Is this correct Baire's Theorem?

  • Thread starter Thread starter julypraise
  • Start date Start date
  • Tags Tags
    Theorem
Click For Summary

Homework Help Overview

The discussion revolves around Baire's Theorem in the context of complete metric spaces. The original poster seeks to verify the correctness of a specific formulation of the theorem, which involves the density of sets derived from closed subsets.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the theorem, discussing the relationship between dense sets and the properties of complete metric spaces. There is a focus on whether the original poster's understanding aligns with established interpretations of Baire's Theorem.

Discussion Status

The conversation is ongoing, with participants providing insights into the nature of Baire's Theorem and its applications. Some guidance has been offered regarding the equivalence of certain statements related to the theorem, but there is no explicit consensus on the correctness of the original poster's formulation.

Contextual Notes

Participants note potential ambiguities in the original poster's understanding, particularly regarding assumptions that may not have been explicitly stated in lectures, such as the necessity for certain sets to be non-empty.

julypraise
Messages
104
Reaction score
0

Homework Statement



Baire's Theorem
Let [itex]X[/itex] be a complete metric space. Suppose [itex]E \subseteq X[/itex] and

[tex]E = \bigcup_{n \in \mathbb{N}} F_{n}[/tex]

where [itex]F_{n} \subseteq X[/itex] is closed in [itex]X[/itex]. If all [itex]X \backslash F_{n}[/itex] are dense then [itex]X \backslash E[/itex] is dense.

Homework Equations


The Attempt at a Solution



Nothing much...

I know there may be a stronger version. But at this stage, all I need to do is to check this theorem is correct.
 
Physics news on Phys.org
Looks like Baire to me. You've got a countable intersection of dense open sets, right? And you are stating the result must be dense since a complete metric space is Baire. Don't you agree?
 
Actually, there are two Baire theorems who state exactly the same thing. One deals with complete metric spaces, the other deals with compact Hausdorff spaces.
 
micromass said:
Actually, there are two Baire theorems who state exactly the same thing. One deals with complete metric spaces, the other deals with compact Hausdorff spaces.

Sure. Being "Baire" is a property of a topological space. Complete metric spaces aren't the only example of Baire spaces. The OP indicated it was probably a special case.
 
Dick said:
Looks like Baire to me. You've got a countable intersection of dense open sets, right? And you are stating the result must be dense since a complete metric space is Baire. Don't you agree?

Yes the statement that you statetd, i.e., the intersection of open dense subsets is also dense, is equivalent to mine.

But is the theorem correct then?

(I've learned this from lectures and the lecturer sometimes does not specify everything like a set should not be empty or etc. So I worry about this theorem too and I've used this in my assignment too.)
 

Similar threads

Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K