Difficulties with Definition of Compact Set

  • Thread starter jeckt
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  • #1
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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!
 

Answers and Replies

  • #2
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You forgot that C is a subset of itself.
 
  • #3
LCKurtz
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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
gb7nash
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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
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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
dextercioby
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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
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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!
 

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