Let's see the definition of a filter with the "$\subset$" order.

Let $X$ be a set. We say a non-empty family $\mathcal{F}$ of subsets of $X$ is a filter if:

  1. $\emptyset \notin \mathcal{F}$
  2. if $A, B \in \mathcal{F}$ then $A \cap B \in \mathcal{F}$
  3. if $A \in \mathcal{F}$ and $A \subset B$, then $B \in \mathcal{F}$

You can't say $$\bigcap_{F \in \mathcal{F}} F \subset \mathcal{F}$$

because not every filter is a principal filter, i.e. the interception above can be the empty set.

A principal filter is a filter generated by a single element.

By filter's statement 2 we know the finite intersection is defined and is an element of the filter. But why can't the interception of all elements of the filter be defined as is the finite intersection?

Maybe my question could be: why there are non-principal filters?


1 Answer 1


Because you can only take finite intersections and be guaranteed that you say in the filter. Once your filter becomes infinite, you can't guarantee that the intersection of all elements in the filter actually lies in the filter.

What your argument does show, however, is that all finite filters are principal. That's why we need to move to infinite filters to get interesting objects.

A common example of a non-principal filter over an infinite set $S$ is the filter $\mathscr{F}=\mathcal{P}_\mathrm{cofin}(S)$ of all cofinite subsets of $S$.

  • $S\in\mathscr{F}$ because $S^\mathsf{c}=\emptyset$ is finite.
  • $\emptyset^\mathsf{c}=S$ is infinite, so $\emptyset\notin \mathscr{F}$
  • If $A,B\in\mathscr{F}$, then $(A\cap B)^\mathsf{c}=A^\mathsf{c}\cup B^\mathsf{c}$ is the union of finite sets and so is finite. Thus, $A\cap B\in\mathscr{F}$
  • If $A\in\mathscr{F}$ and $A\subset B\subset S$, then $B^\mathsf{c}\subset A^\mathsf{c}$ is a subset of a finite set, and so finite. Thus, $B\in\mathscr{F}$.

This gives a second reason why non-principal filters exist: because we can give explicit examples.

EDIT: This is non-principal because if $\mathscr{F}$ is generated by a set $A$, then we know that $A$ must be cofinite so taking any element $a$ of $A$ shows us that $A\setminus\{a\}$ is also cofinite and so is in $\mathscr{F}$. But this doesn't lie in the filter generated by $A$, giving a contradiction.

  • $\begingroup$ Oh, now I see... The only exemple I could think myself was the "standart simple non-creative" subset of the powerset generated by a non-empty subset C of X, which is a principal filter (by definition)... XD $\endgroup$
    – Figurinha
    Apr 22, 2016 at 5:25
  • 1
    $\begingroup$ One way to intuit $\mathcal{P}_\mathrm{cofin}(S)$ as an example is by thinking about what would happen if you had a finite set in your filter. You could show that if there's a finite set in your filter, then it must be principal. This motivates one to look at $\mathcal{P}_\mathrm{cofin}(S)$ (since it has no finite sets). $\endgroup$
    – Hayden
    Apr 22, 2016 at 5:27

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