Difficulties with Definition of Compact Set

[jeckt]

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 $$\mathbb{R}^{n}$$

the set $$S = [0,1]$$ 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 $$C = \{ (-1,2) \}$$ covers $$S$$ 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 $$C_{0} \subset C$$ since there is only one element in $$C$$

Thanks for the help guys!

 

[lugita15]

You forgot that C is a subset of itself.

 

[LCKurtz]

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 $$\mathbb{R}^{n}$$

the set $$S = [0,1]$$ 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 $$C = \{ (-1,2) \}$$ covers $$S$$ 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 $$C_{0} \subset C$$ since there is only one element in $$C$$

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.

 

[gb7nash]

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 $$\mathbb{R}$$, if you show that a set is closed and bounded, then you have shown that it is compact.

 

[jeckt]

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 $$\mathbb{R}$$ 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.

 

[dextercioby]

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.

 

[jeckt]

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!