---
tag: 3MKZST79MK
kind: statement
role: theorem
name: Ultrafilter lemma
msc: ["03E25", "06E05"]
uses: [0XADJKM699, CQM84EW10G, 04RXTQA5ZA]
about: [CQM84EW10G]
sources:
  - url: https://en.wikipedia.org/wiki/Ultrafilter_on_a_set
---

## Statement

Let $B$ be a [[Boolean algebra|0XADJKM699]], and let $F$ be a proper filter of $B$.
Then there is an [[ultrafilter|CQM84EW10G]] $U$ of $B$ with $F \subseteq U$.

## Remarks

Proof idea: consider the proper filters that contain $F$, ordered by inclusion.
This family is nonempty, since it contains $F$.
Every nonempty chain of them has an upper bound, its union, which is still a proper filter.
By [[Zorn's lemma|04RXTQA5ZA]] there is a maximal one, and it is an ultrafilter.
Applied to the power set of a set $X$, the lemma says that every filter on $X$ lies in an [[ultrafilter|J64Y3JC0RE]] on $X$.
For an infinite $X$ and the filter of cofinite sets, this gives non-principal ultrafilters.
Applied to the filter generated by an element $a \ne 0$, it gives an ultrafilter containing $a$, which is the key step of [[Stone's representation theorem|S4QJ395PA5]].
The lemma is not provable in ZF alone, but it is strictly weaker than the [[axiom of choice|0HTGB13FE2]] over ZF; it is equivalent to the Boolean prime ideal theorem.

## Explanations

### L1 human

A filter is a partial, consistent way of saying which elements are true.
The lemma says that such a partial decision can always be completed to a full decision about every element, without contradiction.
Its proof uses a form of the axiom of choice, because the completion is built by infinitely many free choices.

### L2 human

The ultrafilter lemma is the form of choice that much of algebra and topology needs.
It proves the compactness theorem of propositional logic, Tychonoff's theorem for compact Hausdorff spaces, and the representation of Boolean algebras as fields of sets.
For a countable Boolean algebra no choice is needed: one lists the elements and decides them one at a time.