4
$\begingroup$

I have 2 questions about filter and (ultra-)filters:

  1. Which relations are there between free filter, principal filter, ultrafilter, Frechet filter, and co-finite filter?

  2. If a filter is free, does it imply that it is a principal ultrafilter?

|cite|improve this question
$\endgroup$
  • $\begingroup$ You probably wanted to write "ultrafilter" instead of "uttra-filter". I guess you meant Fréchet filter not frecheh. $\endgroup$ – Martin Sleziak Aug 4 '13 at 7:57
  • $\begingroup$ is not important that our space is finite or infinite? $\endgroup$ – freind Aug 4 '13 at 8:09
  • $\begingroup$ All ultrafilters on a finite set are principal. $\endgroup$ – Martin Sleziak Aug 4 '13 at 8:11
5
$\begingroup$

A filter $\mathcal F$ is called free if $\bigcap \mathcal F=\emptyset$. Filter, which is not free is called principal. Hence every filter is either free or principal and the same is true for ultrafilters.1

A principal filter is a filter consisting of supersets of some fixed set. It is clear that a filter cannot be at the same time free and principal.

Principal ultrafilters are very simple; any principal ultrafilter has the form $$\mathcal F_a=\{A\subseteq X; a\in A\}$$ for some $a\in X$.

To show the existence of free ultrafilters, some form of Axiom of Choice is needed. The proof is often based on ultrafilter lemma.

Fréchet filter (a.k.a. cofinite filter) on a set $X$ is the filter consisting of all cofinite set, i.e., it is equal to $$\mathcal F_{F}=\{A\subseteq X; X\setminus A\text{ is finite}\}.$$ An ultrafilter is free if and only if it contains the Fréchet filter.

1 It was pointed out that this is not usual meaning of the term principal filter. I was using "principal" and "non-free" as equivalent, it seems that they are not. (However, this usage is still correct for ultrafilters.)

|cite|improve this answer
$\endgroup$
1
$\begingroup$

By definition, a free ultrafilter is one that is not principal. So any ultrafilter is either principal or free, and never both. The Frechet filter is a filter, not an ultrafilter.

|cite|improve this answer
$\endgroup$

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.