Are filters compact?

Question

Suppose that we are given a filter $F$ of subsets of some set $X$. We may identify each element of $F$ with a function in $\{0,1\}^X$, which is compact under the product topology. Is $F$ a compact subset of $\{0,1\}^X$?

Answer 1

A filter is compact iff it is principal. Indeed, if $F$ is principal (generated by some set $A$), then it is closed in $\{0,1\}^X$ since it is the set of functions $f:X\to\{0,1\}$ such that $f(a)=1$ for all $a\in A$, and for each individual $a$ this defines a basic closed set and the overall condition is just the intersection of all these basic closed sets.

Conversely, if $F$ is not principal, then the set $A=\bigcap_{B\in F} B$ is not an element of $F$ (otherwise $F$ would be principal, generated by $A$). But given any finite subset $Y$ of $X$, there is an element of $F$ that agrees with $A$ on every element of $Y$. Indeed, every element of $F$ contains every element of $A$, so agrees with $A$ on the elements of $Y\cap A$. For each $y\in Y\setminus A$, there exists some $B_y\in F$ which does not contain $y$ (since $A=\bigcap_{B\in F} B$), and the intersection of $B_y$'s for each $y\in Y\setminus A$ is then an element of $F$ that agrees with $A$ on all of $Y$.

This says exactly that $F$ intersects every open neighborhood of $A$, so $A$ is in the closure of $F$. Thus $F$ is not closed in $\{0,1\}^X$, and hence not compact.