10.11 Baire’s Theorem and categories
The following is Baire’s category theorem; there are several equivalent statements.
[0VV]Suppose \((X,d)\) is complete.
Given countably many sets \(A_ n\) that are open and dense in \(X\), we have that \(⋂_ n A_ n\) is dense.
Given countably many sets \(C_ n\) closed with empty interior in \(X\), we have that \(⋃_ n C_ n\) has empty interior.
- E301
[0VX]A complete metric space \(X\) is second category in itself. Hidden solution: [UNACCESSIBLE UUID ’0VY’]
- E301
[0VZ]Given \(X=ℝ\), the set of irrational numbers is second category in \(ℝ\). Hidden solution: [UNACCESSIBLE UUID ’0W0’]
- E301
[0W1]Reflect on the statements:
A closed set \(C\) inside a complete metric space \((X,d)\) is complete (when viewed as a metric space \((C,d)\)).
The set \(C=\{ 0\} ∪\{ 1/n : n∈ℕ\} \) is closed in \(ℝ\), so \(C\) is complete with distance \(d(x,y)=|x-y|\).
\(C\) is composed of countably many points.
A singleton \(\{ x\} \) is a closed set with an empty internal part.
Why is there no contradiction?
Hidden solution: [UNACCESSIBLE UUID ’0W2’]
- E301
[0W3]Topics:perfect set.Prerequisites:1,117.
Suppose \((X,d)\) is a complete metric space. A closed set without isolated points, i.e. consisting only of accumulation points, is called a perfect set. Show that a non-empty perfect set \(E\) contained in \(X\) must be uncountably infinite. (Find a simple direct proof, using Baire’s Theorem 300.)
Hidden solution: [UNACCESSIBLE UUID ’2DZ’]
The Cantor set is a perfect set, see 1.