Encyclopedia > Ultrafilter lemma

  Article Content

Ultrafilter Lemma

Redirected from Ultrafilter lemma

An ultrafilter is a maximal filter -- for every set, either that set or its complement is in the filter. The Ultrafilter Lemma says that every filter is a subset of some ultrafilter.

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 non-trivial 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 Zermelo-Fraenkel axioms) alone, and it cannot be used to prove the axiom of choice, so it is properly weaker.



All Wikipedia text is available under the terms of the GNU Free Documentation License

 
  Search Encyclopedia

Search over one million articles, find something about almost anything!
 
 
  
  Featured Article
List of Quercus species

... - Live oak[?] Quercus wislizenii - Interior live oak[?] Section: Leucobalanus, the white oaks Subsection: white oaks Quercus alba - White oak Quercus arizonica - ...