Is Compactness Necessary for Countable Objects?

  • Context: Undergrad 
  • Thread starter Thread starter kidsasd987
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the necessity of compactness for countable objects, particularly in the context of open coverings and the implications of the compactness theorem. Participants explore the definitions and properties of open sets, epsilon-intervals, and the relationship between compact sets and their coverings.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question how the sets ##H_i## are defined and assert that they are part of an open covering.
  • There is a discussion about the number of ##ε##-intervals needed for coverage around a fixed point ##x_0##, depending on the value of ##ε##.
  • One participant expresses confusion regarding the compactness theorem, suggesting that if epsilon converges to 0, it would require infinitely many sets to cover the set S, which seems to contradict the theorem.
  • Another participant emphasizes that for any given open coverage of a compact set, there exists a finite sub-coverage, regardless of how small ##ε## is chosen.
  • There is a proposal that no matter how small ##ε## is, a smaller ##ε'## can be found, leading to the conclusion that a finite number of open sets will suffice to cover the set S.
  • One participant clarifies that the term "countable" should not be used to describe a finite number of open sets, as countable typically refers to an infinite quantity.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the compactness theorem and the necessity of finite coverings. While some agree on the existence of finite sub-coverings for compact sets, others raise questions about the conditions under which these apply, particularly when epsilon approaches zero.

Contextual Notes

There are unresolved assumptions regarding the definitions of open sets and the implications of epsilon values on the coverage of sets. The discussion also highlights the potential confusion surrounding terminology related to countability.

kidsasd987
Messages
142
Reaction score
4
The question is stated in the attatched file.
 

Attachments

  • Compactness_Q.png
    Compactness_Q.png
    19.4 KB · Views: 544
Physics news on Phys.org
How are the ##H_i## defined?
 
fresh_42 said:
How are the ##H_i## defined?
It is just an open set because it is just a part of open covering.
 
kidsasd987 said:
It is just an open set because it is just a part of open covering.
Ok. Now, if you take ##ε = 2(β-α)##, how many ##ε##-intervals around a fixed ##x_0## would you need for coverage? How many if ##ε = \frac{β-α}{2}##, how many for ##ε = \frac{β-α}{3}## or ##ε = \frac{β-α}{n}## with a fixed ##n##? The limit is essentially the union of all points of ##[α,β]## but for a given open coverage, finitely many will be enough.
 
fresh_42 said:
Ok. Now, if you take ##ε = 2(β-α)##, how many ##ε##-intervals around a fixed ##x_0## would you need for coverage? How many if ##ε = \frac{β-α}{2}##, how many for ##ε = \frac{β-α}{3}## or ##ε = \frac{β-α}{n}## with a fixed ##n##? The limit is essentially the union of all points of ##[α,β]## but for a given open coverage, finitely many will be enough.

Thanks for your reply.

I wanted to confirm the theorem, "if a set is compact then its open covering will contain a subcovering that consists of finitely many open sets." I believe H is an open covering, since it contains S and consists of open sets. However, because epsilon is converging to 0, it would require infinitely many sets to fully cover the set S, which contradicts to the theorem.

Each open set overlaps as much as epsilon, but it is shrinking to 0. Then would we need finitely many open sets Hi to cover the closed set? Somehow I feel there has to be infinitely many.
 
Last edited:
kidsasd987 said:
Thanks for your reply.

I wanted to confirm the theorem, "if a set is compact then its open covering will contain a subcovering that consists of finitely many open sets." I believe H is an open covering, since it contains S and consists of open sets. However, because epsilon is converging to 0, it would require infinitely many sets to fully cover the set S, which contradicts to the theorem.

Each open set overlaps as much as epsilon, but it is shrinking to 0. Then would we need finitely many open sets Hi to cover the closed set? Somehow I feel there has to be infinitely many.
The point is: For any given open coverage of a compact set (in a metric space - although I think Hausdorff is enough) there is a finite sub-coverage. Again: a given coverage!

If you take ##ε=0## then you simply have ##S= \{x | x \in S\} = \bigcup_{x \in S} \{x\}##. This is no open coverage anymore! But as soon as you chose a ##ε > 0##, how small it ever maybe, then finitely many will be sufficient. However, the number will depend on your choice of ##\epsilon##.
 
fresh_42 said:
The point is: For any given open coverage of a compact set (in a metric space - although I think Hausdorff is enough) there is a finite sub-coverage. Again: a given coverage!

If you take ##ε=0## then you simply have ##S= \{x | x \in S\} = \bigcup_{x \in S} \{x\}##. This is no open coverage anymore! But as soon as you chose a ##ε > 0##, how small it ever maybe, then finitely many will be sufficient.
Hmm, can I understand this way.

no matter how small ε is, we can find ε'< ε. which means ε is bounded below by some ε'. Then we can find a sufficiently small integer n which satisfies 1/n < ε.

Also 1/n'<ε'.

since 1/n is bounded by 1/n' and countable therefore finite number of open sets will be enough to cover the set S?
 
kidsasd987 said:
Hmm, can I understand this way.εεε

no matter how small ε is, we can find ε'< ε. which means ε is bounded below by some ε'. Then we can find an integer n which satisfies 1/n < ε.

Also 1/n'<ε'.

since 1/n is bounded by 1/n' and countable therefore finite number of open sets will be enough?
I don't see why you need a ##ε'##, but yes, this is correct. Perhaps you shouldn't say countable because this usually refers to an infinite number of things like the amount of natural numbers, since a finite number of things are always countably many.
 
  • Like
Likes   Reactions: kidsasd987
fresh_42 said:
I don't see why you need a ##ε'##, but yes, this is correct. Perhaps you shouldn't say countable because this usually refers to an infinite number of things like the amount of natural numbers, since a finite number of things are always countably many.

Thanks. My approach and terminology might be coarse because I am an engineering student. Thank you! It helepd a lot
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 18 ·
Replies
18
Views
4K