}_{A}$ are distinct (one contains $A$, the other does not) proper extensions of $\mathcal{F}$. We can extend both of them to ultrafilters, that are thus also distinct. 

We conclude by contraposition that if $\mathcal{F}$ has a unique ultrafilter extension, then for all sets $A \subseteq X$, either $A \in \mathcal{F}$ or $X\setminus A \in \mathcal{F}$, and this property characterises ultrafilters (which follows from the observation that if $\mathcal{G} \supsetneq \mathcal{F}$ for some filter, any $B \in \mathcal{G}\setminus \mathcal{F}$ will fail this criterion, as $B$ is not in $\mathcal{F}$ and $X\setminus B$ isn't either (or it would be in $\mathcal{G}$ which cannot be)).
If T is not an ultrafilter, then there is some A subset X with
A not in T and X - A not in T.  Thus T can be extended to an
ultrafilter F with A in F and to an ultrafilter G with X - A in G.