Ultrafilter on $[0,1]$ consisting of closed sets Today we learned about filters and ultrafilters in the General Topology course. I am trying to play around with these definitions. I wish to ask a question that I am unsure about. Let us say, we have an ultrafilter $\mathcal{F}$ on the closed interval $[0, 1]$. Construct $\overline{\mathcal{F}} = \{ \overline{A} : A \in \mathcal{F}\}$, where the overline denotes closure of a set. My question is, is $\overline{\mathcal{F}}$ an ultrafilter? Is it a filter at all? The intersection property is clear, but it is the superset property of filters that is confusing me. Thanks for any help! 
 A: As Brian said, it depends. $\overline{\mathcal{F}}$ consists of only closed sets, so the superset property will fail in general: if $[0,\frac{1}{2}] \in \overline{\mathcal{F}}$ and $[0,\frac{1}{2}] \subset [0,\frac{3}{4})$ but the latter set is not closed, so not of the form $\overline{A}$ for some $A \in \mathcal{F}$. So it's not a filter over all subsets of $[0,1]$. 
It it a filter base: if $\overline{A_1}, \overline{A_2} \in \overline{\mathcal{F}}$, with $A_1,A_2 \in \mathcal{F}$, then $\emptyset \neq \overline{A_1 \cap A_2} \subseteq (\overline{A_1} \cap \overline{A_2})$, so $\overline{\mathcal{F}}$ has the finite intersection property (which can be used in proofs involving compactness: the property that every family of closed sets with the finite intersection property has non-empty intersection, is equivalent to compactness), and so generates a filter by also inclusing all supersets of elements.
The notion of (ultra)filter can be defined in a lattice (a partially ordered set (a.k.a poset) where every two members $p,q$ has an infimum denoted $p \land q$).
A filter $F$ is then a subfamily of the poset that is closed under $\land$ and if $p \in F$ and $p \le q$ then $q \in F$, so closed under larger elements.
The powerset of a set $X$ is a poset under inclusion and $A \cap B$ is the infimum of $A$ and $B$, but there is also a poset $\mathscr{C}(X)$ of closed subsets of a topological space $X$, with the same operations, but restricted to closed sets only. 
If $\mathcal{F} \subseteq \mathscr{P}(X)$ is a filter, then $\overline{\mathcal{F}}$ is a filter in $\mathscr{C}(X)$: it's in fact the intersection of $\mathscr{C}(X)$ with $\mathcal{F}$: trivially if $A$ is closed and in $\mathcal{F}$, then $A = \overline{A} \in \overline{\mathcal{F}}$, and if $\overline{A} \in \overline{\mathcal{F}}$, then $A \in \mathcal{F}, A \subseteq \overline{A}$, so $\overline{A} \in  \mathscr{C}(X) \cap \mathcal{F}$. From this the property follows easily. 
So your set is a filter, but in another poset. But it is a filterbase of closed sets and this is often useful in proofs involving compactness.
