Editing Meagre set (section)

==Examples==

The empty set is always a closed nowhere dense (and thus meagre) subset of every topological space. 

In the nonmeagre space <math>X=[0,1]\cup([2,3]\cap\Q)</math> the set <math>[2,3]\cap\Q</math> is meagre.  The set <math>[0,1]</math> is nonmeagre and comeagre.

In the nonmeagre space <math>X=[0,2]</math> the set <math>[0,1]</math> is nonmeagre.  But it is not comeagre, as its complement <math>(1,2]</math> is also nonmeagre.

A countable [[T1 space|T<sub>1</sub> space]] without [[isolated point]] is meagre.  So it is also meagre in any space that contains it as a subspace.  For example, <math>\Q</math> is both a meagre subspace of <math>\R</math> (that is, meagre in itself with the subspace topology induced from <math>\R</math>) and a meagre subset of <math>\R.</math>

The [[Cantor set]] is nowhere dense in <math>\R</math> and hence meagre in <math>\R.</math>  But it is nonmeagre in itself, since it is a [[complete metric space]].

The set <math>([0,1]\cap\Q)\cup\{2\}</math> is not nowhere dense in <math>\R</math>, but it is meagre in <math>\R</math>.  It is nonmeagre in itself (since as a subspace it contains an isolated point).

The line <math>\R\times\{0\}</math> is meagre in the plane <math>\R^2.</math>  But it is a nonmeagre subspace, that is, it is nonmeagre in itself.

The set <math>S = (\Q \times \Q) \cup (\Reals \times \{0\})</math> is a meagre sub{{em|set}} of <math>\R^2</math> even though its meagre subset <math>\Reals \times \{0\}</math> is a nonmeagre sub{{em|space}} (that is, <math>\R</math> is not a meagre topological space).{{sfn|Narici|Beckenstein|2011|pp=371-423}} 
A countable Hausdorff space without [[isolated point]]s is meagre, whereas any topological space that contains an isolated point is nonmeagre.{{sfn|Narici|Beckenstein|2011|pp=371-423}} 
Because the [[rational numbers]] are countable, they are meagre as a subset of the reals and as a space—that is, they do not form a [[Baire space]]. 

Any topological space that contains an [[isolated point]] is nonmeagre{{sfn|Narici|Beckenstein|2011|pp=371-423}} (because no set containing the isolated point can be nowhere dense). In particular, every nonempty [[discrete space]] is nonmeagre.

There is a subset <math>H</math> of the real numbers <math>\R</math> that splits every nonempty open set into two nonmeagre sets.  That is, for every nonempty open set <math>U\subseteq \mathbb{R}</math>, the sets <math>U\cap H</math> and <math>U \setminus H</math> are both nonmeagre.

In the space <math>C([0,1])</math> of continuous real-valued functions on <math>[0,1]</math> with the topology of [[uniform convergence]], the set <math>A</math> of continuous real-valued functions on <math>[0,1]</math> that have a derivative at some point is meagre.<ref>{{cite journal|author=Banach, S.|title=Über die Baire'sche Kategorie gewisser Funktionenmengen|journal=[[Studia Math.]]|volume=3|issue=1|year=1931|pages=174–179|doi=10.4064/sm-3-1-174-179|url=https://eudml.org/doc/217560|doi-access=free}}</ref>{{sfn|Willard|2004|loc=Theorem 25.5}}  Since <math>C([0,1])</math> is a complete metric space, it is nonmeagre.  So the complement of <math>A</math>, which consists of the continuous real-valued [[nowhere differentiable function]]s on <math>[0,1],</math> is comeagre and nonmeagre.  In particular that set is not empty.  This is one way to show the existence of continuous nowhere differentiable functions.

On an infinite-dimensional Banach space, there exists a [[discontinuous linear functional]] whose kernel is nonmeagre.<ref name="subspace of Banach">{{cite web | url=https://mathoverflow.net/questions/3188/are-proper-linear-subspaces-of-banach-spaces-always-meager | title=Are proper linear subspaces of Banach spaces always meager? }}</ref> Also, under [[Martin's axiom]], on each separable Banach space, there exists a discontinuous linear functional whose kernel is meagre (this statement disproves the Wilansky–Klee conjecture<ref> https://www.ams.org/journals/bull/1966-72-04/S0002-9904-1966-11547-1/S0002-9904-1966-11547-1.pdf</ref>).<ref name="subspace of Banach" />