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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Ultrafilter ： ウィキペディア英語版 Ultrafilter In the mathematical field of set theory, an ultrafilter is a maximal filter, that is, a filter that cannot be enlarged. Filters and ultrafilters are special subsets of partially ordered sets. Ultrafilters can also be defined on Boolean algebras and sets: * An ultrafilter on a poset ''P'' is a maximal filter on ''P''. * An ultrafilter on a Boolean algebra ''B'' is an ultrafilter on the poset of non-zero elements of ''B''. * An ultrafilter on a set ''X'' is an ultrafilter on the Boolean algebra of subsets of ''X''. Rather confusingly, an ultrafilter on a poset ''P'' or Boolean algebra ''B'' is a subset of ''P'' or ''B'', while an ultrafilter on a set ''X'' is a collection of subsets of ''X''. Ultrafilters have many applications in set theory, model theory, and topology. An ultrafilter on a set ''X'' has some special properties. For example, given any subset ''A'' of ''X'', the ultrafilter must contain either ''A'' or its complement . In addition, an ultrafilter on a set ''X'' may be considered as a finitely additive measure. In this view, every subset of ''X'' is either considered "almost everything" (has measure 1) or "almost nothing" (has measure 0). ==Formal definition for ultrafilter on a set == Given a set ''X'', an ultrafilter on ''X'' is a set ''U'' consisting of subsets of ''X'' such that #The empty set is not an element of ''U'' #If ''A'' and ''B'' are subsets of ''X'', ''A'' is a subset of ''B'', and ''A'' is an element of ''U'', then ''B'' is also an element of ''U''. #If ''A'' and ''B'' are elements of ''U'', then so is the intersection of ''A'' and ''B''. #If ''A'' is a subset of ''X'', then either ''A'' or ''X'' \ ''A'' is an element of ''U''. (Note: axioms 1 and 3 imply that ''A'' and cannot ''both'' be elements of ''U''.) A characterization is given by the following theorem. A filter ''U'' on a set ''X'' is an ultrafilter if any of the following conditions are true: #There is no filter ''F'' finer than ''U'', i.e., $U\backslash subseteq\; F$ implies ''U'' = ''F''. #$A\backslash cup\; B\backslash in\; U$ implies $A\backslash in\; U$ or $B\backslash in\; U$. #$\backslash forall\; A\backslash subseteq\; X\backslash colon\; A\backslash in\; U$ or $X\backslash setminus\; A\; \backslash in\; U$. Another way of looking at ultrafilters on a set ''X'' is to define a function ''m'' on the power set of ''X'' by setting ''m''(''A'') = 1 if ''A'' is an element of ''U'' and ''m''(''A'') = 0 otherwise. Such a function is called a 2-valued morphism. Then ''m'' is a finitely additive measure on ''X'', and every property of elements of ''X'' is either true almost everywhere or false almost everywhere. Note that this does not define a measure in the usual sense, which is required to be ''countably additive''. For a filter ''F'' that is not an ultrafilter, one would say ''m''(''A'') = 1 if ''A'' ∈ ''F'' and ''m''(''A'') = 0 if ''X'' \ ''A'' ∈ ''F'', leaving ''m'' undefined elsewhere. A simple example of an ultrafilter is a ''principal ultrafilter'', which consists of subsets of ''X'' that contain a given element ''x'' of ''X''. All ultrafilters on a finite set are principal. ==Completeness== The completeness of an ultrafilter ''U'' on a set is the smallest cardinal κ such that there are κ elements of ''U'' whose intersection is not in ''U''. The definition implies that the completeness of any ultrafilter is at least $\backslash aleph\_0$. An ultrafilter whose completeness is ''greater'' than $\backslash aleph\_0$—that is, the intersection of any countable collection of elements of ''U'' is still in ''U''—is called countably complete or $\backslash sigma$-complete. The completeness of a countably complete nonprincipal ultrafilter on a set is always a measurable cardinal. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Ultrafilter」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.