# Using Zorn's Lemma to show that a filter is contained in an ultrafilter

Background: I am trying to use Zorn's lemma to show the existence of ultrafilters containing an arbitrary filter on a set $X$. My argument goes as follows:

Let $\mathcal{F}_0$ be a filter on $X$. If $\mathcal{F}_0$ is an ultrafilter, we are done. If $\mathcal{F}_0$ is not an ultrafilter, then we have another filter $\mathcal{F}_1$ which contains $\mathcal{F}_0$. Again, if $\mathcal{F}_1$ is an ultrafilter we are done. If not, we produce another filter $\mathcal{F}_2$ containing $\mathcal{F}_1$ and so on and so forth. Assuming this process continues on ad infinitum, we obtain the following chain of filters ordered by inclusion

$$\mathcal{F}_0\subseteq \mathcal{F}_1\subseteq \mathcal{F}_2\subseteq ...\subseteq \mathcal{F}_n \subseteq ...$$ Let $\mathcal{F}=\bigcup_{i=0} ^{\infty} \mathcal{F}_i$. Then $\mathcal{F}$ is a filter. First, the empty set is not in $\mathcal{F}$ because it is not contained in any $\mathcal{F}_i$. If $A \in \mathcal{F}$ and $B$ is a superset of $A$, then there exists some $\mathcal{F}_i$ such that $A \in \mathcal{F}_i$, which implies $B \in \mathcal{F}_i$ since $\mathcal{F}_i$ is a filter. Thus $B \in \mathcal{F}$. Assume $A,B \in \mathcal{F}$. Then there exists some $\mathcal{F}_j$ such that $A,B \in \mathcal{F}_j$, which implies $A \cap B \in \mathcal{F}_j$. This in turn implies $A \cap B \in \mathcal{F}$, completing the proof. Note that $\mathcal{F}$ is an upper bound of this chain. Using Zorn's lemma, we see that this chain contains a maximal element $\mathcal{G}$. Since $\mathcal{G}$ contains $\mathcal{F}_0$, we see that there exists an ultrafilter containing $\mathcal{F}_0$.

Question: Is it true that $\mathcal{G}=\mathcal{F}$? More generally, is it the case that the upper bound in your chain is the maximal element you are looking for? If not, are there any intuitive reason why or examples of this failure?

Fixed Proof

Let $\mathcal{F}$ be a filter on $X$. Let $S$ be the set of filters containing $\mathcal{F}_0$. Consider an arbitrary chain $C$ in $S$. Let $\mathcal{F}=\bigcup_{i \in I} \mathcal{F}_i$, where $\mathcal{F}_i \in C$ for all $i \in I$. Then $\mathcal{F}$ is a filter. First, the empty set is not in $\mathcal{F}$ because it is not contained in any $\mathcal{F}_i$. If $A \in \mathcal{F}$ and $B$ is a superset of $A$, then there exists some $\mathcal{F}_i$ such that $A \in \mathcal{F}_i$, which implies $B \in \mathcal{F}_i$ since $\mathcal{F}_i$ is a filter. Thus $B \in \mathcal{F}$. Assume $A,B \in \mathcal{F}$. Then there exists some $\mathcal{F}_j$ such that $A,B \in \mathcal{F}_j$, which implies $A \cap B \in \mathcal{F}_j$. This in turn implies $A \cap B \in \mathcal{F}$, completing the proof. Note that $\mathcal{F}$ is an upper bound of this chain. Thus every chain in $S$ has an upper bound. Using Zorn's lemma, we see that $S$ contains a maximal element $\mathcal{G}$. Since $\mathcal{G}$ contains $\mathcal{F}_0$, we see that there exists an ultrafilter containing $\mathcal{F}_0$.

• You seem to be confused about what Zorn's lemma says and how to use it. It doesn't say that chains have upper bounds – rather, that is a hypothesis! Dec 6 '12 at 13:23
• The union of an arbitrary chain of a filters is again a filter, so without appealing to Zorn's lemma you have that $\cal F$ is a filter containing each $\cal F_i$. However, $\cal F$ need not be an ultrafilter, i.e., it may be possible to find a chain of uncountably many filters between $\cal F_0$ and an ultrafilter containing it. All you need to apply Zorn's lemma here is the fact that the union of an arbitrary chain of filters is a filter, which you have proved above (your argument still works if the indices $i$ range over an arbitrary ordinal). Dec 6 '12 at 13:32
• You have a typo on the first sentence you meant to write $\cal F_0$. Secondly what if the chain is longer than you can index with the natural numbers? Dec 6 '12 at 14:05
• @asafkaragila I have to be honest and say that I do not know how to deal with that situation. How does one? I suppose I could just skip this approach altogether and use the axiom of choice to give a function which decides whether $A \in P(X)$ is used to extend $\mathcal{F}_0$ or its complement. I believe that gets around all the size considerations. Dec 6 '12 at 14:14
• @asafkaragila couldn't I just use some arbitrary indexing set $I$ to take the union over? Dec 6 '12 at 14:15

There is a problem with your proposed proof. The notation, at least, seem to imply that you think about countable chains. It need not be the case, especially if $X$ is arbitrary.
The supremum of that particular chain is $\bigcup_{i=0}^\infty\cal F_i$ so it would also be the maximal element of that chain. So, Zorn's lemma wouldn't bring anything new to the proof. Since you didn't prove that the union is an ultrafilter, there might be another bigger filter such that $\bigcup_i\cal F_i\subset\cal F_{\omega+1}$ and this continues ad infinitum and you get a new chain. Zorn's lemma says that if every chain (and they are not necessarily countable) has an upper bound then there is a maximal element in the set of all filters containing $\cal F_0$. Since it uses the Axiom of Choice, you usually won't know anything about the element itself, except that it exists.