TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a set and let $F_{0}$ be a filter on $X$ (Filter on a set). Then there is an ultrafilter $U$ on $X$ (Ultrafilter) with $F_{0} \subseteq U$ .

The hypothesis is spent exactly once, through Zorn's lemma at step 4.1; the rest of the argument is a theorem of ZF.

In particular, every set that carries a filter carries an ultrafilter. The proof uses Zorn's lemma (Zorn's lemma) and therefore the Axiom of Choice. That some choice principle is unavoidable here, if ZF is consistent, is an external independence result, not proved in this library; see the remarks below.

Facts & Assumptions

Given: A set $X$ , a filter $F_{0}$ on $X$ , and Zorn's lemma.

[A1]

$F_{0}$ is a filter on $X$ : $X \in F_{0}$ , $\emptyset \notin F_{0}$ , and $F_{0}$ is closed under pairwise intersection and upward in $X$ (Filter on a set).

[L1]

Zorn's lemma, which assumes the Axiom of Choice: a nonempty poset in which every chain has an upper bound has a maximal element; the hypothesis is about all chains, the empty chain included (Zorn's lemma, The Axiom of Choice).

[L2]

The union of a nonempty inclusion-chain of filters on $X$ is a filter on $X$ and an upper bound of the chain (The union of a nonempty chain of filters is a filter).

[L3]

An ultrafilter on $X$ is a filter that is a maximal element of $(F i l t (X), \subseteq)$ , and $m$ is maximal exactly when $m \leq x$ forces $x = m$ (Ultrafilter, Maximal element and greatest element).

[L4]

Inclusion partially orders any set of sets, and a chain is a subset any two of whose members are comparable, the empty set included (Partial order and partially ordered set, Chain in a poset); an element $u$ of a poset is an upper bound of a subset $S$ when $s \leq u$ for every $s \in S$ (Upper bound, least upper bound, and strict upper bound).

Proof

technique · direct

1.1

Let $F i l t (X)$ be the set of filters on $X$ , a subset of $P (P (X))$ , partially ordered by inclusion.

L3L4construct

1.2

$F_{0} \in F i l t (X)$ .

2.1

Let $P = {F \in F i l t (X) : F_{0} \subseteq F}$ , partially ordered by inclusion as a subset of $F i l t (X)$ .

step 1.1construct

2.2

$P \neq \emptyset$ , since $F_{0} \subseteq F_{0}$ puts $F_{0}$ in $P$ .

step 1.2L4

3.1

The empty chain of $P$ has an upper bound in $P$ : every element of $P$ is vacuously above all of its members, and $P$ is nonempty, so $F_{0}$ is such an upper bound.

step 2.1step 2.2L4

3.2

A nonempty chain $D \subseteq P$ has an upper bound in $P$ : $⋃ D$ is a filter on $X$ and an upper bound of $D$ ; it contains $F_{0}$ because some member of $D$ does and every member is contained in the union; hence $⋃ D \in P$ .

step 2.1L2L4

4.1

Every chain of $P$ has an upper bound in $P$ , and $P$ is a nonempty poset, so Zorn's lemma yields a maximal element $U$ of $P$ .

step 2.2step 3.1step 3.2L1

5.1

$U$ is a filter on $X$ with $F_{0} \subseteq U$ , since $U \in P$ .

step 4.1step 2.1

6.1

Let $G$ be any filter on $X$ with $U \subseteq G$ ; then $F_{0} \subseteq U \subseteq G$ , so $G \in P$ , and maximality of $U$ in $P$ forces $G = U$ .

step 4.1step 5.1step 2.1L3

7.1

So no filter on $X$ strictly contains $U$ : $U$ is a maximal element of $F i l t (X)$ , that is an ultrafilter on $X$ , and it contains $F_{0}$ .

step 5.1step 6.1L3∎

Remarks

The empty chain is the load-bearing case. Zorn's lemma requires an upper bound for every chain of $P$ , and the empty chain has one exactly when $P$ is nonempty. Step 3.1 discharges it with $F_{0}$ itself, and step 2.2 is what makes that possible. This is not a formality: the conclusion of The union of a nonempty chain of filters is a filter fails for the empty chain, since $⋃ \emptyset = \emptyset$ is not a filter, which is exactly why that lemma assumes its chain nonempty; so a proof that says "the union of a chain is a filter" without the case split has a genuine gap at the one chain the hypothesis of Zorn's lemma is easiest to forget.
Why the poset is $P$ and not $F i l t (X)$ . Applying Zorn to $F i l t (X)$ would produce a maximal filter unrelated to $F_{0}$ . Restricting to the filters above $F_{0}$ costs nothing, because step 6.1 transfers maximality back: a filter above $U$ is automatically above $F_{0}$ , so maximality in $P$ already means maximality among all filters. That transfer is the only step where the shape of $P$ is used.
The conclusion is an ultrafilter, not the ultrafilter. Zorn's lemma delivers a maximal element and no construction, so nothing in the proof distinguishes the ultrafilter it produces from any other filter above $F_{0}$ ; the statement asserts existence and no uniqueness whatever. FALSE, once the ultrafilter lemma is available: every ultrafilter is principal runs the argument on the filter of tails of $N$ and obtains a single free ultrafilter, which it can describe no further. How many ultrafilters extend a given filter is a separate counting question, and this library neither states nor uses an answer to it.
Combined with A family lies in a filter exactly when it has the finite intersection property, the lemma takes its most usable form: every family of subsets of $X$ with the finite intersection property is contained in an ultrafilter on $X$ . That is the version applied in topology and in model theory.
What this costs. The proof buys the conclusion with the full Axiom of Choice, through Zorn's lemma and The Axiom of Choice and Zorn's lemma are equivalent, but the statement is, if ZF is consistent, strictly weaker than the Axiom of Choice: it is then neither provable in ZF nor strong enough to recover choice. Both of those are external metamathematical results, conditional on the consistency of ZF, and are recorded, with references and without a claim to prove them, in The proved choice cost of the ultrafilter lemma.

Depends on

Used by

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ultrafilter lemma (Wikipedia) (standard reference, not scraped)
Boolean prime ideal theorem (Wikipedia) (standard reference, not scraped)
Zorn's lemma (Wikipedia) (standard reference, not scraped)
Ultrafilter (Wikipedia) (standard reference, not scraped)
N. Strickland, Notes on Ultrafilters (standard reference, not scraped)