Ultrafilters Are Maximal

Ultra filter
A filer $\mathcal{F}$ containing either $Y$ or $Y^\complement$ for any $Y \subseteq X$.

Two days ago, I spent an afternoon to understand Dudley’s proof of this little result.

A filter is contained in some ultrafilter. A filter is an ultrafilter iff it’s maximal.

At the first glance, I didn’t even understand the organisation of the proof! I’m going to rephrase it for future reference.

  • only if: let $\mathcal{F}$ be an ultrafilter contained in another filter $\mathcal{G}$. If $\mathcal{F}$ isn’t maximal, let $Y \in \mathcal{G} \setminus \mathcal{F}$. Since $\mathcal{F}$ is an ultrafilter, either $Y \in \mathcal{F}$ or $Y^\complement \in \mathcal{F}$. By construction of $Y$, only the later option is possible, so $Y^\complement \in \mathcal{G}$ by hypothesis, but this contradicts our assumption $Y \in \mathcal{G}$: $\varnothing = Y \cap Y^\complement \in \mathcal{G}$, which is false since $\mathcal{G}$ is a filter.

  • if: let $Y \subseteq X$. “Extend” $\mathcal{F}_{\max}$ to $\mathcal{G}$ so that $Y \in \mathcal{G}_1$ or $Y^\complement \in \mathcal{G}_2$. Apply maximality so that $\mathcal{G}_i = \mathcal{F}_{\max}$ for each $i \in \lbrace 1,2 \rbrace$. To construct such $\mathcal{G}$, make a key observation: either one of the following is true.

    1. $\forall G \in \mathcal{F}_{\max}, G \cap Y \ne \varnothing$
    2. $\forall F \in \mathcal{F}_{\max}, F \cap Y^\complement \ne \varnothing$

    To understand this, suppose $(1)$ is false. There exists $G \in \mathcal{F}_{\max}$ such that $G \cap Y = \varnothing$. (i.e. $G \subseteq Y^\complement$) For any $F \in \mathcal{F}_{\max}$, $F \cap G \ne \varnothing$ since $\mathcal{F}_{\max}$ is a filter. Take intersection with $F$ in the first set inclusion to obtain $F \cap G \subseteq F \cap Y^\complement$, so $F \cap Y^\complement \ne \varnothing$. We have just proved $(2)$.

    To visualise this, take, for example, the filter to be the collection of open neighbourhoods of any point in $Y^\complement$.

  • Since $\mathcal{F}$ can be uncountably infinite, Zorn’s lemma is needed. We apply this on the collection $\mathcal{X}$ of filters in $X$. For any chain $\mathcal{C}$ of filters in $X$, check that $\mathcal{U} = \cup \mathcal{C}$ is a filter by definition. Then invoke the lemma to get a maximal element $\mathcal{F}_{\max}$ and we’re done.

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