[proofplan] We order all proper filters on $I$ that contain $\mathcal F$ by inclusion. The key point is that the union of a nonempty chain of such filters is again a proper filter: closure under finite intersections is preserved because any two sets in the union already lie together in one member of the chain. Thus every chain has an upper bound, so [Zorn's lemma](/theorems/1226) gives a maximal proper filter extending $\mathcal F$. By the definition of an ultrafilter as a maximal proper filter, this maximal element is the desired ultrafilter. [/proofplan]

[step:Build the partially ordered family of proper filter extensions] Let $\mathcal P(I)$ denote the power set of $I$. Define \begin{align*} \mathscr P := \{\mathcal G \subseteq \mathcal P(I) : \mathcal G \text{ is a proper filter on } I \text{ and } \mathcal F \subseteq \mathcal G\}. \end{align*} We order $\mathscr P$ by set inclusion. Since $\mathcal F$ is a proper filter on $I$ and $\mathcal F \subseteq \mathcal F$, we have $\mathcal F \in \mathscr P$, so $\mathscr P$ is nonempty. [/step]

[step:Show that every chain has an upper bound]Let $\mathscr C \subseteq \mathscr P$ be a chain with respect to inclusion. If $\mathscr C = \varnothing$, then $\mathcal F \in \mathscr P$ is an upper bound for $\mathscr C$. Assume now that $\mathscr C \neq \varnothing$. Define \begin{align*} \mathcal H := \bigcup_{\mathcal G \in \mathscr C} \mathcal G. \end{align*} We prove that $\mathcal H \in \mathscr P$. First, $\mathcal F \subseteq \mathcal H$: choose some $\mathcal G_0 \in \mathscr C$, which exists because $\mathscr C$ is nonempty; since $\mathcal G_0 \in \mathscr P$, we have $\mathcal F \subseteq \mathcal G_0 \subseteq \mathcal H$. Second, $\mathcal H$ is a proper filter on $I$. Since each $\mathcal G \in \mathscr C$ is a filter, $I \in \mathcal G$ for every $\mathcal G \in \mathscr C$, hence $I \in \mathcal H$. Also $\varnothing \notin \mathcal H$, because if $\varnothing \in \mathcal H$, then $\varnothing \in \mathcal G$ for some $\mathcal G \in \mathscr C$, contradicting that $\mathcal G$ is proper. Let $A, B \in \mathcal H$. Then there exist $\mathcal G_A, \mathcal G_B \in \mathscr C$ such that $A \in \mathcal G_A$ and $B \in \mathcal G_B$. Since $\mathscr C$ is a chain, either $\mathcal G_A \subseteq \mathcal G_B$ or $\mathcal G_B \subseteq \mathcal G_A$. In the first case, $A, B \in \mathcal G_B$, so $A \cap B \in \mathcal G_B \subseteq \mathcal H$. In the second case, $A, B \in \mathcal G_A$, so $A \cap B \in \mathcal G_A \subseteq \mathcal H$. Thus $\mathcal H$ is closed under finite intersections. Finally, let $A \in \mathcal H$ and let $B \subseteq I$ satisfy $A \subseteq B$. Choose $\mathcal G_A \in \mathscr C$ with $A \in \mathcal G_A$. Since $\mathcal G_A$ is a filter on $I$, upward closure gives $B \in \mathcal G_A \subseteq \mathcal H$. Hence $\mathcal H$ is upward closed. Therefore $\mathcal H$ is a proper filter on $I$ containing $\mathcal F$, so $\mathcal H \in \mathscr P$. Since every $\mathcal G \in \mathscr C$ satisfies $\mathcal G \subseteq \mathcal H$, the family $\mathcal H$ is an upper bound for $\mathscr C$ in $\mathscr P$.[/step]

[guided]We must verify the chain condition needed for Zorn's lemma. The ordered set is $\mathscr P$, whose elements are proper filters extending the original filter $\mathcal F$. Let $\mathscr C \subseteq \mathscr P$ be a chain. If $\mathscr C$ is empty, then any element of $\mathscr P$ is an upper bound for it; in particular $\mathcal F \in \mathscr P$ is an upper bound. Now suppose $\mathscr C$ is nonempty. The natural candidate for an upper bound is the union of all filters in the chain: \begin{align*} \mathcal H := \bigcup_{\mathcal G \in \mathscr C} \mathcal G. \end{align*} This union certainly contains every $\mathcal G \in \mathscr C$, so it will be an upper bound once we prove that it is itself an element of $\mathscr P$. First, $\mathcal H$ contains $\mathcal F$. Choose $\mathcal G_0 \in \mathscr C$. Since $\mathcal G_0 \in \mathscr P$, it contains $\mathcal F$, and since $\mathcal G_0 \subseteq \mathcal H$, we get $\mathcal F \subseteq \mathcal H$. Next we check the filter axioms. Since every member of $\mathscr C$ is a filter, every member contains $I$, and therefore $I \in \mathcal H$. The union is also proper: if $\varnothing \in \mathcal H$, then $\varnothing$ belongs to some $\mathcal G \in \mathscr C$, contradicting that $\mathcal G$ is a proper filter. The only point that requires the chain hypothesis is closure under finite intersections. Let $A, B \in \mathcal H$. Then $A$ lies in some filter $\mathcal G_A \in \mathscr C$, and $B$ lies in some filter $\mathcal G_B \in \mathscr C$. Since $\mathscr C$ is linearly ordered by inclusion, one of these filters contains the other. Hence both $A$ and $B$ lie in a single proper filter from the chain. That filter is closed under finite intersections, so $A \cap B$ lies in that filter, and therefore $A \cap B \in \mathcal H$. Finally, upward closure is inherited from the filter that contains the smaller set. If $A \in \mathcal H$, $B \subseteq I$, and $A \subseteq B$, then $A \in \mathcal G_A$ for some $\mathcal G_A \in \mathscr C$. Since $\mathcal G_A$ is a filter, $B \in \mathcal G_A$, hence $B \in \mathcal H$. Thus $\mathcal H$ is a proper filter on $I$ containing $\mathcal F$, so $\mathcal H \in \mathscr P$. It contains every element of the chain, hence it is an upper bound for $\mathscr C$.[/guided]

[step:Apply Zorn's lemma to obtain a maximal proper filter extension] Every chain in $\mathscr P$ has an upper bound in $\mathscr P$. Therefore, by Zorn's lemma (citing a result not yet in the wiki: Zorn's Lemma), there exists a maximal element $\mathcal U \in \mathscr P$ with respect to inclusion. By the definition of $\mathscr P$, the family $\mathcal U$ is a proper filter on $I$ and satisfies $\mathcal F \subseteq \mathcal U$. [/step]

[step:Identify the maximal proper filter as an ultrafilter] An ultrafilter on $I$ is a maximal proper filter on $I$ with respect to inclusion. The maximality of $\mathcal U$ in $\mathscr P$ means that there is no proper filter $\mathcal V$ on $I$ such that \begin{align*} \mathcal U \subsetneq \mathcal V. \end{align*} Indeed, if such a proper filter $\mathcal V$ existed, then $\mathcal F \subseteq \mathcal U \subsetneq \mathcal V$, so $\mathcal V \in \mathscr P$, contradicting the maximality of $\mathcal U$ in $\mathscr P$. Therefore $\mathcal U$ is a maximal proper filter on $I$, hence an ultrafilter on $I$. Since $\mathcal F \subseteq \mathcal U$, this is the required ultrafilter extension of $\mathcal F$. [/step]