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

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

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

Proof.

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

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

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

Proof.

We prove by contradiction that at least one of ${𝒰}_1$ or ${𝒰}_2$ must have the finite intersection property. If neither has the finite intersection property, then for some $Z_1,\mathrm{\dots }{Z}_k$ we must have

$$\mathrm{\varnothing }=\bigcap _{1\le i\le k}{Z}_i\cap Y=\bigcap _{1\le i\le k}{Z}_i\cap (X\backslash \backslash Y).$$

But then

$$\mathrm{\varnothing }=\left(\bigcap _{1\le i\le k}{Z}_i\cap Y\right)\cup \left(\bigcap _{1\le i\le k}{Z}_i\cap (X\backslash \backslash Y)\right)=\bigcap _{1\le i\le k}{Z}_i,$$

and so $𝒰$ does not have the finite intersection property either. This cannot be, since $𝒰$ is a filter.∎

Now, by Claim 2, if $𝒰$ were not an ultrafilter, i.e., if for some $Y$ subset of $X$ we would have neither $Y$ nor $X\backslash \backslash Y$ in $𝒰$, then the filter generated ${𝒰}_1$ or ${𝒰}_2$ would be finer than $𝒰$, and then $𝒰$ would not be maximal.

So $𝒰$ is an ultrafilter containing $ℱ$, as intended.