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Topic: Ultrafilter lemma

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	Ultrafilter
		Equivalently, an ultrafilter F on a set S is a filter on S with the additional property that for every subset A of S, either A is in F or S \ A is in F.
		One can show that every filter is contained in an ultrafilter (see Ultrafilter Lemma) and that free ultrafilters therefore exist, but the proofs involve the axiom of choice in the form of Zorn's Lemma, so explicit examples of free ultrafilters cannot be given.
		Ultrafilters are useful in topology, especially in relation to compact Hausdorff spaces.
	www.ebroadcast.com.au /lookup/encyclopedia/ul/Ultrafilter.html (264 words)

	Ultrafilter Lemma
		An ultrafilter is a maximal filter -- for every set, either that set or its complement is in the filter.
		Proving the lemma from the axiom of choice is an application of Zorn's Lemma, and is fairly standard as these things go.
		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.
	www.ebroadcast.com.au /lookup/encyclopedia/ul/Ultrafilter_lemma.html (225 words)

	Ultrafilter - Biocrawler (Site not responding. Last check: 2007-10-20)
		In this case, ultrafilters are characterized by containing, for each element a of the Boolean algebra, exactly one of the elements a and ¬a (the latter being the Boolean complement of a).
		Ultrafilters on sets are useful in topology, especially in relation to compact Hausdorff spaces, and in model theory in the construction of ultraproducts and ultrapowers.
		Ultrafilters on sets are used in the ultrapower construction of certain fields of hyperreal numbers.
	www.biocrawler.com /encyclopedia/Ultrafilter (756 words)

	Reference.com/Encyclopedia/Ultrafilter
		In the mathematical field of set theory, an ultrafilter on a set X is a collection of subsets of X that can be considered so "large" that they contain "almost all" elements of X.
		In order theory, an ultrafilter is a subset of a partially ordered set (a poset) which is maximal among all proper filters.
		An important special case of the concept occurs if the considered poset is a Boolean algebra, as in the case of an ultrafilter on a set (defined as a filter of the corresponding powerset).
	www.reference.com /browse/wiki/Ultrafilter (1159 words)

	Ultrafilter: Definition and Links by Encyclopedian.com (via CobWeb/3.1 planetlab2.cs.unc.edu) (Site not responding. Last check: 2007-10-20)
		...Ultrafilter Ultrafilter In mathematics, an ultrafilter is a maximal filter....Equivalently, an ultrafilter F on a set S is a filter on S with the additional property that for...
		Each...words, the fixed ultrafilter at x converges only to the points that x is topologically...
		Then an ultrafilter is picked in this Boolean algebra, which assigns values true/false...contains this ultrafilter, which can be understood as a model obtained by extending the old...
	www.encyclopedian.com.cob-web.org:8888 /ul/Ultrafilter.html (464 words)

	Boolean prime ideal theorem (via CobWeb/3.1 planetlab2.cs.unc.edu) (Site not responding. Last check: 2007-10-20)
		It is worth noting that for the special case where the Boolean algebra under consideration is a powerset with the subset ordering, the "maximal filter theorem" is called the #ultrafilter lemma.
		The ultrafilter lemma states that every filter on a set X is a subset of some ultrafilter on X (a maximal filter of nonempty subsets of X.) This lemma is most often used in the study of topology.
		The ultrafilter lemma is equivalent to the Boolean prime ideal theorem, with the equivalence provable in ZF set theory without the axiom of choice.
	www.danceage.com.cob-web.org:8888 /biography/sdmc_Ultrafilter_lemma (1882 words)

	Ultrafilter - Wikipedia, the free encyclopedia
		In the mathematical field of set theory, an ultrafilter on a set X is a collection of subsets of X that is a filter, that can not be enlarged (as a filter).
		If A is a subset of X, then either A or X\A is an element of the ultrafilter (here X\A is the relative complement of A in X; that is, the set of all elements of X that are not in A).
		is contained in an ultrafilter (see ultrafilter lemma) and that free ultrafilters therefore exist, but the proofs involve the axiom of choice in the form of Zorn's Lemma.
	en.wikipedia.org /wiki/Ultrafilter (1332 words)

	[No title] (Site not responding. Last check: 2007-10-20)
		It is customary to denote ultrafilters by small letters u, v, etc. Note that two z-ultrafilters u and v differ iff there are z-sets A 2 u and B 2 v with A \ B = ;.
		Lemma 7 Let F be a filter on a perfectly normal space X. Then a closed set C belongs to F00if and only if every closed neighborhood of C belongs to F.
		By Lemma 7, any closed neighborhood of an element of H00must be an element of H. We will obtain our contradiction by producing an open set W, whose closure is contained in U, such that W cannot be covered by finitely many H-antipolar sets.
	hopf.math.purdue.edu /Feldman-Wilce/fibdegen.txt (5513 words)

	Ultrafilter Lemma (via CobWeb/3.1 planetlab2.cs.unc.edu) (Site not responding. Last check: 2007-10-20)
		In mathematics, the Ultrafilter Lemma states that every filter is a subset of some ultrafilter, i.e.
		This satement is in fact an easy consequence of the Boolean prime ideal theorem that is commonly used in order theory.
		In contrast, this article will only treat the special case of ultrafilters on a set S, which can be characterized as those filters which, for every subset of S, contain either that set or its complement.
	publicliterature.org.cob-web.org:8888 /en/wikipedia/u/ul/ultrafilter_lemma.html (354 words)

	Amazon.com: "ultrafilter lemma": Key Phrase page (Site not responding. Last check: 2007-10-20)
		See all pages with references to ultrafilter lemma.
		The family of all supersets of { I }is an ultrafilter.
		Lemma 5.12 If W satisfy the properties 1-4, then it is dictatorial.
	www.amazon.com /phrase/ultrafilter-lemma (412 words)

	[No title]
		Extend to a filter and then an ultrafilter, D. Let { F } be a finite subset of G, and let’s see that prod(@b, b in I) / D = { F }.
		The trick is to take the quotient with the ultrafilter D which kills all elements of prod(@i, i) with infinitely many zeros, and sets 1/0 to 0 for all elements with finitely many zeros.
		We define an ultrafilter to be a principal ultrafilter if it is generated by a point.
	www.media.mit.edu /physics/pedagogy/babbage/texts/rt.doc (4909 words)

	Ultrafilter Lemma: Definition and Links by Encyclopedian.com (via CobWeb/3.1 planetlab2.cs.unc.edu) (Site not responding. Last check: 2007-10-20)
		The Ultrafilter Lemma says that every filter is a subset of some ultrafilter....filter.
		...is contained in an ultrafilter (see Ultrafilter Lemma) and that free ultrafilters therefore exist, but the proofs...examples of free ultrafilters cannot be given.
		A standard application of Zorn 's lemma shows that every filter...
	www.encyclopedian.com.cob-web.org:8888 /ul/Ultrafilter-lemma.html (369 words)

	Theory FreeUltrafilter (Isabelle repository version) (Site not responding. Last check: 2007-10-20)
		lemma (in FreeUltrafilter) Ultrafilter: "Ultrafilter F" by (rule Ultrafilter.intro) subsection {* Maximal filter = ultrafilter } text { A filter F is an ultrafilter iff it is a maximal filter, i.e.
		qed subsubsection {* Existence of free ultrafilter } lemma* (in UFT) max_cofinite_Filter_Ex: "∃U∈superfrechet.
		Ultrafilter F; - A ∉ F ] ==> A ∈ F
	www.cse.ogi.edu /~brianh/isabelle/Star/FreeUltrafilter.html (544 words)

	Dynamics = Algebra
		in Section 1 is just the ultrafilter iteration of the shift map, i.e., of the universal dynamical system.
		The property of ultrafilters, ``uniformly recurrent and proximal to 0,'' which played a key role in the proof of Theorem 5, has alternative algebraic descriptions that will be useful later.
		When we refer to an idempotent ultrafilter as minimal, we mean with respect to this ordering.
	at.yorku.ca /b/a/a/f/13.l2h/node3.htm (723 words)

	Topology MAT 530
		An ultrafilter is a filter such that any subset is either "big" or "small" (i.e., the complement is "big").
		Although we do not have any examples of nonprincipal ultrafilters, using the Zorn lemma, it is not hard to prove that nontrivial ultrafilters exist on every infinite set.
		The Urysohn lemma states that for a normal topological space X and two disjoint closed subsets A and B of it, there exists a continuous function from X to [0,1] that is 0 on A and 1 on B.
	www.math.sunysb.edu /~azinger/mat530/fall04/index.htm (2907 words)

	An Extension of the Lemma of Rasiowa and Sikorski (ResearchIndex) (Site not responding. Last check: 2007-10-20)
		An Extension of the Lemma of Rasiowa and Sikorski (ResearchIndex)
		An Extension of the Lemma of Rasiowa and Sikorski (1997)
		An ultrafilter U preserves a = V E, if a = 2 U implies e = 2 U for some e 2 E: Since a = E implies 0 = fe " a j e 2 Eg; the Lemma of Rasiowa and Sikorski (cf.
	citeseer.ist.psu.edu /flum97extension.html (390 words)

	Theory Filter (Isabelle2005: October 2005)
		Q n} ∈ F)" by (subst Collect_conj_eq, rule Int_iff) lemma (in ultrafilter) Collect_not: "({n.
		P n} ∉ F)" by (subst Collect_neg_eq, rule Compl_iff) lemma (in ultrafilter) Collect_disj: "({n.
		Q n} ∈ F)" by (subst Collect_disj_eq, rule Un_iff) lemma (in ultrafilter) Collect_all: "({n.
	www.cl.cam.ac.uk /Research/HVG/Isabelle/dist/library/HOL/HOL-Complex/Filter.html (725 words)

	Dr Benedikt Loewe: Advanced Topics in Set Theory (1st Semester 2006/2007)
		Existence of nonprincipal ultrafilters on ω and σ-complete ultrafilters.
		In class we proved that in the 'ultrafilter game' (U is a nonprincipal ultrafilter on ω, players play disjoint finite sets of natural numbers, if I's set is in U then I wins), if player I has a winning strategy then player II has a winning strategy.
		Keep in mind that we already proved a lemma that said that if player II has a winning strategy, then player II has a strategy that can enforce that his move is in U.
	staff.science.uva.nl /~bloewe/2006-07-I/ATST.html (819 words)

	Vector space information - Search.com
		Using Zorn’s Lemma (which is equivalent to the axiom of choice), it can be proved that every vector space has a basis.
		Using the ultrafilter lemma (which is strictly weaker than the axiom of choice), one can show that all bases for a given vector space have the same cardinality.
		Thus vector spaces over a given field are fixed up to isomorphism by a single cardinal number (called the dimension of the vector space) representing the size of the basis.
	domainhelp.search.com /reference/Vector_space (1666 words)

	Theory HyperDef (Isabelle repository version) (Site not responding. Last check: 2007-10-20)
		Y ∈ FreeUltrafilterNat" by (auto, ultra) subsection{Properties of @{term hyprel}} text{Proving that @{term hyprel} is an equivalence relation} lemma hyprel_iff: "((X,Y) ∈ hyprel) = ({n.
		X n - Y n})" by (simp add: hypreal_diff_def hypreal_minus hypreal_add) lemma hypreal_add_minus [simp]: "z + -z = (0::hypreal)" by (cases z, simp add: hypreal_zero_def hypreal_minus hypreal_add) lemma hypreal_add_minus_left: "-z + z = (0::hypreal)" by (simp add: hypreal_add_commute) subsection{Hyperreal Multiplication} lemma hypreal_mult_congruent2: "congruent2 hyprel hyprel (%X Y. hyprel``{%n.
		Use assumption that member @{term FreeUltrafilterNat} is not finite.} text{A few lemmas first} lemma* lemma_omega_empty_singleton_disj: "{n::nat.
	www.cse.unsw.edu.au /~kleing/Isabelle-Library/HOL-Complex/HyperDef.html (1469 words)

	Vector space - ExampleProblems.com (Site not responding. Last check: 2007-10-20)
		A linearly independent set whose span is the whole space is called a basis.
		All bases for a given vector space have the same cardinality by the ultrafilter lemma (a weakened version of the axiom of choice).
		Using Zorn’s Lemma, it can be proved that every vector space has a basis, and vector spaces over a given field are fixed up to isomorphism by a single cardinal number (called the dimension of the vector space) representing the size of the basis.
	www.exampleproblems.com /wiki/index.php/Vector_space (1115 words)

	An uncountable well-ordered subset of the hyperreals
		I have to prove that there exists an uncountable well-ordered subset of the hyperreals.
		An ultrafilter on N is called free if it contains no singleton.
		R is the set of functions from the naturals to the reals, or equivalently, the set of real-valued (denumerable) sequences.
	www.physicsforums.com /showthread.php?threadid=145222 (722 words)

	Axiom of choice - Wikipedia (via CobWeb/3.1 planetlab2.cs.unc.edu) (Site not responding. Last check: 2007-10-20)
		There are also a remarkable number of concepts that are equivalent to the axiom of choice, among them the well-ordering principle and Zorn's lemma.
		It is important to note most results which seem to need the axiom of choice really need a weaker version -- the Ultrafilter Lemma.
		It follows from the axiom of choice, but is properly weaker: there are models of ZFC with the Ultrafilter Lemma but where there are sets which do not have choice functions.
	nostalgia.wikipedia.org.cob-web.org:8888 /wiki/AxiomOfChoice (929 words)

	Ultrafilter Lemma (Site not responding. Last check: 2007-10-20)
		In mathematics, the Ultrafilter Lemma states that every filter (mathematics) is a subset of some ultrafilter, i.e.
		This statement is in fact an easy consequence of the Boolean prime ideal theorem that is commonly used in order theory.
		The Ultrafilter Lemma cannot be proven from Zermelo-Fraenkel set theory (the Zermelo-Fraenkel axioms) alone, and it cannot be used to prove the axiom of choice, so it is properly weaker.
	read-and-go.hopto.org /Order-theory/Ultrafilter-Lemma.html (333 words)

	First-order Model Theory (Stanford Encyclopedia of Philosophy)
		A variant of this lemma is used in the proof of the elementary amalgamation theorem.
		If we have an ultrafilter U over I, then we can construct a reduced product from C by making two elements of C equivalent if and only if the set of indices at which they are equal is a set in the ultrafilter U.
		If U is an ultrafilter then the diagonal map from A to U-prod A is an elementary embedding.
	plato.stanford.edu /entries/modeltheory-fo (6168 words)

	Amazon.com: "generic ultrafilter": Key Phrase page (Site not responding. Last check: 2007-10-20)
		G whenever X E M and X C G. A routine verification (see Exercise 14.10) shows that G is a generic ultrafilter if and only if G is a generic filter on B+.
		Let H be a M-generic ultrafilter on B. It is well-known that M[H] K w1 < b = 2W.
		It appears that we have found a D-generic ultrafilter in P: Clearly, G n D, 54 0 for each a < A. If p E G and p <...
	www.amazon.com /phrase/generic-ultrafilter (588 words)

	[No title]
		Proof: If X is compact then every ultrafilter can be extended to a convergent filter, but ultrafilters don't have proper extensions so it must have converged.
		If every ultrafilter converges, you can extend any F to an ultrafilter, so every filter has a convergent extension.
		Lemma: Let G, H be topological groups and f : G -> H be a continuous group homomorphism.
	br.endernet.org /~loner/topology/kitfiltersandtopology.txt (4404 words)

	Seminar on Filters and Topology - EFnetMath
		The existence of free ultrafilters is independent of ZFC.
		[14:03] (Kit) Proof: If X is compact then every ultrafilter can be extended to a convergent filter, but ultrafilters don't have proper extensions so it must have converged.
		Let F be an ultrafilter which isn't cauchy.
	www.efnet-math.org /w/Seminar_on_Filters_and_Topology (5287 words)

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