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  "body": "\n## Statement\n\nLet $B$ be a [[Boolean algebra|0XADJKM699]], and let $F$ be a proper filter of $B$.\nThen there is an [[ultrafilter|CQM84EW10G]] $U$ of $B$ with $F \\subseteq U$.\n\n## Remarks\n\nProof idea: consider the proper filters that contain $F$, ordered by inclusion.\nThis family is nonempty, since it contains $F$.\nEvery nonempty chain of them has an upper bound, its union, which is still a proper filter.\nBy [[Zorn's lemma|04RXTQA5ZA]] there is a maximal one, and it is an ultrafilter.\nApplied to the power set of a set $X$, the lemma says that every filter on $X$ lies in an [[ultrafilter|J64Y3JC0RE]] on $X$.\nFor an infinite $X$ and the filter of cofinite sets, this gives non-principal ultrafilters.\nApplied 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]].\nThe 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.\n\n## Explanations\n\n### L1 human\n\nA filter is a partial, consistent way of saying which elements are true.\nThe lemma says that such a partial decision can always be completed to a full decision about every element, without contradiction.\nIts proof uses a form of the axiom of choice, because the completion is built by infinitely many free choices.\n\n### L2 human\n\nThe ultrafilter lemma is the form of choice that much of algebra and topology needs.\nIt proves the compactness theorem of propositional logic, Tychonoff's theorem for compact Hausdorff spaces, and the representation of Boolean algebras as fields of sets.\nFor a countable Boolean algebra no choice is needed: one lists the elements and decides them one at a time.\n",
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