Redirected from Ultrafilter lemma
Proving the lemma from the axiom of choice is an application of Zorn's Lemma, and is fairly standard as these things go. The partial ordering is simply that of a subset. The nontrivial part is proving that a maximal filter contains every set or its complement. Let us say F contains neither A nor X \ A. From maximality, that means there is a set B in F such that the intersection of A and B is empty (otherwise, the union of F and {A} would generate a filter). Likewise, there is a C such that the intersection of C and X \ A is empty. The intersection of C and B (let us call it D) is in F. D has empty intersection with both A and X \ A, so it has an empty intersection with X, so it is empty. But a filter cannot contain an empty set. QED
This proof uses Zorn's Lemma, which is equivalent to the axiom of choice. The Ultrafilter Lemma cannot be proven from ZF (the ZermeloFraenkel axioms) alone, and it cannot be used to prove the axiom of choice, so it is properly weaker.
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