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In set theory, a subset of a Polish space is ∞-Borel if it can be obtained by starting with the open subsets of , and transfinitely iterating the operations of complementation and wellordered union. Note that the set of ∞-Borel sets may not actually be closed under wellordered union; see below. ==Formal definition== More formally: we define by simultaneous transfinite recursion the notion of ∞-Borel code, and of the interpretation of such codes. Since is Polish, it has a countable base. Let enumerate that base (that is, is the basic open set). Now: * Every natural number is an ∞-Borel code. Its interpretation is . * If is an ∞-Borel code with interpretation , then the ordered pair is also an ∞-Borel code, and its interpretation is the complement of , that is, . * If is a length-α sequence of ∞-Borel codes for some ordinal α (that is, if for every β<α, is an ∞-Borel code, say with interpretation . Now a set is ∞-Borel if it is the interpretation of some ∞-Borel code. The axiom of choice implies that ''every'' set can be wellordered, and therefore that every subset of every Polish space is ∞-Borel. Therefore the notion is interesting only in contexts where AC does not hold (or is not known to hold). Unfortunately, without the axiom of choice, it is not clear that the ∞-Borel sets ''are'' closed under wellordered union. This is because, given a wellordered union of ∞-Borel sets, each of the individual sets may have ''many'' ∞-Borel codes, and there may be no way to choose one code for each of the sets, with which to form the code for the union. The assumption that every set of reals is ∞-Borel is part of AD+, an extension of the axiom of determinacy studied by Woodin. 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「Infinity-Borel set」の詳細全文を読む スポンサード リンク
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