# proof of every filter is contained in an ultrafilter (alternate proof)

Let $\mathfrak{U}$ be the family of filters over $X$ which are finer than $\mathcal{F}$, under the partial order of inclusion.

Claim 1. Every chain in $\mathfrak{U}$ has an upper bound also in $\mathfrak{U}$.

###### Proof.

Take any chain $\mathfrak{C}$ in $\mathfrak{U}$, and consider the set $\mathcal{C}=\cup\mathfrak{C}$. Then $\mathcal{C}$ is also a filter: it cannot contain the empty set, since no filter in the chain does; the intersection of two sets in $\mathcal{C}$ must be present in the filters of $\mathfrak{C}$; and $\mathcal{C}$ is closed under supersets because every filter in $\mathfrak{C}$ is. Obviously $\mathcal{C}$ is finer than $\mathcal{F}$. β

So we conclude, by Zornβs lemma, that $\mathfrak{U}$ must have a maximal filter say $\mathcal{U}$, which must contain $\mathcal{F}$. All we need to show is that $\mathcal{U}$ is an ultrafilter. Now, for any filter $\mathcal{U}$, and any set $Y\subseteq X$, we must have:

Claim 2. Either $\mathcal{U}_{1}=\{Z\cap Y:Z\in\mathcal{U}\}$ or $\mathcal{U}_{2}=\{Z\cap(X\backslash Y):Z\in\mathcal{U}\}$ (or both) are a filter subbasis.

###### Proof.

We prove by contradiction that at least one of $\mathcal{U}_{1}$ or $\mathcal{U}_{2}$ must have the finite intersection property. If neither has the finite intersection property, then for some $Z_{1},\ldots Z_{k}$ we must have

 $\varnothing=\bigcap_{1\leq i\leq k}Z_{i}\cap Y=\bigcap_{1\leq i\leq k}Z_{i}% \cap(X\backslash Y).$

But then

 $\varnothing=\left(\bigcap_{1\leq i\leq k}Z_{i}\cap Y\right)\cup\left(\bigcap_{% 1\leq i\leq k}Z_{i}\cap(X\backslash Y)\right)=\bigcap_{1\leq i\leq k}Z_{i},$

and so $\mathcal{U}$ does not have the finite intersection property either. This cannot be, since $\mathcal{U}$ is a filter. β

Now, by Claim 2, if $\mathcal{U}$ were not an ultrafilter, i.e., if for some $Y$ subset of $X$ we would have neither $Y$ nor $X\backslash Y$ in $\mathcal{U}$, then the filter generated $\mathcal{U}_{1}$ or $\mathcal{U}_{2}$ would be finer than $\mathcal{U}$, and then $\mathcal{U}$ would not be maximal.

So $\mathcal{U}$ is an ultrafilter containing $\mathcal{F}$, as intended.

Title proof of every filter is contained in an ultrafilter (alternate proof) ProofOfEveryFilterIsContainedInAnUltrafilteralternateProof 2013-03-22 17:52:54 2013-03-22 17:52:54 brunoloff (19748) brunoloff (19748) 8 brunoloff (19748) Proof msc 54A20