Is this correct Baire's Theorem?

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In summary, Baire's Theorem states that in a complete metric space, if a set E is equal to the union of closed subsets F_n and all complements of F_n are dense, then the complement of E is dense. This theorem has two versions, one for complete metric spaces and one for compact Hausdorff spaces. This is a property of topological spaces and not just limited to complete metric spaces. The statement of this theorem may vary depending on the lecturer, so it is important to be aware of the specific assumptions and conditions in each case.
  • #1
julypraise
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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.
 
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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.)
 

FAQ: Is this correct Baire's Theorem?

1. What is Baire's Theorem?

Baire's Theorem is a mathematical theorem that states that if a complete metric space is expressed as a countable union of closed sets with empty interior, then at least one of the sets has to have a non-empty interior.

2. How is Baire's Theorem used in mathematics?

Baire's Theorem is used in various areas of mathematics, including functional analysis, topology, and measure theory. It is often used to prove the existence of solutions to certain equations or to show that certain spaces are not separable.

3. Is Baire's Theorem difficult to understand?

Baire's Theorem can be challenging to understand, especially for those who are not familiar with advanced mathematical concepts. However, with the right background knowledge and practice, it can be comprehended and applied effectively.

4. Are there any variations of Baire's Theorem?

Yes, there are several variations of Baire's Theorem, including the Baire Category Theorem and the Baire-Hausdorff Theorem. These variations have slightly different conditions and conclusions but are all related to the concept of countable intersections of dense sets.

5. Can Baire's Theorem be extended to other types of spaces?

Yes, Baire's Theorem can be extended to other types of spaces, such as Banach spaces, Polish spaces, and locally compact spaces. However, the conditions and conclusions may vary depending on the type of space.

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