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...

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...

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:...

ideals. A variation of this statement for filters on sets is known as the ultrafilter lemma. Other theorems are obtained by considering different mathematical...

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...

filter F {\displaystyle {\mathcal {F}}} on X {\displaystyle X} is an ultrafilter if, for every A ⊆ X , {\displaystyle A\subseteq X,} either A ∈ F {\displaystyle...

principal ultrafilter on X . {\displaystyle X.} Moreover, every principal ultrafilter on X {\displaystyle X} is necessarily of this form. The ultrafilter lemma...

element i ∈ I {\displaystyle i\in I} (all of the same signature), and an ultrafilter U {\displaystyle {\mathcal {U}}} on I . {\displaystyle I.} For any two...

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)...

ultrafilter is called the ultrafilter lemma and cannot be proven in Zermelo–Fraenkel set theory (ZF), if ZF is consistent. Within ZF, the ultrafilter...

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...

million 2002 - Acquired industrial parts maker Ultrafilter International AG for $68 million. Ultrafilter had sales of $100 million and operated in 30 countries...

configurations in the X n {\displaystyle X_{n}} spaces employing an ultrafilter to bypass the need for repeated consideration of subsequences to ensure...

the set of all ultrafilters on X , {\displaystyle X,} with the elements of X {\displaystyle X} corresponding to the principal ultrafilters. The topology...

Countable Uncountable Empty Inhabited Singleton Finite Infinite Transitive Ultrafilter Recursive Fuzzy Universal Universe constructible Grothendieck Von Neumann...

former is equivalent in ZF to Tarski's 1930 ultrafilter lemma: every filter is a subset of some ultrafilter. One of the most interesting aspects of the...

where the property holds is a member of the ultrafilter, i.e. has measure 1.) Also equivalent, the ultrafilter (set of sets of measure 1) is closed under...

} However, an ultrafilter (and any other non-degenerate filter) is free if and only if it includes the Fréchet filter. The ultrafilter lemma states that...

dual to a filter is an order ideal. Special cases of filters include ultrafilters, which are filters that cannot be enlarged, and describe nonconstructive...

proving that certain perfect sets are uncountable, and the construction of ultrafilters. Let X {\textstyle X} be a set and A {\textstyle {\mathcal {A}}} a nonempty...

to a point or a set, and many others. Special types of filters called ultrafilters have many useful technical properties and they may often be used in place...

Countable Uncountable Empty Inhabited Singleton Finite Infinite Transitive Ultrafilter Recursive Fuzzy Universal Universe constructible Grothendieck Von Neumann...

{\displaystyle X} is open. For every x ∈ X , {\displaystyle x\in X,} the fixed ultrafilter at x {\displaystyle x} converges only to x . {\displaystyle x.} For every...

size κ with A ∈ M, and a nonprincipal ultrafilter U on the Boolean algebra P(κ) ∩ M such that: U is an M-ultrafilter: for any sequence ⟨Xβ : β < κ⟩ ∈ M of...

provably equivalent to a weak form of the axiom of choice known as the ultrafilter lemma. In particular, no theory extending ZF can prove either the completeness...

Y} has a neighborhood on which f {\displaystyle f} is constant. Every ultrafilter U {\displaystyle {\mathcal {U}}} on a non-empty set X {\displaystyle...

choice, the uniqueness of the cardinality of the basis requires only the ultrafilter lemma, which is strictly weaker (the proof given below, however, assumes...

direction, a Boolean algebra A {\displaystyle A} has a unique non-principal ultrafilter (that is, a maximal filter not generated by a single element of the algebra)...

generally require the axiom of choice or a weaker form of it, such as the ultrafilter lemma. If V is a vector space over a field F, then: If L is a linearly...

{\displaystyle V_{j(f)(\kappa )}\subset N} . A Shelah cardinal has a normal ultrafilter containing the set of weakly hyper-Woodin cardinals below it. Ernest...