Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a list of questions concerning the properties of filters:

(1) If a finite subset of a poset is downward directed is it necessarily closed under finite intersection? At the very least, if said subset is downward directed, is the intersection of any two subsets of the subset contained in the subset?

(2) If a set is directed does it have a unique bound? For example, do all downward directed sets have a unique lower bound?

(3) Can a filter of neighborhoods of x, where x \in R, be conceived of as a family of intervals containing x and in which x is the only element in the intersection of all subsets in said family?

Thank you for your time.

share|improve this question
What is "intersection" in general posets? –  Asaf Karagila Jul 3 '13 at 18:17
Is this rhetorical? I'm referring to a subset of a poset that is a potential filter. Some manifestations of the definition of a filter imply that all subsets of a filter contain a common element and I am inquiring about the uniqueness of said common element. –  user84815 Jul 3 '13 at 18:27
Strange that you wrote this comment after Andreas wrote his answer and pointed out that there is no notion of intersection for general posets. –  Asaf Karagila Jul 3 '13 at 18:47
I did not see his reply until after I added the comment. My apologies for the confusion. Strictly speaking about filters though (where intersection applies...right?), does every filter have a greatest least bound? If (A,≤) is a filter, and it must be the case that for every x,y∈A, there exists a z∈A such that z≤x and z≤y, then it seems impossible to select an x,y∈A such that both are minimal elements of A because then there can be no z∈A that is less than both x and y. –  user84815 Jul 3 '13 at 19:00
Of course not. Consider the filter of co-finite subsets of $\Bbb N$. Every set in the filter has a strictly smaller set in the filter. –  Asaf Karagila Jul 3 '13 at 19:01
show 1 more comment

1 Answer

up vote 1 down vote accepted

(1) No. In the first place, in a general poset, there is no notion of intersection. Even if your poset is the set $P(X)$ of all subsets of a set $X$, so that "intersection" makes sense, downward-directedness does not guarantee anything about closure under intersections, even binary intersections.

(2) No. Anything below a lower bound of a set is another lower bound of that set. The greatest lower bound of a set is unique if it exists, but in general posets it need not exist. In $P(X)$, greatest lower bounds exist, but the greatest lower bound of a downward-directed set need not be a member of that set.

(3) A filter, being closed under supersets, won't consist only of intervals. In the real line, the neighborhood filter of a point $x$ consists of the open intervals containing $x$ and all the supersets of those intervals. $x$ is indeed the only point in the intersection of all those intervals.

share|improve this answer
Thank you, this is very helpful. I have a questions regarding your response: do filters, being downward directed and closed under intersection, have a greatest least bound? –  user84815 Jul 3 '13 at 18:33
A filter $F$ in $P(X)$ has a greatest lower bound that is an element of $P(X)$ but usually not an element of $F$. For example, consider any infinite set $X$, and let $F$ consist of those subsets of $X$ whose complements are finite. Then the greatest lower bound of $F$ is the empty set, which is not in $F$. –  Andreas Blass Jul 3 '13 at 19:01
@user84815 Also, in your comment, you wrote "closed under intersection"; filters are closed under finite intersection. That is, the intersection of finitely many sets from a filter $F$ is again in $F$. The same does not in general hold for the intersection of infinitely many sets from $F$. –  Andreas Blass Jul 3 '13 at 19:03
I see. Thank you for your help. –  user84815 Jul 3 '13 at 19:08
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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