# Every free filter is contained in a free ultrafilter?

Let $$X$$ be an infinite set. A nonempty collection of subsets, say $$F$$, of $$X$$ is called a filter if

$$\emptyset \notin F$$

$$A, B \in F$$ implies $$A \cap B \in F$$

$$A \in F$$, $$A \subseteq B$$ implies $$B \in F$$

and is called an ultrafilter if in addition,

for all $$A \subseteq X$$, $$A \in F$$ or $$A^c \in F$$

holds.

A (ultra)filter is called free if for all $$a \in X$$, $$\{a\} \notin F$$.

I want to prove that every free filter $$F$$ is contained in a free ultrafilter.

My try: I tried to use Zorn's lemma to the collection $$\mathcal C := \{A: A$$ is a free filter containing $$F \}$$. However, I can't prove that the maximal element $$U$$ is an ultrafilter.

If $$U$$ were not an ultrafilter, there will exist an $$A \neq \emptyset$$ such that $$A$$, $$A^c \notin U$$. Then we will be able to extend $$U \cup \{ A \}$$ into a filter. However, I cannot prove that this filter is free, so that it contradicts the maximality of $$U$$.

Thank you!

• Actually, that is not the definition of free in the general case of filters. A filter $F$ is free if $\bigcap F\notin F$. In the ultrafilter case it coincides with your definition. Jan 31, 2020 at 9:01
• Without changed definition of free, this is false: a "free" filter that contains a finite subset, does not have such an extension. See my answer for that. Feb 2, 2020 at 12:16

First of all, as Asaf Karaglia wrote, a filter $$\mathscr F$$ is free if $$\,\bigcap \mathscr F \not\in\mathscr F$$. Clearly, a free filter does not possess finite sets.

It is not hard to show that, if a filter $$\mathscr{F}$$ is free, then adding to it all the complements of finite sets of $$X$$, and taking all the the finite intersections of its sets, you produce another free filter $$\mathscr{F}'\supset\mathscr{F}$$.

Using the Zorn's Lemma, on the filters, bigger than $$\mathscr F'$$, you obtain an ultrafilter which does not contain finite sets, and hence it is a free ultrafilter.

• Thanks for your elaboration. Now I am confused since a different answer and the Wikipedia('filter' page) says that a filter is free if its intersection is empty. Is this equivalent to your and Asaf's definition? Jan 31, 2020 at 13:35

In the poset of all filters (extending $$\mathcal{F}$$) under inclusion, a maximal element is an ultrafilter in your sense:

Suppose $$\mathcal{M}$$ is maximal and extends $$\mathcal{F}$$, and $$A \subseteq X$$ and $$A \notin \mathcal{M}$$, then suppose there some $$B \in \mathcal{M}$$ exists with $$A \cap B = \emptyset$$. Then $$B \subseteq A^\complement$$ and so $$A^\complement \in \mathcal{M}$$ and we're done. If this is not the case, i.e. $$\forall B \in \mathcal{M}: B \cap A \neq \emptyset\tag{1}$$

then we can define $$\mathcal{M}' = \{B \subseteq X: \exists C \in \mathcal{M}: C \cap A \subseteq B\}\tag{2}$$ and $$(2)$$ defines a filter that extends $$\mathcal{M}$$ (so $$\mathcal{F}$$ too) and contains $$A$$, so $$\mathcal{M} \subsetneq \mathcal{M}'$$, contradicting maximality of $$\mathcal{M}$$. So the second case does not occur and we're done showing $$\mathcal{M}$$ is an ultrafilter.

Now, the question of being free is a definitional matter: usually this is defined to mean (for any filter) that $$\bigcap \mathcal{F} \notin \mathcal{F}$$ and for your notion of ultrafilter this is equivalent: if $$\mathcal{M}$$ is an ultrafilter, $$\{a\} \in \mathcal{M}$$ iff $$\{a\} = \bigcap \mathcal{M}$$ iff $$\mathcal{M} = \{A \subseteq X: a \in M\}=: \mathcal{M}_a$$.

If $$\mathcal{F} \subseteq \mathcal{M}$$ and $$\{a\} \in \mathcal{M}$$ this implies that $$A \in \mathcal{F} \to A \in \mathcal{M} \to \{a\} \cap A \neq \emptyset \to a \in A$$, so $$a \in \bigcap \mathcal{F}$$. But this is not a direct contradiction with being free in your sense. In fact if $$\mathcal{F}$$ are all supersets of the even integers in the set of all integers $$X$$, then it is free in your sense but one possible extension that is maximal is all subsets that contain $$0$$, which is not free in your sense. So there is no insurance directly that we'll get a free ultrafilter from a free starting filter. But this can be helped if $$\mathcal{F}$$ contains no finite subset: in that case define $$\mathcal{F}' = \{C \subseteq X: \exists F \in \mathcal{F} \exists P \subseteq X \text { finite }: (F\setminus P) \subseteq C\}\tag{3}$$

and check that $$\mathcal{F}'$$ defines a filter on $$X$$ that extends $$\mathcal{F}$$ and then check that any maximal filter that extends $$\mathcal{F}'$$ is a free ultrafilter extending $$\mathcal{F}$$.

if however $$\mathcal{F}$$ contains a finite subset there can be no free ultrafilter extending it:

Let $$A \in \mathcal{F}$$ be finite, say $$A=\{a_1, a_2, \ldots, a_n\}$$. If $$\mathcal{M}$$ extends $$\mathcal{F}$$ and is free, $$\{a_i\}^\complement \in \mathcal{M}$$ for $$i=1,\ldots,n$$ and so $$A^\complement = \bigcap_{i=1}^n \{a_i\}^\complement \in \mathcal{M}$$ as filters are closed under finite intersections, contradicting $$A \in \mathcal{M}$$ (as $$\mathcal{M}$$ extends $$\mathcal{F}$$). So $$\mathcal{F}= \{A \subseteq \Bbb N: \{0,1,2\} \subseteq A\}$$

is a counterexample to your statement. Unless you redefine "free" of course.

With Axiom of choice yes, without Axiom of choice is this undecidable. For a filter $$\mathcal{F}$$, the system of filters $$\mathcal{F}\subseteq \mathcal{F}'$$ not containing finite set satisfies that for any chain $$\mathcal{F}\subseteq \mathcal{F}_1\subseteq \mathcal{F}_2\subseteq \mathcal{F}_3\text{...}$$

(may be uncountable)

the union of them is also filter not containing finite set. So, by Zorn's lemma (equivalent to Axiom of choice!) there exist a maximal element of all filters $$\mathcal{F}\subseteq \mathcal{F}'$$ and must be ultrafilter.

A filter is free iff it contains the Fréchet filter, so every filter extending a free filter is free.

Alternatively a filter $$G$$ is free iff $$\bigcap G=\varnothing$$. Let $$F$$ be the filter generated by $$A\cup U$$. Since $$\bigcap U=\varnothing$$ and $$U\subseteq F$$ we also have $$\bigcap F=\varnothing$$.

Also note that your definition of free works fine for ultrafilters, since an ultrafilter containing a finite set contains a singleton, but not for filters, for example the filter $$\{A\subseteq\omega\mid\{3,4\}\subseteq A\}$$ contains no singletons, but is not free.

• What is the most standard definition, which works for both filter and ultrafilters, of being free? I am totally new to the concept and am confused by the various definitions provided in the literature. Jan 31, 2020 at 8:58
• @JohnDoe usually one of the two characterizations I used in my answer is given as definition and the other is proved to be equivalent Jan 31, 2020 at 9:11