• Thumbnail for Ultrafilter
    In the mathematical field of order theory, an ultrafilter on a given partially ordered set (or "poset") P {\textstyle P} is a certain subset of P , {\displaystyle...
    20 KB (2,956 words) - 14:53, 25 July 2024
  • Thumbnail for Ultrafilter on a set
    In the mathematical field of set theory, an ultrafilter on a set X {\displaystyle X} is a maximal filter on the set X . {\displaystyle X.} In other words...
    47 KB (7,377 words) - 04:20, 9 April 2024
  • Thumbnail for Almost everywhere
    property true almost everywhere is sometimes defined in terms of an ultrafilter. An ultrafilter on a set X is a maximal collection F of subsets of X such that:...
    9 KB (1,261 words) - 12:11, 1 July 2024
  • ideals. A variation of this statement for filters on sets is known as the ultrafilter lemma. Other theorems are obtained by considering different mathematical...
    15 KB (2,257 words) - 03:04, 29 November 2023
  • principal ultrafilter on X . {\displaystyle X.} Moreover, every principal ultrafilter on X {\displaystyle X} is necessarily of this form. The ultrafilter lemma...
    6 KB (823 words) - 05:16, 4 October 2024
  • to the following criterion: assuming the ultrafilter lemma, a space is compact if and only if each ultrafilter on the space converges. With this in hand...
    15 KB (2,094 words) - 09:49, 19 July 2024
  • configurations in the X n {\displaystyle X_{n}} spaces employing an ultrafilter to bypass the need for repeated consideration of subsequences to ensure...
    14 KB (2,462 words) - 09:48, 17 May 2024
  • Thumbnail for Hyperreal number
    be extended to an ultrafilter, but the proof uses the axiom of choice. The existence of a nontrivial ultrafilter (the ultrafilter lemma) can be added...
    33 KB (4,899 words) - 05:38, 27 September 2024
  • then the ultrafilter U witnessing that κ is measurable will be in Vκ+2 and thus in M. So for any α < κ, we have that there exist an ultrafilter U in j(Vκ)...
    2 KB (271 words) - 06:29, 4 March 2024
  • Thumbnail for Boolean algebra (structure)
    ultrafilter is called the ultrafilter lemma and cannot be proven in Zermelo–Fraenkel set theory (ZF), if ZF is consistent. Within ZF, the ultrafilter...
    49 KB (3,356 words) - 02:25, 17 September 2024