I have 2 questions about filter and (ultra-)filters:
Which relations are there between free filter, principal filter, ultrafilter, Frechet filter, and co-finite filter?
If a filter is free, does it imply that it is a principal ultrafilter?
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$\begingroup$ 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
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$\begingroup$ is not important that our space is finite or infinite? $\endgroup$ – freind Aug 4 '13 at 8:09
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$\begingroup$ All ultrafilters on a finite set are principal. $\endgroup$ – Martin Sleziak Aug 4 '13 at 8:11
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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.)
answered Aug 4 '13 at 8:04
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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.
answered Aug 4 '13 at 7:59
