Proving Interior of S is NOT Empty: Baire's Theorem

  • Thread starter robertdeniro
  • Start date
  • Tags
    Analysis
In summary, we are given a vector space V and a subset S of V with the property that for all v in V, there exists a positive integer p where v/p is in S. We are asked to prove that the interior of S is not empty, given that S is closed and has this property. To do so, we can define closed sets S1, S2, etc. such that the interior of their union is not equal to 0. However, defining Sn as {v/n|v in S} may not be valid as every element of Sn is also in S, resulting in the union being equal to S and the interior of S being 0.
  • #1
robertdeniro
39
0

Homework Statement


let V be a vector space and S be a subset of V with the following property
for all v in V, there exist some positive integer p such that v/p is in S

given: S is closed and S has the property as described above. prove interior of S is NOT empty


Homework Equations





The Attempt at a Solution


since S is closed, we can define other closed sets S1, S2... such that the int(Sn)=0 and int(S U S1 U S2 U...)=/=0. once we have this we can apply baire's theorem.

problem is how do i define S1, S2,...
 
Physics news on Phys.org
  • #2
? i don't think it is legal to define Sn={v/n|v in S} because every element of Sn is also in S, which would make S U S1 U S2 U...=S and int(S)=0
 

Related to Proving Interior of S is NOT Empty: Baire's Theorem

What is Baire's Theorem?

Baire's Theorem is a fundamental theorem in topology that states that in a complete metric space, a countable intersection of dense open sets is also dense. This theorem is often used in analysis and functional analysis to prove the existence of certain mathematical objects.

How does Baire's Theorem relate to proving the interior of a set is not empty?

Baire's Theorem can be used to prove that the interior of a set is not empty by showing that the complement of the set is not dense. If the complement is not dense, then the set must have non-empty interior.

Can Baire's Theorem be used to prove that a set is open?

No, Baire's Theorem only guarantees the existence of a non-empty interior. It does not guarantee that the interior is open. Other techniques such as showing that a set is the countable intersection of open sets can be used to prove that a set is open.

Is Baire's Theorem a necessary condition for a set to have non-empty interior?

No, Baire's Theorem is not a necessary condition. There are other techniques and theorems that can be used to prove that a set has non-empty interior. However, Baire's Theorem is a powerful tool that is frequently used in mathematical proofs.

What are some examples of using Baire's Theorem to prove the interior of a set is not empty?

One common example is using Baire's Theorem to prove the existence of a nowhere-dense set. Another example is using Baire's Theorem to prove the existence of a nowhere-continuous function. These examples demonstrate the power of Baire's Theorem in proving the non-emptiness of a set's interior.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
594
  • Calculus and Beyond Homework Help
Replies
1
Views
772
  • Calculus and Beyond Homework Help
Replies
4
Views
834
  • Calculus and Beyond Homework Help
Replies
1
Views
545
  • Calculus and Beyond Homework Help
Replies
0
Views
466
  • Calculus and Beyond Homework Help
Replies
1
Views
619
  • Calculus and Beyond Homework Help
Replies
3
Views
541
  • Calculus and Beyond Homework Help
Replies
1
Views
668
Replies
2
Views
413
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top