Difficulties with Definition of Compact Set

In summary, the conversation discusses the concept of compact sets and how to show that a set is compact. The definition of a subcover is also mentioned, with some confusion about whether it must be a strict subset or not. Ultimately, it is clarified that a subcover does not necessarily have to be a strict subset.
  • #1
jeckt
19
0
Hi All,

This is really a stupid question...I can't seem to get my head around it and it's making me depressed just thinking about it. Anyways, let's consider [tex] \mathbb{R}^{n} [/tex]

the set [tex] S = [0,1] [/tex] is not compact (I know it is but I can't see the flaw in my argument which seems it should be blatantly obvious.) Since [tex] C = \{ (-1,2) \} [/tex] covers [tex] S [/tex] but has no finite subcover (it does and this was pointed out in another thread but they didn't go into detail, I guess because it was so easy). When I say no finite subcover I mean [tex] C_{0} \subset C [/tex] since there is only one element in [tex] C [/tex]

Thanks for the help guys!
 
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  • #2
You forgot that C is a subset of itself.
 
  • #3
jeckt said:
Hi All,

This is really a stupid question...I can't seem to get my head around it and it's making me depressed just thinking about it. Anyways, let's consider [tex] \mathbb{R}^{n} [/tex]

the set [tex] S = [0,1] [/tex] is not compact (I know it is but I can't see the flaw in my argument which seems it should be blatantly obvious.) Since [tex] C = \{ (-1,2) \} [/tex] covers [tex] S [/tex] but has no finite subcover (it does and this was pointed out in another thread but they didn't go into detail, I guess because it was so easy). When I say no finite subcover I mean [tex] C_{0} \subset C [/tex] since there is only one element in [tex] C [/tex]

Thanks for the help guys!

An open covering is a collection U of open sets that covers S. U typically might have an infinite number of open sets which overlap. A finite subcover from U would be finitely many of the open sets from U which still cover S. In your example, U has only one member which is given to cover S. It is automatically a finite subcover because there is one element in the subcover.
 
  • #4
What's your question? I'm assuming you're just having difficulty in understanding the concept of compact sets.

A compact set is defined to be a set in which for every open cover, there exists a finite subcover. Certainly, (-1,2) is a finite cover of the set. This is a red herring though, as it doesn't really tell us anything. In order to show that a set is compact, you must show that for every open cover, there exists a finite subcover. In this fashion, it is not a trivial task to show an arbitrary set is compact. However, for [tex] \mathbb{R}[/tex], if you show that a set is closed and bounded, then you have shown that it is compact.
 
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  • #5
This makes more sense now, thanks guys! the reply is so quick too. In class, the definition I had was that a subcover is a strict subset of a cover. So in this case the subcover would be the cover and thus it would be finite.

@gb7nash - yeah I know that to show a set is compact i need to show that every open cover of the set contains a finite subcover. In [tex] \mathbb{R} [/tex] like you said, we just have to show that the set is closed and bounded. This is the Heine Borel theorem if I remember correclty.
 
  • #6
The subcover (Kelley's term) or the subcollection (Munkres' term) doesn't necessary mean a strict inclusion in terms of sets when speaking of covers and subcovers.
 
  • #7
Thanks dextercioby, yeah I checked some online resources and textbooks. Just that in my analysis class the lecturer used a strict subset or I wrote it down wrong but it's all good now and I get where the issue is.

Thanks for the help guys!
 

1. What is a compact set?

A compact set is a subset of a metric space that is closed and bounded. In simpler terms, it is a set where every sequence of elements has a converging subsequence within the set.

2. How do you determine if a set is compact?

To determine if a set is compact, there are a few criteria that need to be met. The set must be closed, meaning it contains all its limit points. It must also be bounded, meaning there is a finite distance between its elements. Lastly, every sequence within the set must have a converging subsequence within the set.

3. Can a set be compact in one metric space but not in another?

Yes, a set can be compact in one metric space but not in another. This is because different metric spaces have different definitions of compactness. For example, a set may be compact in a Euclidean space but not in a discrete space.

4. What is the relationship between compact sets and closed sets?

Compact sets and closed sets are closely related, as a compact set must be closed. However, a closed set is not necessarily compact. A closed set is one where all its limit points are contained within the set, while a compact set also has the additional criterion of being bounded.

5. Are there any practical applications of compact sets?

Yes, compact sets have many practical applications in mathematics and other fields such as physics and engineering. They are used in optimization problems, topological spaces, and in proving the existence of solutions for certain equations. Compact sets also have applications in computer science, specifically in algorithms and data compression.

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