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## Summary

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 ${\textstyle \bigcup _{n}A_{n}}$ 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 intersection of countably many dense open sets in $X$  is dense in $X$ ;
2. Every union of countably many closed subsets of $X$  with empty interior has empty interior;
3. Every non-empty open subset of $X$  is a nonmeager subset of $X$ ;
4. Every comeagre subset of $X$  is dense in $X$ ;
5. 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;
6. 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 balanced 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 $\mathbb {R}$ .
• Another large class of Baire spaces are algebraic varieties with the Zariski topology. For example, the space $\mathbb {C} ^{n}$  of complex numbers whose open sets are complements of the vanishing sets of polynomials $f\in \mathbb {C} [x_{1},\ldots ,x_{n}]$  is an algebraic variety with the Zariski topology. Usually this is denoted $\mathbb {A} ^{n}$ .
• 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 $0$ :
$\bigcap _{m=1}^{\infty }\bigcup _{n=1}^{\infty }\left(r_{n}-\left({\tfrac {1}{2}}\right)^{n+m},r_{n}+\left({\tfrac {1}{2}}\right)^{n+m}\right)$
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.