Baire space

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In mathematics, a topological space ${\displaystyle X}$ is said to be a Baire space, if for any given countable collection ${\displaystyle \{A_{n}\}}$ of closed sets with empty interior in ${\displaystyle X}$, their union ${\displaystyle \cup A_{n}}$ also has empty interior in ${\displaystyle X}$.^[1] Equivalently, a locally convex space which is not meagre in itself is called a Baire space.^[2] According to Baire category theorem, compact Hausdorff spaces and complete metric spaces are examples of a Baire space.^[3] Bourbaki coined the term "Baire space".^[4]

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 ${\displaystyle \mathbb {R} ,}$ 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 ${\displaystyle X}$ is called a Baire space if it satisfies any of the following equivalent conditions:

1. Every non-empty open subset of ${\displaystyle X}$ is a nonmeager subset of ${\displaystyle X}$;^[5]
2. Every comeagre subset of ${\displaystyle X}$ is dense in ${\displaystyle X}$;
3. The union of any countable collection of closed nowhere dense subsets (i.e. each closed subset has empty interior) has empty interior;^[5]
4. Every intersection of countably many dense open sets in ${\displaystyle X}$ is dense in ${\displaystyle X}$;^[5]
5. Whenever the union of countably many closed subsets of ${\displaystyle X}$ has an interior point, then at least one of the closed subsets must have an interior point;
6. Every point in ${\displaystyle X}$ has a neighborhood that is a Baire space (according to any defining condition other than this one).^[5]
   • So ${\displaystyle X}$ 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) Every complete pseudometric space is a Baire space.^[5] More generally, every topological space that is homeomorphic to an open subset of a complete pseudometric space is a Baire space. In particular, every completely metrizable space is a Baire space.
• (BCT2) Every locally compact Hausdorff space (or more generally every locally compact sober space) is a Baire space.

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

• The space ${\displaystyle \mathbb {R} }$ of real numbers
• The space of irrational numbers, which is homeomorphic to the Baire space ${\displaystyle \omega ^{\omega }}$ of set theory
• Every compact Hausdorff space is a Baire space.
  • In particular, the Cantor set is a Baire space.
• Indeed, every Polish 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.^[5]
• A topological vector space is nonmeagre if and only if it is a Baire space,^[5] which happens if and only if every closed absorbing subset has non-empty interior.^[6]

Examples[]

• The space ${\displaystyle \mathbb {R} }$ 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 ${\displaystyle \mathbb {R} }$.
• The Cantor set is a Baire space, and so is of second category in itself, but it is of first category in the interval ${\displaystyle [0,1]}$ with the usual topology.
• Here is an example of a set of second category in ${\displaystyle \mathbb {R} }$ with Lebesgue measure ${\displaystyle 0}$:

  ${\displaystyle \bigcap _{m=1}^{\infty }\bigcup _{n=1}^{\infty }\left(r_{n}-({\tfrac {1}{2}})^{n+m},r_{n}+({\tfrac {1}{2}})^{n+m}\right)}$

  where ${\displaystyle \left(r_{n}\right)_{n=1}^{\infty }}$ 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 ${\displaystyle \mathbb {Q} }$ inside of the real line ${\displaystyle \mathbb {R} }$ with the standard euclidean topology. Given an indexing of the rationals by the natural numbers ${\displaystyle \mathbb {N} }$ so a bijection ${\displaystyle f:\mathbb {N} \to \mathbb {Q} ,}$ and let ${\displaystyle {\mathcal {A}}=\left(A_{n}\right)_{n=1}^{\infty }}$ where ${\displaystyle A_{n}:=\mathbb {Q} \setminus \{f(n)\},}$ which is an open, dense subset in ${\displaystyle \mathbb {Q} .}$ Then, because the intersection of every open set in ${\displaystyle {\mathcal {A}}}$ is empty, the space ${\displaystyle \mathbb {Q} }$ 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 ${\displaystyle X}$ 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 ${\displaystyle [0,1].}$
• Every open subspace of a Baire space is a Baire space.
• Given a family of continuous functions ${\displaystyle f_{n}:X\to Y}$= with pointwise limit ${\displaystyle f:X\to Y.}$ If ${\displaystyle X}$ is a Baire space then the points where ${\displaystyle f}$ is not continuous is a meagre set in ${\displaystyle X}$ and the set of points where ${\displaystyle f}$ is continuous is dense in ${\displaystyle X.}$ 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[]

• Baire space (set theory)
• Banach–Mazur game
• Barrelled space
• Descriptive set theory
• Meagre set
• Nowhere dense set
• Property of Baire
• Blumberg theorem

Citations[]

References[]

External links[]

• Encyclopaedia of Mathematics article on Baire space
• Encyclopaedia of Mathematics article on Baire theorem