# Meagre set

In the mathematical fields of general topology and descriptive set theory, a meagre set (also called a meager set or a set of first category) is a set that, considered as a subset of a (usually larger) topological space, is in a precise sense small or negligible. A topological space T is called meagre if it is a meager subset of itself; otherwise, it is called nonmeagre.

The meagre subsets of a fixed space form a σ-ideal of subsets; that is, any subset of a meagre set is meagre, and the union of countably many meagre sets is meagre. General topologists use the term Baire space to refer to a broad class of topological spaces on which the notion of meagre set is not trivial (in particular, the entire space is not meagre). Descriptive set theorists mostly study meagre sets as subsets of the real numbers, or more generally any Polish space, and reserve the term Baire space for one particular Polish space.

The complement of a meagre set is a comeagre set or residual set. A set that is not meagre is called nonmeagre and is said to be of the second category. Note that the notions of a comeagre set and a nonmeagre set are not equivalent.

## Definition

Throughout, ${\displaystyle X}$ will be a topological space.

A subset ${\displaystyle B\subseteq X}$ of a topological space ${\displaystyle X}$ is called nowhere dense and rare in ${\displaystyle X}$ if its closure has empty interior. Equivalently, ${\displaystyle B}$ is nowhere dense in ${\displaystyle X}$ if for each open set ${\displaystyle U\subseteq X,}$ the set ${\displaystyle B\cap U}$ is not dense in ${\displaystyle U.}$

A closed subset of ${\displaystyle X}$ is nowhere dense in ${\displaystyle X}$ if and only if its topological interior in ${\displaystyle X}$ is empty.

A subset of a topological space ${\displaystyle X}$ is said to be meagre in ${\displaystyle X,}$ a meagre subset of ${\displaystyle X,}$ and of the first category in ${\displaystyle X}$ if it is a countable union of nowhere dense subsets of ${\displaystyle X.}$ A subset that is not of first category in ${\displaystyle X}$ is said to be nonmeagre in ${\displaystyle X,}$ a nonmeagre subset of ${\displaystyle X,}$ and of the second category in ${\displaystyle X.}$

A topological space is said to be meagre (respectively, nonmeagre) if it is a meagre (respectively, nonmeagre) subset of itself.

Warning: If ${\displaystyle S\subseteq X}$ is a subset of ${\displaystyle X}$ then ${\displaystyle S}$ being a "meagre subspace" of ${\displaystyle X}$ means that when ${\displaystyle S}$ is endowed with the subspace topology (induced on it by ${\displaystyle X}$) then ${\displaystyle S}$ is a meagre topological space (that is, ${\displaystyle S}$ is a meagre subset of ${\displaystyle S}$). In contrast, ${\displaystyle S}$ being a "meagre subset" of ${\displaystyle X}$ means that ${\displaystyle S}$ is equal to a countable union of nowhere dense subsets of ${\displaystyle X.}$ The same warning applies to nonmeagre subsets versus nonmeagre subspaces. More details on how to tell these notions apart (and why the slight difference in these terms is reasonable) are given in this footnote.[note 1]

For example, if ${\displaystyle S:=\mathbb {N} }$ is the set of all positive integers then ${\displaystyle S}$ is a meager subset of ${\displaystyle \mathbb {R} }$ but not a meager subspace of ${\displaystyle \mathbb {R} .}$ If ${\displaystyle x\in X}$ is not an isolated point of a T1 space ${\displaystyle X}$ (meaning that ${\displaystyle \{x\}}$ is not an open subset of ${\displaystyle X}$) then ${\displaystyle \{x\}}$ is a meager subspace of ${\displaystyle X}$ but not a meager subset of ${\displaystyle X.}$

A subset ${\displaystyle A\subseteq X}$ is called a residual subset of ${\displaystyle X}$ and is said to be comeagre in ${\displaystyle X}$ if its complement ${\displaystyle X\setminus A}$ is meagre in ${\displaystyle X}$ (this use of the prefix "co" is consistent with its use in other terms such as "cofinite"). A subset is comeagre in ${\displaystyle X}$ if and only if it is equal to an intersection of countably many sets, each of whose topological interior is a dense subset of ${\displaystyle X.}$

Importantly, being of the second category is not the same as being comeagre — a set may be neither meagre nor comeagre (in this case it will be of second category).

## Sufficient conditions

Every Baire space is nonmeagre but there exist nonmeagre spaces that are not Baire spaces.[1] Since complete (pseudo)metric spaces as well as Hausdorff locally compact spaces are Baire spaces, they are also nonmeagre spaces.[1]

Any subset of a meagre set is a meagre set, as is the union of countably many meagre sets.[2] If ${\displaystyle h:X\to X}$ is a homeomorphism then a subset ${\displaystyle S\subseteq X}$ is meagre if and only if ${\displaystyle h(S)}$ is meagre.[2]

Every nowhere dense subset is a meagre set.[2] Consequently, any closed subset of ${\displaystyle X}$ whose interior in ${\displaystyle X}$ is empty is of the first category of ${\displaystyle X}$ (that is, it is a meager subset of ${\displaystyle X}$). Thus a closed subset of ${\displaystyle X}$ that is of the second category in ${\displaystyle X}$ must have non-empty interior in ${\displaystyle X.}$[3]

Any topological space that contains an isolated point (such as any non-empty discrete space) is nonmeagre.[1]

Comeagre subset

Any superset of a comeagre set is comeagre, as is the intersection of countably many comeagre sets (because countable union of countable sets is countable).

## Examples

Every non-empty discrete space is nonmeagre because this is true of any topological space that contains one or more isolated points.[1]

Meagre subsets and subspaces

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

A singleton subset ${\displaystyle \{x\}\subseteq X}$ is always a nonmeagre subspace of ${\displaystyle X}$ (that is, it is a nonmeagre topological space). If ${\displaystyle x}$ is an isolated point of ${\displaystyle X}$ (meaning that ${\displaystyle \{x\}}$ is an open subset) then ${\displaystyle \{x\}}$ is also a nonmeagre subset of ${\displaystyle X}$; the converse holds if ${\displaystyle X}$ is a T1 space.

The set ${\displaystyle S=(\mathbb {Q} \times \mathbb {Q} )\cup \mathbb {R} }$ is a meagre subset of ${\displaystyle \mathbb {R} ^{2}}$ even though ${\displaystyle \mathbb {R} }$ is a nonmeagre subspace (that is, ${\displaystyle \mathbb {R} }$ is not a meagre topological space).[1] A countable Hausdorff space without isolated points is meagre, whereas any topological space that contains an isolated point is nonmeagre.[1] 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.

The Cantor set is meagre as a subset of the reals, but not as a subset of itself, since it is a complete metric space and is thus a Baire space, by the Baire category theorem.

The Smith–Volterra–Cantor set is a closed nowhere dense (and thus meagre) subset of the unit interval ${\displaystyle [0,1]}$ that has positive Lebesgue measure.

Function spaces

The set of functions that have a derivative at some point is a meagre set in the space of all continuous functions.[4]

## Properties

Banach category theorem[5] — In any space ${\displaystyle X,}$ the union of any countable family of open sets of the first category is of the first category.

If ${\displaystyle B\subseteq X}$ is of the second category in ${\displaystyle X}$ and if ${\displaystyle S_{1},S_{2},\ldots }$ are subsets of ${\displaystyle X}$ such that ${\displaystyle B\subseteq S_{1}\cup S_{2}\cup \cdots }$ then at least one ${\displaystyle S_{n}}$ is of the second category in ${\displaystyle X.}$

A closed subset of ${\displaystyle X}$ that is of the second category in ${\displaystyle X}$ must have non-empty interior in ${\displaystyle X}$[3] (because otherwise it would be nowhere dense and thus of the first category).

A nonmeagre locally convex topological vector space is a barreled space.[1]

### Meagre subsets and Lebesgue measure

A meagre set need not have measure zero. There exist nowhere dense subsets (which are thus meagre subsets) that have positive Lebesgue measure.[1]

### Relation to Borel hierarchy

Just as a nowhere dense subset need not be closed, but is always contained in a closed nowhere dense subset (viz, its closure), a meagre set need not be an ${\displaystyle F_{\sigma }}$ set (countable union of closed sets), but is always contained in an ${\displaystyle F_{\sigma }}$ set made from nowhere dense sets (by taking the closure of each set).

Dually, just as the complement of a nowhere dense set need not be open, but has a dense interior (contains a dense open set), a comeagre set need not be a ${\displaystyle G_{\delta }}$ set (countable intersection of open sets), but contains a dense ${\displaystyle G_{\delta }}$ set formed from dense open sets.

## Banach–Mazur game

Meagre sets have a useful alternative characterization in terms of the Banach–Mazur game. Let ${\displaystyle Z}$ be a topological space, ${\displaystyle {\mathcal {W}}}$ be a family of subsets of ${\displaystyle Z}$ that have nonempty interiors such that every nonempty open set has a subset belonging to ${\displaystyle {\mathcal {W}},}$ and ${\displaystyle Z}$ be any subset of ${\displaystyle Z.}$ Then there is a Banach–Mazur game corresponding to ${\displaystyle X,{\mathcal {W}},Z.}$ In the Banach–Mazur game, two players, ${\displaystyle P}$ and ${\displaystyle Q,}$ alternately choose successively smaller elements of ${\displaystyle {\mathcal {W}}}$ to produce a sequence ${\displaystyle W_{1}\supseteq W_{2}\supseteq W_{3}\supseteq \cdots .}$ Player ${\displaystyle P}$ wins if the intersection of this sequence contains a point in ${\displaystyle X}$; otherwise, player ${\displaystyle Q}$ wins.

Theorem — For any ${\displaystyle {\mathcal {W}}}$ meeting the above criteria, player ${\displaystyle Q}$ has a winning strategy if and only if ${\displaystyle X}$ is meagre.

1. ^ This distinction between "subspace" and "subset" is a consequence of the fact that in general topology, the word "space" means "topological space", which is a pair ${\displaystyle (X,\tau )}$ consisting of a set and topology, and (similarly) the word "subspace" means "topological subspace"; consequently, "subspace of ${\displaystyle X}$" refers to the pair consisting of the subset together with the subspace topology that it inherits from ${\displaystyle X}$ whereas "subset of ${\displaystyle X}$" refers only to the set. Consequently, if the subset ${\displaystyle S}$ lacks any topology then "${\displaystyle S}$ is meagre of subset of ${\displaystyle S}$" is not well-defined, leaving "${\displaystyle S}$ is a meagre subset of ${\displaystyle X}$" as the only possible meaning of "${\displaystyle S}$ is meagre". But if ${\displaystyle S}$ is endowed with a topology then (by definition) "${\displaystyle S}$ is meagre" means "${\displaystyle S}$ is a meagre subset of ${\displaystyle S.}$" Saying "${\displaystyle S}$ is a meagre subspace of ${\displaystyle X}$" is just a combination of the following two statements: (1) "${\displaystyle S}$ is a subspace of ${\displaystyle X}$", which by definition means that ${\displaystyle S}$ is endowed with a topology that is equal to the subspace topology induced by on it by ${\displaystyle X}$ (denote this topology by ${\displaystyle \tau _{S}}$), and (2) "${\displaystyle S}$ is a meagre space", which by definition means "${\displaystyle S}$ is a meagre subset of ${\displaystyle \left(S,\tau _{S}\right)}$". However, if ${\displaystyle S}$ happens to be endowed with a topology (say ${\displaystyle \tau _{S}}$) then the statement "${\displaystyle S}$ is a meagre subset of ${\displaystyle X}$" does not mean "${\displaystyle S}$ is a meagre subset of ${\displaystyle \left(S,\tau _{S}\right)}$" because in this statement, ${\displaystyle S}$ is being considered as a set (and not as a topological space). The same is true of a statement such as "let ${\displaystyle S}$ be a subspace of ${\displaystyle X}$ that is a meagre subset of ${\displaystyle X}$" and its more succinct equivalent "let ${\displaystyle S}$ be a subspace that is meagre in ${\displaystyle X}$" (note that the meaning is completely changed without the words "in ${\displaystyle X}$").