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Meagre set
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==Characterizations and sufficient conditions== Every nonempty [[Baire space]] is nonmeagre. In particular, by the [[Baire category theorem]] every nonempty [[complete metric space]] and every nonempty [[locally compact Hausdorff]] space is nonmeagre. Every nonempty [[Baire space]] is nonmeagre but there exist nonmeagre spaces that are not Baire spaces.{{sfn|Narici|Beckenstein|2011|pp=371-423}} Since [[Complete metric space|complete]] [[Pseudometric space|(pseudo)]][[metric space]]s as well as [[Hausdorff space|Hausdorff]] [[locally compact]] spaces [[Baire category theorem|are Baire spaces]], they are also nonmeagre spaces.{{sfn|Narici|Beckenstein|2011|pp=371β423}} Any subset of a meagre set is a meagre set, as is the union of countably many meagre sets.{{sfn|Rudin|1991|p=43}} If <math>h : X \to X</math> is a [[homeomorphism]] then a subset <math>S \subseteq X</math> is meagre if and only if <math>h(S)</math> is meagre.{{sfn|Rudin|1991|p=43}} Every nowhere dense subset is a meagre set.{{sfn|Rudin|1991|p=43}} Consequently, any closed subset of <math>X</math> whose interior in <math>X</math> is empty is of the first category of <math>X</math> (that is, it is a meager subset of <math>X</math>). The {{visible anchor|'''Banach category theorem'''}}{{sfn|Oxtoby|1980|p=62}} states that in any space <math>X,</math> the union of any family of open sets of the first category is of the first category. All subsets and all countable unions of meagre sets are meagre. Thus the meagre subsets of a fixed space form a [[Sigma-ideal|Ο-ideal]] of subsets, a suitable notion of [[negligible set]]. Dually, all [[superset]]s and all countable intersections of comeagre sets are comeagre. Every superset of a nonmeagre set is nonmeagre. Suppose <math>A \subseteq Y \subseteq X,</math> where <math>Y</math> has the [[subspace topology]] induced from <math>X.</math> The set <math>A</math> may be meagre in <math>X</math> without being meagre in <math>Y.</math> However the following results hold:{{sfn|Bourbaki|1989|p=192}} * If <math>A</math> is meagre in <math>Y,</math> then <math>A</math> is meagre in <math>X.</math> * If <math>Y</math> is open in <math>X,</math> then <math>A</math> is meagre in <math>Y</math> if and only if <math>A</math> is meagre in <math>X.</math> * If <math>Y</math> is dense in <math>X,</math> then <math>A</math> is meagre in <math>Y</math> if and only if <math>A</math> is meagre in <math>X.</math> And correspondingly for nonmeagre sets: * If <math>A</math> is nonmeagre in <math>X,</math> then <math>A</math> is nonmeagre in <math>Y.</math> * If <math>Y</math> is open in <math>X,</math> then <math>A</math> is nonmeagre in <math>Y</math> if and only if <math>A</math> is nonmeagre in <math>X.</math> * If <math>Y</math> is dense in <math>X,</math> then <math>A</math> is nonmeagre in <math>Y</math> if and only if <math>A</math> is nonmeagre in <math>X.</math> In particular, every subset of <math>X</math> that is meagre in itself is meagre in <math>X.</math> Every subset of <math>X</math> that is nonmeagre in <math>X</math> is nonmeagre in itself. And for an open set or a dense set in <math>X,</math> being meagre in <math>X</math> is equivalent to being meagre in itself, and similarly for the nonmeagre property. A topological space <math>X</math> is nonmeagre if and only if every countable intersection of dense open sets in <math>X</math> is nonempty.{{sfn|Willard|2004|loc=Theorem 25.2}}
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