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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.
+