proof for one equivalent statement of Baire category theorem

First, let’s assume Baire’s category theoremMathworldPlanetmath and prove the alternative statement.
We have B=n=1Bn, with int(Bk¯)=k𝐍.


Then X-Bk¯ is dense in X for every k. Besides, X-Bk¯ is open because X is open and Bk¯ closed. So, by Baire’s Category Theorem, we have that


is dense in X. But Bn=1Bn¯X-n=1Bn¯X-B, and then X=X-n=1Bn¯¯X-B¯=X-int(B)int(B)=.

Now, let’s assume our alternative statement as the hypothesisMathworldPlanetmathPlanetmath, and let (Bk)kN be a collectionMathworldPlanetmath of open dense sets in a complete metric space X. Then int(X-Bk¯)=int(X-int(Bk))=int(X-Bk)=X-Bk¯= and so X-Bk is nowhere dense for every k.

Then X-n=1Bn¯=int(X-n=1Bn)=int(n=1X-Bn)= due to our hypothesis. Hence Baire’s category theorem holds.

Title proof for one equivalent statement of Baire category theorem
Canonical name ProofForOneEquivalentStatementOfBaireCategoryTheorem
Date of creation 2013-03-22 14:04:52
Last modified on 2013-03-22 14:04:52
Owner gumau (3545)
Last modified by gumau (3545)
Numerical id 8
Author gumau (3545)
Entry type Proof
Classification msc 54E52