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## From Wikipedia, the free encyclopedia

In mathematics, a topological space $X$ is said to be a Baire space, if for any given countable collection $\{A_{n}\}$ of closed sets with empty interior in $X$ , their union $\cup A_{n}$ also has empty interior in $X$ . Equivalently, a locally convex space which is not meagre in itself is called a Baire space. According to Baire category theorem, compact Hausdorff spaces and complete metric spaces are examples of a Baire space. Bourbaki coined the term "Baire space".

## Motivation

In an arbitrary topological space, the class of closed sets with empty interior consists precisely of the boundaries of dense open sets. These sets are, in a certain sense, "negligible". Some examples are finite sets in $\mathbb {R} ,$ smooth curves in the plane, and proper affine subspaces in a Euclidean space. If a topological space is a Baire space then it is "large", meaning that it is not a countable union of negligible subsets. For example, the three-dimensional Euclidean space is not a countable union of its affine planes.

## Definition

The precise definition of a Baire space has undergone slight changes throughout history, mostly due to prevailing needs and viewpoints. A topological space $X$ is called a Baire space if it satisfies any of the following equivalent conditions:

1. Every non-empty open subset of $X$ is a nonmeager subset of $X$ ;
2. Every comeagre subset of $X$ is dense in $X$ ;
3. The union of any countable collection of closed nowhere dense subsets (i.e. each closed subset has empty interior) has empty interior;
4. Every intersection of countably many dense open sets in $X$ is dense in $X$ ;
5. The interior (taken in $X$ ) of every union of countably many closed nowhere dense sets is empty;
6. Whenever the union of countably many closed subsets of $X$ has an interior point, then at least one of the closed subsets must have an interior point;
7. The complement in $X$ of every meagre subset of $X$ is dense in $X$ ;
8. Every point in $X$ has a neighborhood that is a Baire space (according to any defining condition other than this one).
• So $X$ is a Baire space if and only if it is "locally a Baire space."

## Sufficient conditions

### Baire category theorem

The Baire category theorem gives sufficient conditions for a topological space to be a Baire space. It is an important tool in topology and functional analysis.

BCT1 shows that each of the following is a Baire space:

BCT2 shows that every manifold is a Baire space, even if it is not paracompact, and hence not metrizable. For example, the long line is of second category.

### Other sufficient conditions

• A product of complete metric spaces is a Baire space.
• A topological vector space is nonmeagre if and only if it is a Baire space, which happens if and only if every closed absorbing subset has non-empty interior.

## Examples

• The space $\mathbb {R}$ of real numbers with the usual topology, is a Baire space, and so is of second category in itself. The rational numbers are of first category and the irrational numbers are of second category in $\mathbb {R}$ .
• The Cantor set is a Baire space, and so is of second category in itself, but it is of first category in the interval $[0,1]$ with the usual topology.
• Here is an example of a set of second category in $\mathbb {R}$ with Lebesgue measure $0$ :
$\bigcap _{m=1}^{\infty }\bigcup _{n=1}^{\infty }\left(r_{n}-({\tfrac {1}{2}})^{n+m},r_{n}+({\tfrac {1}{2}})^{n+m}\right)$ where $\left(r_{n}\right)_{n=1}^{\infty }$ is a sequence that enumerates the rational numbers.
• Note that the space of rational numbers with the usual topology inherited from the real numbers is not a Baire space, since it is the union of countably many closed sets without interior, the singletons.

### Non-example

One of the first non-examples comes from the induced topology of the rationals $\mathbb {Q}$ inside of the real line $\mathbb {R}$ with the standard euclidean topology. Given an indexing of the rationals by the natural numbers $\mathbb {N}$ so a bijection $f:\mathbb {N} \to \mathbb {Q} ,$ and let ${\mathcal {A}}=\left(A_{n}\right)_{n=1}^{\infty }$ where $A_{n}:=\mathbb {Q} \setminus \{f(n)\},$ which is an open, dense subset in $\mathbb {Q} .$ Then, because the intersection of every open set in ${\mathcal {A}}$ is empty, the space $\mathbb {Q}$ cannot be a Baire space.

## Properties

• Every non-empty Baire space is of second category in itself, and every intersection of countably many dense open subsets of $X$ is non-empty, but the converse of neither of these is true, as is shown by the topological disjoint sum of the rationals and the unit interval $[0,1].$ • Every open subspace of a Baire space is a Baire space.
• Given a family of continuous functions $f_{n}:X\to Y$ = with pointwise limit $f:X\to Y.$ If $X$ is a Baire space then the points where $f$ is not continuous is a meagre set in $X$ and the set of points where $f$ is continuous is dense in $X.$ A special case of this is the uniform boundedness principle.
• A closed subset of a Baire space is not necessarily Baire.
• The product of two Baire spaces is not necessarily Baire. However, there exist sufficient conditions that will guarantee that a product of arbitrarily many Baire spaces is again Baire.

## See also

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