I am very new to filters. Wikipedia says that principal ultrafilters are filters containing least element.

However, then it says principal ultrafilters look like $F_a = \{x: a \leq x \}$ for some elements $a$ of the given poset.

How can the $F_a$ contain the least element, if it consists of $x$ bigger than the $a$?

Also, how can principal ultrafilter have the condition that the intersection of all its subsets is the filter itself?

I think I generally have trouble understanding the definition of principal ultrafilters, since I just cannot see how this makes sense.

Thank you for any help.

  • 2
    $\begingroup$ No, it is a poset ultrafilter, and has nothing to do with intersection but only $\leq$ and possible meet (if lattice). $\endgroup$ Sep 29, 2021 at 10:00
  • $\begingroup$ I'm not sure what you mean with "the intersection of all its subsets is the filter itself". This cannot be correct, since $\varnothing$ is a subset of any set, and any intersection containing $\varnothing$ is empty. $\endgroup$
    – Vsotvep
    Sep 29, 2021 at 11:06
  • $\begingroup$ @user10354138 So what is the difference between "ultrafilter" and "poset ultrafilter"? $\endgroup$ Sep 29, 2021 at 15:50
  • $\begingroup$ @TerezaTizkova there is no real difference: filters on sets use "$\subseteq$" and "intersection" etc. as terms, while in posets the terms "$\leq$" and "meet" etc. are commonly used. Every power set algebra $(\mathcal P(X),\subseteq)$ is a poset, but not every poset can be modelled as a power set algebra. So terms like "intersection" make no sense when talking about ultrafilters on posets in general. $\endgroup$
    – Vsotvep
    Sep 29, 2021 at 15:58

2 Answers 2


This question seems to be about posets in general, but since your other question was about filters on sets specifically, I will preamble this answer that you could regard $(\mathcal P(X),\subseteq)$ as a partial order with minimal element $\varnothing$. Simply replace $P$ with $\mathcal P(X)$, $\leq$ with $\subseteq$ and $0$ with $\varnothing$ below if you're only interested in sets.

Let's fix some arbitrary poset $(P,\leq)$ and consider filters on $P$.

A filter $F$ is principal if it contains some least element, not the least element. What this means to say, is that there is some element in $F$ that is smaller than all other elements of $F$. For instance, $F_a=\{x\in P\mid a\leq x\}$ is principal, but it is not necessarily the case that $a$ is the least element of $P$, and if in fact $a$ happens to be the least element of $P$, then $F_a$ would not be an interesting filter: it would contain every element of $P$.

The filter $F_a$ contains all elements $x$ such that $x\geq a$, thus $a$ is the least element of $F_a$: first, $a\in F_a$ since $a\leq a$, and second, if $x\in F_a$, then $a\leq x$.

If $F$ is a principal filter, then it is not necessarily an ultrafilter. For example, if $P$ has a minimal element $0$, and furthermore $0<a<b$ are two elements of the poset, then $F_a\supset F_b$, but $a\notin F_b$. Both $F_a$ and $F_b$ are principal filters, but since $F_b$ is not maximal, it is not an ultrafilter.

An element $a\in P$ is called atomic if there is a minimal element $0\in P$ for which $0<a$ and for every $x\in P$, if $x<a$, then $x=0$. A principal filter $F$ is an ultrafilter if and only if $F=F_a$ for some atomic $a$:

Suppose $F=F_b$ for some non-atomic $b$, then there is some $0<x<b$, and thus $F_x\supset F_b$, which shows that $F_b$ is not maximal.

Conversely if $a$ is atomic and $F\supset F_a$ is a filter containing some $x\notin F_a$, then by filters being downward directed, there is some $y\leq x$ and $y\leq a$. Now $y\neq a$, since if $y=a$, then $a\leq x$ contradicts that $x\notin F_a$. But $y\neq a$ implies that $y=0$, since $a$ is atomic, hence $F_y=F_0=P$ is not a proper filter.

Not every ultrafilter has to be principal, but it is hard to give concrete examples of nonprincipal ultrafilters. It generally requires some form of the Axiom of Choice to prove their existence in the first place.

  • $\begingroup$ Thank you for your answer! So filters are usually put either on sets, or on partially ordered sets? Or on anything else that goes with the axioms? I got quite confused, sorrry, only worked with filters on sets. $\endgroup$ Sep 29, 2021 at 16:00
  • 1
    $\begingroup$ @TerezaTizkova Both. Partially ordered sets are more general than power set algebras, so anything provable for partially ordered sets also holds for power sets, but not vice versa. E.g. for power sets, atomic elements always exist, but for general partially ordered sets they don't necessarily, thus there aren't necessarily principal ultrafilters on every partially ordered set. As an example, consider $(\Bbb Z,\leq)$ as a partial order, then a poset filter would be just an upward closed set of integers, such as $\{n\in\Bbb Z\mid n\geq 3\}$, but no such filter can be maximal. $\endgroup$
    – Vsotvep
    Sep 29, 2021 at 16:04
  • $\begingroup$ Don't confuse this with filters over the set $\Bbb Z$, though, which would be a subset $F\subset \mathcal P(\Bbb Z)$. Of course here there do exist principal ultrafilters, such as the filter $\{X\subset \Bbb Z\mid 0\in X\}$. $\endgroup$
    – Vsotvep
    Sep 29, 2021 at 16:05

$F_a = \{x\mid a \le x\}$ is always a filter in any lattice: If $b' \ge b$ and $b \in F_a$ then $a \le b \le b'$ so $a \le b'$ and so $b' \in F_a$ too. And if $b,b' \in F_a$ then $a \le b$ and $a \le b'$ so $a$ is a lower bound for $\{b,b'\}$ and so $a \le b \land b'$ and $b \land b' \in F_a$ by maximality of the meet. (it's quite obvious when we're in a lattice of sets (powerset, e.g.) and $\le$ means $\subseteq$).

In some cases (as the Wikipedia entry says) this filter is in fact maximal, e.g. when we're in a powerset and $a= \{c\}$ for some element of the set.

In either case the meet/intersection of $F_a$ is indeed $a$ as can be seen from the definitions.

A boring case is $P(X)$ and $F_X=\{X\}$ e.g. which will almost never be an ultrafilter.


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