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----
-title: Baire Spaces
-parent: General Topology
-nav_order: 1
-description: >
-  A Baire space is a topological space with the property that the intersection
-  of countably many dense open subsets is still dense. One version of the Baire
-  Category Theorem states that complete metric spaces are Baire spaces. We give
-  a self-contained proof of Baire's Category Theorem by contradiction.
-# spellchecker:words
----
-
-# 
-
-
-A topological space is said to be a *Baire space*,
-if any of the following equivalent conditions holds:
->
-- The intersection of countably many dense open subsets is still dense.
-- The union of countably many closed subsets with empty interior has empty interior.
-
-
-Note that
-a set is dense in a topological space
-if and only if
-its complement has empty interior.
-
-Any sufficient condition
-for a topological space to be a Baire space
-constitutes a *Baire Category Theorem*,
-of which there are several.
-Here we give one
-that is commonly used in functional analysis.
-
-
-Baire Category Theorem
-
->
-Complete metric spaces are Baire spaces.
-
-**Proof:**
-Let C-C-C be a metric space
-with complete metric D-D-D.
-Suppose that F-F-F is not a Baire space.
-Then there is a countable collection G-G-G of dense open subsets of B-B-B
-such that the intersection C-C-C is not dense in D-D-D.
-
-In a metric space, any nonempty open set contains an open ball.
-It is also true, that any nonempty open set contains a closed ball,
-since F-F-F if G-G-G.
-
-We construct a sequence B-B-B of open balls C-C-C satisfying
-
-V-V-V
-as follows: By hypothesis,
-the interior of D-D-D is not empty (otherwise F-F-F would be dense in G-G-G),
-so we may choose an open ball B-B-B with C-C-C
-such that D-D-D.
-Given F-F-F,
-the set G-G-G is nonempty, because B-B-B is dense in C-C-C,
-and it is open, because D-D-D and F-F-F are open.
-This allows us to choose an open ball G-G-G as desired.
-
-Note that by construction B-B-B for C-C-C,
-thus D-D-D.
-Therefore, the sequence F-F-F is Cauchy
-and has a limit point G-G-G by completeness.
-In the limit B-B-B, we obtain C-C-C (strictness is lost),
-hence D-D-D for all F-F-F.
-This shows that G-G-G for all B-B-B, that is C-C-C.
-On the other hand, D-D-D,
-in contradiction to the preceding statement.
-