3
$\begingroup$

Preliminaries. Let $S$ be an infinite set. We are only going to consider filters in the powerset algebra $\mathcal P(S)$. In this setting, a filter over $S$ is a subset $\mathcal F \subset \mathcal P(S)$ such that

  1. (Closed under intersections) whenever $A,B \in \mathcal F$, then $A \cap B \in \mathcal F$,
  2. (Monotonicity) whenever $A \in \mathcal F$ and $A \subset B\subset S$, then $B \in \mathcal F$ ,
  3. (Properness) $\emptyset \in \mathcal F$ and $F \neq \emptyset$.

A filter $\mathcal U$ is called an ultrafilter, if it satisfies in addition

  1. (Maximality) Whenever $A \subset S$, either $A \in \mathcal U$ or $S\setminus A \in \mathcal U$.

An ultrafilter $\mathcal U$ is called free if $\bigcap \mathcal U = \emptyset$.

The set $\mathcal P_{\text{cof}}(S)$ of cofinite subsets of $S$ is called the Fréchet filter. The Fréchet filter is a filter, but not an ultrafilter (since $S$ is infinite). It turns out that

  • Any ultrafilter containing the Fréchet filter is free,
  • Any free ultrafilter contains the Fréchet filter.

Question. Let $\mathcal F$ be a filter over an infinite set $S$ satisfying the following properties.

  1. Any ultrafilter containing $\mathcal F$ is free.
  2. Any free ultrafilter contains $\mathcal F$.
  3. Any filter $\mathcal G$ strictly containing $\mathcal F$ does not satisfy property 2. In other words, there exists an ultrafilter $\mathcal U$ such that $\mathcal G$ is not a subset of $\mathcal U$.

Does it follow that $\mathcal F$ is the Fréchet filter? Does the Fréchet filter even satisfy property 3.?

$\endgroup$

1 Answer 1

5
$\begingroup$

The Fréchet filter is actually the only filter that satisfies both (1) and (2). More precisely a filter satisfies (1) iff it contains the Frechét filter, and satisfies (2) iff it is contained in the Fréchet filter. (It follows that the Fréchet filter also satisfies (3).)

Clearly any filter containing the Fréchet filter satisfies (1). Conversely, suppose a filter $\mathcal{F}$ does not contain the Fréchet filter, so there is a cofinite set $A\subseteq S$ which is not in $\mathcal{F}$. We can then enlarge $\mathcal{F}$ to the filter $\mathcal{G}$ generated by $\mathcal{F}$ and $S\setminus A$, and then extend $\mathcal{G}$ to an ultrafilter $\mathcal{U}$. Since $S\setminus A\in\mathcal{U}$ and $S\setminus A$ is finite, $\mathcal{U}$ is not free. Thus $\mathcal{F}$ does not satisfy (1).

Similarly, clearly any filter contained in the Fréchet filter satisfies (2). Conversely, suppose a filter $\mathcal{F}$ is not contained in the Fréchet filter, so there is a set $A\in\mathcal{F}$ whose complement is infinite. We can then take the filter $\mathcal{G}$ generated by the Fréchet filter together with $S\setminus A$, and extend it to an ultrafilter $\mathcal{U}$. This ultrafilter is free since it contains the Fréchet filter, but does not contain $A$ and so does not contain $\mathcal{F}$. Thus $\mathcal{F}$ does not satisfy (2).

More generally, similar arguments show that every filter is equal to the intersection of all the ultrafilters that contain it, so a filter is determined by the ultrafilters that contain it. If $\beta S$ denotes the set of ultrafilters on $S$, this gives an inclusion-reversing bijection between the set of filters on $S$ and the set of closed subsets of $\beta S$ with respect to the product topology, considering $\beta S$ as a subset of $\{0,1\}^{\mathcal{P}(S)}$ in the obvious way. (We map each filter to the set of ultrafilters that contain it, and the inverse is given by taking the intersection of the ultrafilters in any closed set.) This is part of the more general theory of Stone duality between Boolean algebras and totally disconnected compact Hausdorff spaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.